text:
Q: Django ForeignKey from external Database I'm working with 2 Databases, 1 external not managed by Django and the other internal.
Thi is the external database now working on models and in Admin site:
class Client(models.Model):# IMPORT FROM MYSQL
code = models.CharField(max_length=3, primary_key=True)
name = models.CharField(max_length=250)
region_id = models.IntegerField(db_column="RegionId")

class Meta:
    managed = False
    db_table = 'client'

And this is the one managed by Django on the other DB:
class ClientJuno(models.Model):
    LABORATORIES = (('SCR','SCREENS'), ('DVR','DVMR'),)
    client = models.ForeignKey(Client, on_delete=models.SET_NULL, null=True)
    profiles = models.ManyToManyField(JunoProfile)
    laboratory = models.CharField(max_length=20, choices=LABORATORIES)

See error on opening this last model. Can't find table Client.
  OperationalError at /admin/settings/clientjules/
no such table: client
Request Method: GET
Request URL:    http://127.0.0.1:8000/admin/settings/clientjules/
Django Version: 2.2.11
Exception Type: OperationalError
Exception Value:
no such table: client
Exception Location: /usr/local/lib/python3.6/site-packages/django/db/backends/sqlite3/base.py in execute, line 383
Python Executable:  /usr/local/bin/python
Python Version: 3.6.10
Python Path:
['/app/mmike',
 '/usr/local/lib/python36.zip',
 '/usr/local/lib/python3.6',
 '/usr/local/lib/python3.6/lib-dynload',
 '/usr/local/lib/python3.6/site-packages']
Server time:    Mon, 20 Apr 2020 08:11:26 +0000

enter image description here

A: It seems this is not supported by Django.
How to use django models with foreign keys in different DBs?
Cross-database limitations
Django doesn't currently provide any support for foreign key or many-to-many relationships spanning multiple databases. If you have used a router to partition models to different databases, any foreign key and many-to-many relationships defined by those models must be internal to a single database.
Django - limitations-of-multiple-databases


text:
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- Hey!
I'm Dr. Matt, and I've got the perfect cure
for that headache, backache, toothache,
heartache, handshake, corn flake, tummyache.
It's called meth! (laughs nervously)
Ride the snake!
All right, well, obviously I'm not a doctor,
and prescribing that sounds nuts.
But, there have been some insane treatments
prescribed by medical experts who,
arguably, didn't know better.
And what's really disturbing is that these were all commonly used in the past.
If a physician used any of the methods
you're going to hear about today, they'd be in jail.
So, what exactly are they?
Here are 10 banned medical practices
you won't believe existed.
Number one is Mercury.
It definitely sounds crazy, but there was a time
when people thought that the best way to treat syphilis was to ingest Mercury.
Side effects included losing your teeth,
damage to internal organs, or simply death.
Using Mercury as a cure began as early
as 1500 B.C. in Egypt, and was believed
to cure ailments, heal wounds, and even prolong life.
It turned out to be the exact opposite.
They also used to use Mercury ointments, pills,
and lotions to treat all sorts of ailments,
attributing the side-effects and deaths to the underlying condition the patient already had.
Nowadays, we're weary about Mercury in fish.
Back then, they were drinkin' it like it was an ale.
Ah, gimmie some Mercury!
Yeah, I feel great! (groans)
Number two is radioactive water.
It's hard to believe, but in the early 1900s,
Radium was all the rage.
Today we know that exposure to Radium causes cancer,
but back then it was used as a cure for mental illness,
diarrhea, malaria, and was even thought to prevent aging
due to increased cell activity.
And what better way to get this stuff into consumers' bodies than to sell it as bottled water?
That's right, Radium-infused water could be purchased in many stores
alongside Radium-infused chocolate, radioactive toothpaste, and,
why not, radioactive suppositories.
Up and at 'em.
(grunts)
Number three is whirling chairs.
Between 1850 and 1900, in an attempt to stop the practice of restraining the mentally ill
and locking them in dark cells, yes, that actually happened,
doctors turned to a method they genuinely thought could help:
Strapping patients into a chair and spinning them until they lost consciousness.
That'll get the voices to stop.
They used a simple chair with a lever system to spin the person around.
The belief was that this would shuffle
the contents of the patient's brain around,
knocking out conditions like schizophrenia.
Used alongside other unbelievable practices like high-dose laxatives, insulin comas,
and ice-water baths, it took far too long for doctors to learn that this really wasn't working.
Just made a whole lot of vomit...
Just...just...buckets.
Buckets of vomit.
Number four is the lobotomy.
Often declared as one of the most barbaric mistakes ever perpetrated by main-stream medicine,
lobotomies involved driving a large ice pick into a patient's head.
Follow me on this.
Entering through the eye socket,
the spike was pressed into the prefrontal cortex of the brain,
and then struck with a hammer.
Doctors truly believed that whatever the outcome,
the patient would still be better off than they were pre-lobotomy.
Side effects included severe brain damage,
or even death, which was the case with
John F. Kennedy's sister who, after a lobotomy
was reduced mentally to an infant, and never recovered.
How...how did they ever think that driving an ice pick
through the eye into the brain was ever a good idea?
Just...
It's scary stuff.
Number five is the electrical impotence cure.
By the 1900s, everybody and their brother had a miracle cure for anything.
And disturbingly, many of them were for male...
(clears throat) issues....
Down there...
Wit' ya dick.
Anywho, as fascination with electricity was on the rise,
medical practitioners turned to it to solve impotence.
By shocking the hell out of a man's...
package.
It was widely believed that a shock provided energy
to a man's body which he could use to...
rise to the occasion.
Of course, what they didn't tell you,
until after they were paid, was that
not only was there no real science involved in it,
but it was incredibly painful!
Dudes these days be buyin' these online enhancement pills,
and all they really needed was a car battery and some jumper cables.
Number six is trepanning.
The twisted cousin of lobotomy,
trepanning was used for many years
and it involved drilling into a mentally ill patient's head.
As one of the oldest methods of treatment,
trepanning dates all the way back to 6500 B.C. in France,
and consisted of drilling burr holes
into a patient's skull to let the evil spirits out.
(groans)
The mentally ill were often mistaken to be possessed
by demons, and, as we all know,
the only way to get a demon out of your head
is to make a little hole for it to get out of.
Long after science recognized mental illness, however,
trepanning was still used when all other treatments failed,
as doctors just drilled a few holes to see what happens.
Uh, science, doctors, if I ever become ill,
please, don't treat my head like a bowling ball.
If there's demons in there, you just let 'em chill.
Just let 'em chill.
I don't want no holes, okay?
Still livin' and then I'll die.
I'll die, that's cool, I just...
I don't wanna be a Matthew Santoro bowling ball.
That's...
That's disturbing.
Number seven is bloodletting.
A treatment used for over 2,000 years until the late 1800s,
bloodletting was the act of draining blood from the body.
Now, while this is still done in a way today,
it's nothing like it used to be.
Many doctors used to believe that
blood would pool in certain places,
and that was why people got sick.
Vast quantities of it would be drained from the body for ridiculous reasons like a stuffy nose,
or even a headache.
Obviously, this caused many deaths,
but, once again, practitioners just chalked up loss of life to the illness itself,
or, state of mind of the patient.
Barbers were the most common blood-letters,
and that's the reason they use red and white poles.
Disturbing, I know.
Think about that next time you walk by a barber.
Nightmares for days.
Number eight is urine therapy.
Urine therapy was the practice of using one's own pee to cure them of ailments.
Sometimes, it would be the patient's own urine,
but other times it was a healthy person's.
Or even the doctor's.
How generous.
It was thought that by drinking urine, or applying it around affected areas,
it could help heal broken bones,
clear up strep throat, cure acne, or even whiten teeth.
Don't get any ideas.
Though technically debunked, there are some people
still in the world that claim to gain the benefits from this truly bizarre form of therapy.
So, careful where you go in the world,
because that's...that lemonade...
may not be lemonade.
Number nine is Ms. Winslow's soothing syrup.
Hey, want a quick way to stop your child from throwing a temper-tantrum?
Well, don't worry, with Ms. Winslow's soothing syrup,
they'll be calm and cool,
and halfway to a coma in no time!
Sold in the early 1900s as a remedy
for children with ailments such as cold,
fever, rambunctiousness, or even if a parent just wanted to catch a quick nap,
Ms. Winslow's remedy was fast-acting
mostly because it contained heroin,
cannabis, powdered opium, morphine,
and an insane slew of other substances.
The cocktail proved deadly to those children
who didn't mellow out, and, to top it off,
it was marketed to parents of babies who were teething,
and thus, always crying.
Mom and Dad can finally relax by putting your child to sleep...
That was a scary time to be alive, medical science.
And, number 10, the tobacco enema.
Let me just say, first of all, that all the remedies
you heard of today are 100% true.
I am not blowing smoke up your ass.
Although, if you lived in the 1700s and
needed emergency, life-saving treatment,
that's exactly what they'd do.
It was actually believed that tobacco smoke
could warm up a person's body just before or immediately after death,
and could jump-start the respiratory system.
Of course, the best place to jump-start the lungs and the heart is through the butt!
Using rubber rectal tubes.
Doctors claimed that this practice
would combat hernias, typhoid fever,
and even headaches.
And then, in 1811, it was discovered
that this is really doing just way more damage
than helping, and tobacco enemas were discontinued.
I think the moral of the story here, friends,
is just don't try to smoke with ya butt.
Don't smoke with your butt.
Leave the anus...
It's a helpful tip.
And those were 10 banned medical practices
that actually existed.
Thank you guys, so much, for watching this episode.
I really hope you enjoyed it.
If you have a recommendation for a topic
of a top ten, please leave a comment below,
and I'll be reading through your comments.
Remember to click that red subscribe button
to be notified of my future videos on this channel.
On the right, you'll find an annotation to my last video,
as well as an annotation to get some merch
if you feel like pickin' up some cool shirts.
As well, you'll also find an annotation
to my second channel, where I make vlogs, unboxings,
and much more, so be sure to subscribe to that,
and I will see all of you next time
with a brand...new...video.
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10 BANNED Medical Practices You Won't Believe Existed!
David Chen posted on 2015/11/15
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Q: During installation can't change directory I try to install mysql db server on windows 10.
Version mysql-installer-community-5.7.16.0.msi
During the installation Install MSI does not offer to change the installation directory.
Do I correctly understand that I can't change the installation directory during the installation process?

