text:
Q: What is "$\exists x \in S , \phi(x)$" shorthand for? I recently learned that "$\forall x \in S, \phi(x)$" is shorthand for "$\forall x \big(x \in S \rightarrow \phi(x)\big)$". Does the same idea apply for the shorthand "$\exists x \in S, \phi(x)$"?
i.e. is "$\exists x \in S, \phi(x)$" logically equivalent to "$\exists x \big(x \in S \rightarrow \phi(x)\big)$"?
I am somewhat tempted to think that it may actually denote "$\exists x\big( x \in S \land \phi(x)\big )$" but am not positive! Any insight is greatly appreciated. Cheers~

A:
I am somewhat tempted to think that it may actually denote "$\exists x\big( x \in S \land \phi(x)\big )$"

Yes, that's what it means.

i.e. is "$\exists x \in S, \phi(x)$" logically equivalent to "$\exists x \big(x \in S \rightarrow \phi(x)\big)$"?

That would not be very useful, because $x \in S \rightarrow \phi(x)$ is true for every $x \notin S$, no matter what $\phi$ is, and so $\exists x \big(x \in S \rightarrow \phi(x)\big)$ is true for any $S$ that is not the whole universe.

A: $\exists x{\in}S~.\phi(x)$ is intended to be read as "There is something, call it $x$, which is in $S$ and satisfies $\phi(x)$," and so is indeed synonymous with $\exists x~.(x\in S\wedge\phi(x))$
Moreover, we want quantifier duality to hold in restricted domains, so will need
$\neg\forall x{\in}S~.\neg\phi(x)$ and $\exists x{\in}S~.\phi(x)$ to be logical equivalences.
Now because $\neg \forall x{\in}S~.\neg\phi(x)$ is synonymous with $\neg\forall x~.(x\in S\to \neg\phi(x))$, which by quantifier duality is equivalent to $\exists x~.\neg(x\in S\to\neg\phi(x))$, and thus with $\exists x~.(x\in S\wedge\phi(x))$, therefore we shall want this to be synonymous with $\exists x{\in}S~.\phi(x)$ .


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text:
Dodge tennis teams host Jefferson and Thompson
Thursday, April 3. 2014
By Frank Erwin
The Dodge County High School (DCHS) Indians tennis teams hosted Jefferson County March 20, 2014. The boys won five to zero and the girls won four to one.
Winners for the boys were number one singles, Jacob Danforth, six to zero in the first set and six to zero in the second set; number two singles, Dallas Hardin, six to zero in the first set and six to two in the second set; number three singles, Dwight Inman, won by forfeit; number one doubles, Vic Mizell and Corvon Parker, six to zero in the first set and six to two in the second set; and number two doubles, CJ Kemp and Brandon Roland, six to three in the first set and six to two in the second set.
Winners for the girls were number one singles, Kylie Erwin, seven to five in the first set and six to three in the second set; number three singles, Brenna Hutcheson, six to three in the first set and six to two in the second set; number one doubles, Lila Coleman and Lillie Douglas, six to zero in the first set and six to zero in the second set; and number two doubles, Faith Manning and Myranda Burch, six to one in the first set and six to zero in the second set.
The Indians tennis teams hosted Thompson on March 21, 2014. The boys won five to zero and the girls won five to zero.
Winners for the boys were number one singles, Danforth, six to two in the first set and six to one in the second set; number two singles, Hardin, six to zero in the first set and six to one in the second set; number three singles, Inman, six to zero in the first set and six to one in the second set; number one doubles, Mizell and Parker, six to one in the first set and six to one in the second set; and number two doubles, Kemp and Roland, won by forfeit.
Winners for the girls were number one singles, Erwin, six to three in the first set and six to zero in the second set; number two singles, Taylor Thomas, six to two in the first set and six to zero in the second set; number three singles, Hutcheson, six to one in the first set and six to zero in the second set; number one doubles, Coleman and Douglas, six to zero in the first set and six to one in the second set; and number two doubles, Manning and Burch, six to zero in the first set and six to zero in the second set.

redpajama_set_name: RedPajamaCommonCrawl


text:
Austria: Because he was not satisfied with the work performance of his employee, a Syrian business owner poured gasoline over him and attempted to set him on fire
Posted on 4. Oct 2020 by medforth
Dramatic scenes took place in the night to Saturday in a firm in the Vienna district of Wieden. A 32-year-old is under strong suspicion of having doused a 28-year-old man with gasoline and tried to set him on fire – which the suspected perpetrator did not succeed in doing. As it turned out, the suspect is the man's boss. The Syrian citizen was arrested. The reason for the crime: The 32-year-old was not satisfied with his work performance.The incident happened in the night on Saturday at about 3.25 am, as the police reported on Saturday morning. After the 32-year-old suspect had doused the 28-year-old victim with gasoline, "he probably tried to set the man on fire with a lighter," according to police spokeswoman Barbara Gass. With the help of a 20-year-old witness, who was also present at the scene of the crime, the victim "was able to stop the suspect from his intention".A scuffle subsequently ensued, it was said further, after which the suspected perpetrator first fled, only to return to the scene of the crime a short time later. There, however, he was already awaited by the police. The 32-year-old suspect was arrested.The interrogation of the suspected perpetrator on Saturday brought some light into the darkness: Thus it came to an argument between the 32-year old and the 28-year old first, because the older one was not satisfied with the work performance of its employee, reported Gass in the afternoon from the first interrogation. The 32-year-old Syrian denied that he intended to kill the employee and also denied having poured gasoline on the person in question. The man was taken into custody. The Vienna State Criminal Investigation Office, Mitte branch office, is investigating.
https://www.krone.at/2243790
1 thought on "Austria: Because he was not satisfied with the work performance of his employee, a Syrian business owner poured gasoline over him and attempted to set him on fire"
boudicabpi on 4. Oct 2020 at 13:24 said:
Reblogged this on Boudica BPI Weblog.


text:
There's One Perennial Complaint About Democracy. Here's How to Solve It.
May 10, 2016 Government Political Science
A perennial complaint about democracy in America is that too large a portion of the electorate is poorly informed about important political issues. This is the problem of the ignorant voter.
This problem was exacerbated in the current presidential election cycle by the large number of candidates vying for the Republican nomination that made keeping track of their various, and often shifting, policy positions extraordinarily difficult. As a result, many of those who voted in the presidential primaries cast their ballots with little idea of where the candidates stood on many important issues.
Wouldn't it be nice if there was some way to nudge voters into making more informed choices?
Well, actually, there is. But in saying this, I am using the word "nudge" advisedly.
What is a "nudge?"
Among contemporary policy analysts, "nudge" is a term of art. It refers to creating a context within which people make choices–a "choice architecture"–that makes it more likely that people will select one option rather than another. The typical example of a nudge is a school cafeteria in which fruits and vegetables are placed in front in easy to reach locations and less healthy fare is placed in less visible and harder to reach locations. No one is forced to select the fruit or vegetables, but the choice architecture makes it more likely that people will.
The key feature of a nudge is that it is not coercive. It is an effort to influence choice, not to impose it. People are always able to "opt out" of the nudge. Thus, to nudge is to design the context in which individuals make decisions so as to influence their choice without eliminating any options.
What does a "nudge" look like in politics?
To see how nudging can be employed to help voters make more informed decisions, consider the following scenario.
A bipartisan good government group creates a list of the most significant contemporary policy issues. It then invites all candidates to state their positions on the issues. In the current campaign, candidates could be invited to state where they stand on gay marriage, immigration, intervention in Syria, climate change, tax reform, the minimum wage, gun control, income inequality, etc.
This information would be collected and fed into the relevant election commission computer. When voters enter the voting booth, they would have the option of electronically recording their policy preferences on the same form that the candidates completed.
The computer would display a ranking of the candidates on the basis of how closely their positions aligned with the voter's. After receiving this information, voters would cast their ballots.
This proposal is a nudge. It is completely non-coercive. No candidate would be required to complete the list of his or her policy positions, although refusing to do so might be viewed negatively by voters. No voter would be required to utilize the option. All would remain free to simply walk into the booth and cast their vote. Even those who utilize the option remain free to disregard its results and vote for whomever they please.
The proposal simply alters the choice architecture of voting to build in access to a source of information about the candidates. Yet, it makes it more likely that citizens will cast more informed votes than they do at present.
Why is nudging voters a good idea?
Nudging voters in this way has much to recommend it. In the first place, it greatly reduces the cost to voters of educating themselves about the candidates. Rather than having to expend the time and effort to research the candidates' policy positions and decipher misleading political advertisements, voters can get the relevant information at the touch of a button precisely when they need it.
Secondly, this type of nudge ameliorates the effect of money in politics. Well-financed candidates or their super-PAC surrogates often use their resources to drown out or distort the message of their less well-endowed rivals during campaigns. But the size of a candidate's war chest cannot affect the ranking the voter receives in the voting booth. This is determined exclusively by the candidates' policy positions, and is insulated from the distortions of political advertising. Thus, this nudge makes it possible for even the most impoverished candidates to command the voters' attention when it matters most.
Finally, this nudge reduces politicians' ability to profit from pandering. Candidates often try to appeal to different segments of the electorate by tailoring their policy prescriptions to fit the audience they are addressing. By requiring candidates to take a definite stand on the issues and basing each voter's ranking on those explicit commitments, the nudge undermines the effectiveness of such pandering.
This proposal is likely to have widespread appeal. Voters should like it because it integrates complex information about the candidates into a user-friendly form that can help them vote intelligently. Candidates should like it because it helps them circumvent the attacks ads and misrepresentations of their positions that are characteristic of contemporary campaigns. Even journalists should like it because it incentivizes candidates to put their policy positions on record, which would facilitate their reporting.
But perhaps most importantly for a posting on this particular blog, libertarians should like it because it is likely to show a significant portion of the population that they agree with libertarian candidates more than they do with Democrat or Republican candidates.
And I suspect that this would be more effective than all the educational outreach efforts in the world.
Tags: Democracy nudge rational voting voters voting
Congestion tax in Manhattan: Readers speak!
March 9, 2018 Economics and Taxing
Learn Liberty Turns 7
February 11, 2018 Uncategorized
The organic industry is a case study in rent-seeking.
February 5, 2018 Economics, Politics & Policy, Regulations, and Subsidies


text:
← Softball Takeaways from Simone Biles
Tip for Learning to Feel the Separation of Hips and Shoulders When Hitting →
Coaching, Bearded Dragons and Relationship Building
It was yet another oppressively hot, very humid day here in the Midwest – the type we used to associate with Mississippi but now is happening here on a regular basis.
My first lesson of the afternoon, a young 10U pitcher we'll call Hermione (for reasons that will become obvious shortly), had the look of someone who wanted to be anywhere but a pitching lesson.
"Jump some rope to warm up," I told her, and she slowly shuffled over to the fence where the rope was hanging before giving it a half-hearted effort. Then she went out to throw with her mom, and could barely get the ball there despite the fact they weren't more than 15 feet apart.
This was the second time in a row she had come out that way. "Maybe she just doesn't want to pitch," I said to her mom. That's ok, I said, but then no sense torturing her with a lesson.
Mom left it up to Hermione and she said she wanted to do it so we continued – after I talked to her for a few minutes.
"I know it's hot," I told her, "but there's nothing we can do about that. But attitude is a choice. You can choose to put the effort in or not – it's totally in your control."
That worked about as well as you'd expect it to work on a 10U player, but we continued anyway.
"She may be tired," her mom offered. "We went to the zoo yesterday."
"Really?" I replied, thinking we might have a different way to approach things. "What zoo?" I asked Hermione.
"Mmph," she said.
"Milwaukee," her mom added.
"I love the Milwaukee Zoo," I said to Hermione. "What are your favorite exhibits?"
"Mmph," she replied again.
Then mom stepped in again. "We went to the small mammals, (something else I don't remember), and reptiles."
"Reptiles?" I said. That's an unusual choice if you're only seeing three.
"I like reptiles." Hermione offered.
Aha! An actual response. I dug into it a little more, and then came the turning point. Her mom said the reptile house didn't have any bearded dragons.
"Oh, you like those?" I asked.
"Yes," she said with some enthusiasm. "I have a bearded dragon in my room."
From then on it was like someone flipped a switch. Instead of looking like someone in a hostage video she became quite animated as we talked about beardies in general (that's what the "in the know" people call them) and hers in particular.
Over the course of the lesson I learned her beardy's name as Phoenix Firebolt (for Dumbledore's pet and Harry Potter's broomstick), which of course meant I also learned she liked Harry Potter. I learned that beardies eat grain and bugs, and that bugs are a lot more expensive than grain.
I learned she liked to dress Phoenix up in costumes, and that Phoenix wasn't exactly a fan.
"I just have a couple," she told me. "A lot of people do it all the time.
I also learned she liked to take Phoenix for walks, and that if she took the leash off outside without an adult present Phoenix would try to escape. A lesson discovered the hard way.
Throughout all of this conversation Hermione continued to pitch. Her throws got harder and more accurate.
I would make corrections here or there, but quickly follow that up with a beardy question. The lesson flew by, with Hermione making tremendous progress.
It was a good reminder for me. There is a saying that players don't care how much you know until they know how much you care.
In other words, your ability as a coach to build a relationship with a player on her terms is critical to your success.
Whether it carries over automatically to next week remains to be seen. But hopefully, at least with this one player, I have a key to unlock her resistance or reluctance and help her enjoy her pitching lessons instead of just enduring them.
That's the challenge for all coaches. Get to know your players, especially at the younger ages. Find out what floats their boats.
Because if you can establish some common ground there, they just might let you in a little faster to help them become the players they're meant to be.
Bearded dragon photo by Saketh Upadhya on Pexels.com
Posted on August 20, 2021, in Coaching, General Thoughts and tagged attitude, bearded dragons, Harry Potter, making a connection, Milwaukee Zoo, relationship building. Bookmark the permalink. Leave a comment.


