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African Youth want more space in government service
by Sam Ibanda Mugabi
Youth on the African continent say the biggest leadership gap is caused by failure of young people to occupy leadership positions at all levels. They blamed this on lack of commitment from governments to provide an enabling environment and policies for youths to meaningfully participate in leadership.
The youth were debating on a motion, "Harnessing the youth in leadership; How can young people play a more transformational role in policies and leadership" during a plenary session of the 4th Commonwealth Youth Parliament, on 17 April, 2018 at the Parliamentary chambers.
Abubaker Khalifa Usman (Nigeria) urged governments to emulate Nigeria which passed a law that allows youths aged 25 to contest for any elective position.
"Uganda has tried in promoting youth leadership through the five youth MPs, but this is not enough. Let us get our governments to give youths more space to lead Africa," said Usman.
David Milamba (Malawi) also cited an example of Australia which voted a 22-year old as a senator in Parliament while Kenya voted a 23-year old as a Member of Parliament.
"Those leaders who have over stayed in power should know that their time is up," said Milamba.
Hellen Bariki Mushi (Tanzania) encouraged African leaders to emulate former South African President, Nelson Mandela. She noted that Mandela set a footprint in transformational leadership while encouraging young people across the world to participate in leadership
"Governments should enhance the leadership capacities of youths by identifying and nurturing them instead of encouraging old people to hold onto power," said Mushi.
Thalente Ndebele (South Africa) advised governments to review their policies and laws to allow more youths to occupy higher positions of influence since they are the majority.
"Youths should be allowed to be ministers and young women should particularly be encouraged to take up such positions. Young people are ready and willing to lead, we only need space," said Ndebele.
Hamis Saidi (Tanzania) noted the importance of succession plans which target young people.
"Youths should be given opportunities to participate in policy development as a way of preparing them to take up leadership," said Saidi.
Prince Matshwarakgole (Botswana) warned against discrimination against youth with disabilities. He said that Africa will not develop if the disabled youth are not involved in all programs.
"In the next Commonwealth Youth Parliament, I would like to see increased representation of youths with disabilities," said Matshwarakgole.
Other youth however, noted the need for attitude change among young people, saying negative attitude has stopped some from becoming leaders.
"You need to show your governments that you love your countries and offer yourselves for elections while participating in election of other leaders," said Pamela Abito.
Maria Natabi (Uganda) challenged youth to use the available platforms to play a transformational role in leadership.
"Can we use the existing students' movement platforms to advance our agenda, but first we need to develop long term clear agenda because without an agenda, we cannot cause any transformation," said
Tags: government serviceUgandanewsYouth parliament
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The Ministry of Health has launched a ten year national health supply chain roadmap which will ensure availability of essential...
The executive director of the Uganda Media Centre Ofwono Opondo has asked the youth of the National Resistance Movement (NRM)...
by Miss D
Heard that the only way to get over someone is to get under someone else? That for you to forget...
Igeme Nabeta runs to court, seeks to annul Mwiru's election