A: You can make the following steps:

*

*Manually create folder C:\Program files\MySQL\MySQL Server 5.7  (this is in 64 bits version)

*Run the installer normally

Create an empty folder, although is empty folder, makes the installer allow you to choose another location.

A: There are two methods available, either use ZIP archive to install or you may have to change the datadir variable in my.ini after installation. You can use following steps

*

*Complete the installation

*Identify the config file and then look for related datadir being used, you may used mysql>show variables like 'datadir' within mysql to identify datadir or the service details for concerned my.ini

*Stop the service if it is already running

*Change the value of datadir in my.ini, it is usually placed under C:\ProgramData\MySQL

*Copy/move the folder contents to newer location

*Make sure permissions are provided to user who owns mysql service

*Start service and verify using mysql>show variables like 'datadir'

Hope it helps

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text:
\section{Introduction}
The theory of $C^*$-crossed product by group partial actions and inverse semigroup actions are very well developed \cite{partial} \cite{sieben}. In this paper, we show that the theory of crossed products by actions of inverse semigroups can be generalized to partial actions of inverse semigroups.\par In section \ref{two} we define a partial action of an inverse semigroup as a partial homomorphism from the inverse semigroup into a symmetric inverse semigroup on some set. We will refer the reader to \cite{EXPAN} for an extensive treatment of partial actions of inverse semigroups. In section \ref{two}, we define the crossed products by partial actions of inverse semigroups.\par It turns out that there is a close connection between crossed products by partial actions of inverse semigroups and crossed products by inverse semigroups actions. In section \ref{five}, we will show that every crossed products by partial action of an inverse semigroup is isomorphic to a crossed product by an inverse semigroup action.
\section{Partial Actions of Inverse Semigroups and Covariant Representations}\label{two}
We will assume that throughout this work $G$ is a unital inverse semigroup with unit element $e$ and $\mathcal{A}$ is a $C^*$-algebra.

We recall from ~\cite{EXPAN} that a partial action of an inverse semigroup $S$ on a set $X$ is a partial
homomorphism $\alpha :S\mapsto \verb"I"(X)$, that is, for each $s,t\in S$ $$\alpha(s^{*})\alpha(s)\alpha(t)=\alpha(s^{*})\alpha(st),\;\;\;\;\alpha(s)\alpha(t)\alpha(t^{*})=\alpha(st)\alpha(t^{*}),$$
where $\verb"I"(X)$ denotes the inverse semigroup of all partial bijections between subsets of $X$.
But we use ~\cite[Proposition 3.4]{EXPAN} to give a definition of a partial action of an inverse semigroup.
\begin{definition}\label{defpar}
Suppose that $S$ is an inverse semigroup and $X$ is a set. By a partial action of $S$ on
$X$ a map  we mean $\alpha :S\mapsto \verb"I"(X)$ satisfied the following conditions:
\begin{enumerate}
    \item [(\textit{i})]$\alpha_{s}^{-1}=\alpha_{s^{*}}$,
    \item
    [(\textit{ii})]$\alpha _{s}(X_{s^{*}}\cap X_t)=X_s\cap
    X_{st}$ for all $s,t\in S $ (where $X_s $ denotes the rang of $\alpha_{s}$ for each $s\in
    S$),
    \item [(\textit{iii})] $\alpha_s(\alpha_t(x))=\alpha_{st}(x)$ for all $x\in X_{t^*}\cap
    X_{t^*s^*}.$
\end{enumerate}
\end{definition}
To define a partial action $\alpha$ of an inverse semigroup $S$ on an
associative $\mathcal{K}$-algebra $\mathcal{A}$, we suppose in Definition \ref{defpar}
that each $X_s$ $(s\in S)$  is an ideal of $\mathcal{A}$ and that every map
$\alpha_s:X_{s^*}\mapsto X_s$ is an algebra isomorphism.
Furthermore, if the inverse semigroup $S$ is unital with unit $e$, we shall
suppose that $X_e=\mathcal{A}$. The next Proposition shows that for such a partial action $\alpha$ we have $\alpha_e$ is the identity map on $\mathcal{A}$.
\begin{proposition}
If $\alpha$ is a partial action of $G$ on a $C^*$-algebra $\mathcal{A}$ then $\alpha_e$ is the identity map $\ell$ on $\mathcal{A}$.
\end{proposition}
\begin{proof}
By definition of partial action, $\alpha_e$ is an invertible map on it's domain, $D_e=\mathcal{A}$. Now, $$\ell=\alpha_e\alpha_e^{-1}=\alpha_e\alpha_{e^*}=\alpha_e\alpha_e=\alpha_e.$$
Note that we have used part (3) of Definition \ref{defpar} in the fourth equality above.
\end{proof}
The following Lemma will be used in the proof of Theorem \ref{theorem1.4}
\begin{lemma}\label{lemma2.5}
If $\alpha$ is a partial action of $G$ on a $C^*$-algebra $\mathcal{A}$, then for all $t,s_1,...,s_n\in G$ $$\alpha_t(D_{t^*}D_{s_1}...D_{s_n})=D_{t}D_{ts_1}...D_{ts_n}.$$
\end{lemma}
\begin{proof}
For $t,s_1,...,s_n\in G$ we have
\begin{eqnarray*}
  \alpha_t(D_{t^*}D_{s_1}...D_{s_n}) &=& \alpha_t(D_{t^*}\cap D_{s_1}\cap ...\cap D_{t^*}\cap D_{s_n}) \\
   &=& \alpha_t(D_{t^*}\cap D_{s_1})\cap...\cap\alpha_t(D_{t^*}\cap D_{s_n}) \\
   &=& \alpha_t(D_t\cap D_{ts_1})\cap...\cap\alpha_t(D_t\cap D_{ts_n}) \\
   &=&  \alpha_t(D_t\cap D_{ts_1}\cap...\cap D_t\cap D_{ts_n})\\
   &=& \alpha_t(D_t D_{ts_1}... D_{ts_n})
\end{eqnarray*}