By WFWP Austria, Katharina Bauer: "On Saturday 23 April, I went to Ukraine, a trip I was looking forward to for the last 2 months. As soon as I stepped off the plane in Ukraine, I received the warmest welcome by the Ukrainian members. From the beginning on, it was also perfectly organized. While there, I heard the most beautiful, touching and wise explanation of Vision 2020: it's the time when we learn how to let go of our selfishness and learn to lead a happy and fulfilling life of living for others through uniting with God's heart. This is the essence and sense of the whole week I was in Ukraine and for my whole life.
Monday was for sightseeing, so that we could see all the historical places as well as how things were developing in Maidan and Kiev, in general. We went to beautiful, old churches with their roofs shining with gold to the sky, we walked through the streets where the first Christians were witnessing, historical places with young and old memories of the nation. We walked through the squares and streets, where just two years ago huge tents stood full of protestors from Ukraine and all over the world. We could still see bullet holes through the thick, iron lampposts. How strong the bullets were to leave such holes. Our tour stopped to pray by the photos of the "Heavenly Hundred" who were shot in the demonstrations. There is a chapel, close to Maidan, where many people lost their lives, where we can come to pray and cry. This is a special, "all religions" chapel. Maybe the first in the world of this type. The tents are now gone, and the streets are breathing, welcoming the sunshine onto the flowers growing there. I felt freedom and peace.
I am grateful that the community in Kiev organized this for us. The next day, 26 April, was the 30th Anniversary of the Chernobyl disaster. In the early morning, we went to a place in Kiev where memorials of the victims stand. Because WFWP Ukraine actively take care of the children, who are victims of the Chernobyl disaster, our representatives were invited officially to offer flowers and pray next to the Ukrainian President, the Mayor of Kiev and other politicians. The weather on this day was extremely cold and windy. In fact, the skies are crying every year on 26 April. The Ukrainian president rang a bell in remembrance as the nation offered a prayer exactly at 10 o'clock.
Wednesday, 27 April, the Kiev City Hall hosted a round table with politicians and NGO's. This was organized together with WFWP Ukraine under the title "Women, Peace, Security." This topic is especially relevant for this nation nowadays. I, along with Angelika Selle, WFWP International, Tina Coombs, WFWP Europe, and Tatiana Kotseba, WFWP Ukraine, gave inspiring and essential speeches on the subject. I felt very grateful to these wonderful sisters, who made such a long journey to support WFWP Ukraine in this excellent and important work.
The main event for WFWP was the Global Women's Peace Network Conference on "Universal Values as a Cornerstone of Global Peace and Harmony," on 28 April. With 500 in attendance, 70 were current WFWP members, and the rest were partners and friends. The conference was an opportunity to honour cultural activities such as the arts and sports in Ukraine. It was successful in inspiring more local branches of WFWP to be founded in at least three areas.
I was so impressed by the numbers and quality of VIPs that the Ukrainian sisters are working with. These are deep, personal, warm connections. Like Angelika said, they all have such a motherly heart. I am proud to be a woman when I listen to them. Tatiana, Anna, Dascha, and many others are doing such a great job and working so hard to improve life in Ukraine. They reach out to the world level and are welcomed and have become important members of the International Women's Council. The conference was amazing. I am so inspired through this event, and grateful for being invited, I can just admire them and learn from these amazing sisters and the amazing Ukrainian community. I share this inspiring experience to encourage more of our WFWP sisters to come and get to know the sisters in WFWP Ukraine and experience this event. I feel that their work is how we all could and should be working.
Dear sisters, dear friends, come, experience the love, and support the excellent, valuable work of our Ukrainian sisters on the national and international level. I feel that God is happy to watch them. Thank you very much. I wish you all the best and send the warmest greetings from Austria.