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The Art of Communicating an Acquisition: M&A in 2016
March 24, 2016 Maureen Wolff Leave a comment
Global merger and acquisition activity set an all-time high last year, breaking the previous record set in 2007. According to an EY survey in October 2015, 59% of executives planned to actively pursue acquisitions in the coming 12 months. Given that this number is significantly higher than the 40% reported in the survey a year ago, we very well could see another record-breaking year for M&A in 2016.
But institutional investors are not nearly as convinced as the executives about the merits of acquisitions. Buy-side analysts and portfolio managers are often skeptical about M&A as a strategic alternative for driving growth, because they believe no corporate event has greater potential for failure than a major deal.
Plan for Success
How to improve the odds for M&A success? The obvious answer - essential in any strategic initiative - is careful planning. Unfortunately, mergers and acquisitions often develop so rapidly that time for preparation is limited. This is why one of the mainstays of investor relations is an ongoing program of skillful relationship building with a company's key stakeholders. Management teams that stay in touch with their priority constituencies, including institutional shareholders, will be in a better position to understand, in advance, the different mindsets these interests will bring to an impending deal.
In addition to keeping your shareholders updated on company developments, management should obtain feedback from your top 25 shareholders on issues that matter. How do your institutional holders feel about M&A as an element in your company's growth strategy? Are they more concerned about dilution or debt? Do they require all deals to be immediately accretive? How does their thinking about industry consolidation align with yours? Do they have concerns about your management team's M&A experience? A comprehensive third-party perception audit of your investor base can help answer these questions.
Develop Your Strategy
Armed with this insight, the company can develop an effective M&A messaging strategy. From the announcement of a deal through the closing, this strategy should be aimed at two outcomes. The first is to support successful completion of the deal. The second is to ensure that, whether the deal goes through or not, your company will emerge from the process with even stronger positioning on Wall Street.
Messaging based on an understanding of your shareholders' perceptions and concerns can convert them into advocates for the deal. Current shareholders will want to know how the transaction benefits them. Assess the transaction through their eyes, particularly when dilution is involved. Whether a deal is strategic or tactical is often crucial in determining how much dilution Wall Street will support. Are you acquiring the target for its products, its technology platform, its customers or the geography of its operations? An explanation of expected synergies and long-term benefits can help offset dilution concerns.
Consider Contingencies
Sound planning for M&A communications always considers "what if" scenarios. Chief among them: What if the deal fails to close? It is imperative to communicate that, as attractive as the deal may be, it is by no means your company's sole option for achieving its long-term goals.
Successful M&A messaging also must reflect the perceptions of stakeholders on the other side of the deal. (Value is in the eyes of the beholder.) Both companies will emphasize that they are getting the best possible price, but they may come from opposite directions in the reasoning and messaging. Your success in M&A communications can hinge on the ability to mediate the contradictory nature of the messaging on opposite sides of the transaction.
Negotiate the Outcome
It often can take as much diplomacy to coordinate messaging strategy as to reach agreement on deal terms. The best M&A communication outcomes result from both parties clearly understanding their respective roles, as well as their IR messaging objectives. Coordinated planning also is crucial. It is imperative for each company to be systematic in maintaining open communications with its institutional shareholders – and to get in touch with these constituents immediately after the transaction is announced.
Overnight, the management teams on both sides of the deal will find themselves encountering new universes of shareholders, often with very different investment objectives and horizons. As closing approaches, the acquiring company should become thoroughly familiar with the target's shareholder base and be prepared for the necessary outreach once the deal is complete. This should include personal phone calls to welcome inherited shareholders or outgoing e-mails with IR contact information. These actions engender goodwill and can make a lasting positive impression.
Cultivate Broad Support
Although CFOs and investor relations officers tend to bear the primary responsibility for spearheading M&A communications, the scope of the effort must extend beyond these functions. Senior management should assemble an internal task force that also consists of the legal department, corporate communications, HR and the business unit where the acquisition will reside. Working together, these groups can cover the entire range of external and internal constituencies.
Too often, communications with non-financial stakeholders get short shrift from management during the frenzy of activity surrounding a deal. Employees will be asking: Will I have a job? What will happen to my pay and benefits? What are the implications for my career, my lifestyle and my security? Customers and suppliers will have their own sets of questions about the transaction and its potential impact on their business relationships with both companies involved.
Coordinate the Details
In addition to crafting messages that respond to these questions, a number of logistical communications issues will need to be addressed. For example, will both companies conduct a joint conference call when announcing the transaction? If so, which members of the executive teams will participate in the call? If each company is conducting its own conference call, answers to potential analyst and media questions should be carefully scripted and rehearsed to ensure consistency.
If the current trend is any indication, M&A communications will be an important tool this year for companies and their executive teams. Communicating effectively before and throughout the M&A process will enhance relationships with all constituencies and possibly even enable a more successful transaction.
Maureen Wolff is president and partner at Sharon Merrill. She is a National Investor Relations Institute Fellow, Senior Roundtable Member and Honorary NIRI Boston Director. She is a trusted advisor to CEOs, CFOs and boards of directors on critical communications issues including corporate governance, shareholder activism and proxy contests, CEO succession planning and disclosure issues.
For more information on Sharon Merrill Associates, please contact us at info@InvestorRelations.com, or at 617-542-5300.
Disclosure, Strategic Messaging, Mergers & Acquisitions, Credibility, CFO, Shareholder Communications, Investor Relations, CEO, Trends, 2016