\end{proof}
\begin{theorem}\label{theorem1.4}
If $\alpha$ is a partial action of $G$ on a $C^*$-algebra $\mathcal{A}$, then for $s_1,...,s_n\in G$ the partial automorphism $\alpha_{s_1}...\alpha_{s_n}$ has domain\\ $D_{s_n^*}D{s_n^*s_{n-1}^*}...D_{s_n^*...s_1}$ and range $D_{s_1}...D_{s_1...s_n}$.
\end{theorem}
\begin{proof}
We will use induction to prove the statement about the domain. For $n=1$
\begin{equation*}
    dom \alpha_{s_1}=ran \alpha_{s^*}=D_{s^*}.
\end{equation*}
Now,
\begin{eqnarray*}
  dom \alpha_{s_1}...\alpha_{s_n} &=& \alpha_{s_n}^{-1}(dom (\alpha_{s_1}...\alpha_{s_{n-1}})\cap ran \alpha_{s_n}) \\
   &=& \alpha_{s_n^*}(D_{s_{n-1}^*}...D_{s_{n-1}^*...s_1^*}\cap D_{s_n}) \\
   &=& \alpha_{s_n^*}(D_{s_n}D_{s_{n-1}^*}...D_{s_{n-1}^*...s_1^*} ) \\
   &=& D_{s_n^*}D{s_n^*s_{n-1}^*}...D_{s_n^*...s_1}.
\end{eqnarray*}
Note that we obtained the last equality by using Lemma \ref{lemma2.5}. For the second statement, we have
\begin{eqnarray*}
  ran \alpha_{s_1}...\alpha_{s_n} &=& dom \alpha_{s_n^*}...\alpha_{s_1^*} \\
  &=& D_{s_1}...D_{s_1...s_n}
\end{eqnarray*}
by the first statement.
\end{proof}
If we consider a group $G$ as an inverse semigroup, then the two definitions of partial actions as a group and as an inverse semigroup are the same. This fact motivates us to define a covariant representation of a partial action of an inverse semigroup.
\begin{definition}\label{def1}
Let $\alpha$ be a partial action of $G$ on an algebra $\mathcal{A}$.  A covariant representation of $\alpha$
is a triple $(\pi, u, \mathcal{H})$, where $\pi:\mathcal{A}\to B(\mathcal{H})$ is a non-degenerate representation of $\mathcal{A}$ on a Hilbert space $\mathcal{H}$ and for each $g\in G$, $u_g$ is a partial isometry on $\mathcal{H}$ with initial space $\pi(D_{g^*})\mathcal{H}$ and final space $\pi(D_g)\mathcal{H}$, such that
\begin{enumerate}
  \item $u_g\pi(a)u_{g^*}=\pi(\alpha_g(a))\hspace{0.5cm}a\in D_{g^*}$,
  \item $u_{st}h=u_s u_th\hspace{0.5cm}$ for all $h\in \pi(D_{t^*}D_{t^*s^*})\mathcal{H}$,
  \item $u_{s^*}=u_s^*$.
\end{enumerate}
\end{definition}
Notice that by the \emph{Cohen-Hewitt factorization} Theorem $\pi(D_g)\mathcal{H}$ is a closed subspace of $\mathcal{H}$ and so the notions of initial and final spaces make sense.\par Now, we show that $u_e=1_{\mathcal{H}}$, where $e$ denotes the unit of $G$.
Since $D_e=\mathcal{A}$, by (2) of Definition \ref{def1} for all $h\in \pi(\mathcal{A})\mathcal{H}=\mathcal{H}$ we have that $$u_eh=u_{ee}h=u_eu_eh.$$ Since $u_e$ is one to one on $\pi(\mathcal{A})\mathcal{H}=\mathcal{H}$, we have $u_eh=h$ for all $h\in\mathcal{H}$ as we claimed.
\begin{definition}
Let $\alpha$ be a partial action of $G$ on a $C^*$-algebra $\mathcal{A}$. For $s\in G$, let $\rho_s$ denote the central projection of $\mathcal{A}^{**}$ which is the identity of $D_s^{**}$.
\end{definition}
Let ($\pi, u, \mathcal{H}$) be a covariant representation of ($\mathcal{A}, G, \alpha$). Since $\pi$ is a non-degenerate representation of $\mathcal{A}$, $\pi$ can be extended to a normal morphism of $\mathcal{A}^{**}$ onto $\pi(\mathcal{A})''$. We will denote this extension also by $\pi$. Note that $\pi(D_{s_1}...D_{s_n})\mathcal{H}=\pi(\rho_{s_1}\ldots\rho_{s_n})\mathcal{H}$ for all $s_1,\ldots ,s_n\in G$, and $u_su_{s^*}=\pi(\rho_s)$ for all $s\in G$.
\begin{theorem}\label{theorem2.9}
Let ($\pi, u, \mathcal{H}$) be a covariant representation of ($\mathcal{A}, G, \alpha$). Then for all $s_1,...,s_n\in G$, $u_{s_1}...u_{s_n}$ is a partial isometry with initial space $$\pi(D_{s_n^*}D_{s_n^*s_{n-1}^*}...D_{s_n^*...s_1^*})\mathcal{H}$$ and final space $$\pi(D_{s_1}...D_{s_1...s_n})\mathcal{H}.$$
\end{theorem}
\begin{proof}
Firstly, we show that $u_{s_1}...u_{s_n}u_{s_n}^*...u_{s_1}^*=\pi(\rho_{s_1}...\rho_{s_1...s_n})$. For $n=1$ we have proved that $u_{s_1}u_{s_1}^*=\pi(\rho_{s_1})$. Now,
\begin{eqnarray*}
  u_{s_1}...u_{s_n}u_{s_n}^*...u_{s_1}^* &=& u_{s_1}u_{s_1}^*u_{s_1}\pi(\rho{s_2}...\rho_{s_2...s_n})ua_{s_1}^* \\
   &=& u_{s_1}u_{s_1^*}u_{s_1}\pi(\rho_{s_2}...\rho_{s_2...s_n})u_{s_1}^* \\
   &=& u_{s_1}\pi(\rho_{s_1^*})\pi(\rho_{s_2}...\rho_{s_2...s_n})u_{s_1}^*\\
   &=& u_{s_1}\pi(\rho_{s_1^*}\rho_{s_2}...\rho_{s_2...s_n})u_{s_1^*} \\
   &=& \pi(\alpha_{s_1}(\rho_{s_1^*}\rho_{s_2}...\rho_{s_2...s_n})) \\
   &=& \pi(\rho_{s_1}\rho_{s_1s_2}...\rho_{s_1s_2...s_n}),
\end{eqnarray*}
so, $u_{s_1}...u_{s_n}u_{s_n}^*...u_{s_1}^*$ is a projection since $\rho_{s_1},...,\rho_{s_1...s_n}$ are commute. Finally, the initial space of $u_{s_1}...u_{s_n}$ is equal to
\begin{eqnarray*}
u_{s_1}...u_{s_n}u_{s_n}^*...u_{s_1}^*\mathcal{H}&=&\pi(\rho_{s_1}...\rho_{s_1...s_n})\mathcal{H}  \\
  &=&\pi(D_{s_n^*}D_{s_n^*s_{n-1}^*}...D_{s_n^*...s_1^*})\mathcal{H}.
   \end{eqnarray*}
   Similarly, we can prove that the final space of $u_{s_1}...u_{s_n}$ is equal to $$\pi(D_{s_1}...D_{s_1...s_n})\mathcal{H}.$$
\end{proof}
\begin{corollary}\label{cor2.10}
If ($\pi, u, \mathcal{H} $) is a covariant representation of ($\mathcal{A}, G, \alpha$), then
\begin{equation*}
u_{s_1...s_n}h=u_{s_1}...u_{s_{n}}h\;\text{ for all}\; h\in\pi(D_{s_n^*}...D_{s_n^*...s_1^*})\mathcal{H},
\end{equation*}
and
\begin{equation*}
  \pi(a)u_{s_1...s_n}=\pi(a)u_{s_1}...u_{s_n}\;\text{for all}\; a\in D_{s_1}D_{s_1s_2}...D_{s_1...s_n}.
\end{equation*}
\end{corollary}
\begin{proof}
For $n=2$, if $h\in\pi(D_{s_2^*}D_{s_2^*s_1^*})$ then by Definition \ref{def1} part (2) we have $u_{s_1s_2}h=u_{s_1}u_{s_2}$.
For $h\in \pi(D_{s_n^*}...D_{s_n^*...s_1^*})\mathcal{H}$ we have
$u_{s_1...s_n}h=u_{s_1...s_{n-1}}u_{s_n}h$ by Definition \ref{def1} part (2). Now, since
\begin{eqnarray*}
  u_{s_n}h\in u_{s_n}\pi(\rho_{s_{n}^*}...\rho_{s_n^*...s_1^*})\mathcal{H} &=& \pi(\rho_{s_n^*}\rho_{s_n^*s_{n-1}^*}...\rho_{s_{n-1}^*...s_1^*})\mathcal{H} \\
   &\subseteq& \pi(\rho_{s_{n-1}^*}...\rho_{s_{n-1}^*...s_1^*})\mathcal{H} \\
   &=& \pi(D_{s_{n-1}^*}...D_{s_{n-1}^*...s_1^*})\mathcal{H},
\end{eqnarray*}
by induction hypothesis we have $u_{s_1...s_{n-1}}u_{s_n}h=u_{s_1}...u_{s_{n-1}}u_{s_n}h$. By the first statement, we have
\begin{equation}\label{eq1}
  u_{s_n^*s_{n-1}^*...s_1^*}\pi(a^*)=u_{s_n^*}...u_{s_1^*}\pi(a^*)
\end{equation}
since $\pi(a^*)\in\pi(D_{s_1}...D_{s_1...s_n})$ for $a\in D_{s_1}...D_{s_1...s_n}$. Taking the conjugate, we have $\pi(a)u_{s_1...s_n}=\pi(a)u_{s_1}...u_{s_n}.$
\end{proof}
\begin{corollary}\label{cor2.11}
If $(\pi, u, \mathcal{H})$ is a covariant representation of ($\mathcal{A}, G, \alpha$), then $S=\{ u_{s_1}...u_{s_n}\;:\; s_1,..., s_n\in G\}$ is a unital inverse semigroup of partial isometries of $\mathcal{H}$.
\end{corollary}
Now, we are able to define an inverse semigroup associated to a covariant representation of a unital inverse semigroup $G$.
\begin{proposition}\label{pro3.3}
Let $\alpha$ be a partial action of an inverse semigroup $G$ on the $C^*$-algebra $\mathcal{A}$, and let ($\pi, u. \mathcal{H}$) be a covariant representation of $\alpha$. Let $S_G=\{(\alpha_{g_1}...\alpha_{g_n}, u_{g_1}...u_{g_n})\;:\;g_1,...,g_n\in G\}$. Then $S_G$ is a unital inverse semigroup with coordinate wise multiplication. For $s=(\alpha_{g_1}...\alpha_{g_n}, u_{g_1}...u_{g_n})\in S_G$ let
\begin{equation*}
  E_s=D_{g_1}...D_{g_1...g_n},
\end{equation*}
and
\begin{equation*}
  \beta_s=\alpha_{g_1}...\alpha_{g_n}:E_{s^*}\to E_s.
\end{equation*}
Then $\beta$ is an action of $S_G$ on $\mathcal{A}$.
\end{proposition}
\begin{proof}
Let $s=(\alpha_{g_1}...\alpha_{g_n}, u_{g_1}...u_{g_n}), t=(\alpha_{h_1}...\alpha_{h_n}, u_{h_1}...u_{h_n})$, then $st=(\alpha_{g_1}...\alpha_{g_n}\alpha_{h_1}...\alpha_{h_n}, u_{g_1}...u_{g_n}u_{h_1}...u_{h_n})\in S_G$, and the unit of $S_G$ is $(\alpha_e, u_e)$. Obviously $E_s$ is a closed ideal of $\mathcal{A}$ and $\beta_s$ is an isomorphism. Now, we define the domain of $\beta_s$.
\begin{eqnarray*}
  dom(\alpha_{g_1}...\alpha_{g_n}) &=& \alpha_{g_n^*}(dom(\alpha_{g_1}...\alpha_{g_{n-1}}) D_{g_n}) \\
   &=& \alpha_{g_n^*}(\alpha_{g_{n-1}^*}(dom(\alpha_{g_1}...\alpha_{g_{n-2}})D_{g_{n-1}}) D_{g_n})) \\
   &\vdots&\\
  &=& D_{g_n^*}...D_{g_{n}^*...g_1^*}=E_{s^*}.
\end{eqnarray*}
Now, let us show that $ran\beta_s=E_s$. To do this, we will use induction. For $n=2$,
\begin{eqnarray*}