text:
\section*{Introduction}
Let $G$ be a locally compact group, and let $\mu$ be a probability measure on
$G$ whose support generates $G$. A function $\phi \in L^\infty(G)$ is called
\emph{$\mu$-harmonic} if $\mu \ast \phi = \phi$. Harmonic functions play
a crucial r\^ole for the investigation of random walks on locally compact
groups and are extensively studied (see \cite{Aze} and \cite{Fur}, for
example). The collection of all $\mu$-harmonic functions is denoted by
${\cal H}_\mu$; it is easily seen to be a $w^\ast$-closed subspace of
$L^\infty(G)$, but is a subalgebra only if it consists of the constants alone.
Nevertheless, ${\cal H}_\mu$ can be equipped with a product---different,
of course, from the one in $L^\infty(G)$---turning it into an abelian von
Neumann algebra (\cite{Aze}).
\par
Recently, the notion of harmonicity has been ``quantized'' in two, seemingly
entirely different directions.
\par
One such quantization was introduced and studied by C.-H.\
Chu and A.\ T.-M.\ Lau in \cite{CL}. Their approach can be considered dual to
the classical concept of a harmonic function. In \cite{Eym}, P.\ Eymard
introduced the so-called Fourier algebra $A(G)$ and Fourier--Stieltjes algebra
$B(G)$ of $G$. If $G$ is abelian with dual group $\hat{G}$, the Fourier and
Fourier--Stieltjes transforms, respectively, yield isometric isomorphisms
$A(G) \cong L^1(\hat{G})$ and $B(G) \cong M(\hat{G})$. Instead of looking
at harmonicity with respect to a probability measure, Chu and Lau consider
harmonicity of functionals on $A(G)$, i.e., of elements of the group von
Neumann algebra $\operatorname{VN}(G)$, with respect to a positive definite function
$\sigma \in B(G)$: an operator $T \in \operatorname{VN}(G)$ is said to be $\sigma$-harmonic
if $\sigma \cdot T = T$, where $\cdot$ is the canonical module action of
$B(G)$ on $\operatorname{VN}(G)$. The collection of all $\sigma$-harmonic functionals is
denoted by ${\cal H}_\sigma$. Even though, this new theory seems to parallel
the classical theory of harmonic functions on the surface, it is, in fact,
strikingly different. For instance, ${\cal H}_\sigma$ is \emph{always} a
von Neumann \emph{subalgebra} of $\operatorname{VN}(G)$ (\cite[Remark 3.2.11]{CL}).
\par
A completely different type of quantization was recently carried out by
W.\ Jaworski and the first-named author (\cite{JN}). Starting point is the
the result by F.\ Ghahramani (\cite{Gha}) that there is a natural isometric
representation $\Theta$ of the measure algebra $M(G)$ on ${\cal B}(L^2(G))$,
such that for $\mu \in M(G)$ and $\phi \in L^\infty(G)$---viewed as a
multiplication operator on $L^2(G)$---we have $\Theta(\mu)\phi =
\mu \ast \phi$. Hence, the authors of \cite{JN} define an operator
$T \in {\cal B}(L^2(G))$ to be $\mu$-harmonic for a probability measure
$\mu$ if $\Theta(\mu)(T) = T$. The collection of all $\mu$-harmonic operators
is denoted by $\tilde{\cal H}_\mu$. Like ${\cal H}_\mu$, the space
$\tilde{\cal H}_\mu$ is a von Neumann algebra, but with a product usually
different from the one in ${\cal B}(L^2(G))$;
in fact, $\tilde{\cal H}_\mu$ can be described as the crossed product
of ${\cal H}_\mu$ with $G$, where the action of $G$ on ${\cal H}_\mu$ is
given by left translation (\cite[Proposition 6.3]{JN}).
\par
In the present paper, we extend Chu's and Lau's notion of $\sigma$-harmonicity from $\operatorname{VN}(G)$ to ${\cal B}(L^2(G))$ in a way that parallels the extension of $\mu$-harmonicity from $L^\infty(G)$ to ${\cal B}(L^2(G))$ in \cite{JN}.
\par
As the predual space of the operator algebra $\operatorname{VN}(G)$, the Fourier algebra
carries a canonical operator space structure. (For the theory of operator
spaces, we refer to \cite{ER}.) By $\mathcal{M}_\mathrm{cb}(A(G))$, we denote the
completely bounded multipliers of $A(G)$, i.e., those functions on $G$
that induce completely bounded multiplication operators on $A(G)$.
Obviously, $\mathcal{M}_\mathrm{cb}(A(G))$ is a commutative Banach algebra, and it
contains $B(G)$ (with equality if and only if $G$ is amenable).
In \cite{NRS}, the first-named author, Z.-J.\ Ruan, and N.\ Spronk
constructed a representation $\hat{\Theta}$ of $\mathcal{M}_\mathrm{cb}(A(G))$ on
${\cal B}(L^2(G))$ which displays a perfect duality with Ghahramani's
representation of $M(G)$ (for details, see \cite{NRS}); in particular, it
extends the action of $B(G)$ on $\operatorname{VN}(G)$ to ${\cal B}(L^2(G))$. For $\sigma
\in \mathcal{M}_\mathrm{cb}(A(G))$, it is then natural to define $T \in {\cal B}(L^2(G))$ to be
$\sigma$-harmonic if $\hat{\Theta}(\sigma)(T) = T$. We denote the collection
of all $\sigma$-harmonic operators by $\tilde{\cal H}_\sigma$.
\par
The aim of this paper is to explore the connections between this setting
and the two quantizations from \cite{CL} and \cite{JN}. For instance,
one of our main results is that---under very mild hypotheses which are
always satisfied if $G$ is amenable or the free group in two
generators---$\tilde{\cal H}_\sigma$ is a von
Neumann subalgebra of ${\cal B}(L^2(G))$, namely the von Neumann subalgebra
of ${\cal B}(L^2(G))$ generated by ${\cal H}_\sigma$ and $L^\infty(G)$.
\par
The paper is organized as follows.
\par
First, we fix our notation and terminology
and review a few basic facts about harmonic functions (see \cite{JN})
and harmonic functionals in $\operatorname{VN}(G)$ (as introduced and studied in \cite{CL}).
We also recall results from \cite{NRS} on the representation of $\mathcal{M}_\mathrm{cb}(A(G))$
on ${\cal B}(L^2(G))$.
\par
Section \ref{adapted} gives a characterization of
adaptedness in terms of the Fourier--Stieltjes transform (in the framework of
locally compact \emph{abelian} groups). This motivates our definition of the
dual notion of adaptedness for positive definite functions.
\par
In Section \ref{support}, we introduce the support of an arbitrary
(bounded linear) operator on $L^2(G)$ such that it generalizes Eymard's
corresponding notion for elements of $\operatorname{VN}(G)$. In order to prepare the ground
for our main results, we investigate the properties of our notion of support
in detail.
\par
Those main results are contained in Section \ref{dualpic}. Using
the representation of $\mathcal{M}_\mathrm{cb}(A(G))$ from \cite{NRS}, we naturally extend the
notion of a $\sigma$-harmonic functional
in $\operatorname{VN} (G)$ (\cite{CL}) to the one of a $\sigma$-harmonic operator in
${\cal B}(L^2(G))$, where $\sigma$ is a positive definite function on $G$.
Assuming either a very mild condition on $G$---the approximation property
(\cite{HK})---or that $\sigma$ belongs to $A(G)$, we show that the space of
$\sigma$-harmonic
operators $\tilde{\cal H}_\sigma$ is always a von Neumann subalgebra of
${\cal B}(L^2(G))$ and, in fact, precisely the von Neumann algebra
generated by $L^\infty(G)$ and the algebra ${\cal H}_\sigma$ of
$\sigma$-harmonic functionals in $\operatorname{VN} (G)$. This may be viewed as a result
``dual'' to the corresponding characterization (see \cite{JN})
of the von Neumann algebra of $\mu$-harmonic operators in ${\cal B}(L^2(G))$
as a crossed product over the algebra of $\mu$-harmonic functions in
$L^\infty(G)$, where $\mu$ is a probability measure on $G$.
Our notion of support for an arbitrary operator on $L^2(G)$ allows for
an alternative characterization of $\tilde{\cal H}_\sigma$---at least if $G$
has the approximation property---, namely as the operators in
${\cal B}(L^2(G))$ with support contained in the subgroup
$G_\sigma = \sigma^{-1} (\{ 1 \})$ of $G$.
\par
Finally, in Section \ref{ideals}, we present an approach to harmonic operators
in ${\cal B}(L^2(G))$ via ideals in the predual ${\cal T}(L^2(G))$ in the
spirit of \cite{Wil} (see also \cite[Section 3]{JN}). This makes it
necessary to endow the space ${\cal T}(L^2(G))$ with a product very different
from the composition of operators; this product arises naturally when
when one regards ${\cal B}(L^2(G))$ as a Hopf--von Neumann algebra, with a
co-multiplication naturally extending the one of
$\operatorname{VN}(G)$ (see \cite{NRS}, \cite{PR}, and \cite{JN}).
\subsubsection*{Acknowledgments}
Most of this paper was written during an RiP ($=$ ``Research in Paradise'')
stay by the authors at the Mathematisches Forschungsinstitut Oberwolfach in
May 2005. We acknowledge this support and the uniquely stimulating atmosphere
at Oberwolfach with gratitude.
\par
We would also like to thank Tony Lau for helping with the proof of Lemma
\ref{wdense}---in particular, for bringing \cite{Lau} to our attention---and
Zhiguo Hu for catching some oversights in an earlier version of this paper. Finally, thanks are due to the referee for his helpful suggestions.
\section{Preliminaries}
Throughout this section, let $G$ denote a locally compact group equipped with
left Haar measure.
We follow standard notation and terminology of abstract harmonic analysis as, e.g., in \cite{HR}.
For $p \in [1,\infty]$, we denote the corresponding $L^p$-space with respect to left Haar measure on $G$ by $L^p(G)$. By $M(G)$,
we denote the algebra of all complex, regular Borel measures; $M^1(G)$
stands for the probability measures in $M(G)$. We note that $L^\infty(G)$
is a von Neumann algebra acting naturally on $L^2(G)$ as multiplication
operators: for $\phi \in L^\infty(G)$, let $M_\phi \in {\cal B}(L^2(G))$
be given by $(M_\phi \xi)(x) = \phi(x) \xi(x)$ for $\xi \in L^2(G)$ and
$x \in G$. (For the sake of notational simplicity, we shall sometimes write
$\phi$ and $M_\phi$ interchangeably.)
\par
The Banach algebra $M(G)$ acts on $L^p(G)$ for $p \in [1,\infty]$ via
convolution from the left. For $\mu \in M(G)$, we define
$\phi \in L^\infty(G)$ as \emph{$\mu$-harmonic} if $\mu \ast \phi = \phi$
and set
\[
  {\cal H}_\mu := \{ \phi \in L^\infty(G) : \text{$\phi$ is $\mu$-harmonic} \}.
\]
(For the motivation for the name ``$\mu$-harmonic function'', see the
introduction of \cite{CL}). Usually, $\mu$-harmonic functions are considered
only for $\mu \in M^1(G)$. To avoid pathologies, we shall assume
$\mu$ to be \emph{adapted}, i.e., $\langle {\operatorname{supp}} \mu \rangle$, the closed
subgroup of $G$ generated by the support of $\mu$, is all of $G$. For abelian $G$ and adapted $\mu \in M^1(G)$, the classical Choquet--Deny theorem asserts that ${\cal H}_\mu$ consists only of the constant functions. For a general locally compact group $G$, a measure $\mu \in M^1(G)$ such that ${\cal H}_\mu \cong {\mathbb C}$ exists if and only if $G$ is amenable and $\sigma$-compact (\cite[Proposition 2.1.3]{CL}). Even though ${\cal H}_\mu$ is not a von Neumann subalgebra of $L^\infty(G)$---except if ${\cal H}_\mu \cong {\mathbb C}$---, there is a natural product on ${\cal H}_\mu$ turning it into an abelian von Neumann algebra in its own right: given $f,g \in {\cal H}_\mu$, one can show that the limit
\[
  \text{$w^\ast$-}\lim_n \int_G \rho(x) (fg) \, d\mu^{\ast n} (x)
\]
exists in $L^\infty(G)$ and is an explicit formula for the multiplication
in ${\cal H}_\mu$ (here, $\mu^{\ast n}$ stands for the $n$-th convolution
power of $\mu$, and $\rho(x)$ denotes right
translation by $x$, i.e., $(\rho(x)\phi)(y) :=\phi(yx)$ for
$\phi \!: G \to {\mathbb C}$ and $x,y \in G$).
For the classical theory of random walks and
harmonic functions, see \cite{Aze}, \cite{Nev}, and \cite{Rev}, for example.
\par
For any $\phi \!: G \to {\mathbb C}$ and $x \in G$, we define $\lambda(x)\phi
\!: G \to {\mathbb C}$ by letting $(\lambda(x)\phi)(y) := \phi(x^{-1}y)$ for
$x \in G$. For fixed $x \in G$, the map $L^2(G) \ni \xi \mapsto \lambda(x)\xi$
is a unitary operator on $L^2(G)$, and
\[
  \lambda \!: G \to {\cal B}(L^2(G)), \quad x \mapsto \lambda(x)
\]
is a unitary representation of $G$ on $L^2(G)$, the \emph{left regular
representation} of $G$. The \emph{group von Neumann algebra} $\operatorname{VN}(G)$ of
$G$ is defined as $\operatorname{VN}(G) := \lambda(G)''$. The \emph{Fourier algebra}
$A(G)$ of $G$---introduced by P.\ Eymard in \cite{Eym}---is the (unique)
predual of $\operatorname{VN}(G)$.
\par
As any operator algebra, $\operatorname{VN}(G)$ is a concrete operator space. (For the
theory of operator spaces, our reference is \cite{ER}). Since the category of
operator spaces allows for a natural duality theory, the dual space
$\operatorname{VN}(G)^\ast$---and thus its subspace $A(G)$---is an operator space in a
canonical manner. In particular, we may speak of completely bounded maps
on $A(G)$. (Following \cite{ER}, we denote the completely bounded maps
on an operator space $E$ by ${\cal CB}(E)$.) A \emph{multiplier} of $A(G)$ is a
function $\sigma$ on $G$ such that $\sigma A(G) \subset A(G)$; a multiplier is
completely bounded if the corresponding multiplication operator is
completely bounded. The collection of all completely bounded multipliers
is denoted by $\mathcal{M}_\mathrm{cb}(A(G))$. It naturally inherits an operator space structure
from ${\cal CB}(A(G))$ which turns it into a commutative, completely contractive
Banach algebra, i.e., a Banach algebra which is an operator space such that
multiplication is completely contractive. An important subalgebra of
$\mathcal{M}_\mathrm{cb}(A(G))$ is the \emph{Fourier--Stieltjes algebra} $B(G)$ of $G$ (introduced also in \cite{Eym}).
It consists of all functions of the form $G \ni x \mapsto \langle \pi(x) \xi, \eta \rangle$,
where $\pi$ is a (always strongly continuous) unitary representation of $G$ on a Hilbert space ${\mathfrak H}$ and
$\xi, \eta \in {\mathfrak H}$. It can be identified with the dual space of
the full group ${C^\ast}$-algebra ${C^\ast}(G)$, i.e., of the enveloping
${C^\ast}$-algebra of $L^1(G)$, and thus also has a canonical operator space
structure. It contains $A(G)$ as a closed ideal and thus canonically embeds
into $\mathcal{M}_\mathrm{cb}(A(G))$; this embedding is always completely contractive, but is
an isometric isomorphism if and only if $G$ is amenable. For more details
and references to the original literature, see \cite{Spr}.
\par
It is immediate from the definition of $\mathcal{M}_\mathrm{cb}(A(G))$
that $A(G)$ is a completely contractive Banach $\mathcal{M}_\mathrm{cb}(A(G))$-module through
pointwise multiplication. Consequently, $\operatorname{VN}(G)$ carries a dual
$\mathcal{M}_\mathrm{cb}(A(G))$-module structure, namely
\[
  \langle \phi, \sigma \cdot T \rangle := \langle \phi \sigma, T \rangle
  \qquad (\sigma \in \mathcal{M}_\mathrm{cb}(A(G)), \, T \in \operatorname{VN}(G), \, \phi \in A(G)).
\]
Slightly generalizing the definition from \cite{CL}, we say that, for
$\sigma \in \mathcal{M}_\mathrm{cb}(A(G))$, a von Neumann algebra element $T \in V(G)$, is
\emph{$\sigma$-harmonic} if $\sigma \cdot T = T$, and define
\[
  {\cal H}_\sigma := \{ T \in \operatorname{VN}(G) : \text{$T$ is $\sigma$-harmonic} \}.
\]
(Chu and Lau consider ${\cal H}_\sigma$ only for
$\sigma \in P^1(G)$, which denotes the positive definite functions of norm one
in $B(G)$.) For abelian $G$ with dual group $\hat{G}$, the Fourier and
Fourier--Stieltjes transforms, respectively, yield isometric isomorphism
$A(G) \cong L^1(\hat{G})$ and $B(G) \cong M(\hat{G})$, and conjugation with
the Plancherel transform yields that
\begin{equation} \label{planschi}
  {\cal H}_\mu \cong {\cal H}_{\hat{\mu}} \qquad (\mu \in M(G)).
\end{equation}
Despite the formal analogies with harmonic functions, the harmonic
functionals according to Chu and Lau display a strikingly different behavior:
for $\sigma \in P^1(G)$, the set
\[
  G_\sigma := \{ x \in G : \sigma(x) = 1 \}
\]
is a closed subgroup of $G$ such that ${\cal H}_\sigma = \lambda(G_\sigma)''$
(\cite[Proposition 3.2.10]{CL}); in particular, ${\cal H}_\sigma$ is
a von Neumann subalgebra of ${\cal B}(L^2(G))$.
\par
Finally, we require a construction from \cite{NRS}: there the first-named
author, Z.-J.\ Ruan, and N.\ Spronk
define complete isometry $\hat{\Theta} \!: \mathcal{M}_\mathrm{cb}(A(G)) \to
{\cal CB}({\cal B}(L^2(G)))$ with the following properties:
\begin{itemize}
\item $\hat{\Theta}$ is a unital, $w^\ast$-$w^\ast$-continuous algebra
homomorphism;
\item $\hat{\Theta}(\mathcal{M}_\mathrm{cb}(A(G)))$ consists precisely of those normal operators
in ${\cal CB}({\cal B}(L^2(G)))$ which are $L^\infty(G)$-bimodule homomorphisms and leave
$\operatorname{VN}(G)$ invariant;
\item for $\phi \in \mathcal{M}_\mathrm{cb}(A(G))$ and $T \in \operatorname{VN}(G)$, we have
$\hat{\Theta}(\phi)(T) = \phi \cdot T$, i.e., the action of $\mathcal{M}_\mathrm{cb}(A(G))$
on ${\cal B}(L^2(G))$ induced by $\hat{\Theta}$ extends the canonical one
of $\mathcal{M}_\mathrm{cb}(A(G))$ on $\operatorname{VN}(G)$.
\end{itemize}
Under many aspects, $\hat{\Theta}$ can be viewed as dual to Ghahramani's
representation of $M(G)$ on ${\cal B} (L^2(G))$ (for details, see Section \ref{dualpic}, and \cite{NRS}).
\par
For later use, we indicate how $\hat{\Theta}$ is defined. Given $\phi
\in \mathcal{M}_\mathrm{cb}(A(G))$, there are a Hilbert space ${\mathfrak H}$ and continuous
functions $\boldsymbol{\xi}, \boldsymbol{\eta} \!: G \to {\mathfrak H}$ such that
\[
  \phi(x^{-1}y) = \langle \boldsymbol{\xi}(y),
                  \boldsymbol{\eta}(x) \rangle \qquad (x,y \in G)
\]
(see \cite{Jol} for an accessible proof). Let $( e_i )_{i \in \mathbb I}$
be an orthonormal basis, and define
\[
  \phi_i(x) := \langle e_i, \boldsymbol{\eta}(x) \rangle
  \quad\text{and}\quad