redpajama_set_name: RedPajamaCommonCrawl


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Home /Water Should Be a Public Good, Not a Commodity. Catalonia Is Showing How
Water Should Be a Public Good, Not a Commodity. Catalonia Is Showing How
Yves here. Some readers have asked for more positive stories on how to move forward with agendas that advance the interests of citizens at large, as opposed to corporations and the super wealthy. Given the ongoing and too often successful efforts to privatize water supplies, Catalonia is an important case study of how to reverse this tendency.
One reasons we don't feature much in the visionary future-oriented category is that a lot of the proposals show a lack of understanding of the complexity of the issues, and address only part of the problem (and then often the easiest part). Or they make a decent enough stab at defining the terrain, but then engage in "Assume a can opener" level remedies.
Having said that, while this post on water is useful, I wish it had spent more time describing how the various activists were able to achieve this win. And I'd very much like to get comments from readers in Spain as to how it works. The administrative structure looks complex, and that may be a deliberate choice to impede capture by the professionals who will run the water works. That may not be a bad thing with a utility, since there should not be that many decisions to make on a daily or even monthly basis.
By Míriam Planas, an activist with Engineering Without Borders in Catalonia and spokesperson for L'Aigua és Vida (Water is Life), an organisation that campaigns to remunicipalise water in the Barcelona region and bring it back into public control, and Juan Martínez, a member of the Taula de l'Aigua de Terrassa and becomes the president of the Terrassa Water Observatory, designated by the Plenary. Originally published at openDemocracy
Despite being a public good, water is often treated as a commodity. While some argue that privatisation of water increases efficiency and improves the quality of water services, the negative effects in terms of loss of accountability, poor performance, and elevated utility costs are well documented. As the corona virus makes its way into poor and vulnerable communities, we are yet again reminded of the devastating impact that these have on people's lives. When your ability to protect yourself from infectious disease is determined by your access to safe and affordable water, water service provision is not a matter of convenience – it is a matter of life and death. Worldwide, civil society organisations (CSOs) and grassroots movements therefore argue that water management should not be handled by private companies. Instead, water services and policy need to be democratised and placed under public control.
This is not always an easy task, but in Catalonia, years of hard work has paid off. The movement to remunicipalise water supply in the region has grown slowly and steadily over the past decade, thanks to the Aigua és Vida(Water is Life) platforms that many local groups set up since 2011. Local platforms such as Aigua és VidaGironaand Tauladel'AiguadeTerrassahave also emerged. After successful remunicipalisation, the Terrassa Water Observatory was established in 2018 – a dedicated citizen organisation that guides strategic decisions affecting Terressa's water management. The Observatory has now become a point of reference for other municipalities that wish to build new forms of water governance.
Recovering Ownership of Water Services in Terrassa
Terrassa is a relatively large municipality (more than 200,000 residents) located some 20 kilometers away from the city of Barcelona. Although its water supply services are now managed by a public enterprise that is 100 per cent owned by the municipal government, the municipality's support of public water ownership is quite recent. Before 2016, the water supply services in Terressa were managed by the private company MinaPúblicadeAguasdeTerrassa,S.A. This company was awarded its concessions in 1941 and had thus run the water services for over seven decades before the pro-public platform Tauladel'AiguadeTerrassastarted to push for a non-renewal of the concessions in 2013.
Together with other local networks,Tauladel'AiguadeTerrassalaunched a 2015 campaign called 'Social Pact for Public Water' that focused on public, participatory management of the entire water cycle. The campaign was a great success. In 2016, the Terrassa City Council decided to reclaim control over the water supply not only by creating a new public enterprise, but also by establishing a whole new water management body – the Terressa Water Observatory (Observatorio del Agua de Terrassa). What is the Observatory? It's for water management to be truly participatory and to ensure citizen decision-making power. Users can discuss their water concerns and make collective recommendations to the public water company – Taigua, Aigua Municipal de Terrassa.
The Observatory as a Forum for Collaboration
The Terrassa Water Observatory was based on the model of the Citizen Parliament of Terrassa, a forum that promotes debate and agreements pushed by citizens for consideration by local government. As an autonomous organisation, the Observatory is responsible for carrying out studies and advising the Terressa City Council on all matters related to municipal water management.
The Observatory's highest governing body – the Plenary – is composed of representatives from each political group in the municipality and representatives of technical staff, business, community groups, and unions. Both the elected president and the Plenary itself can invite people from outside the Observatory (with no voting rights) to enrich the debate and discussion on any specific issue. In addition to this, there are six working groups and three collaborative boards that work to implement the Plenary's work plan. The working groups are open to everyone from civil society who wishes to contribute. The three boards (on education, research and citizenship) act as interest groups and help strengthen the Observatory's role and influence by being the central point of communication between community groups, universities, and the Observatory. Although its organisational structure is quite complicated, it also enhances its consensus-building capacity and gives it a high level of legitimacy.
The organisational structure of Terrassa Water Observatory
Moving Forward – A Model for Management?
Since the Observatory water management model is relatively new, there is still a lot to tease out. There is significant public support for the Observatory in Terressa, but also ongoing debates about the different actors' involvement: what are the roles and responsibilities of the local government and the public water operator? How should information be managed? Where and how does coordination take place?
Answering these questions will require time and effort coupled with a good portion of patience. Building collaborative water governance is not done in a trice. However, in a region where eighty-four per cent of people depend on a private company for their water supply, the Terrassa Water Observatory represents more than a body that ensures that citizens have a say in the production of policy. It also represents a new political culture, where recovering public water management means actively fostering social empowerment. This is clear when looking at the tangible results such as training proposals for educational activities within schools that the Observatory has produced in only a matter of months.
Evidently, the Observatory's work both depends on and strengthens people's level of engagement. Promoting active citizenship, this provides yet another reason why we should look to Terrassa. In a time of crisis, we are now more than ever in need of good examples of how citizens can take the lead in the management of common goods.


Audit of hardware equipment within the organisation.
Procedures for data and software security.
Employment and deployment and terms of co-location and hosting for web services.
Students taking this course must have completed the Advanced Programming at the Advanced Diploma level or an equivalent course.
NOTE: This course does not require a formal examination. Students taking this course have to deliver, at an agreed date, a completed and documented working website.
Students taking this course have to be registered to enable their work to be uploaded to our servers for their development and for our assessment.
Embedding PHP into HTML Pages.
Using Data from a Form.
Passing Information through a Form and Links.
PHP with Forms and Cookies.
Checking User-Entered Data for Required Fields.
Setting up PHP with MySQL.
Putting Content into Database with PHP.
To provide the student with the requisite skills of strategic analysis, development of strategic options and choices and the understanding of techniques for strategic decision making.
The overall objective of this Case Study subject is to integrate the theories and knowledge attained from the different subjects contained in the Advanced Diploma Level.
The Case Study integrates the courses that the student has completed. Students are expected to suggest practical solutions to the problems that they have identified.
Students taking this examination will be provided with the Case Study and other background information and notes at least 30 days before the examination. Students are expected to analyse the case and to identify key issues and managerial problems which may require solutions.
At the examination, students will be given questions relating to the Case Study concerned. Students are then expected to suggest their solutions to the questions posed and to support their arguments with their appropriate rationale and reasons.
Examiners will be paying close attention to the reasoning and that has been applied in deriving the students' recommendations and advice.