 ran\beta_s&=&ran \alpha_{g_1}\alpha_{g_2} \\
   &=& \alpha_{g_1}(D_{g_2}D_{g_1^*}) \\
    &=& D_{g_1}D_{g_1g_2}=E_s.
\end{eqnarray*}
On the other hand, for $s=(\alpha_{g_1}...\alpha_{g_n}, u_{g_1}...u_{g_n})$,
\begin{eqnarray*}
  ran\beta_s &=& ran\alpha_{g_1}...\alpha_{g_n} \\
   &=& \alpha_{g_1}(ran(\alpha_{g_2}...\alpha_{g_n})D_{g_{1}^*}) \\
   &=& \alpha_{g_1}(D_{g_2}D_{g_2g_3}...D_{g_2...g_n}D_{g_{1}^*})\\
   &=& D_{g_1}D_{g_1g_2}...D_{g_1...g_n}=E_s.
\end{eqnarray*}
So, $\beta_s:E_{s^*}\to E_s$ is an isomorphism, and clearly for $s,t\in S$ we have $\beta_s\beta_t=\beta_{st}$.
\end{proof}
The following Proposition shows that there exists a relation between covariant representation of $(\mathcal{A}, S_G, \beta)$ and covariant representation of $(\mathcal{A}, G, \alpha).$
\begin{proposition}\label{pro3.5}
keeping the notation of Proposition \ref{pro3.3}, define $\nu:S_G\to B(\mathcal{H})$ by $\nu_s=u_{g_1}...u_{g_n}$, where
$s=(\alpha_{g_1}...\alpha_{g_n}, u_{g_1}...u_{g_n})$. Then $(\pi, \nu, \mathcal{H})$ is a covariant representation of $(\mathcal{A}, S_G, \beta)$. Conversely, if ($\rho, z, \mathcal{K} $) is a covariant representation of $(\mathcal{A} , S_G , \beta)$, then the function $\omega : G\to B(\mathcal{K})$ defined by $\omega_g=z(\alpha_g,u_g)$ gives a covariant representation $(\rho, \omega, \mathcal{K})$ of $(\mathcal{A}, G, \alpha)$.
\end{proposition}
\begin{proof}
Let $s=(\alpha_{g_1}...\alpha_{g_n}, u_{g_1}...u_{g_n})\in S$. By Theorem \ref{theorem2.9}, $\nu_s=u_{g_1}...u_{g_n}$ is a partial isometry with initial space $\pi(E_{s^*})\mathcal{H}$ and final space $\pi(E_s)\mathcal{H}$. Obviously, $\nu$ is multiplicative. Let $a\in E_{s^*}= D_{g_n^*}...D_{g_n^*...g_1^*}$, then
\begin{eqnarray*}
  \nu_s\pi(a)\nu_{s^*} &=& u_{g_1}...u_{g_n}\pi(a)u_{g_n^*}...u_{g_1^*} \\
  &=& u_{g_1}...u_{g_{n-1}}\pi(\alpha_{g_n}(a))u_{g_{n-1}^*}...u_{g_1^*}  \\
  &\vdots&\\
   &=& \pi(\alpha_{g_1}...\alpha_{g_n}(a))=\pi(\beta_s(a)).
\end{eqnarray*}
Conversely, suppose that ($\rho, z, \mathcal{K}$) is a covariant representation of $(\mathcal{A}, S_G, \beta)$. We want to show that ($\rho, \omega, \mathcal{K}$) is a covariant representation of ($\mathcal{A}, G, \alpha$). By the definition of $\omega_g$, $g\in G$, $\omega_g$ is a partial isometry with initial space $\pi(D_{g^*})\mathcal{K}$ and final space $\pi(D_g)\mathcal{K}$. For $g_1, g_2\in G$, put $s=(\alpha_{g_1g_2}, u_{g_1g_2})$, $s_1=(\alpha_{g_1}, u_{g_1})$, and
$s_2=(\alpha_{g_2}, u_{g_2})$. By the definition of partial action, if $x\in D_{g_2^*}D_{g_1^*}$ then $\alpha_{g_1}\alpha_{g_2}(x)=\alpha_{g_1g_2}(x)$. But,
\begin{equation*}
  ran\alpha_{g_2^*}\alpha_{g_1^*}=\alpha_{g_2^*}(D_{g_1^*}D_{g_2})=D_{g_2^*}D_{g_2^*g_1^*}.
\end{equation*}
Consequently,
\begin{equation}\label{eq2}
  \alpha_{g_1g_2}(\alpha_{g_1}\alpha_{g_2})^*=\alpha_{g_1g_2}\alpha_{g_2^*}\alpha_{g_1^*}=\alpha_{g_1}\alpha_{g_2}(\alpha_{g_1}\alpha_{g_2}).
\end{equation}
By Definition \ref{def1} part (2), $u_{g_1g_2}=u_{g_1}u_{g_2}$ on $\pi (D_{g_2^*}D_{g_2^*g_1^*})\mathcal{H}$. On the other hand, by Theorem \ref{theorem2.9} final space of $(u_{g_1}u_{g_2})^*$ is $\pi(D_{g_2^*}D_{g_2^*g_1^*})\mathcal{H}$. Thus
\begin{equation}\label{eq3}
  u_{g_1g_2}(u_{g_1}u_{g_2})^*=u_{g_1}u_{g_2}(u_{g_1}u_{g_2})^*.
\end{equation}
Hence
\begin{eqnarray}\label{eq4}
\nonumber  s(s_1s_2)^* &=& (\alpha_{g_1g_2}, u_{g_1g_2})[(\alpha_{g_1}, u_{g_1})(\alpha_{g_{2}}, u_{g_2}))]^* \\
 \nonumber  &=& (\alpha_{g_1}\alpha_{g_2}(\alpha_{g_1}\alpha_{g_2})^*, u_{g_1}u_{g_2}(u_{g_1}u_{g_2})^*) \\
  &=& s_1s_2(s_1s_2)^*,
\end{eqnarray}
note that we have used equations \ref{eq2} and \ref{eq3} in the second equality above. So,
\begin{eqnarray}\label{eq5}
  \nonumber z_sz_{(s_1s_2)^*}&=& z_{s(s_1s_2)^*} \\
  \nonumber &=& z_{s_1s_2(s_1s_2)^*} \\
  &=& z_{s_1s_2}z_{(s_1s_2)^*}
\end{eqnarray}
by equality \ref{eq4}. Now, for $h\in\rho(D_{g_2^*}D_{g_2^*g_1^*})\mathcal{K}$
\begin{eqnarray*}
  z_s h&=&z_{s_1s_2}h  \\
   &=& z_{s_1}z_{s_2}h
\end{eqnarray*}
note that we have used \ref{eq5} and the fact that $\rho(D_{g_2^*}D_{g_2^*g_1^*})\mathcal{K}$ is the final space of $z_{(s_1s_2)^*}$ in the first equality. Hence $\omega_{g_1g_2}=\omega_{g_1}\omega_{g_2}$ on $\rho(D_{g_2^*}D_{g_2^*g_1^*})\mathcal{K}$. For $g\in G$,
\begin{equation*}
  \omega_{g^*}=z_{(\alpha_{g^*}, u_{g^*})}=z_{s^*}=z^*_s=\omega^*_g.
\end{equation*}
Consequently, $(\mathcal{A}, G, \omega)$ is a covariant representation of $(\mathcal{A}, G, \alpha)$.
\end{proof}
\section{Crossed Products}\label{sec3}
Mc Calanahan defines the partial crossed product $\mathcal{A}\ltimes _{\alpha}G$ of the $C^*$-algebra $\mathcal{A}$ and the group $G$ by the partial action $\alpha$ as the enveloping $C^*$-algebra of $L=\{x\in \ell^1(G,\mathcal{A})\;:\;x(g)\in D_g\}$ with the multiplication and involution
\begin{equation*}
  (x\ast y)(g)=\sum_{h\in G}\alpha_h[\alpha_{h^{-1}}(x(h))y(h^{-1}g)],
\end{equation*}
and
\begin{equation*}
  x^*(g)=\alpha_g(x(g^{-1})^*).
\end{equation*}
He shows that there is a bijective correspondence $(\pi, u, \mathcal{H})\leftrightarrow (\pi\times u, \mathcal{H})$ between covariant representations of $(\mathcal{A}, G, \alpha)$ and non-degenerate representations of $\mathcal{A}\ltimes_{\alpha} G$, where $\pi\times u$ defined by $x\mapsto\sum_{g\in G}\pi(x(g))u_g$. We are going to follow his footsteps constructing the crossed product of a $C^*$-algebra and a unital inverse semigroup by a partial action.\par Let $\alpha$ be a partial action of the unital inverse semigroup $G$ on the $C^*$-algebra $\mathcal{A}$. Consider the subset $L=\{x\in\ell^1(G,\mathcal{A})\;:\;x(g)\in E_g\}$ of $\ell^1(G,\mathcal{A})$ with the multiplication and involution as follows:
\begin{eqnarray*}
  (x\ast y)(g) &=& \sum_{hk=g}\beta_{h}(\beta_{h^*}(x(h))y(k)), \\
  x^*(g) &=& \beta_s(x(g^*)^*).
\end{eqnarray*}
Notice that by Definition \ref{defpar} part (ii), $(x\ast y)(g)\in E_g$. It is easy to see that for $x,y\in L$ we have $x\ast y, x^*\in L$, and
\begin{equation*}
  \| x\ast y \|\leq \|x\|\|y\|,
\end{equation*}
and
\begin{equation*}
  \|x^*\|=\|x\|.
\end{equation*}
Obviously, $L$ is a closed subset of $\ell^1(G, \mathcal{A})$, so, $L$ is a Banach space. Easily one can shows that
\begin{itemize}
  \item [(\textit{i})] $(x+y)^*=x^*+ y^*,$
  \item [(\textit{ii})] $(ax)^*=\bar{a}x^*,$
  \item [(\textit{iii})] $(x\ast y)^*=y^*\ast x^*.$
\end{itemize}
\begin{proposition}\label{pro4.1}
$L$ is a Banach *-algebra.
\end{proposition}
\begin{proof}
By the argument above, $L$ is a Banach space closed under multiplication and involution. To show that $L$ is a Banach *-algebra, it is enough check the associativity of multiplication. It suffices to show this for $x=a_r\delta_r, y=a_s\delta_s,$ and $a_t\delta_t$. Let $\{u_{\lambda}\}$ be an approximate identity for $E_{s^*}$, then
\begin{eqnarray*}
  (a_r\delta_r\ast a_s\delta_s)\ast a_t\delta_t &=& \beta_r(\beta_{r^*}(a_r)a_s)\delta_{rs}\ast a_t\delta_t \\
  &=& \beta_{rs}(\beta_{s^*r^*}(\beta_r(\beta_{r^*}(a_r)a_s))a_t)\delta_{rst} \\
   &=&  \beta_{rs}(\beta_{s^*}\beta_{r^*}(\beta_r(\beta_{r^*}(a_r)a_s))a_t)\delta_{rst} \\
   &=& \beta_{rs}(\beta_{s^*}(\beta_{r^*}(a_r)a_s)a_t)\delta_{rst} \\
   &=&  \lim_{\lambda}\beta_{rs}(\beta_{s^*}(\beta_{r^*}(a_r)a_s)u_{\lambda}a_t)\delta_{rst}\\
   &=&  \lim_{\lambda}\beta_{r}\beta{s}(\beta_{s^*}(\beta_{r^*}(a_r)a_s)a_t)\delta_{rst}\\
   &=&\lim_{\lambda}\beta_r(\beta_{r^*}(a_r)a_s\beta_{s}(u_{\lambda}a_t))\delta_{rst}\\
   &=&\lim_{\lambda}\beta_r(\beta_{r^*}(a_r)\beta_{s}(\beta_{s^*}(a_s)u_{\lambda}a_t))\delta_{rst}\\
   &=&\beta_r(\beta_{r^*}(a_r)\beta_{s}(\beta_{s^*}(a_s)a_t))\delta_{rst}\\
   &=& a_r\delta_r\ast(\beta_{s}(\beta_{s^*}(a_s)a_t)))\delta_{st}\\
   &=& a_r\delta_r\ast(a_s\delta_s\ast a_t\delta_t).
\end{eqnarray*}
\end{proof}
Note that authors in \cite{skew} prove the associativity of $L$ in a general case, where $\mathcal{A}$ is just an algebra.
\begin{definition}\label{def4.2}
If $(\pi, \nu. \mathcal{H})$ is a covariant representation of ($\mathcal{A}, S, \beta$), then define
  $\pi\times\nu: L\mapsto B(\mathcal{H})$ by $(\pi\times\nu)(x)=\sum_{s\in S}\pi(x(s))\nu_s.$
\end{definition}
\begin{proposition}\label{pro4.3}
$\pi\times\nu$ is a non-degenerate representation of $L$.
\end{proposition}
\begin{proof}
clearly $\pi\times\nu$ is a linear map from $L$ into $B(\mathcal{H})$. As for multiplicativity, it suffices to verify this for elements of the form $a_s\delta_s$. For such elements, we have
\begin{eqnarray*}
  \pi\times\nu(a_s\delta_s\ast a_t\delta_t) &=& \pi\times\nu(\beta_s(\beta_{s^*}(a_s)a_t)\delta_{st}) \\
  &=& \pi(\beta_s(\beta_{s^*}(a_s)a_t)\nu_{st}.
\end{eqnarray*}
We also have
\begin{eqnarray*}
  \pi\times\nu(a_s\delta_s)\pi\times\nu(a_t\delta_t) &=& \pi(a_s)\nu_s\pi(a_t)\nu_t \\
   &=& \nu_s\nu_{s^*}\pi(a_s)\nu_s\pi(a_t)\nu_t \\
  &=& \nu_s\pi(\beta_{s^*}(a_s))\pi(a_t)\nu_t \\
   &=& \nu_s\pi(\beta_{s^*}(a_s)a_t)\nu_t \\
   &=& \nu_s\pi(\beta_{s^*}(a_s)a_t)\nu_{s^*}\nu_s\nu_t \\
   &=& \pi(\beta_s(\beta_{s^*}(a_s)a_t))\nu_s\nu_t\\
\end{eqnarray*}
We have used the fact that $\nu_s\nu_{s^*}\pi(a_s)=\pi(a_s)\nu_{s^*}\nu_s=\pi(a_s)$ for $a\in E_s$ in the second and fifth equalities above. Since $\beta_s(\beta_{s^*}(a_s)a_t)$ is in $\beta_s(E_{s^*}E_{t})=E_sE_{st}$, it follows from Definition \ref{def1} that
\begin{equation*}
  \pi(\beta_s(\beta_{s^*}(a_s)a_t))\nu_s\nu_t=\pi(\beta_s(\beta_{s^*}(a_s)a_t))\nu_{st},
\end{equation*}
 so, the multiplicativity of $\pi\times\nu$ follows. The following computations verify that $\pi\times\nu$ preserves the $*$-operation.
\begin{eqnarray*}
  \pi\times\nu((a_s\delta_s)^*) &=& \pi\times\nu(\beta_{s^*}(a_s^*)\delta_{s^*}) \\
  &=& \pi(\beta_{s^*}(a_s^*))\nu_{s^*} \\
   &=& \nu_{s^*}\pi(a_{s}^*)\nu_s\nu_{s^*} \\
   &=& \nu_{s^*}\pi(a_s^*) \\
   &=& (\pi(a_s)\nu_s)^*=(\pi\times\nu(a_s\delta_s))^*.
\end{eqnarray*}
If $\{u_{\lambda}\}$ is a bounded approximate identity for $\mathcal{A}$, then $\{u_{\lambda}\delta_e\}$ is a bounded approximate identity for $L$ since for $a\in E_s$ we have
\begin{equation*}
  \lim_{\lambda}u_{\lambda}\delta_e\ast a\delta_s=\lim_{\lambda}u_{\lambda}a\delta_s=a\delta_s,
\end{equation*}
and
\begin{equation*}
  \lim_{\lambda}a\delta_s\ast u_{\lambda}\delta_e=\lim_{\lambda}\beta_s(\beta_{s^*}(a)u_{\lambda})\delta_s=a\delta_s.
\end{equation*}
Since $\pi$ is a non-degenerate representation, $\pi\times\nu(u_{\lambda}\delta_e)=\pi(u_{\lambda})$ converges strongly to $1_{B(\mathcal{H})}$ and so $\pi\times\nu$ is non-degenerate.
\end{proof}
\begin{definition}
Let $\mathcal{A}$ be a $C^*$-algebra and $\beta$ be a partial action of the unital inverse semigroup $G$ on $\mathcal{A}$. Define a seminorm $\|.\|_1$ on $L$ by
\begin{equation*}
  \|x\|_1=\sup\{\|\pi\times\nu(x)\|\;:\;(\pi,\nu)\textsl{is a covariant representation of } (\mathcal{A}, G, \beta)\},
\end{equation*}
and let $N=\{x\in L\;:\;\|x\|_1=0\}$
\end{definition}
The crossed product $\mathcal{A}\ltimes_{\beta} G$ is the $C^*$-algebra obtained by completing the quotient $\frac{L}{N}$ with respect to $\|.\|_1.$
\begin{lemma}\label{lem4.5}
If $s\leq t$ in $G$, then $\Phi(a\delta_s)=\Phi(a\delta_t)$ for all $a\in E_s$, where $\Phi$ is the quotient map of $L$ onto $\frac{L}{N}$.
\end{lemma}
\begin{proof}
Notice that since $s\leq t$  there is an idempotent $f$ in $G$ such that $s=ft$, and we have $E_s\subseteq E_t$ by \cite[Proposition 3.8]{EXPAN}, so, $a\in E_t$. If ($\pi, \nu$) is a covariant representation of ($\mathcal{A}, G, \beta$), then
\begin{eqnarray*}
  \pi\times\nu(a\delta_s- a\delta_t) &=& \pi(a)\nu_s - \pi(a)\nu_t \\
   &=& \pi(a)\nu_{ft} - \pi(a)\nu_t \\
   &=& \pi(a)\nu_f\nu_t- \pi(a)\nu_t.
\end{eqnarray*}
We have used the fact that for $a\in E_s$ $\pi(a)\nu_{ft}=\pi(a)\nu_f\nu_t$ in the the third equality. Since $f$ is an idempotent, $\nu_f$ is identity on $\pi(E_f)\mathcal{H}$. Now for $h\in \mathcal{H}$ if $\nu_t(h)\in \pi(E_f)\mathcal{H}$, then
\begin{equation*}
  \pi(a)\nu_f\nu_t(h) - \pi(a)\nu_t(h)=0.
\end{equation*}
If $\nu_t(h)\in (\pi(E_f)\mathcal{H})^\bot= Ker\hspace{0.1cm} \nu_f$, then
\begin{equation*}
  \pi(a)\nu_f\nu_t(h)=0.
\end{equation*}
On the other hand, $\pi(a)\nu_t(h)=0$ because if $k\in\mathcal{H}$ then
\begin{equation*}
  <\pi(a)\nu_t(h), k>= <\nu_t(h), \pi(a^*)k>=0
\end{equation*}
since $a^*\in E_s=E_{ft}\subseteq E_f$. Hence, $\Phi(a\delta_s - a\delta_t)=0$. Note that the fact that $E_{ft}\subseteq E_f$ follows from \cite[Corollary 2.21]{EXPAN} and the fact that
\begin{equation*}
  E_{ft}= ran \beta_{ft}=\beta_f\beta_t=\beta_f(E_tE_f)=E_{ft}\cap E_f.
\end{equation*}