  \psi_i(x) := \langle \boldsymbol{\xi}(x), e_i \rangle
  \qquad (i \in \mathbb{I}, \, x \in G).
\]
Then we have
\[
  \hat{\Theta}(\phi)(T) := \sum_{i \in \mathbb I} M_{\phi_i(x)} T M_{\psi_i(x)}.
\]
For more details, see \cite{NRS}.
\section{A dual characterization of adapted probability measures} \label{adapted}
Let $G$ be a locally compact group, and let $\mu \in M^1(G)$. It is fair to
say that almost nothing can be said about ${\cal H}_\mu$ unless
$\mu$ is adapted. In order to have a dual notion of adaptedness, i.e.,
in the context of harmonic functionals as in \cite{CL}, we first
characterize the adapted probability measures on a locally
compact \emph{abelian} group in terms of their Fourier--Stieltjes transforms.
\begin{proposition} \label{adapprop}
Let $G$ be a locally compact abelian group. Then the following are equivalent
for $\mu \in M^1(G)$:
\begin{items}
\item $\mu$ is adapted;
\item if $\gamma \in \hat{G}$ is such that $\gamma |_{{\operatorname{supp}} \, \mu} \equiv 1$,
then $\gamma = 1$;
\item $\{ \gamma \in \hat{G} : \hat{\mu}(\gamma) = 1 \} = \{ 1 \}$.
\end{items}
\end{proposition}
\begin{proof}
(i) $\Longleftrightarrow$ (ii) and (iii) $\Longrightarrow$ (i) are straightforward.
\par
(ii) $\Longrightarrow$ (iii): Let $\gamma \in \hat{G}$ be such that
\[
  \hat{\mu}(\gamma) = \int_G \overline{\gamma(x)} \, d\mu(x) = 1,
\]
so that
\[
  1 = \overline{\hat{\mu}(\gamma)} = \overline{\int_G
  \overline{\gamma(x)} \, d\mu(x)} =
  \int_G \gamma(x) \, d\mu(x)
\]
and thus
\[
  \int_G {\operatorname{Re}} \, \gamma(x) \,d \mu(x) = \frac{1}{2} \left(
  \int_G \overline{\gamma(x)} \, d\mu(x) + \int_G \gamma(x) \, d\mu(x)
  \right) = 1.
\]
Since $\gamma(G) \subset {\mathbb T}$, we have $({\operatorname{Re}} \, \gamma)(G) \subset [-1,1]$,
so that
\[
  0 = \int_G 1 \, d\mu(x) - \int_G {\operatorname{Re}} \, \gamma(x) \, d\mu(x) =
  \int_{{\operatorname{supp}} \, \mu} \underbrace{(1 - {\operatorname{Re}} \, \gamma(x))}_{\geq 0} \, d\mu(x).
\]
By the continuity of $\gamma$, this means that ${\operatorname{Re}} \, \gamma |_{{\operatorname{supp}} \, \mu} \equiv 1$,
thus $\gamma |_{{\operatorname{supp}} \, \mu} \equiv 1$, and therefore $\gamma = 1$ by (ii).
\end{proof}
\par
Using (\ref{planschi}) and \cite[Proposition 3.2.10]{CL}, we obtain a
dual approach to the Choquet--Deny theorem:
\begin{corollary}
Let $G$ be a locally compact abelian group, and let $\mu \in M^1(G)$ be
adapted. Then ${\cal H}_\mu \cong {\mathbb C}$ holds.
\end{corollary}
\par
In view of Proposition \ref{adapprop}, we
define for general locally compact groups:
\begin{definition} \label{adapdef}
Let $G$ be a locally compact group. Then we call $\sigma \in P^1(G)$
\emph{adapted} if $G_\sigma = \{ e \}$.
\end{definition}
\begin{remark}
In terms of Definition \ref{adapdef}, Proposition \ref{adapprop} can be
reformulated as follows: a probability measure on a locally compact abelian group
is adapted if and only if its Fourier--Stieltjes transform is adapted.
\end{remark}
\par
It is well known that there is an adapted probability measure on a locally
compact group $G$ if and only if $G$ is $\sigma$-compact. Since a locally
compact abelian group is $\sigma$-compact if and only if its dual is
first countable, it is immediate that there is an adapted positive definite
function on such a group if and only if it is first countable.
\par
The following proposition extends this to general locally compact groups:
\begin{proposition} \label{adapprop2}
The following are equivalent for a locally compact group:
\begin{items}
\item there is an adapted $\sigma \in P^1(G)$;
\item $G$ is first countable;
\item there is an adapted $\sigma \in A(G) \cap P^1(G)$.
\end{items}
\end{proposition}
\begin{proof}
(i) $\Longrightarrow$ (ii): Let $\sigma \in P^1(G)$ be adapted, fix a compact
neighborhood $U$ of $e$, and define,
for $n \in {\mathbb N}$,
\[
  U_n := \left\{ x \in U : | \sigma(x) - 1 | < \frac{1}{n} \right\}.
\]
From the continuity of $\sigma$, it is clear that $\{ U_n : n \in {\mathbb N} \}$
consists of neighborhoods of $e$. Let $V$ be a neighborhood
of $e$, and suppose without loss of generality that $V$ is open and contained
in $U$. Since $\sigma$ is continuous and adapted, and since $U \setminus V$
is compact, we have that
\[
  \epsilon_0 := \inf \{ | \sigma(x) - 1 | : x \in U \setminus V \} > 0.
\]
Choose $n_0 \in {\mathbb N}$ so large that $\frac{1}{n_0} \leq \epsilon_0$.
It follows that $U_{n_0} \subset V$. Consequently, $\{ U_n : n \in {\mathbb N} \}$
is a base of neighborhoods of $e$.
\par
Through translation, we see that every point of $G$ has a countable
base of neighborhoods.
\par
(ii) $\Longrightarrow$ (iii): Let $\{ U_n : n \in {\mathbb N} \}$ be a base
of neighborhoods of $e$, and suppose without loss of generality that
$U_{n+1} \subset U_n$ for $n \in {\mathbb N}$. For each $n \in {\mathbb N}$,
there is $\sigma_n \in A(G) \cap P^1(G)$ with ${\operatorname{supp}} \, \sigma_n \subset U_n$.
Define
\[
  \sigma := \sum_{n=1}^\infty \frac{1}{2^n} \sigma_n
\]
so that, clearly, $\sigma \in A(G) \cap P^1(G)$. Let $x \in G \setminus \{ e \}$.
Since $\{ U_n : n \in {\mathbb N} \}$ is a base of neighborhoods of $e$, there
is $n_0 \in {\mathbb N}$ such that $x \notin U_{n_0}$ and thus $x \notin U_n$
for $n \geq n_0$. It follows that
\[
  |\sigma(x)| \leq \sum_{n=1}^{n_0-1} \frac{1}{2^n} |\sigma_n(x)|
  \leq  \sum_{n=1}^{n_0-1} \frac{1}{2^n} < 1.
\]
This proves (iii).
\par
Finally, (iii) $\Longrightarrow$ (i) is trivial.
\end{proof}
\section{The support of an operator on $L^2(G)$} \label{support}
In \cite{Eym}, P.\ Eymard introduced the notion of support for elements
of a group von Neumann algebra: if $G$ is a locally compact group and $T
\in \operatorname{VN}(G)$, then the \emph{support} ${\operatorname{supp}} \, T$ of $T$ is defined to
consist of those $x \in G$ such that $\phi(x) = 0$ for all $\phi \in A(G)$
with $\phi \cdot T = 0$.
\par
Using $\hat{\Theta}$, this notion can naturally be extended to arbitrary
operators on $L^2(G)$:
\begin{definition} \label{suppdef}
Let $G$ be a locally compact group, and let $T \in {\cal B}(L^2(G))$.
Then the \emph{support} ${\operatorname{supp}} \, T$ of $T$ is defined to be
\[
  {\operatorname{supp}} \, T := \{ x \in G: \text{$\phi(x) = 0$ for all $\phi \in A(G)$
  with $\hat{\Theta}(\phi)(T) = 0$} \}.
\]
\end{definition}
\begin{remarks}
\item For operators in $\operatorname{VN}(G)$, this notion of support coincides with the
one from \cite{Eym}.
\item The support of an operator is obviously a closed subset of $G$.
\end{remarks}
\par
We first prove a few general assertions on the support of an operator.
\begin{proposition} \label{supportgeneral}
Let $G$ be a locally compact group. Then, for $\phi \in A(G)$ and $T \in {\cal B}(L^2(G))$, we have
\[
  {\operatorname{supp}} \, \hat{\Theta}(\phi)(T) \subseteq {\operatorname{supp}} \, \phi \cap {\operatorname{supp}} \, T.
\]
\end{proposition}
\begin{proof}
Fix $\phi \in A(G)$ and $T \in {\cal B}(L^2(G))$.
\par
Since $\hat{\Theta}$ is multiplicative and $A(G)$ is commutative, it is
straightforward that any $\psi \in A(G)$ with $\hat{\Theta}(\psi)(T) = 0$
satisfies
$\hat{\Theta}(\psi)(\hat{\Theta}(\phi)(T)) = 0$ as well. From Definition
\ref{supportgeneral}, it is then immediate that ${\operatorname{supp}} \, \hat{\Theta}(\phi)(T)
\subset {\operatorname{supp}} \, T$.
\par
To see that ${\operatorname{supp}} \, \hat{\Theta}(\phi)(T) \subset {\operatorname{supp}} \, \phi$ as well, let
$x \in G \setminus {\operatorname{supp}} \, \phi$, and assume towards a
contradiction that $x \in {\operatorname{supp}} \, \hat{\Theta}(\phi)(T)$.
Since $x \notin {\operatorname{supp}} \, \phi$, there is a neighborhood $V$ of $x$ such that
$\phi |_V \equiv 0$. Take $\psi \in A(G)$ such that ${\operatorname{supp}} \, \psi \subseteq V$
and $\psi (x) = 1$. Then $\psi \phi \equiv 0$ holds, so that
$\hat{\Theta}(\psi) (\hat{\Theta}(\phi)(T)) =
\hat{\Theta}(\psi \phi) (T) = 0$.
Since $x \in {\operatorname{supp}} \, \hat{\Theta}(\phi)(T)$, this implies that
$\psi (x) = 0$, which is a contradiction.
\end{proof}
\par
The following is an alternative description of the support
of an operator:
\begin{proposition} \label{suppideal}
Let $G$ be a locally compact group, and let $T \in {\cal B}(L^2(G))$.
Then
\begin{equation} \label{ideal}
  \{ \phi \in A(G) : \hat{\Theta}(\phi)(T) = 0 \}
\end{equation}
is an ideal of $A(G)$ whose hull equals ${\operatorname{supp}} \, T$.
\end{proposition}
\begin{proof}
Let the ideal (\ref{ideal}) be denoted by $I$. Then we have for $x \in G$ that
\begin{align*}
  x \notin {\operatorname{supp}} \, T & \iff \text{there is $\phi \in A(G)$ such that $\phi(x) \neq 0$ and
  $\hat{\Theta}(\phi)(T) = 0$} \\
  & \iff \text{there is $\phi \in I$ with $\phi(x) \neq 0$} \\
  & \iff x \notin {\operatorname{hull}}(I).
\end{align*}
This completes the proof.
\end{proof}
\par
Since $A(G)$ is Tauberian for any locally compact group $G$, the following
is clear:
\begin{corollary} \label{suppcor0}
Let $G$ be a locally compact group, and let $T \in {\cal B}(L^2(G))$ be
such that ${\operatorname{supp}} \, T = \varnothing$. Then $\hat{\Theta}(\phi)(T) = 0$ holds
for all $\phi \in A(G)$.
\end{corollary}
\par
Recall (\cite{HK}) that a locally compact group $G$ is said to have the
\emph{approximation property} if the constant function $1$ lies in the
$w^\ast$-closure of $A(G)$ in $\mathcal{M}_\mathrm{cb}(A(G))$. Clearly, every amenable, locally
compact group has the approximation property (by Leptin's theorem), but so
does also every weakly amenable group in the sense of \cite{dCH}, such as
${\mathbb F}_2$, the free group in two generators. Nevertheless, the approximation
property is weaker than weak amenability: by \cite[Corollary 1.17 and
Remark 3.10]{HK}, the group ${\mathbb Z}^2 \rtimes \operatorname{SL}(2,{\mathbb Z})$ has the
approximation property, but is not weakly amenable.
\begin{proposition} \label{supprop0}
Let $G$ be a locally compact group with the approximation property. Then
the following are equivalent for $T \in {\cal B}(L^2(G))$:
\begin{items}
\item ${\operatorname{supp}} \, T = \varnothing$;
\item $T = 0$.
\end{items}
\end{proposition}
\begin{proof}
Of course, only (i) $\Longrightarrow$ (ii) needs proof.
\par
Let $(e_\alpha)_\alpha$ be a net in $A(G)$ that converges to $1$ in the
$w^\ast$-topology of $\mathcal{M}_\mathrm{cb}(A(G))$. Since $\hat{\Theta}$ is unital and
$w^\ast$-$w^\ast$-continuous (\cite[Theorem 4.5]{NRS}), it follows that
${\mathrm{id}}_{{\cal B}(L^2(G))} = \text{$w^\ast$-}\lim_\alpha \hat{\Theta}(e_\alpha)$
and thus, by Corollary \ref{suppcor0},
\[
  T = \hat{\Theta}(1)(T) =
  \text{$w^\ast$-}\lim_\alpha \hat{\Theta}(e_\alpha)(T) = 0.
\]
This proves (ii).
\end{proof}
\begin{remark}
It is well possible that Proposition \ref{supprop0} is true for \emph{every}
locally compact group.
\end{remark}
\par
Let $G$ be a locally compact group, and let $F \subset G$ be closed. As
is customary, we write
\[
  I(F) := \{ \phi \in A(G) : \phi |_F \equiv 0 \}.
\]
We also define
\[
  {\cal B}_F(L^2(G)) := \{ T \in {\cal B}(L^2(G)) : {\operatorname{supp}} \, T \subset F \}.
\]
\begin{lemma} \label{supplem2}
Let $G$ be a locally compact group, and let $F \subset G$ be a set of
synthesis for $A(G)$. Then $\hat{\Theta}(\phi)(T) = 0$ holds for all
$\phi \in I(F)$ and $T \in {\cal B}_F(L^2(G))$.
\end{lemma}
\begin{proof}
Let $T \in {\cal B}_F(L^2(G))$, and set
\[
  I := \{ \phi \in A(G) : \hat{\Theta}(\phi)(T) = 0 \}.
\]
By Proposition \ref{suppideal}, we have
\[
  {\operatorname{hull}}(I) = {\operatorname{supp}} \, T \subset F.
\]
Since $F$ is a set of synthesis, this means that $I(F) \subset I$.
\end{proof}
\par
Since the multiplication of $\mathcal{M}_\mathrm{cb}(A(G))$ is separately $w^\ast$-continuous
(see \cite{FRS}, for instance), we obtain immediately (from the
$w^\ast$-$w^\ast$-continuity of $\hat{\Theta}$):
\begin{corollary} \label{suppcor1}
Let $G$ be a locally compact group with the approximation property,
let $F \subset G$ be a set of synthesis, and let $T \in {\cal B}_F(L^2(G))$.
Then $\hat{\Theta}(\phi)(T) =0$ for all $\phi \in \mathcal{M}_\mathrm{cb}(A(G))$ with
$\phi |_F \equiv 0$.
\end{corollary}
\begin{proof}
Let $\phi \in \mathcal{M}_\mathrm{cb}(A(G))$ be such that $\phi |_F \equiv 0$, and let
$( e_\alpha )_\alpha$ be a net in $A(G)$ converging to $1$ in the
$w^\ast$-topology of $\mathcal{M}_\mathrm{cb}(A(G))$. Then $( e_\alpha \phi )_\alpha$ is
a net in $I(F)$ that is $w^\ast$-convergent to $\phi$. Lemma \ref{supplem2}
and the $w^\ast$-$w^\ast$-continuity of $\hat{\Theta}$ then yield the claim.
\end{proof}
\par
Let $G$ be a locally compact group, let $F \subset G$ be a set of synthesis,
and let $T \in {\cal B}_F(L^2(G))$.
Then it is clear from Lemma \ref{supplem2} that $\hat{\Theta}(\phi)(T)$
for $\phi \in A(G)$ depends only on $\phi |_F$ (if $G$ has the approximation
property, this is even true for all $\phi \in \mathcal{M}_\mathrm{cb}(A(G))$ by
Corollary \ref{suppcor1}).
\begin{proposition} \label{supprop1}
Let $G$ be a locally compact group, let $H$ be a closed subgroup of $G$,
and let $T \in {\cal B}_H(L^2(G))$. Then:
\begin{items}
\item if there is $\sigma \in A(G)$ with $\sigma |_H = 1$ and
$\hat{\Theta}(\sigma)(T) = T$, then $\hat{\Theta}(\phi)(T) = \phi(e) T$
for all $\phi \in A(G)$ which are constant on $H$;
\item if $G$ has the approximation property, then
$\hat{\Theta}(\phi)(T) = \phi(e) T$ holds for all $\phi \in \mathcal{M}_\mathrm{cb}(A(G))$
which are constant on $H$.
\end{items}
\end{proposition}
\begin{proof}
First, recall that $H$, as a closed subgroup, is a set of synthesis for $G$.
(\cite[Theorem 3]{TT2}).
\par
Suppose that there is $\sigma \in A(G)$ with $\sigma |_H = 1$ and
$\hat{\Theta}(\sigma)(T) = T$. Let $\phi \in A(G)$ be constant on
$H$. Then $\phi(e) \sigma - \phi$ vanishes on $H$, so that, by Lemma
\ref{supplem2} we have,
\[
  0 = \hat{\Theta}(\phi(e) \sigma - \phi)(T) = \phi(e) \hat{\Theta}(\sigma)(T)
  - \hat{\Theta}(\phi)(T) = \phi(e) T - \hat{\Theta}(\phi)(T).
\]
This proves (i).
\par
For (ii), just note that, if $\phi \in \mathcal{M}_\mathrm{cb}(A(G))$ is constant on $H$,
then $\phi(e) - \phi$ vanishes on $H$. An application of Corollary
\ref{suppcor1} then yields (ii).
\end{proof}
\par
As we already noted, our notion of support coincides with the one from
\cite{Eym} for operators in $\operatorname{VN}(G)$. We now compute the support of
multiplication operators:
\begin{example}
Let $G$ be a locally compact group, and let $f \in L^\infty(G)$.
For any $\phi \in A(G)$, we have
\begin{equation} \label{Linfty}
  \hat{\Theta}(\phi)(M_f) = M_f \hat{\Theta}(\phi)(1) = \phi(e) M_f ,
\end{equation}
where the first equality is due to the fact that $\hat{\Theta}(\phi)$ is an $L^\infty(G)$-bimodule homomorphism.
If $x \in G \setminus \{ e \}$, we can find $\phi \in A(G)$ with $\phi(x)
\neq 0 = \phi(e)$, so that $\hat{\Theta}(\phi)(M_f) = 0$ by (\ref{Linfty}).
Hence, $x$ cannot lie in ${\operatorname{supp}} \, M_f$. Since $x \in G \setminus \{ e \}$
was arbitrary, this means that $M_f \in {\cal B}_e(L^2(G))$. Moreover,
it is also clear from (\ref{Linfty}) that $e \in {\operatorname{supp}} \, M_f$ whenever
$f \neq 0$, i.e., ${\operatorname{supp}} \, M_f = \{ e \}$.
\end{example}
\section{Harmonic operators: the dual picture} \label{dualpic}
Let $G$ be a locally compact group, and let $\Theta \!: M(G) \to
{\cal CB}({\cal B}(L^2(G)))$ be the completely isometric representation from
\cite{Gha} (see also \cite{Neu}, \cite{Neu1}, and---for abelian
$G$---\cite{Sto}) given by
\[
  \langle \Theta (\mu) T , \omega \rangle := \int_G \langle \rho(t)
   T \rho(t^{-1}) , \omega \rangle \, d\mu (t)
  \qquad (\mu \in M(G), \,  T \in {\cal B}(L^2(G)), \,
   \omega \in {\cal T}(L^2(G))) .
\]
For $\mu \in M^1(G)$, W.\ Jaworski and the first-named author define
$T \in {\cal B}(L^2(G))$ to be \emph{$\mu$-harmonic} if $\Theta(\mu)(T) = T$
(\cite{JN}). Since
\[
  \Theta(\mu)M_\phi = M_{\mu \ast \phi} \qquad (\phi \in L^\infty(G)),
\]
this generalizes the notion of a $\mu$-harmonic function. More precisely,
denoting the space of $\mu$-harmonic functions and operators
by ${\cal H}_\mu$ and $\tilde{\cal H}_\mu$, respectively,
this shows that ${\cal H}_\mu \subseteq \tilde{\cal H}_\mu$. Moreover,
$\tilde{\cal H}_\mu$ carries a natural multiplication extending the one of
${\cal H}_\mu$---and turning $\tilde{\cal H}_\mu$ into a non-commutative von
Neumann algebra (\cite{JN}): given $S,T \in \tilde{\cal H}_\mu$, their product
in $\tilde{\cal H}_\mu$ is explicitly given by
\[
  \text{$w^\ast$-}\lim_n \int_G \rho(x) (ST) \rho (x^{-1}) \,\mu^{\ast n} (x).
\]
One of the main results obtained in \cite{JN}---affirmatively answering a
question by M.\ Izumi (\cite{Izu})---consists of a precise structural
description of $\tilde{\cal H}_\mu$: provided
that $G$ is second countable, $\tilde{\cal H}_\mu$ is exactly the crossed
product of ${\cal H}_\mu$ with $G$ acting by left translation
(\cite[Theorem 6.3]{JN}). In particular, it shows that the algebra $\operatorname{VN}(G)$,
as a subalgebra of $\tilde{\cal H}_\mu$, plays the same r\^ole as the scalars
for the classical algebra of harmonic functions ${\cal H}_\mu$.
\par
Using $\hat{\Theta} \!: \mathcal{M}_\mathrm{cb}(A(G)) \to {\cal CB}({\cal B}(L^2(G)))$ from
\cite{NRS}, we can extend the notion of a $\sigma$-harmonic functional