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Schröding ist ein Gemeindeteil der Gemeinde Lengdorf im Landkreis Erding in Oberbayern.

Geografie
Der Weiler liegt vier Kilometer nördlich von Lengdorf entfernt. 800 Meter westlich entspringt die Große Vils.

Verkehr
Der Bahnhof Thann-Matzbach liegt vier Kilometer südlich. Die Bundesautobahn 94 verläuft sechs Kilometer südlich.

Weblinks

Einzelnachweise

Weiler in Oberbayern
Ort im Landkreis Erding
Geographie (Lengdorf)

redpajama_set_name: RedPajamaWikipedia


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\subsection{Same conductors}
For a trait $T$
with generic geometric point $\eta$ and closed point $t$ such that the residue field of $t$ is algebraically closed
and for a constructible complex $\mathcal{F}$ of $\Lambda$-modules on $T$,
the Artin conductor $a(\mathcal{F})$ is defined by
$a(\mathcal{F})=\rk(\mathcal{F}_{\eta})-\rk(\mathcal{F}_{t})+\Sw(\mathcal{F}_{\eta})$.

For a regular scheme $X$ of dimension 1 whose closed points have perfect residue fields,
a constructible complex $\mathcal{F}$ of $\Lambda$-modules on $X$,
and a geometric point $x$ over a closed point of $X$,
the Artin conductor $a_x(\mathcal{F})$ at $x$ is defined by $a_x(\mathcal{F})=a(\mathcal{F}|_{X_{(x)}})$.

For a geometric point $x$ over a closed point of $X$,
we denote the generic point of $X_{(x)}$ by $\eta_x$
and a geometric point over $\eta_x$ by $\bar{\eta_x}$.
For a constructible complex $\mathcal{G}$ on a dense open subscheme $U$ of $X$
the Swan conductor $\Sw_x(\mathcal{G})$ at $x$ is defined to be
the Swan conductor of the virtual representation $[\mathcal{G}_{\bar{\eta_x}}]$
of the absolute Galois group $\Gal(\bar{\eta_x}/\eta_x)$.

Let $S$ be an excellent noetherian scheme.
By Zariski's main theorem, for a regular $S$-curve $C$,
there exists a dense open immersion $j:C\to\overline{C}$ over $S$
to a regular scheme $\overline{C}$ proper over $S$.
We call such an open immersion a canonical regular compactification of $C$ over $S$.
It is unique up to unique isomorphism.

We assume that every closed point of $S$ has perfect residue field.
\begin{Def}
\label{sc}
Let
$X$ be an $S$-scheme separated of finite type
and $\mathcal{F}$ and $\mathcal{F}'$ constructible complexes of $\Lambda$-modules and $\Lambda'$-modules respectively on $X$.
We say $\mathcal{F}$ and $\mathcal{F}'$ have {\it universally the same conductors} over $S$
if for every regular $S$-curve $C$,
every $S$-morphism $g:C\to X$,
every canonical regular compactification $j:C\to\overline{C}$ of $C$ over $S$
and every geometric point $v\to \overline{C}$ over a closed point,
we have $a_v(j_!g^*\mathcal{F})=a_v(j_!g^*\mathcal{F}')$.
\end{Def}

\begin{Lem}
\label{char of sc}
$\mathcal{F}$ and $\mathcal{F}'$ have universally the same conductors over $S$
if and only if the following two conditions hold;
\begin{enumerate}[{\rm (i)}]
\item
\label{rk}
for every geometric point $x$ of $X$,
we have $\rk(\mathcal{F}_x)=\rk(\mathcal{F}'_x)$,

\item
\label{swc}
for every regular $S$-curve $C$,
every $S$-morphism $g:C\to X$,
every canonical compactification $C\to\overline{C}$,
and every geometric point $v\to \overline{C}$ over a closed point,
we have $\Sw_v(g^*\mathcal{F})=\Sw_v(g^*\mathcal{F}')$.
\end{enumerate}
\end{Lem}

\begin{proof}
The sufficiency is clear.
We prove the necessity.
Assume $\mathcal{F}$ and $\mathcal{F}'$ have universally the same conductors over $S$.
Let $x$ be a geometric point of $X$.
Take a stratification $X=\coprod_i X_i$
	such that $\mathcal{F}|_{X_i}$ and $\mathcal{F}'|_{X_i}$ are smooth for every $i$
	and a covering $W\to X_j$ trivializing $\mathcal{F}|_{X_j}$ and $\mathcal{F}'|_{X_j}$,
		where $X_j$ is the subset over which $x$ lies.
Then, for every regular $S$-curve $C$ which is not proper and an $S$-morphism $g:C\to W$
and every geometric point $v$ of $\overline{C}\setminus C$,
where $j:C\to\overline{C}$ is a canonical compactification over $S$,
we have
$a_v(j_!g^*(\mathcal{F}|_W))=\rk(\mathcal{F}_x)$, and similarly for $\mathcal{F}'$.
Thus, by the assumption we get $\rk(\mathcal{F}_x)=\rk(\mathcal{F}'_x)$.
By the definition of the Artin conductor
	we have $\Sw_v(g^*\mathcal{F})=a_v(j_!g^*\mathcal{F})-\rk(\mathcal{F}_{g(v)})$
	and (ii) also follows.
\end{proof}

\subsection{Same wild ramification}
In this subsection, we do not use the assumption
that closed points of $S$ have perfect residue fields.