\end{proof}
\begin{corollary}\label{cor4.6}
If $G$ is a semilattice, then $\mathcal{A}\ltimes_\beta G$ is isomorphic to $\mathcal{A}$.
\end{corollary}
\begin{proof}
Let $e$ be the identity element of $G$, then $g\leq e$ for each $g\in G$. Define $\psi_1: \mathcal{A}\to \mathcal{A}\ltimes_\beta G$ by $a\mapsto \Phi(a\delta_e)$. Obviously $\psi_1$ is a $*$-homomorphism. Now, define $\psi_2: \frac{L}{N}\to \mathcal{A}$ by $\Phi(a\delta_g)\mapsto a$. Now, we will show that  $\psi_2$ is well-defined. If $\Phi(a\delta_e)=\Phi(b\delta_e)$, then for each covariant representation ($\pi, \nu, \mathcal{H}$) we have
\begin{equation*}
  \pi\times\nu(a\delta_e - b\delta_e)= \pi(a - b)=0,
\end{equation*}
so, $a - b=0$ since $\mathcal{A}$ has a universal representation. This shows that $\psi_1$ is well-defined  since  $\Phi(a\delta_g)=\Phi(a\delta_e)$ for each $g\in G$. Clearly, $\psi_2$ is a $*$-homomorphism that can be extended to $\mathcal{A}\ltimes_\beta G$. Finally, it is easy to see that $\psi_1\circ\psi_2$ and $\psi_2\circ\psi_1$ are identity maps on $\mathcal{A}\ltimes_\beta G$ and $\mathcal{A}$ respectively.
\end{proof}
\begin{proposition}\label{pro4.7}
Let $(\Pi,\mathcal{H})$ be a non-degenerate representation of $\mathcal{A}\ltimes_\beta G$. Define a representation $\pi$ of $\mathcal{A}$ on $\mathcal{H}$ and a map $\nu:S\to B(\mathcal{H})$ by \begin{equation*}
  \pi(a)=\Pi(a\delta_e),\hspace{0.5cm} \nu_s=\lim_\lambda \Pi(u_\lambda\delta_s)\rho_{s^*},
\end{equation*}
where $\{u_\lambda\}$ is an approximate identity of $E_s$, limit is the strong limit, and $\rho_{s^*}$ is the orthogonal projection onto $\pi(E_{s^*})\mathcal{H}$. Then $(\pi, \nu, \mathcal{H})$ is a covariant representation of $(\mathcal{A}, G, \beta)$.
\end{proposition}
\begin{proof}
Clearly $\pi$ is a representation of $\mathcal{A}$ on $\mathcal{H}$. Now, let $\{u_\lambda \}$ be an approximate identity for $E_s$, and let $h\in\mathcal{H}$. We will consider two cases:\\
If $h\in\pi(E_{s^*})\mathcal{H}$: then there exist elements $a\in E_{s^*}$ and $h'\in \mathcal{H}$  such that  $h=\pi(a)h'$. So,
\begin{eqnarray*}
  \nu_s(h) &=& \lim_\lambda\Pi(u_\lambda\delta_s)(\Pi(a\delta_e)h') \\
   &=& \lim_\lambda\Pi(u_\lambda\delta_s\ast a\delta_e)h' \\
   &=&\lim_\lambda \Pi(\beta_s(\beta_{s^*}(u_{\lambda})a)\delta_s)h' \\
  &=& \Pi(\beta_s(a)\delta_s)h'.
\end{eqnarray*}
If $h\in (\pi(E_{s^*})\mathcal{H})^\bot$: by  the definition we have
 \begin{eqnarray*}