on $A(G)$ from \cite{CL} to that of a $\sigma$-harmonic operator in
a way analogous to the passage from ${\cal H}_\mu$ to $\tilde{\cal H}_\mu$ via $\Theta$:
\begin{definition}
Let $G$ be a locally compact group, and let $\sigma \in \mathcal{M}_\mathrm{cb}(A(G))$.
Then an operator $T \in {\cal B}(L^2(G))$ is called
\emph{$\sigma$-harmonic} if $\hat{\Theta}(\sigma)(T) = T$. We denote
the collection of all $\sigma$-harmonic operators on $L^2(G)$ by
$\tilde{\cal H}_\sigma$.
\end{definition}
\begin{remarks}
\item Obviously, $\tilde{\cal H}_\sigma$ is a $w^\ast$-closed subspace
of ${\cal B}(L^2(G))$.
\item Trivially, $\tilde{\cal H}_\sigma$ contains ${\cal H}_\sigma$, and since,
\[
  \hat{\Theta}(\phi)(M_f) = M_f \, \hat{\Theta}(\phi)(1) = M_f \qquad (f \in L^\infty(G)),
\]
it contains $L^\infty(G)$ as well.
\end{remarks}
\par
As proven in \cite{CL}, ${\cal H}_\sigma$ is a von Neumann subalgebra
of $\operatorname{VN}(G)$ for $\sigma \in P^1(G)$, which stands in marked contrast to
${\cal H}_\mu$ with $\mu \in M^1(G)$. In the remainder of this section,
we shall see that a similar statement is true for $\tilde{\cal H}_\sigma$:
under some, fairly mild, additional hypotheses, it is a von Neumann
subalgebra of ${\cal B}(L^2(G))$; in fact, we shall prove that
\begin{equation} \label{generated}
  \tilde{\cal H}_\sigma = ({\cal H}_\sigma \cup L^\infty(G))''.
\end{equation}
\begin{remark}
The description (\ref{generated}) may be
viewed as a dual version of the central structural result
\cite[Theorem 6.3]{JN}; in our setting, it is the algebra $L^\infty(G)$ which
plays the r\^ole of the scalars. Noting that $L^\infty(G)$ and $\operatorname{VN}(G)$ are
Kac algebras dual to each other (see \cite{ES}), we are
inclined to believe that one may find one single structure result that
unifies those descriptions of $\tilde{\cal H}_\sigma$ and $\tilde{\cal H}_\mu$
in the general framework of Kac algebras (or, even more generally,
locally compact quantum groups).
\end{remark}
\par
We proceed through a series of lemmas and propositions.
\begin{proposition} \label{prop1}
Let $G$ be a locally compact group, and let $\sigma \in P^1(G)$. Then
\[
  ({\cal H}_\sigma \cup L^\infty(G))'' \subset \tilde{\cal H}_\sigma
\]
holds.
\end{proposition}
\begin{proof}
By von Neumann's double commutant theorem, $({\cal H}_\sigma \cup L^\infty(G))''$ is the von
Neumann subalgebra of ${\cal B}(L^2(G))$ generated by ${\cal H}_\sigma$
and $L^\infty(G)$. Since $\tilde{\cal H}_\sigma$ is a $w^\ast$-closed
subspace of ${\cal B}(L^2(G))$, it is sufficient to show that finite
products of operators from ${\cal H}_\sigma \cup L^\infty(G)$ belong to
$\tilde{\cal H}_\sigma$. Moreover, since ${\cal H}_\sigma \cong \operatorname{VN}(G_\sigma)$, it is enough to consider products of the form
\begin{equation} \label{product}
  \prod_{j=1}^n \lambda(x_j) M_{f_j},
\end{equation}
where $x_1, \ldots, x_n \in G_\sigma$ and $f_1, \ldots, f_n \in L^\infty(G)$. Since
\[
  \lambda(x) M_f = M_{\lambda(x) f} \lambda(x) \qquad (x \in G, \, f \in L^\infty(G)),
\]
a simple induction on the number of factors in (\ref{product}) shows that (\ref{product}) is, in fact, of the form $M_f \lambda(x)$ with $f \in L^\infty(G)$ and $x \in G_\sigma$. Since $\hat{\Theta}(\sigma)$ is an $L^\infty(G)$-bimodule homomorphism and
since $\hat{\Theta}(\sigma)$ fixes $\lambda(x)$ if $x \in G_\sigma$, we have
\[
  \hat{\Theta}(\sigma)(M_f \lambda(x)) =
  M_f \hat{\Theta}(\sigma)(\lambda(x)) =  M_f \lambda(x)
  \qquad (f \in L^\infty(G), \, x \in G_\sigma).
\]
i.e., $M_f \lambda(x) \in \tilde{\cal H}_\sigma$.
\par
In view of the foregoing remarks, this completes the proof.
\end{proof}
\par
While Proposition \ref{prop1} provides an estimate for $\tilde{\cal H}_\sigma$ ``from below'', we now give one ``from above''.
\par
Extending our previous notation, we set $G_\sigma := \{ x \in G : \sigma(x) = 1 \}$ for \emph{any} $\sigma \in \mathcal{M}_\mathrm{cb}(A(G))$. Note that
$G_\sigma$ need not be a subgroup of $G$ unless $\sigma \in P^1(G)$, but that it is still a closed subset of $G$, so that the conclusion of the following proposition still makes sense.
\begin{proposition} \label{prop2}
Let $G$ be a locally compact group, and let $\sigma \in \mathcal{M}_\mathrm{cb}(A(G))$. Then
we have $\tilde{\cal H}_\sigma \subset {\cal B}_{G_\sigma}(L^2(G))$.
\end{proposition}
\begin{proof}
Let $T \in \tilde{\cal H}_\sigma$, and let $x \in G \setminus G_\sigma$, i.e.,
$\sigma(x) \neq 1$. Let $\phi \in A(G)$ be such that $\phi(x) \neq 0$, so that
$\psi(x) := \phi(x)\sigma(x) - \phi(x) \neq 0$. On the other hand, we have
\[
  \hat{\Theta}(\psi)(T) =
  \hat{\Theta}(\phi\sigma)(T) - \hat{\Theta}(\phi)(T) =
  \hat{\Theta}(\phi) (\hat{\Theta}(\sigma)(T)) - \hat{\Theta}(\phi)(T)
  = \hat{\Theta}(\phi)(T) - \hat{\Theta}(\phi)(T) =0.
\]
This means that $x \notin {\operatorname{supp}} \, T$.
\end{proof}
\par
Let $G$ be a locally compact group, and let $H$ be a closed subgroup.
We define
\[
  L^\infty(G:H) := \{ \phi \in L^\infty(G) :
                   \text{$\lambda(x)\phi = \phi$ for all $x \in H$} \}.
\]
It is obvious that $L^\infty(G:H)$ is $w^\ast$-closed in $L^\infty(G)$.
Slightly deviating from \cite{For}, we set
\[
  B(G:H) := \{ \phi \in B(G) :
                   \text{$\lambda(x)\phi = \phi$ for all $x \in H$} \}.
\]
(Note that in \cite{For}, right instead of left translates are considered.)
We can canonically embed $B(G:H)$ into $L^\infty(G:H)$. Also, we set
\[
  P^1_H(G) := \{ \sigma \in P^1(G) : \sigma |_H \equiv 1 \}.
\]
As is observed in \cite{KL}, the functions in
$P^1_H(G)$ are constant on both left and right cosets of $H$ and thus
are contained in $B(G:H)$.
Following \cite{KL}, we say that $G$ has the \emph{$H$-separation property}
if, for each $x \in G \setminus H$, there is $\sigma \in P^1_H(G)$ such that
$\sigma(x) \neq 1$. For instance, whenever $H$ is open, compact, or
neutral---this includes all normal subgroups as well as all closed subgroups
of $[\operatorname{SIN}]$-groups---, $G$ has the $H$-separation property (see \cite{KL}).
\begin{lemma} \label{wdense}
Let $G$ be a locally compact group, and let $H$ be a closed subgroup of $G$
such that $G$ has the $H$-separation property.
Then $B(G:H)$ is $w^\ast$-dense in $L^\infty(G:H)$.
\end{lemma}
\begin{proof}
Let ${\mathfrak M}$ denote the $w^\ast$-closure of the linear span of all
right translates of all functions in $P^1_H(G)$ in $L^\infty(G)$.
Then ${\mathfrak M}$ is a von Neumann subalgebra of $L^\infty(G)$ and invariant
under right translation. Set
\[
  L := \{ x \in G : \text{$\lambda(x)\phi = \phi$ for all $\phi \in {\mathfrak M}$} \}.
\]
Then $L$ is a closed subgroup of $G$ containing $H$, and from
\cite[Theorem 2]{TT} (see also \cite[Lemma 3.2]{Lau}), we conclude that
${\mathfrak M} = L^\infty(G:L)$.
\par
Assume that there is $x \in L \setminus H$. Since $G$ has the $H$-separation
property, there is $\sigma \in P^1_H(G) \subset L^\infty(G:L)$ such that
$\sigma(x) \neq 1$ and thus $\lambda(x^{-1}) \sigma \neq \sigma$. This is
a contradiction, so that $H = L$.
\end{proof}
\par
For any locally compact group $G$ and $\sigma \in P^1(G)$, it is clear by
definition that $G$ has the $G_\sigma$-separation property. Hence, we obtain:
\begin{corollary} \label{wdensecor}
Let $G$ be a locally compact group, and let $\sigma \in P^1(G)$.
Then $B(G:G_\sigma)$ is $w^\ast$-dense in $L^\infty(G:G_\sigma)$.
\end{corollary}
\begin{lemma} \label{invlem}
Let $G$ be a locally compact group, let $H$ be a closed subgroup, and let
$\phi \in B(G:H)$. Then there are a unitary representation $\pi$ of $G$
on a Hilbert space ${\mathfrak H}$ as well as $\xi, \eta \in {\mathfrak H}$ such that
\[
  \phi(x) = \langle \pi(x) \xi, \eta \rangle \qquad (x \in G)
\]
and $\pi(y)\eta = \eta$ for all $y \in H$.
\end{lemma}
\begin{proof}
By definition of $B(G)$, there are a unitary representation $\pi$ of
$G$ on a Hilbert space ${\mathfrak H}$ as well as $\xi, \eta_0 \in {\mathfrak H}$
such that
\[
  \phi_{\xi,\eta_0}(x) := \langle \pi(x) \xi, \eta_0 \rangle
  = \phi(x) \qquad (x \in G).
\]
Without loss of generality, suppose that $\| \xi \| = 1$. Set
\[
  C := \{ \eta \in {\mathfrak H} : \text{$\| \eta \| \leq \| \eta_0 \|$
  and $\phi_{\xi,\eta} = \phi$} \}.
\]
Then $C$ is a non-empty, convex, weakly compact subset of ${\mathfrak H}$.
Let $\eta \in C$ and $y \in H$. Then we have
\[
  \phi_{\xi,\pi(y)\eta}(x) = \langle \pi(x) \xi, \pi(y)\eta \rangle \\
  = \langle \pi(y^{-1} x) \xi, \eta \rangle = (\lambda(y)\phi)(x)
  = \phi(x) \qquad (x \in G),
\]
so that $\pi(y)\eta \in C$ again. From the Ryll-Nardzewski
fixed point theorem (\cite[(9.6) Theorem]{GD}), we
conclude that there is $\eta \in C$ with $\pi(y)\eta = \eta$ for $y \in H$.
\end{proof}
\begin{proposition} \label{supprop}
Let $G$ be a locally compact group, let $H$ be a closed subgroup of
$G$, let $T \in {\cal B}_H(L^2(G))$, and suppose that one of the following
holds:
\begin{alphitems}
\item there is $\sigma \in A(G)$ with $\sigma |_H = 1$ and
$\hat{\Theta}(\sigma)(T) = T$;
\item $G$ has the approximation property and the $H$-separation property.
\end{alphitems}
Then $T$ lies in $(\operatorname{VN}(H) \cup L^\infty(G))''$.
\end{proposition}
\par
For the proof, we first establish some conventions for (possibly infinite) matrices. The extensive use of matrix calculations may seem like an unnecessary complication at the first glance, but appears to be unavoidable in view of how $\hat{\Theta} \!: \mathcal{M}_\mathrm{cb}(A(G)) \to {\cal CB}({\cal B}(L^2(G))$ is defined in \cite{NRS}, namely via the extended Haagerup tensor product of operator spaces, an object that itself is defined in terms of arbitrarily large matrices.
\par
Given a linear space $E$ and index sets $\mathbb I$ and $\mathbb J$,
we denote by $M_{\mathbb{I} \times \mathbb{J}}(E)$ the matrices $[ x_{i,j}
]_{i \in \mathbb{I} \atop j \in \mathbb J}$ with $x_{i,j} \in E$ for
$(i,j) \in \mathbb{I} \times \mathbb{J}$. If $\mathbb{I} = \mathbb J$, we
write $M_{\mathbb J}(E)$, and if $E = {\mathbb C}$, we simply use the symbols
$M_{\mathbb{I} \times \mathbb{J}}$ and $M_{\mathbb J}$ instead of
$M_{\mathbb{I} \times \mathbb{J}}({\mathbb C})$ and $M_{\mathbb J}({\mathbb C})$,
respectively. For $x \in E$ and $[ \alpha_{i,j}
]_{i \in \mathbb{I} \atop j \in \mathbb J}$, we set
$[ \alpha_{i,j} ]_{i \in \mathbb{I} \atop j \in \mathbb J} \otimes x =
[ \alpha_{i,j} x ]_{i \in \mathbb{I} \atop j \in \mathbb J}$. We convene
to interpret families $[ x_j ]_{j \in \mathbb J}$ as row vectors, and write
$[ x_j ]_{j \in \mathbb J}^t$ for the corresponding column vector.
Finally, we denote by $1_\mathbb{J}$ the matrix $[ \delta_{j,k} ]_{j,k \in
\mathbb J}$, and set $0_{\mathbb{I} \times \mathbb{J}} =
[ \alpha_{i,j}]_{i \in \mathbb{I} \atop j \in \mathbb J}$ with $\alpha_{i,j}
= 0$ for $(i,j) \in \mathbb{I} \times \mathbb{J}$.
\begin{proof}
If (a) holds, $H$ is necessarily compact, so that in both case (a) and
case (b), $G$ has the $H$-separation property. By Lemma
\ref{wdense}, $B(G:H)$ is therefore $w^\ast$-dense in $L^\infty(G:H)$.
We will show that $M_\phi T = T M_\phi$ for
all $\phi \in B(G:H)$, so that $T$ lies in $L^\infty(G:H)'$. Since,
as is easily checked,
\[
  (\operatorname{VN}(H) \cup L^\infty(G))' = L^\infty(G:H),
\]
this will prove the proposition.
\par
We adapt part of the proof of \cite[Theorem 5.1]{NRS} to our situation.
\par
Let $\phi \in B(G:H)$. By Lemma \ref{invlem}, there thus are a unitary
representation $\pi$ of $G$ on some Hilbert space ${\mathfrak H}$ as well as
$\xi, \eta \in {\mathfrak H}$ such that
\[
  \phi(x) = \langle \pi(x) \xi, \eta \rangle \quad (x \in G)
\]
and $\pi(y) \eta = \eta$ for $y \in H$. Let
\[
  \mathfrak{K} := \{ \zeta \in {\mathfrak H} : \text{$\pi(y) \zeta = \zeta$
  for $y \in H$} \},
\]
so that, in particular, $\eta \in \mathfrak K$.
Let $( e_i )_{i \in \mathbb I}$ be an orthonormal basis for $\mathfrak K$,
and extend it to an orthonormal basis $( e_i )_{i \in \mathbb J}$ for ${\mathfrak H}$ (so that, in particular, $\mathbb{I} \subset \mathbb{J}$). For $(i,k) \in \mathbb{J} \times \mathbb J$ set
\[
  \phi_{i,k}(x) := \langle \pi(x) e_k, e_i \rangle
  \quad\text{and}\quad
  \check{\phi}_{i,k}(x) := \phi_{i,k}(x^{-1})
  \qquad (x \in G)
\]
Since $e_i \in \mathfrak K$, it follows that $\phi_{i,k} \in B(G:H)$ for all $(i,k) \in \mathbb{I} \times \mathbb J$. By the definition
of $\hat{\Theta}$, we have
\begin{equation} \label{fart1}
  \hat{\Theta}(\phi_{i,k})(T) =
  \sum_{j \in \mathbb J} M_{\phi_{i,j}} T M_{\check{\phi}_{j,k}}
  \qquad ((i,k) \in \mathbb{I} \times \mathbb{J});
\end{equation}
By Proposition \ref{supprop1}(i) or (ii)---depending on whether (a) or (b) is satisfied---, the left hand side of (\ref{fart1})
equals $\phi_{i,k}(e)T$ for all $(i,k) \in \mathbb{I} \times \mathbb{J}$, so that (\ref{fart1}) becomes
\begin{equation} \label{fart+}
  \phi_{i,k}(e)T = \sum_{j \in \mathbb J} M_{\phi_{i,j}} T M_{\check{\phi}_{j,k}}
  \qquad ((i,k) \in \mathbb{I} \times \mathbb{J}).
\end{equation}
Interpreting (\ref{fart+}) as a matrix identity---with matrices indexed
over $\mathbb{J} \times \mathbb{J}$---, we obtain
\begin{equation} \label{fart2}
  \phi_{i,k}(e)T  = [ \phi_{i,j}]_{j \in \mathbb J} \,
  (1_\mathbb{J} \otimes T) \,   \left[ \check{\phi}_{j,k} \right]_{j \in \mathbb J}
  \qquad ((i,k) \in \mathbb{I} \times \mathbb{J}).
\end{equation}
We view $\Phi := [ \phi_{i,k} ]_{i,k \in \mathbb J}$ as an element of
$M_\mathbb{J}({\cal C}_b(G))$ and set $\check{\Phi} :=
\left[ \check{\phi}_{i,k} \right]_{i,k \in \mathbb J}$. Then
$\check{\Phi}$ also
lies in $M_\mathbb{J}({\cal C}_b(G))$ and satisfies
\begin{equation} \label{unitary}
  \Phi \, \check{\Phi} = 1_\mathbb{J} \otimes 1 = \check{\Phi} \, \Phi.
\end{equation}
Furthermore, set $\Psi :=
[ \phi_{i,k} ]_{i \in \mathbb{I} \atop k \in \mathbb J} \in
M_{\mathbb{I} \times \mathbb J}({\cal C}_b(G))$.
Since $\phi_{i,k}(e) = \delta_{i,k}$ for $(i,k)\in \mathbb{I}
\times \mathbb J$, we obtain
from (\ref{fart2}) that
\[
  \begin{split}
  \left(1_\mathbb{I} \oplus
  0_{\mathbb{I} \times (\mathbb{J} \setminus \mathbb{I})} \right)
  \otimes T &
  = [ \phi_{i,k}(e)T ]_{i \in \mathbb{I} \atop k \in \mathbb J} \\
  & = \left[ [ \phi_{i,j}]_{j \in \mathbb J} \,
  (1_\mathbb{J} \otimes T) \,
  \left[ \check{\phi}_{j,k} \right]_{j \in \mathbb J}
  \right]_{i \in \mathbb{I} \atop k \in \mathbb J} \\
  & = \Psi \, ( 1_\mathbb{J} \otimes T) \, \check{\Phi},
  \end{split}
\]
and thus, by (\ref{unitary})
\begin{equation} \label{commutes}
  \Psi \, ( 1_\mathbb{J} \otimes T) =
  \left( \left(1_\mathbb{I} \oplus
  0_{\mathbb{I} \times (\mathbb{J} \setminus \mathbb{I})}\right)
  \otimes T \right) \, \Phi.
\end{equation}
Let $[ \alpha_j ]_{j \in \mathbb J}$ and $[ \beta_i ]_{i \in \mathbb I}$ be
in $\ell^2(\mathbb{J})$ and $\ell^2(\mathbb{I})$, respectively, such that
$\xi = \sum_{j \in \mathbb J} \alpha_j e_j$
and $\eta = \sum_{i \in \mathbb I} \beta_i e_i$; it follows that
\[
  \phi(x) = \langle \pi(x)\xi, \eta \rangle =
  \sum_{j \in \mathbb{J} \atop i \in \mathbb I}
  \alpha_j \langle \pi(x)e_j,e_i \rangle \bar{\beta}_i=
  \sum_{j \in \mathbb{J} \atop i \in \mathbb I}
  \alpha_j \, \phi_{i,j}(x) \, \bar{\beta}_i
  \qquad (x \in G)
\]
or, in matrix notation,
\[
  \phi = \left[ \bar{\beta}_i \right]^t_{i \in \mathbb I}
  \, \Psi \, [ \alpha_j ]_{j \in \mathbb J}.
\]
Hence, we obtain eventually:
\[
  \begin{split}
  M_\phi T & =  \left[ \bar{\beta}_i \right]_{i \in \mathbb I}^t \, \Psi \,
  (1_\mathbb{J} \otimes T) \, [ \alpha_j ]_{j \in \mathbb J} \\
  & = \left[ \bar{\beta}_i \right]_{i \in \mathbb I}^t \,
  \left( \left(1_\mathbb{I} \oplus
  0_{\mathbb{I} \times (\mathbb{J} \setminus \mathbb{I})} \right)
  \otimes T \right) \, \Phi
  \,  [ \alpha_j ]_{j \in \mathbb J} \\
  & = TM_\phi.
  \end{split}
\]
This proves the claim.
\end{proof}
\par
We can now prove the main result of this section:
\begin{theorem}
Let $G$ be a locally compact group, and let $\sigma \in P^1(G)$.
Then the following hold:
\begin{items}
\item if $\sigma \in A(G)$, then $\tilde{\cal H}_\sigma =
({\cal H}_\sigma \cup L^\infty(G))''$;
\item if $G$ has the approximation property, then
\[
  \tilde{\cal H}_\sigma = ({\cal H}_\sigma \cup L^\infty(G))''
  = {\cal B}_{G_\sigma}(L^2(G)).
\]
\end{items}
In either case, $\tilde{\cal H}_\sigma =  ({\cal H}_\sigma \cup L^\infty(G))''$
is a von Neumann subalgebra of ${\cal B}(L^2(G))$.
\end{theorem}
\begin{proof}
By Propositions \ref{prop1} and \ref{prop2}
\[
  ({\cal H}_\sigma \cup L^\infty(G))'' \subset \tilde{\cal H}_\sigma
  \subset {\cal B}_{G_\sigma}(L^2(G))
\]
holds without any additional hypotheses.
\par
In case (i), we conclude from Proposition \ref{supprop} (with condition
(a)) that $\tilde{\cal H}_\sigma \subset
({\cal H}_\sigma \cup L^\infty(G))''$.
For case (ii), recall that, as we remarked before Corollary
\ref{wdensecor}, $G$ has the $G_\sigma$-separation property.
Hence, by Proposition \ref{supprop}  (with condition (b)), we even
have ${\cal B}_{G_\sigma}(L^2(G))
\subset ({\cal H}_\sigma \cup L^\infty(G))''$. This proves the theorem.
\end{proof}
\begin{remark}
We believe, but have been unable to prove, that
\[
  \tilde{\cal H}_\sigma = ({\cal H}_\sigma \cup L^\infty(G))''
  = {\cal B}_{G_\sigma}(L^2(G))
\]
holds without any additional hypotheses on $\sigma$ or $G$.
\end{remark}
\par
For adapted $\sigma$, we obtain as a special case:
\begin{corollary}
Let $G$ be a locally compact group, and let $\sigma \in P^1(G)$ be
adapted. Then the following hold:
\begin{items}
\item if $\sigma \in A(G)$, then $\tilde{\cal H}_\sigma =
L^\infty(G)$;
\item if $G$ has the approximation property, then
\[
  \tilde{\cal H}_\sigma = L^\infty(G) = {\cal B}_e(L^2(G)).
\]
\end{items}
In either case, $\tilde{\cal H}_\sigma = L^\infty(G)$ is a von
Neumann subalgebra of ${\cal B}(L^2(G))$.
\end{corollary}
\section{Harmonic operators through ideals in the predual} \label{ideals}
An alternative, very fruitful approach to the classical space ${\cal H}_\mu$ of
$\mu$-harmonic functions (where $\mu$ is a probability measure) has been carried out
by G.\ A.\ Willis (\cite{Wil}). There, the study of ${\cal H}_\mu$ is
transferred to its pre-annihilator
\[
  J_\mu := \overline{\{ f - f * \mu : f \in L^1(G) \}}^{L^1(G)}
\]
in $L^1(G)$, which (obviously) forms a left ideal in the group algebra. In
order to investigate the non-commutative analogue
$\tilde{\cal H}_\mu$ of ${\cal H}_\mu$ in a
similar fashion, W.\ Jaworski and the first-named author use the ``quantized''
convolution product in the space ${\cal T}(L^2(G)) = {\cal B}(L^2(G))_\ast$
of trace class operators, as introduced and studied in \cite{Neu} and
\cite{Neu2} (see also \cite{Pir} for further results on this product). Indeed,
they show that the pre-annihilator
\[
  \tilde{J}_\mu := \overline{\{ \omega - \Theta(\mu)_\ast (\omega) :
  \omega \in {\cal T}(L^2(G)) \}}^{{\cal T}(L^2(G))}
\]
of $\tilde{\cal H}_\mu$ in ${\cal T}(L^2(G))$ forms a left ideal with respect
to this convolution (\cite[Proposition 3.3]{JN}).
\par
We shall first indicate how to equip ${\cal T}(L^2(G))$ with a product
that turns it into a completely contractive Banach algebra, and may be thought
of as a ``non-commutative pointwise product'', in other words, a Fourier
algebra type product, instead of a convolution type product, as described
above. Just as the latter, the multiplication we shall consider here is also
very different from the ordinary composition of operators. From a
Hopf--von Neumann algebraic point of view, both the convolution type and the
Fourier algebra type product on the space ${\cal T}(L^2(G))$ are constructed
in exactly the same way: the first one is derived from a co-multiplication
on ${\cal B}(L^2(G))$ that canonically extends the one of $L^\infty(G)$
whereas the second one is based on a canonical extension of the
co-multiplication of $\operatorname{VN}(G)$. In this sense,
both products on ${\cal T}(L^2(G))$ are ``dual'' to each other.
\par
Let $G$ be a locally compact group. We define a unitary operator
$W \in {\cal B}(L^2(G \times G))$ by letting
\[
  (W \xi)(x,y) := \xi(x,xy) \qquad (\xi \in L^2(G), \, x,y \in G).
\]
Identifying $L^2(G \times G)$ with $L^2(G) \tilde{\otimes}_2 L^2(G)$ (Hilbert space
tensor product), we denote the flip map on $L^2(G \times G)$ by $\sigma$.
Then $\hat{W} := \sigma W^\ast \sigma$ is a again unitary, and
\[
  \hat{\Gamma} \!: {\cal B}(L^2(G)) \to {\cal B}(L^2(G \times G)),
  \quad T \mapsto \hat{W}(1 \otimes T) \hat{W}^\ast
\]
is a \emph{co-multiplication}, i.e., a normal, unital, injective
$^\ast$-homomorphism satisfying
\[
  (\hat{\Gamma} \otimes {\mathrm{id}}) \circ \hat{\Gamma} =
  ({\mathrm{id}} \otimes \hat{\Gamma}) \circ \hat{\Gamma}.
\]
The pre-adjoint
\[
  \hat{\Gamma} \!: {\cal T}(L^2(G)) \hat{\otimes} {\cal T}(L^2(G)) \to
  {\cal T}(L^2(G)),
\]
where $\hat{\otimes}$ denotes the projective tensor product of operator spaces,
is a complete contraction and turns ${\cal T}(L^2(G))$ into a
completely contractive
Banach algebra. In order to tell this product on ${\cal T}(L^2(G))$
apart from the usual composition of operators, we denote it by $\bullet$,
so that
\[
  \langle \omega \bullet \rho, T \rangle := \langle \omega \otimes \rho,
  \hat{\Gamma}(T) \rangle \qquad (\omega, \rho \in {\cal T}(L^2(G)), \,
  T \in {\cal B}(L^2(G))).
\]
Furthermore, the pre-adjoint of the inclusion $\operatorname{VN}(G) \subset {\cal B}(L^2(G))$
is an algebra homomorphism $\pi \!: {\cal T}(L^2(G)) \to A(G)$
(and necessarily a complete quotient map), and ${\cal B}(L^2(G))$ becomes a completely
contractive ${\cal T}(L^2(G))$-bi\-mo\-dule through
\begin{multline*}
  \omega \cdot T := ({\mathrm{id}} \otimes \omega)(\hat{\Gamma}(T))
  \quad\text{and}\quad
  T \cdot \omega := (\omega \otimes {\mathrm{id}})(\hat{\Gamma}(T)) \\
  \qquad (\omega \in {\cal T}(L^2(G)), \, T \in {\cal B}(L^2(G))).
\end{multline*}
(These module actions have been studied in \cite{NRS} and \cite{PR}.)
\par
Since $\hat{W} \in \operatorname{VN}(G) \bar{\otimes} L^\infty(G)$, we have
\[
  \omega \cdot T = ({\mathrm{id}} \otimes \omega)(\hat{W}(1 \otimes T)\hat{W}^\ast) \in \operatorname{VN}(G)
  \qquad (\omega \in {\cal T}(L^2(G)), \, T \in {\cal B}(L^2(G))).
\]
Hence, the module action induces a complete contraction $\hat{\mathcal{S}} \!: {\cal T}(L^2(G))
\hat{\otimes} {\cal B}(L^2(G)) \to \operatorname{VN}(G)$. In \cite{PR},
$\mathrm{UCQ}(G)$ is defined to be the range of $\hat{\mathcal{S}}$ in
$\operatorname{VN}(G)$, equipped with the quotient operator space structure.
The dual space $\mathrm{UCQ}(G)^\ast$ then naturally
becomes a completely contractive Banach algebra, which
contains $\mathcal{M}_\mathrm{cb} A(G)$ as a closed subalgebra, and the adjoint
$\hat{\mathcal{S}}^\ast$ extends
the representation $\hat{\Theta}$ from $\mathcal{M}_\mathrm{cb}(A(G))$ to $\mathrm{UCQ}(G)^\ast$.
More precisely, if, for $n \in \mathrm{UCQ}(G)^\ast$ and
$T \in {\cal B}(L^2(G))$, one defines $n \diamond T \in {\cal B}(L^2(G))$
through
\[
  \langle n \diamond T , \omega \rangle :=
  \langle n , \omega \cdot T \rangle \qquad
  (\omega \in {\cal T}(L^2(G)))
\]
then, for each $n \in \mathrm{UCQ}(G)^\ast$, the map
\[
  {\cal B}(L^2(G)) \to {\cal B}(L^2(G)), \quad T \mapsto n \diamond T
\]
is a completely bounded operator on ${\cal B}(L^2(G))$, which we denote by
$\tilde{\hat{\Theta}}(n)$. It is easy to check that $\tilde{\hat{\Theta}} =
\hat{\mathcal{S}}^\ast$. For more information, see \cite{PR}
(and \cite{NRS} for the amenable case).
\par
In the sequel, we shall use the fact that $\tilde{\hat{\Theta}}$ and
$\hat{\Theta}$ coincide on $A(G)$, which follows from \cite{PR}
(or \cite{NRS} if $G$ is amenable). For the reader's convenience, we include
a different (and short) proof:
\begin{lemma} \label{thetatheta}
Let $G$ be a locally compact group. Then we have
\[
  \tilde{\hat{\Theta}} (\phi) = \hat{\Theta} (\phi) \qquad (\phi \in A(G)).
\]
\end{lemma}
\begin{proof}
Let $\phi \in A(G)$, and note that both $\tilde{\hat{\Theta}} (\phi)$ and
$\hat{\Theta} (\phi)$ are normal, which follows easily from the definition
(see also \cite[Theorem 4.3]{NRS} for $\hat{\Theta} (\phi)$).
\par
Both $\tilde{\hat{\Theta}} (\phi)$ and $\hat{\Theta} (\phi)$ are
$L^\infty(G)$-bimodule maps: we already know this for $\hat{\Theta} (\phi)$, and it follows for $\tilde{\hat{\Theta}} (\phi)$
from \cite[Theorem 2.3]{PR}. Furthermore,  $\tilde{\hat{\Theta}} (\phi)$ and $\hat{\Theta} (\phi)$ coincide on all operators
$\lambda(x)$ with $x \in G$: we know that $\hat{\Theta} (\phi)$ on $\operatorname{VN}(G)$
is nothing but the canonical action of $\phi$ on $\operatorname{VN}(G)$, and the same follows for $\tilde{\hat{\Theta}} (\phi)$ from its definition.
Consequently, we have
\begin{multline*}
  \tilde{\hat{\Theta}} (\phi)(M_f \lambda(x)) = M_f \tilde{\hat{\Theta}} (\phi)(\lambda(x)) \\
  = M_f \hat{\Theta} (\phi)(\lambda(x)) = \hat{\Theta} (\phi)(M_f \lambda(x)) \qquad (f \in L^\infty(G), \, x \in G).
\end{multline*}
\par
As in the proof of Proposition \ref{prop1}, we conclude that $\tilde{\hat{\Theta}} (\phi) = \hat{\Theta} (\phi)$.
\end{proof}
Following \cite{CL}, where the analogous questions for $\operatorname{VN}(G)$ and
${\cal H}_\sigma$ were considered, we now study the pre-annihilator
of $\tilde{\cal H}_\sigma$ in ${\cal T}(L^2(G))$
for any $\sigma \in \mathcal{M}_\mathrm{cb}(A(G))$.
\begin{lemma} \label{rightmod}
For any locally compact group $G$, we have
\[
  T \cdot \omega = \hat{\Theta}(\pi(\omega))(T)
  \qquad (\omega \in {\cal T}(L^2(G)), \, T \in {\cal B}(L^2(G)).
\]
where $\pi \!: {\cal T}(L^2(G)) \to A(G)$ is the canonical quotient map.
\end{lemma}
\begin{proof}
Let $\omega \in {\cal T}(L^2(G))$, and let $T \in {\cal B}(L^2(G))$.
Then we have
\begin{equation*} \label{modid}
  \langle \rho, T \cdot \omega \rangle = \langle \omega\bullet\rho, T \rangle
  = \langle \omega, \rho \cdot T \rangle
  \qquad (\rho \in {\cal T}(L^2(G))).
\end{equation*}
Since $\rho \cdot T \in \operatorname{VN}(G)$ for
$\rho \in {\cal T}(L^2(G))$, the evaluation of $\omega$ at
$\rho \cdot T$ depends only on $\pi(\omega)$,
so that $\langle \rho, T \cdot \omega \rangle = \langle \pi(\omega), \rho \cdot T \rangle$
for each $\rho \in {\cal T}(L^2(G))$. Hence we obtain
\[
  \begin{split}
  \langle \rho, T \cdot \omega \rangle
  & = \langle \pi(\omega), \rho \cdot T \rangle \\
  & = \langle \pi(\omega), \hat{\mathcal{S}} (\rho \otimes T) \rangle \\
  & = \langle \rho \otimes T , \hat{\mathcal{S}}^* (\pi(\omega)) \rangle \\
  & = \langle \rho \otimes T , \tilde{\hat{\Theta}} (\pi(\omega)) \rangle \\
  & = \langle \rho , \tilde{\hat{\Theta}} (\pi(\omega)) (T) \rangle \\
  & = \langle \rho , \hat{\Theta} (\pi(\omega)) (T) \rangle, \qquad\text{by Lemma \ref{thetatheta}},
  \end{split}
\]
for all $\rho \in {\cal T}(L^2(G))$, as desired.
\end{proof}
The following is a consequence of \cite[Lemma 2.1]{PR}, using the fact that
$\tilde{\hat{\Theta}} \mid_{\mathcal{M}_\mathrm{cb}((A(G))} = \hat{\Theta}$. We include our own (short) proof for the sake of completeness:
\begin{lemma} \label{leftmod}
For any locally compact group $G$, we have
\[
  \hat{\Theta}(\phi)(\omega \cdot T) = \omega \cdot \hat{\Theta}(\phi)(T)
  \qquad (\phi \in \mathcal{M}_\mathrm{cb}((A(G)), \,
  \omega \in {\cal T}(L^2(G)), \, T \in {\cal B}(L^2(G))).
\]
\end{lemma}
\begin{proof}
Let $\phi \in \mathcal{M}_\mathrm{cb}((A(G))$, let $\omega, \rho \in {\cal T}(L^2(G))$,
and let $T \in {\cal B}(L^2(G))$.
\par
We first note that
\[
  \begin{split}
  \langle \pi(\hat{\Theta}(\phi)_\ast(\rho)), S \rangle
  & = \langle \hat{\Theta}(\phi)_\ast(\rho), S \rangle \\
  & = \langle \rho, \hat{\Theta}(\phi)(S) \rangle \\
  & = \langle \rho, \phi \cdot S \rangle \\
  & = \langle \pi(\rho), \phi \cdot S \rangle \\
  & = \langle \pi(\rho) \phi, S \rangle \qquad (S \in \operatorname{VN}(G)),
  \end{split}
\]
so that
\begin{equation} \label{yai}
  \pi(\hat{\Theta}(\phi)_\ast(\rho)) = \pi(\rho) \phi.
\end{equation}
Now we obtain:
\[
  \begin{split}
  \langle \rho, \hat{\Theta}(\phi)(\omega \cdot T) \rangle
  & = \langle \hat{\Theta}(\phi)_\ast(\rho), \omega \cdot T \rangle \\
  & = \langle \hat{\Theta}(\phi)_\ast(\rho) \bullet \omega , T \rangle \\
  & = \langle \omega , T \cdot \hat{\Theta}(\phi)_\ast(\rho)\rangle \\
  & = \langle \omega, \hat{\Theta}(\pi(\hat{\Theta}(\phi)_\ast(\rho)))(T)
      \rangle, \qquad\text{by Lemma \ref{rightmod}}, \\
  & = \langle \omega, \hat{\Theta}(\pi(\rho)\phi)(T) \rangle,
      \qquad\text{by (\ref{yai})}, \\
  & = \langle \omega, \hat{\Theta}(\pi(\rho))(\hat{\Theta}(\phi)(T)) \rangle \\
  & = \langle \omega, \hat{\Theta}(\phi)(T) \cdot \rho \rangle, \qquad
      \text{by Lemma \ref{rightmod} again}, \\
  & = \langle \rho \bullet \omega, \hat{\Theta}(\phi)(T) \rangle \\
  & = \langle \rho, \omega \cdot \hat{\Theta}(\phi)(T) \rangle.
  \end{split}
\]
Since $\rho \in {\cal T}(L^2(G))$ is arbitrary, this yields that
\[
  \hat{\Theta}(\phi)(\omega \cdot T) = \omega \cdot \hat{\Theta}(\phi)(T),
\]
as claimed.
\end{proof}
\begin{theorem}
Let $G$ be a locally compact group, and let $\sigma \in \mathcal{M}_\mathrm{cb}(A(G))$.
Then
\[
  \tilde{I}_\sigma := \varcl{\{ \hat{\Theta}(\sigma)_\ast(\omega) - \omega :
  \omega \in {\cal T}(L^2(G)) \}}^{{\cal T}(L^2(G))}
\]
is the pre-annihilator of $\tilde{\cal H}_\sigma$ in ${\cal T}(L^2(G))$---so that $\tilde{\cal H}_\sigma \cong ({\cal T}(L^2(G)) / \tilde{I}_\sigma)^\ast$---and a two-sided ideal of $({\cal T}(L^2(G)), \bullet)$.
\end{theorem}
\begin{proof}
It is straightforward to see that
\[
  \omega \in \tilde{I}_\sigma \iff \text{$\langle \omega, T \rangle = 0$ for all $T \in \tilde{\cal H}_\sigma$}
  \qquad (\omega \in {\cal T}(L^2(G))).
\]
Hence, $\tilde{I}_\sigma$ is indeed the pre-annihilator of $\tilde{\cal H}_\sigma$ in ${\cal T}(L^2(G))$, and $({\cal T}(L^2(G)) / \tilde{I}_\sigma)^\ast = \tilde{I}_\sigma^\perp \cong \tilde{\cal H}_\sigma$ holds.
\par
We first show that $\tilde{I}_\sigma$ is a left ideal in ${\cal T}(L^2(G))$.
To this end, let $\omega, \rho \in {\cal T}(L^2(G))$, and note that
\[
  \begin{split}
  \langle \omega \bullet \hat{\Theta}(\sigma)_\ast(\rho), T \rangle & =
  \langle \hat{\Theta}(\sigma)_\ast(\rho), T \cdot \omega \rangle \\
  & = \langle \rho, \hat{\Theta}(\sigma)(T \cdot \omega) \rangle \\
  & = \langle \rho, \hat{\Theta}(\sigma)(\hat{\Theta}(\pi(\omega))(T)) \rangle,
  \qquad\text{by Lemma \ref{rightmod}}, \\
  & = \langle \rho, \hat{\Theta}(\pi(\omega)\sigma)(T) \rangle \\
  & =\langle \rho, \hat{\Theta}(\pi(\omega))(\hat{\Theta}(\sigma)(T)) \rangle \\
  & = \langle \rho, \hat{\Theta}(\sigma) (T) \cdot \omega \rangle, \qquad
  \text{again by Lemma \ref{rightmod}}, \\
  & = \langle \omega \bullet \rho, \hat{\Theta}(\sigma)(T) \rangle \\
  & = \langle \hat{\Theta}(\sigma)_\ast(\omega \bullet \rho), T \rangle.
  \qquad (T \in {\cal B}(L^2(G))),
  \end{split}
\]
so that
\[
  \omega \bullet \hat{\Theta}(\sigma)_\ast(\rho) =
  \hat{\Theta}(\sigma)_\ast(\omega \bullet \rho)
\]
and therefore
\begin{equation} \label{ofi}
  \omega \bullet (\hat{\Theta}(\sigma)_\ast(\rho) - \rho) =
  \omega \bullet \hat{\Theta}(\sigma)_\ast(\rho) - \omega \bullet \rho =
  \hat{\Theta}(\sigma)_\ast(\omega \bullet \rho) - \omega \bullet \rho
  \in \tilde{I}_\sigma
\end{equation}
Since $\omega, \rho \in {\cal T}(L^2(G))$ were arbitrary, it follows that
$\tilde{I}_\sigma$ is a left ideal as claimed.
\par
To see that $\tilde{I}_\sigma$ is also a right ideal, let $\omega, \rho
\in {\cal T}(L^2(G))$, so that
\[
  \begin{split}
  \langle (\hat{\Theta}(\sigma)_\ast(\rho)) \bullet \omega, T \rangle
  & = \langle (\hat{\Theta}(\sigma)_\ast(\rho)), \omega \cdot T \rangle \\
  & = \langle \rho, \hat{\Theta}(\sigma)(\omega \cdot T) \rangle \\
  & = \langle \rho, \omega \cdot \hat{\Theta}(\sigma)(T) \rangle,
    \qquad\text{by Lemma \ref{leftmod}}, \\
  & = \langle \rho \bullet \omega, \hat{\Theta}(\sigma)(T) \rangle \\
  & = \langle \hat{\Theta}(\sigma)_\ast(\rho\bullet\omega), T \rangle
  \qquad (T \in {\cal B}(L^2(G)))
  \end{split}
\]
and thus
\[
  (\hat{\Theta}(\sigma)_\ast(\rho)) \bullet \omega =
  \hat{\Theta}(\sigma)_\ast(\rho\bullet\omega).
\]
A calculation similar to (\ref{ofi}) then lets us conclude that
$\tilde{I}_\sigma$ is a right ideal, too.
\end{proof}
\begin{remark}
The above result mirrors both \cite[Definition 3.2.1]{CL} and
\cite[Proposition 3.3]{JN} in our setting.
\end{remark}
\par
Finally, we consider another ideal of $({\cal T}(L^2(G)), \bullet)$.
\begin{proposition}
Let $G$ be a locally compact group, and let $\sigma \in \mathcal{M}_\mathrm{cb}(A(G))$. Then
\[
  L^\infty(G)_\perp := \{ \omega \in {\cal T}(L^2(G)) :
  \text{$\langle \omega, M_\phi \rangle = 0$ for all $\phi \in L^\infty(G)$} \}
\]
is a two-sided ideal in $({\cal T}(L^2(G)), \bullet)$ which is contained in
the \emph{augmentation ideal}
\[
  {\cal T}_0(L^2(G)) := \{ \omega \in {\cal T}(L^2(G)) :
                           \langle \omega, 1 \rangle = 0 \}
\]
and, if $\sigma(e) = 1$, contains $\tilde{I}_\sigma$.
\end{proposition}
\begin{proof}
Trivially, $L^\infty(G)_\perp \subset {\cal T}_0(L^2(G))$ holds.
\par
Moreover, if $\sigma(e) = 1$, then
\[
  \langle \hat{\Theta}(\sigma)_\ast(\rho) - \rho, M_\phi \rangle
  = \langle \rho,  \hat{\Theta}(\sigma)(M_\phi) \rangle -
    \langle \rho, M_\phi \rangle
  = \langle \rho, M_\phi \rangle - \langle \rho, M_\phi \rangle
\]
holds for all $\rho \in {\cal T}(L^2(G))$ and $\phi \in L^\infty(G)$,
so that $\tilde{I}_\sigma \subset L^\infty(G)_\perp$.
\par
It remains to be shown that $L^\infty(G)_\perp$ is indeed an ideal of
${\cal T}(L^2(G))$.
\par
Let $\rho , \omega \in {\cal T}(L^2(G))$.
Then we see that
\begin{equation} \label{yup}
  \begin{split}
  \langle \rho \bullet \omega, M_\phi \rangle & =
  \langle \omega , M_\phi \cdot \rho \rangle \\
  & = \langle \omega, \hat{\Theta}(\pi(\rho))(M_\phi) \rangle,
  \qquad\text{by Lemma \ref{rightmod}}, \\
  & = \langle \omega, M_\phi \hat{\Theta}(\pi(\rho))(1) \rangle \\
  & = \langle \omega, M_\phi (\pi(\rho) \cdot 1) \rangle \\
  & = \langle \rho, 1 \rangle \langle \omega, M_\phi \rangle
  \qquad (\phi \in L^\infty(G))
  \end{split}
\end{equation}
holds. From (\ref{yup}), it is immediate that $L^\infty(G)_\perp$ is indeed a
two-sided ideal of ${\cal T}(L^2(G))$.
\end{proof}
\begin{remark}
Since $L^\infty(G)_\perp$ is a two-sided ideal of $({\cal T}(L^2(G)),
\bullet)$, the product $\bullet$ induces a product---likewise denoted by
$\bullet$---on the quotient algebra ${\cal T}(L^2(G)) / L^\infty(G)_\perp
\cong L^1(G)$. This product, however, is \emph{not} the usual convolution
product on $L^1(G)$: from (\ref{yup}), it is clear that
\[
  f \bullet g = \langle f, 1 \rangle g \qquad (f,g \in L^1(G)).
\]
\end{remark}