We note that by Nagata's compactification theorem,
separated scheme of finite type over a noetherian scheme have a compactification.

We say an element $\sigma$ of a pro-finite group $G$ is {\it pro-p}
if the closed subgroup $\overline{\langle\sigma\rangle}$ of $G$ is pro-$p$.
For a pro-$p$ element $\sigma\in G$, we put $\Gamma^\sigma=\Gamma^{\overline{\langle\sigma\rangle}}$.
For the definition of $\Gamma^{\overline{\langle\sigma\rangle}}$, see Section \ref{swbr}.

For the spectrum $x=\Spec F$ of a field $F$,
we denote the characteristic of $F$ by $p_x$.

\begin{Def}
Let $\overline{X}$ be a noetherian normal connected scheme
and $X$ a dense open subscheme of $\overline{X}$.
For smooth complexes $\mathcal{F}$ and $\mathcal{F}'$
of $\Lambda$-modules and $\Lambda'$-modules respectively on $X$,
we say $\mathcal{F}$ and $\mathcal{F}'$ have {\it the same virtual wild ramification} along $\overline{X}\setminus X$
if for any geometric point $x\to \overline{X}$
and any pro-$p_x$ element $\sigma$
of $\pi_1(\overline{X}_{(x)}\times_{\overline{X}}X,y)$,
where $y$ is a geometric point,
we have $\Gamma ^{\sigma}([\mathcal{F}_y])=\Gamma ^{\sigma}([\mathcal{F}'_y])$.
\end{Def}

To show $\mathcal{F}$ and $\mathcal{F}'$ have the same virtual wild ramification along $\overline{X}\setminus X$,
it suffices to check the above condition for a geometric point over every closed point,
i.e. it suffices to check that
for a geometric point $x\to \overline{X}$ {\it over every closed point}
and every pro-$p_x$ element $\sigma$
of $\pi_1(\overline{X}_{(x)}\times_{\overline{X}}X,y)$,
where $y$ is a geometric point,
we have $\Gamma ^{\sigma}([\mathcal{F}_y])=\Gamma ^{\sigma}([\mathcal{F}'_y])$.
Note that
for a geometric point $x'$ of $\overline{X}$ and a geometric point $x$ which specializes $x'$,
if we choose a geometric point $y'$ of $\overline{X}_{(x')}\times_{\overline{X}}X$,
the specialization morphism $\overline{X}_{(x')}\to\overline{X}_{(x)}$
induces a homomorphism
$\pi_1(\overline{X}_{(x')}\times_{\overline{X}}X,y')\to\pi_1(\overline{X}_{(x)}\times_{\overline{X}}X,y)$,
where $y$ is the image of $y'$.

\begin{Def}[cf. Definition 5.1 in \cite{SY}]
\label{svwr}
Let $X$ be an $S$-scheme separated of finite type
and $\mathcal{F}$ and $\mathcal{F}'$ constructible complexes
of $\Lambda$-modules and $\Lambda '$-modules respectively on $X$.
\begin{enumerate}[(i)]
\item
\label{sm case}
Assume that $X$ is normal and connected
and that $\mathcal{F}$ and $\mathcal{F}'$ are smooth.
We say $\mathcal{F}$ and $\mathcal{F}'$ have the {\it same virtual wild ramification} over $S$
if there exist
a normal compactification $\overline{X}$ of $X$ over $S$
such that $\mathcal{F}$ and $\mathcal{F}'$ have the same virtual wild ramification along $\overline{X}\setminus X$.
\item
We say $\mathcal{F}$ and $\mathcal{F}'$ have the {\it same virtual wild ramification} over $S$
if there exists a decomposition $X=\coprod_i X_i$
such that
$X_i$ is locally closed connected normal subscheme
and that $\mathcal{F}|_{X_i}$ and $\mathcal{F}'|_{X_i}$ are smooth complexes
and have the same virtual wild ramification in the sense of (\ref{sm case}) for every $i$.
\end{enumerate}
\end{Def}
If a $G$-torsor is given,
we use the terminology same virtual wild ramification
also for virtual representations of $G$.