   \nu_s&=& \lim_\lambda\Pi(u_\lambda\delta_s)\rho_{s^*}h=0. \\
 \end{eqnarray*}
 This show that $\nu_s$ is independent of the choice of approximate identity of $E_s$, so $\nu$ is well-defined. Now, we want to show that $\nu^*_s=\nu_{s^*}$ for $s\in S$. First we remark that for $a_s\in E_s$ we have $\Pi(a_s\delta_s)\rho_{s^*}=\rho_s\Pi(a_s\delta_s)$. Let $\{u_\lambda \}$ be an approximate identity for $E_s$. It follows that
 \begin{eqnarray*}

   (\nu_s)^* &=& \lim_\lambda(\Pi(u_\lambda\delta_s)\rho_{s^*})^* \\
    &=& \lim_\lambda\rho_{s^*}\Pi(\beta_{s^*}(u_\lambda)\delta_{s^*}) \\
    &=&\lim_\lambda \Pi(\beta_{s^*}(u_\lambda)\delta_{s^*})\rho_{s}\\
    &=&\nu_{s^*}
 \end{eqnarray*}
 since $\{\beta_{s^*}(u_\lambda)\}$ is an approximate identity for $E_{s^*}$. As for the covariance condition, let $x\in E_{s^*}$ and observe that
 \begin{eqnarray*}

   \nu_s\pi(x)\nu_{s^*} &=& \lim_{\lambda,\mu}\rho_s\Pi(u_\mu\delta_s)\Pi(x\delta_e)\Pi(\beta_{s^*}(u_\lambda)\delta_{s^*})\rho_s \\
    &=& \lim_{\mu,\lambda}\rho_s\Pi(u_\mu\delta_s\ast x\delta_e\ast \beta_{s^*}(u_\lambda)\delta_{s^*}) \rho_s\\
    &=& \lim_{\mu,\lambda}\rho_s\Pi(u_\mu\beta_s(x)u_\lambda\delta_{ss^*})\rho_s \\
    &=& \lim_{\mu,\lambda}\rho_s\Pi(u_\mu\beta_s(x)u_\lambda\delta_{e})\rho_s \\
    &=& \rho_s\pi(\beta_s(x))\rho_s \\
    &=& \pi(\beta_s(x)).
 \end{eqnarray*}
It should be noted that we have used the fact that $\Pi\equiv0$ on $N$ in the forth equality above. As for property (2) of Definition \ref{def1}, notice that for $a_s\in E_s$ we have
 \begin{eqnarray*}

   \Pi(a_s\delta_s) &=& \lim_\lambda\Pi(a_su_\lambda\delta_s) \\
    &=& \lim_\lambda\Pi(a_s\delta_e\ast u_\lambda\delta_s) \\
    &=& \pi(a_s)\rho_s\lim_\lambda\Pi(u_\lambda\delta_s) \\
    &=& \pi(a_s)\lim_\lambda\Pi(u_\lambda\delta_s)\rho_{s^*} \\
    &=& \pi(a_s)\nu_s.
 \end{eqnarray*}
 Thus,
 \begin{eqnarray*}

   \Pi(a_s\delta_s)\Pi(a_t\delta_t) &=& \pi(a_s)\nu_s\pi(a_t)\nu_t \\
    &=& \nu_s\nu_{s^*} \pi(a_s)\nu_s\pi(a_t)\nu_t \\
    &=& \nu_s\pi(\beta_{s^*}(a_s)a_t)\nu_t \\
    &=& \nu_s\pi(\beta_{s^*}(a_s)a_t)\nu_{s^*}\nu_s\nu_t \\
    &=& \pi(\beta_s(\beta_{s^*}(a_s)a_t))\nu_s\nu_t.
 \end{eqnarray*}
 Because $\Pi$ is multiplicative, the above expression is the same as
 \begin{eqnarray*}

   \Pi(a_s\delta_s\ast a_t\delta_t) &=& \Pi(\beta_s(\beta_{s^*}(a_s)a_t)\delta_{st}) \\
   &=& \pi(\beta_s(\beta_{s^*}(a_s)a_t))\nu_{st}.
 \end{eqnarray*}
 Elements of the form $\beta_{s^*}(a_s)a_t$ generate $E_{s^*}E_t$. Since $\beta_s$ maps $E_{s^*}E_t$ onto $E_sE_{st}$, it follows that elements of the form $\beta_s(\beta_{s^*}(a_s)a_t)$ generate $E_sE_{st}$ and so property (2) of Definition \ref{def1} follows. Clearly, $\pi$ is a non-degenerate representation of $\mathcal{A}$. Thus ($\pi, \nu, \mathcal{H}$) is a covariant representation of ($\mathcal{A}, S, \beta$).
 \end{proof}
 \begin{proposition}\label{pro4.8}
 The correspondence $(\pi, \nu , \mathcal{H})\leftrightarrow (\pi\times\nu, \mathcal{H})$is a bijection between covariant representations of ($\mathcal{A}, S, \beta$) and non-degenerate representations of $\mathcal{A}\ltimes_{\beta} S$.
 \end{proposition}
 \begin{proof}
 We will show that the correspondences $(\pi, \nu, \mathcal{H})\mapsto(\pi\times \nu, \mathcal{H})$ and $(\Pi,\mathcal{H})\mapsto(\pi, \nu, \mathcal{H})$ are inverses of each other. Let ($\pi', \nu', \mathcal{H}$) be a covariant representation of ($\mathcal{A}, S, \beta$). Let ($\pi, u, \mathcal{H}$) be a covariant representation of ($\mathcal{A}, S, \beta$) induced by $\pi'\times\nu'$. Then for $a\in\mathcal{A}$ and $s\in S$ we have
 \begin{equation*}
   \pi(a)=\pi'\times\nu'(a\delta_e)=\pi'(a)
 \end{equation*}
 and
 \begin{eqnarray*}
   u_s&=&\lim_\lambda\rho_s\pi'\times\nu'(\omega_\lambda\delta_s)\\
   &=&\lim_\lambda\rho_s\pi'(\omega_\lambda)\nu'_s\\
   &=&\lim_\lambda\pi'(\omega_\lambda)\nu'_s=\nu'_s.
 \end{eqnarray*}
 We have used the fact that $\rho_s\pi'(\omega_\lambda)\nu'_s=\pi'(\omega_\lambda)\nu'_s$ since $\rho_s$ is the orthogonal projection onto $\pi'\times\nu'(E_s)\mathcal{H}=\bar{\textsf{Span}}\{\pi'(a_s)\nu'_s\;:\;a_s\in E_s\}$. Let $\Pi$ be a non-degenerate representation of $\mathcal{A}\ltimes_\beta S$ on $\mathcal{H}$. Let $(\pi, \nu, \mathcal{H})$ be a covariant representation of ($\mathcal{A}, S, \beta$) induced by $\Pi$. Then if $a_s\in E_s$ we have
 \begin{eqnarray*}