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Ty Dolla $ign and Wiz Khalifa come together for a music video to their track 'Brand New' off of the TGOD Vol. 1 compilation mixtape. Directed by Gabriel Hart, Ty and Wiz enjoy the Los Angeles scenery with plenty of gorgeous women to accompany them. Watch the video to 'Brand New' above.


RTGS -The acronym "RTGS" stands for Real Time Gross Settlement. The RTGS system facilitates transfer of funds from accounts in one bank to another on a "real time" and on "gross settlement" basis. The RTGS system is the fastest possible inter bank money transfer facility available through secure banking channels in India.
What is the minimum / maximum amount for RTGS /NEFT transactions?
When does the beneficiary get the credit for a RTGS payment? Under normal circumstances the beneficiary bank's branch receives the funds in real time as soon as funds are transferred by the remitting bank.
When does the beneficiary get the credit for a NEFT payment? The funds will be sent to the RBI within three hours of the transaction. The actual time taken to credit the beneficiary depends on the time taken by the beneficiary bank to process the payment.
If an RTGS transaction is not credited to a beneficiary account, does the Remitter get back the money? Yes. If it is unable to credit the amount of the remittance to the account of the beneficiary for any reason, the beneficiary's bank has to return the money to the remitting bank within 2 hours. Once this amount is received back by the remitting bank, the amount is credited to the Remitter's account by the remitting bank branch.
If an NEFT transaction is not credited to a beneficiary account, does the Remitter get back the money? Yes. If the money cannot be credited for any reason, the beneficiary's bank has to return the money to the remitting bank . Once this amount is received back by the remitting bank, the amount is credited to the Remitter's account by the remitting bank branch.
At what time during the day /week is the RTGS & NEFT service is available?Presently, NEFT is settled in six batches at 09:30 A.M., 10:30 A.M., 12:00 P.M., 1:00 P.M., 3:00 P.M. and 4:00 P.M. hours on weekdays and 09:30 A.M., 10:30 A.M. and 12:00 P.M. hours on Saturdays in RBI. OnlineSBI sends the transaction requests upto 3:30 P.M. on weekdays and upto 11:30 A.M. on Saturdays.
RTGS transaction requests will be sent to RBI upto 2:30 PM. on weekdays and upto 11:30 AM on Saturdays.
How to find the IFSC code of the beneficiary branch? Based on simple search option- Location, Bank & Branch. The IFSC code will automatically be updated or if you know the IFSC code of the beneficiary bank, you can provide it directly.
Do all bank branches in India provide RTGS & NEFT service? No. RTGS facility is enabled only in specific bank branches across India. You can view a list of these branches in the RBI website: www.rbi.org.in/Scripts/Bs_viewRTGS.aspx. NEFT is enabled only in specific bank branches across India. You can view a list of these branches in the RBI website: http://www.rbi.org.in/scripts/neft.aspx.
How do I go about using this service?
You should be an active Internet Banking user with transaction rights.
Log on to www.onlinesbi.com by using your SBI Internet Banking ID and Password.
Click on the 'Manage Interbank Payee' link in the 'Profile' tab.
Select the 'Add' option and provide the Beneficiary Name, Beneficiary Account Number, Beneficiary Bank and transfer limit to register the Beneficiary.
Click the IFSC Code option and a textbox is displayed where you can enter the 11 digit IFSC Code of the Beneficiary Bank.
Click the Location option, the following are retrieved and displayed.
The confirm button will be enabled only after clicking the accept button in the terms and conditions window.
After providing all the details, the beneficiary will be added. It will be shown whether the added beneficiary bank is RTGS or NEFT enabled.
After adding the Beneficiary you will receive a high security password in your mobile number. This is done basically to double check your identity. Provide the password to authorize the Beneficiary After a Beneficiary is authorized you can start transferring funds.
You can proceed to make payments by clicking the 'Inter bank Transfer' link in the 'Payments/Transfers' tab.
According to the transaction type selected (RTGS/NEFT), the credit account details will be displayed depending on whether the branch is RTGS or NEFT enabled or both.
Select the Beneficiary from the list of registered Beneficiaries.
You can either confirm or cancel the transaction.
This move will further consolidate SBI's position against ICICI. In my opinion, ICICI's online banking service was their main benefit as customers could transfer money electronically to any branch .. Now SBI's customers can have this feature too with no extra cost .. This competition is opening up new initiatives for the benefit of customers .. My verdict : The best just got better… !! Yay ..!
how long it will takes to approve mobile num?
contact me if u can..
my name is ramujagar yadav.
Next PostNext Last.fm Client for Symbian phones at last ..!