\begin{Rem}
	\begin{enumerate}[(i)]
\item
Our definition of ``same virtual wild ramification'' is weaker than that in \cite{I} and \cite{V}.
Consider the case where $S$ is a spectrum of a field of characteristic $p>0$
or that of a discrete valuation ring with residue field of characteristic $p>0$.
In \cite[D\'efinition 2.3.1]{V},
for smooth complexes $\mathcal{F}$ and $\mathcal{F}'$ of $\Lambda$-modules
on a normal connected scheme $X$ separated of finite type over $S$,
the property that $\mathcal{F}$ and $\mathcal{F}'$ have the same virtual wild ramification
is defined by the following condition:
There exists a normal compactification $\overline{X}$ of $X$ over $S$
such that for every geometric point $x$ of $\overline{X}$
and every pro-$p$ element $\sigma\in\pi_1(\overline{X}_{(x)}\times_{\overline{X}}X,y)$,
we have
$\Trbr(\sigma,[\mathcal{F}_y])=\Trbr(\sigma,[\mathcal{F}'_y])$.
Thus our ``same virtual wild ramification'' is weaker than that in \cite{V}.
Indeed, for a finite group $G$,
an element $\sigma\in G$ of order a power of $p$,
and an element $M\in K(\Lambda[G])$,
we have
$$\Gamma^\sigma(M)=\frac{1}{|\langle\sigma\rangle|}\sum_{g\in\langle\sigma\rangle}\Trbr(g,M)$$
(c.f. \cite[1.4.7]{I}).

\item
If $S$ is a spectrum of a field and
$\mathcal{F}$ and $\mathcal{F}'$ are constructible sheaves,
our ``same virtual wild ramification'' coincides with
``same wild ramification'' in \cite[Definition 5.1]{SY}.
\end{enumerate}
\end{Rem}

\begin{Lem}
\label{pullback}
Let $S'\to S$ be a morphism of excellent noetherian schemes.
Consider a commutative diagram
$$
\begin{CD}
X'@>{g}>>X\\
@VVV@VVV\\
S'@>>>S
\end{CD}
$$
such that the vertical arrows are separated of finite type.
Let $\mathcal{F}$ and $\mathcal{F}'$ be constructible complexes of sheaves of $\Lambda$-modules and $\Lambda'$-modules respectively on $X$.
If $\mathcal{F}$ and $\mathcal{F}'$ have the same virtual wild ramification over $S$,
then $g^*\mathcal{F}$ and $g^*\mathcal{F}'$ have the same virtual wild ramification over $S'$.
\end{Lem}

\begin{proof}
We note that the normal locus of an excellent scheme is open.
By devissage, we may assume
$X$ and $X'$ are normal and connected
and $\mathcal{F}$ and $\mathcal{F}'$ are smooth complexes.
Then we take a normal compactification $\overline{X}$ of $X$ over $S$
such that $\mathcal{F}$ and $\mathcal{F}'$ have the same virtual wild ramification along $\overline{X}\setminus X$.
We can take a normal compactification $\overline{X'}$ of $X'$ over $S'$
with a morphism $\bar{g}:\overline{X'}\to\overline{X}$ extending $g$.
Then,
for a geometric point $x'\to\overline{X'}$,
a geometric point $y'$ of $\overline{X'}_{(x')}\times_{\overline{X'}}X'$,
and a pro-$p_{x'}$ element $\sigma'$ of $\pi_1(\overline{X'}_{(x')}\times_{\overline{X'}}X',y')$,
we have $\Gamma^{\sigma'}([(g^*\mathcal{F})_{y'}])=\Gamma^{\bar{g}_*(\sigma')}([\mathcal{F}_{\bar{g}(y')}])$
and the same equality for $\mathcal{F}'$,
where $\bar{g}_*$ is the induced homomorphism
$\pi_1(\overline{X'}_{(x')}\times_{\overline{X'}}X',y')\to\pi_1(\overline{X}_{(\bar{g}(x'))}\times_{\overline{X}}X,\bar{g}(y'))$.
Here since $\bar{g}_*(\sigma')$ is a pro-$p_{\bar{g}(x')}$ element of
$\pi_1(\overline{X}_{(\bar{g}(x'))}\times_{\overline{X}}X,\bar{g}(y'))$,
we have $\Gamma^{\sigma'}([\mathcal{F}_{y'}])=\Gamma^{\sigma'}([\mathcal{F}'_{y'}])$.
\end{proof}

\section{Same wild ramification and Same conductors}
Let $S$ be an excellent noetherian scheme.
We always assume that
closed points of the base scheme $S$ have perfect residue fields.
We define the terminologies {\it same virtual wild ramification} and {\it same conductors}
for complexes on a scheme over $S$.
The terminology {\it same virtual wild ramification}
is a modification of the terminology {\it same wild ramification}
in \cite[Definition 5.1]{SY}.