   \pi\times\nu(a_s\delta_s) &=& \pi(a_s)\nu_s \\
    &=& \Pi(a_s\delta_e)\lim_\lambda\Pi(u_\lambda\delta_s)\rho_{s^*} \\
    &=& \Pi(a_s\delta_e)\rho_s\lim_\lambda\Pi(u_\lambda\delta_s) \\
    &=& \Pi(a_s\delta_e)\lim_\lambda\Pi(u_\lambda\delta_s)\\
   &=&\lim_\lambda\Pi(a_su_\lambda\delta_s)=\Pi(a_s\delta_s).
 \end{eqnarray*}
 Thus the correspondence is bijective.
  \end{proof}
 \section{Conection Between Croossed Products}\label{five}
 Throughout this section we will assume that $G$ is an inverse semigroup with unit element $e$.
 \begin{lemma}\label{lem5.1}
 Let ($\mathcal{A}, G, \alpha$) and ($\mathcal{A}, S_G, \beta$) be as in Proposition \ref{pro3.3}. Let ($\rho, z, \mathcal{K}$) be a covariant representation of ($\mathcal{A}, S_G, \beta$), and define a covariant representation ($\rho, \omega, \mathcal{K}$) of ($\mathcal{A}, G, \alpha$) by $\omega_g=z(\alpha_g, u_g)$ as in Proposition \ref{pro3.5}. Then $(\rho\times z)(\mathcal{A}\ltimes_\beta S)= (\rho\times\omega)(\mathcal{A}\times_\alpha G)$.
 \end{lemma}
\begin{proof}
For $g\in G$, let $s=(\alpha_g, u_g)\in S$, then $E_s=D_g$, so, $\rho(D_g)\omega_g=\rho(E_s)z_s$. Thus,
\begin{equation}\label{eq}
  \sum_{g\in G} \rho(D_g)\omega_g\subseteq\sum_{s\in S}\rho(E_s)z_s.
\end{equation}
On the other hand, if
\begin{equation*}
  s=(\alpha_{g_1}...\alpha_{g_n}, u_{g_1}...u_{g_n})\;\text{and}\;a\in E_s=D_{g_1}D_{g_1g_2}...D_{g_1...g_n}
\end{equation*}
then by Corollary \ref{cor2.10} we have
\begin{eqnarray}\label{eqn}
  \nonumber \rho(a)z_s&=&\rho(a)z_{(\alpha_{g_1}, u_{g_1})...(\alpha_{g_n}, u_{g_n})}  \\
  \nonumber &=& \rho(a)z_{(\alpha_{g_1}, u_{g_1})}...z_{(\alpha_{g_n}, u_{g_n})}\\
 &=& \rho(a)\omega_{g_1}...\omega_{g_n}.
\end{eqnarray}
Let $\Phi(\sum a_g\delta_g)\in\frac{L}{N}$. Then by \ref{eq} we have
\begin{equation*}
  \rho\times\omega(\Phi(\sum a_g\delta_g))=\sum\rho(a_g)\omega_g\subseteq (\rho\times z)(\mathcal{A}\ltimes_\beta S),
\end{equation*}
 so, $(\rho\times\omega)(\mathcal{A}\ltimes_\alpha G)\subseteq (\rho\times z)(\mathcal{A}\ltimes_\beta S)$. If $\Phi(\sum a_s\delta_s)\in \mathcal{A}\ltimes _\beta S$, then
\begin{equation*}
  \rho\times z(\Phi(\sum a_s\delta_s))=\sum \rho(a_s)z_s\in \rho\times\omega(\mathcal{A}\ltimes_\alpha G)
\end{equation*}
by \ref{eqn}.
\end{proof}
\begin{theorem}\label{theorem5.2}
Let $\alpha$ be a partial action of a unital inverse semigroup $G$ on a $C^*$-algebra $\mathcal{A}$ such that the representation $\pi\times u$ of $\mathcal{A}\ltimes G$ is faithful. Define an inverse semigroup $S_G$ by $S_G=\{(\alpha_{g_1}...\alpha_{g_n}, u_{g_1}...u_{g_n})\;:\;g_1,...,g_n\in G\}$ and an action $\beta$ of $S_G$ by $\beta_s=\alpha_{g_1}...\alpha_{g_n}$ for $s=(\alpha_{g_1}...\alpha_{g_n}, u_{g_1}...u_{g_n})$, as in Proposition \ref{pro3.3}. Then the crossed product $\mathcal{A}\ltimes_\alpha G$ and $\mathcal{A}\ltimes_\beta S$ are isomorphic.
\end{theorem}
\begin{proof}
Let $\nu_s=u_{g_1}...u_{g_n})$ for $s=(\alpha_{g_1}...\alpha_{g_n}, u_{g_1}...u_{g_n})$. We know from Proposition \ref{pro3.5} $(\pi, \nu, \mathcal{H})$ is a covariant representation of $(\mathcal{A}, S, \beta)$. If we show that $\pi\times\nu$ is a faithful representation of $\mathcal{A}\ltimes_\beta S$, then $(\pi\times\nu)^{-1}\circ\pi\times\nu$ is an isomorphism. Consider the universal representation of $\mathcal{A}\ltimes_\beta S$, which by proposition \ref{pro4.8} must be in the form $\rho\times z$ for some covariant representation $(\rho, z)$ of $(\mathcal{A}, S, \beta)$. By Proposition \ref{pro3.5} the definition $\omega_g= z_{(\alpha_g, u_g)}$ gives a covariant representation $(\rho, \omega, \mathcal{K})$ of $(\mathcal{A}, G, \alpha)$ and we have $(\rho\times\omega)(\mathcal{A}\ltimes_\alpha G)= (\rho\times z)(\mathcal{A}\ltimes_\beta S)$ by Lemma \ref{lem5.1}. Put $\Theta(x)=(\rho\times\omega)(\pi\times u)^{-1}(x)$, thus, $\Theta\circ\pi\times u=\rho\times\omega$. We will show that $\Theta\circ(\pi\times\nu)=\rho\times z$. It suffices to check this on generators $a\delta_s$, where $s=(\alpha_{g_1}...\alpha_{g_n}, u_{g_1}...u_{g_n})$ and $a\in E_s=D_{g_1}...D_{g_1...g_n}$.
\begin{eqnarray*}
  \Theta((\pi\times\nu)(a\delta_s)) &=& \Theta(\pi(a)\nu_s) \\
   &=& (\rho\times\omega)(\pi\times u)^{-1}(\pi(a)\nu_s)\\
   &=& (\rho\times\omega)(\pi\times u)^{-1}(\pi(a)u_{g_1}...u_{g_n}) \\
   &=&  (\rho\times\omega)(\pi\times u)^{-1}(\pi(a)u_{g_1...g_n})\\
   &=& (\rho\times\omega)(a\delta_{g_1...g_n})\\
   &=&  \rho(a)\omega_{g_1...g_n}\\
   &=&  \rho(a)\omega_{g_1}...\omega_{g_n}\\
   &=& \rho(a)z_{(\alpha_{g_1},u_{g_1})}...z_{(\alpha_{g_n}, u_{g_n})}\\
   &=&\rho(a)z_{(\alpha_{g_1}...\alpha_{g_n},u_{g_1}... u_{g_n})}\\
   &=&\rho\times z (a\delta_s)
\end{eqnarray*}
where we have appealed to Corollary \ref{cor2.10} twice more.
\end{proof}

redpajama_set_name: RedPajamaArXiv


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Welcome to the Forum. Some of us here are in to the very old Studebakers. The majority are hyped by hydraulic brakes and electric windshield wipers.
You may need patience for specifics on a car from the twenties but the assistance you need is available.
I would highly recommend you join the Antique Studebaker Club. We cater to those who own or admire pre-war Studebakers, and the bi-monthly ANTIQUE STUDEBAKER REVIEW is your best resource for parts and people who share your interest.
It is edited by Richard Quinn, who posts on this Forum as Studebakerwheel.


Today I worked to get the web page in order by adding a new Alpaca page. It is short but it tells the story of my getting them. Then I got all the work done on the business that I was putting off. All I have to do is get things mailed and I am caught up with everything.
Okay, so there is one thing that I am putting off and that is laying out my booth for my very first fiber festival. Mom said she will help me but that just gives me another reason to wait. I do have until July to figure it out so it isn't like it is a rush or anything.
This afternoon I had my Rodeo at the flying cow. I really didn't do much good in the team roping but I placed 2nd in the tie down and 3rd in the ribbon roping. Oh and I ended up buying another rope. Maybe it will work better for me and will get more calves caught.
I can't wait to see the jackets on the sheep! I think they look so silly wearing them, I can't help but find them adorable.


In a recent conversation, the idea of an analytics tool was discussed. As usual, Tableau is the clear leader in the space. However, it does pose a challenge for those not willing (or at least timid) to learn its vast capabilities. For some, the learning curve was just too steep. While that sounds crazy to most of us, it still remains that some will not make the effort to learn a new tool; rather, they just want to benefits of the tool presented to them without any serious cognitive dissonance.
Imagine your elite Tableau developers create a set of template/standard workbooks. These can contain anything from time series charts to scatter plots (and everything between). They're deep and elaborate and complex. They're the essence of Tableau.
Not a problem, Tableau Working Wax is an automated platform which enables users to choose a report, chart type, filter (via url parameters), frequency of delivery and delivery method. That's it. After that, the report is sent to them in seconds at an interval of their choosing (daily, weekly or monthly).
I'll start putting code segments here.
The key is to have your Tableau team create a template workbook(s) first. Once that's done, you're going to leverage all of the neat functionality in Tableau (tabcmd, rest api, url filtering, and visualizations).
The config file. Users enter in their order and Tableau is delivered.
Bob the Builder thinks it would be really neat to have a report about his construction inventory sent to him. He knows what it looks like in his head; he wants a timeline and a cross-tab of inventory data. Suddenly, he remembers there is a document he can fill out that will simply let him choose a few configurations and then, viola, it will be sent to him. He didn't need to learn all about Tableau (but he knows he should have).
Nothing. We've literally scaled this to 100s of distinct reports and all users have to do is adjust some configurations on a file.
For a demo, please just let me know!