text:
Q: Cluster analysis in matlab or python - k-means clustering on three dimensional data I have 3dimensional data. So for 705 frequencies i have 24*2 data. 24 rows represent hours and columns represent mean and coefficient of variance. This 24*2*705 matrix data.
I would like to perform k-means on this data.
How should i do this either in matlab or python?
Any suggestions.
Vaishali

A: K-means does not correlated or mix components. The results are invariant to any permutation of axes.
Hence it is safe to reshaoe your data into a n.p format as used by k-means.
The bigger question is whether SSR is the right objective for your problem. Are you solving the right problem, or just using k-means because it is cool.

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text:
About Harley
The Wollie Series
Keeper of the Moon
Narrating
Where's Harley?
I wanted to act from a very early age. Ever since I saw The Parent Trap. The first one, not the 1998 remake. I mean, I liked the remake quite a lot, but it didn't make me want to hack my way through the screen and leap into that fictional world like the first one did, although that might've had something to do with being 40, instead of 4.
And then, when I was 5, my mom got me my first onstage acting role, in Valley City, North Dakota. It was the opera Dido & Aneas and my mom was the Sorceress. I was an angel. Thrilling stuff. Staying up past my bedtime, wearing red lipstick, walking into blindingly bright lights, tossing rose petals on a corpse. My fate was sealed.
At least, it was sealed for the next 35 or 40 years (see resumé). Then I took a 15-year maternity leave, during which I turned my creative ambitions to writing novels (see other tab) and raising kids, and people would say, "will you ever go back to acting?" and I'd say, "I don't really know." But whenever I'd drive past a film shoot, and see the camera trucks and the honey wagons, my heart would skip a beat and I'd realized I did in fact miss it. But like I said, I was busy. Who isn't?
And then one night I dreamed of being on a set and I was so happy, having such fun with my co-stars and hair & makeup and the camera crew and I woke up and thought, "time to go back."
So I'm back. Fifteen years is a long time, especially in Hollywood years, so there were people in the industry who thought I'd died. My agent actually did die. I found myself in another age category altogether after my "comeback" and the number of auditions and roles is tiny compared to those available in my 20's and 30's, but that's not exactly news. And there's a lot to be said for being older and wiser and not being the one on the set having to act in her underwear.
View Harley's Acting Resume
The Accolade Global Film Competition
Award of Merit Special Mention: Supporting Actress for More Beautiful for Having Been Broken
Patty Woo at Seven Summits (310) 550-6777 or patty@pattywoo.com
Talent Agent:
Talent Works (818) 972-4300
Commercial Agent:
AKA Talent Agency (323) 965-5600
Acting Studio I love: The BGB Studio
The BGB studio is my home away from home. Check out their website, where you can see, among zillions of other things,
The 8-minute film I wrote and directed, called Bouillabaisse
The podcast I'm in, Episode 5: "Wisdom of Women"
The blog I wrote, Second Act
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