\input{wildram.comparison-svwr.tex}
\input{wildram.comparison-sc.tex}

\section{The Euler-Poincar\'e characteristic}
We prove Theorem \ref{A to X}.
The following proposition is a consequence of the result of Deligne-Illusie.
\begin{Prop}[cf. Proposition 0.2 in \cite{SY}]
\label{W to X}
Let
$k$ be a perfect field
and $\bar{k}$ an algebraic closure of $k$.
Let $X$ be a scheme separated of finite type over $k$
and $\mathcal{F}$ and $\mathcal{F}'$ constructible complexes of
$\Lambda$-modules and $\Lambda'$-modules respectively on $X$.
Assume that $\mathcal{F}$ and $\mathcal{F}'$ have the same virtual wild ramification $(\text{over }k)$.
Then, we have $\chi _c(X_{\bar{k}},\mathcal{F})=\chi _c(X_{\bar{k}},\mathcal{F}')$.
\end{Prop}
\begin{Prop}
\label{EP and brtr}
Let $X$ be a normal connected scheme separated of finite type over an algebraically closed field,
$\mathcal{F}$ be a smooth complex of $\Lambda$-module on $X$,
and $W\to X$ a $G$-torsor, for a finite group $G$, trivializing $\mathcal{F}$.
We denote the virtual representation of $G$ corresponding to $\mathcal{F}$ by $M$.
\begin{enumerate}[{\rm (i)}]
\item {\rm \cite[Lemme 2.3 and Remarque 2.4]{I}}
We have
$$
\chi_c(X,\mathcal{F})=\frac{1}{|G|}\sum_{\sigma\in G}\Tr(\sigma,H_c^*(W,\mathbb{Q}_l))\cdot\Trbr(\sigma,M),
$$
where $l$ is the characteristic of $\Lambda$
and for $\sigma\in G$ of order divided by $l$ we put $\Trbr(\sigma,-)=0$.
\item {\rm \cite[3.3]{DL}}
$\Tr(\sigma,H_c^*(W,\mathbb{Q}_l))$ is an integer independent of $l\neq p$.

\item
Let $\overline{X}$ be a normal compactification of $X$
and $\overline{W}$ a normalization of $\overline{X}$ in $W$.
For an element $\sigma\in G$, if $\Tr(\sigma,H_c^*(W,\mathbb{Q}_l))\neq0$
then $\sigma$ is of order a power of $p$ and fixes some point of $\overline{W}$.
\end{enumerate}
\end{Prop}
\begin{proof}
For (iii), see the beginning of the proof of \cite[Lemme 2.5]{I}.
\end{proof}

\begin{proof}[Proof of Proposition \ref{W to X}]
The way of the proof is the same as that of \cite[Proposition 0.2]{SY}.
We may assume $k$ is algebraically closed.
By taking a decomposition $X=\coprod_i X_i$ as in Definition \ref{svwr} (ii),
we may assume that $X$ is normal and connected and that $\mathcal{F}$ and $\mathcal{F}'$ are smooth complexes.
Then by Proposition \ref{EP and brtr} (i)(iii),
we have
$$
\chi_c(X,\mathcal{F})=\frac{1}{|G|}\sum_{\sigma\in S}\Tr(\sigma,H_c^*(W,\mathbb{Q}_l))\cdot\Trbr(\sigma,M),
$$
where $S=\{\sigma\in G\mid\sigma\text{ is of }p\text{-power order and fixes some point of }\overline{W}\}$.
Since $\chi_c(X,\mathcal{F})$ is an integer, by Lemma \ref{brtr},
$$
\chi_c(X,\mathcal{F})=\frac{1}{|G|}\sum_{\sigma\in S}\Tr(\sigma,H_c^*(W,\mathbb{Q}_l))\cdot
\frac{p\cdot\Gamma^\sigma(M)-\Gamma^{\sigma^p}(M)}{p-1}.
$$
Hence the assertion follows from Proposition \ref{EP and brtr} (ii).
\end{proof}

We use the following lemma to reduce Theorem \ref{A to X} to the surface case.
\begin{Lem}
\label{red to surface case}
Let $f:X\to Y$ be a morphism
of schemes separated of finite type
over an algebraically closed field
and $\mathcal{F}$ and $\mathcal{F}'$ constructible complexes
of $\Lambda$-modules and $\Lambda '$-modules respectively on $X$.
Assume that
for every smooth curve $C$
and every morphism $g:C\to Y$,
we have $\chi_c(X\times_YC,g'^*\mathcal{F})=\chi_c(X\times_YC,g'^*\mathcal{F}')$,
where $g'$ is the morphism $X\times_YC\to X$.
Then, $Rf_!\mathcal{F}$ and $Rf_!\mathcal{F}'$ have universally the same conductors.
\end{Lem}
\begin{proof}
By the assumption and the proper base change theorem,
for every smooth curve $C$ over $k$
and every morphism $g:C\to Y$,
we have $\chi_c(C,g^*Rf_!\mathcal{F})=\chi_c(C,g^*Rf_!\mathcal{F}')$.
Hence we may assume that $f=\id$.
Then the assertion follows from the proof of \cite[Lemma 3.3]{SY}.
\end{proof}

\begin{proof}[Proof of Theorem \ref{A to X}]
We may assume $k$ is algebraically closed.
We prove the assertion by induction on $\dim X$.
First we assume $\dim X\leq2$.
Then by Theorem \ref{A to W},
having universally the same conductors
implies having the same virtual wild ramification.
Hence the assertion follows Proposition \ref{W to X}.