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Need one good reason to get outside and embrace the cold? Snow tubing. The slick rides are like sledding—on steroids. No special skills are required. And the colder it gets, the faster you'll fly down the slopes: at speeds of up to 20 miles per hour. "As the snow freezes more, it gets a little icier," says Alex Cole, a manager at Nashoba Valley Tubing Park, in Littleton, Massachusetts. There, 18 tubing lanes, each nearly a quarter of a mile long, are packed with man-made (and some natural) snow and extend along two sides of the hill, at a 100-foot vertical drop.
The park—the largest snow-tubing venue in New England—is part of the family-run Nashoba Valley Ski Area, founded in 1964. Tubing lanes opened on an adjacent hill in 2001, and quickly became a popular intergenerational winter activity. Families and kids come in droves, especially during school-vacation weeks, but so do groups of young adults and "the occasional older couple," Cole adds. "We have thousands of people every week for tubing; in a season, we could see 50,000." Tips: go early or late in the day to avoid crowds; two-hour individual tickets are $35; group-rate discounts are available; and night-time tubing, until 10 p.m., ups the thrill factor.
Other snow-tubing sites close to Boston include Ski Ward, in Shrewsbury, Massachusetts, and McIntyre Ski Area, in Manchester, New Hampshire. Those are smaller and perhaps tamer, but nonetheless offer fresh air and exercise, human contact, and a fun day out: all helpful combatants against winter doldrums.


text:
Is Decorum Gone In Washington?
12 Aug, 2010 by	William Teach
That's a question that ABC News asks on the front page of their website. And, if guessed that they would simply blame Republicans, nope, sorry, you owe me a shot of Tequila
There was an extraordinary moment during the final senate debate on Elena Kagan's Supreme Court confirmation. While Senator Mitch McConnell (R-KY) delivered remarks, Senator Al Franken (D-MN) was gasping, shaking his head, and commenting under his breath. Franken's displeasure with McConnell's speech was so blatant that the Kentucky Republican felt compelled to confront him afterwards.
According to Republican aides who overheard, Republican leader McConnell admonished Senator Franken saying, "this is not Saturday Night Live."
Franken later apologized, but the incident certainly raises the question: is the tradition, and indeed, the rule of decorum dying in Congress?
Remember Senator Carl Levin's generous use of the "s"-word during a Senate Subcommittee on Investigations hearing with Goldman Sachs? Originally reading from an e mail and then questioning Goldman exec Daniel Sparks, the Senator used a word we can't even say on TV (rhymes with witty) 21 times in a single hearing.
Also mentioned is the confrontation between Patrick Leahy and Dick Cheney (the "F-bomb" one), Anthony Weiner's unhinged rant aimed at Peter King, Patrick Kennedy losing it on the House floor, and Joe Wilson's "you lie". Missing, of course, was Obama's slaps thrown at the SCOTUS members sitting right in front of him during has SOTU speech.
But……is it really that bad?
According to the office of the House Historian, the first case of violence on the House floor happened when Congress was meeting in Philadelphia in 1798. Representative Matthew Lyon of Vermont verbally assaulted and spat on Representative Roger Griswold of Connecticut on the House floor. Griswold retaliated by striking Lyon with a cane. Then Lyon found a pair of fireplace tongs and began attacking Griswold.
Also mention: one Congress member shooting another in a duel, and another beating the snot out of a fellow member with a cane. So, it appears that, perhaps, ABC News is attempting to cover for the unhinged and ranting members of Congress, usually Democrats, by saying "hey, at least they aren't fighting! And dueling!"
So, has decorum been lost? Well, there's always been conflict in Congress, and in American politics. Remember, then VP Aaron Burr engaged in a duel with former Treasury Sec. Alexander Hamilton. William Howard Taft dueled an attendee of a speech in hand to hand cage match no holds barred combat, used a choke hold, then crushed his opponents spine, killing him. Andrew Jackson dueled another man. But, these incidents, and the incidents in Congress, were generally accompanied by flowing, gracious language, and a protocol method long established. Back then, there were only newspapers to report the incidents. Now, with TV and the Internet, we have politicians and their peeps saying the most amazing things, sometimes on purpose, sometimes by accident, all caught for posterity (or notoriety). And you have a Party with a completely unhinged base, allowing for incidents like Harry Reid saying the Iraq war is lost, and calling the President a loser. You have a president who demeans private citizens and blames everyone.
So, has decorum been lost? You betcha. Are Republicans blameless? No, but, the issues are generally from Democrats. Heck, you have unhinged Alan Grayson calling Robert Gibbs "Bozo The Spokesman." Nor does it stop with the federal crazies. In New Hampshire, you have Keith Halloran, a Democrat candidate for the State House, wishing death on Sarah Palin. So, maybe I'm wrong, and it's not so much that decorum has been lost, just that one Party has gone stark raving bat guano.
Crossed at Pirate's Cove. Follow me on Twitter @WilliamTeach


text:
The checks and balances of Charlie Baker
PUBLISHED: February 12, 2010 at 12:00 a.m. | UPDATED: July 13, 2019 at 12:00 a.m.
Simpler and smaller.
That's the way Republican Charlie Baker sees state government.
Hardly were these words out of his mouth than Gov. Deval Patrick and his new friend, Senate President Therese Murray, who once said the governor was "irrelevant," teamed up to propose a streamlining of state agencies to help small businesses create jobs and stimulate economic development.
"I'm thrilled to see this come about," Baker said. "It confirms what I've been saying all along. We've got to cut the size of government, period. We are spending at an unsustainable rate. We've got to make government simpler and smaller."
Baker is running for governor and he believes he can defeat troubled Democratic incumbent Patrick, who is seeking re-election. But first Baker must get by gadfly Republican businessman Christy Mihos in the GOP primary. Mihos is running for governor for the second time.
Should that occur, as is expected, Baker will face Patrick and outgoing state Treasurer Timothy Cahill in the November election. Cahill left the Democratic Party and is running for governor as an independent candidate.
Baker shrugs off Cahill's candidacy, which has signs of being a spoiler candidacy that could enable Patrick to squeak back into office in the three-way battle.
"I don't spend too much time on the horse-race issue," Baker said. "If I do what I need to do, and do it well, I have as good a shot at being elected."
Baker, once a Swampscott selectman, left his job as CEO of Harvard Pilgrim Health Care to make the run for governor. Prior to working in the health-care industry, Baker served as secretary of Administration and Finance in the administrations of Republican Govs. William Weld and Paul Cellucci. In that capacity, Baker put together more state budgets than the incumbent governor has. With state Sen. Richard Tisei as his running mate, the pair bring to the campaign more political and business experience than the rest of the field of candidates for governor and lieutenant governor combined.
Liberals have attempted to link Baker's conservative fiscal outlook with that of former Republican Gov. Mitt Romney, as though that were a bad thing. Romney was a penny-pinching conservative when it came to spending taxpayers' money. You can be critical of Romney for a lot of things — like using the governor's office as a steppingstone to run for president — but when he was governor, Romney did not raise taxes, did away with political patronage and refused to make deals with lobbyists or with anyone else. In fact, he didn't even take the salary.
When there was a Republican governor of Massachusetts — as there was for 16 years before Patrick was elected — you had a good system of checks and balances between a Republican governor and a Democratic Legislature. With the election of Patrick, that balance disappeared. Patrick expected the Democratic House and Senate to roll over for him. It did not work out that way. With a Republican in the governor's office, the Democratic leaders in the House and Senate are big wheels in their own right. With a Democrat in the Corner Office, their roles are lessened, and they resent it. The truth of the matter is that the Democratic leadership in the House and Senate would prefer dealing with a Republican governor, although they would never admit it.
Baker would bring back that constructive balance. He would need all the cooperation he could get if he were to cut back Patrick's increase in the sales tax from 6.25 percent to 5 percent, as well as pare back the 5.3 percent income tax to 5 percent.
Baker knows government finances, and his knowledge concerns Patrick, who eventually will have to face Baker in a debate and defend all those failed campaign promises he made four years ago.
Which probably explains the churlish remarks by Lt. Gov. Timothy Murray, who said that listening to Baker t alk about fiscal responsibility was "like getting lectured on abstinence from Paris Hilton."
Patrick, who warned voters to be on guard against "fear mongering" from his opponents, apparently has no problem with Lt. Gov. Murray's "smear mongering" remarks, complimenting Murray for his "way with words."
OK. So here's a quote from Paris Hilton in the L.A. County Jail to Barbara Walters that Murray might consider using: "I used to act dumb. That act is no longer cute."
Peter Lucas' political column appears Tuesday and Friday. E-mail comments to plucas@lowellsun.com


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Posted on Wednesday, 04 March 2015 19:48 .
In this premiere edition our resident guitar guru lays out his vision for a biweekly helping of guitar lore and offers a little personal history.
Posted on Sunday, 31 August 2014 15:43 .
The huge projection and reverberant bass notes that pour out of the Martin Custom Sinker Mahogany Dreadnought are largely the result of the extraordinary wood used for the back and sides. But before we get to the guitar itself, the story behind that mahogany is a fascinating one worth telling.