Assume $\dim X>2$.
By the induction hypothesis, it suffices to show the assertion after replacing $X$ by a dense open subscheme.
We can find a flat morphism $f:X\to Y$ to a surface $Y$ over $k$ by shrinking $X$ if necessary
(for example, take an \'etale morphism $X\to \mathbb{A}^d$, by shrinking $X$,
and compose with some projection $\mathbb{A}^d\to \mathbb{A}^2$).
By the assumption, for each smooth curve $C$ and each morphism $g:C\to Y$,
the pullbacks $g'^*\mathcal{F}$ and $g'^*\mathcal{F}'$
by the canonical morphism $g':X\times_YC\to X$
have universally the same conductors.
Here $\dim X\times_YC<\dim X$ by the flatness of $f$
and we can apply the induction hypothesis to $g'^*\mathcal{F}$ and $g'^*\mathcal{F}'$.
Hence we have
$\chi_c(X\times_YC,g'^*\mathcal{F})=\chi_c(X\times_YC,g'^*\mathcal{F}')$.
Therefore by Lemma \ref{red to surface case},
$Rf_!\mathcal{F}$ and $Rf_!\mathcal{F}$ have the universally the same conductors.
By the $\dim X=2$ case, we have $\chi_c(X,\mathcal{F})=\chi_c(X,\mathcal{F}')$
and the assertion follows.
\end{proof}

\begin{Def}[Definition 3.1 in \cite{SY}]
\label{def of X}
\upshape
Let $X$ be a separated scheme of finite type over a perfect field $k$.
For constructible complexes $\mathcal{F}$ and $\mathcal{F}'$
of $\Lambda$-modules and $\Lambda'$-modules respectively on $X$,
we say $\mathcal{F}$ and $\mathcal{F}'$ have {\it universally the same Euler-Poincar\'e characteristics}
if for any scheme $Z$ separated of finite type over $k$
and any morphism $g:Z\to X$,
we have
$\chi_c(Z_{\bar{k}},g^*\mathcal{F})=\chi_c(Z_{\bar{k}},g^*\mathcal{F}')$.
\end{Def}

\begin{Cor}
\label{sc and sx}
Let $X$ be a separated scheme of finite type over a perfect field $k$
and $\mathcal{F}$ and $\mathcal{F}'$ constructible complexes
of $\Lambda$-modules and $\Lambda'$-modules respectively on $X$.
Consider the following three conditions.

\begin{enumerate}[{\rm (i)}]
\item\label{have the svwr}
$\mathcal{F}$ and $\mathcal{F}'$ have the same virtual wild ramification.

\item\label{have the sc}
$\mathcal{F}$ and $\mathcal{F}'$ have universally the same conductors.

\item\label{have the sx}
$\mathcal{F}$ and $\mathcal{F}'$ have universally the same Euler-Poincar\'e characteristics.
\end{enumerate}

Then, the implications
{\rm (\ref{have the svwr})}$\Rightarrow${\rm (\ref{have the sc})}$\Leftrightarrow${\rm (\ref{have the sx})} hold.
Moreover, if $\dim X\leq2$, then these three conditions are equivalent.
\end{Cor}

\begin{proof}
(\ref{have the svwr})$\Rightarrow$(\ref{have the sc}):
This is nothing but Proposition \ref{W to A}.

(\ref{have the sc})$\Rightarrow$(\ref{have the sx}):
Since the property having universally the same conductors is preserved by pullbacks,
this follows from Theorem \ref{A to X}.

(\ref{have the sx})$\Rightarrow$(\ref{have the sc}):
This is a special case of Lemma \ref{red to surface case} (Take $f=\id$).

Finally, if $\dim X\leq2$, then the implication (\ref{have the sc})$\Rightarrow$(\ref{have the svwr})
is nothing but Theorem \ref{A to W}.
\end{proof}

\begin{Cor}
\label{functoriality over a field}
Let $f:X\to Y$ be a morphism of schemes separated of finite type over $k$
and $\mathcal{F}$ and $\mathcal{F}'$ constructible complexes
of $\Lambda$-modules and $\Lambda'$-modules respectively on $X$.
Then,
\begin{enumerate}[{\rm (i)}]
\item
{\rm \cite[Lemma 3.3.2]{SY}}
If $\mathcal{F}$ and $\mathcal{F}'$ have universally the same Euler-Poincar\'e characteristics
then so do $Rf_!\mathcal{F}$ and $Rf_!\mathcal{F}'$.
\item
If $\mathcal{F}$ and $\mathcal{F}'$ have universally the same conductors
then so do $Rf_!\mathcal{F}$ and $Rf_!\mathcal{F}'$.
\item
Assume $\dim Y\leq2$.
Then, if $\mathcal{F}$ and $\mathcal{F}'$ have the same virtual wild ramification
then so do $Rf_!\mathcal{F}$ and $Rf_!\mathcal{F}'$.
\end{enumerate}
\end{Cor}
\begin{proof}
\begin{description}
\item{(ii)}
This follows from (i) and Corollary \ref{sc and sx} (ii)$\Leftrightarrow$(iii).

\item{(iii)}
By (i)$\Rightarrow$(iii) of Corollary \ref{sc and sx},
$\mathcal{F}$ and $\mathcal{F}'$ have universally the same Euler-Poincar\'e characteristics.
Then by (i), $Rf_!\mathcal{F}$ and $Rf_!\mathcal{F}'$ have universally the same Euler-Poincar\'e characteristics
and the assertion follows from (iii)$\Rightarrow$(i) of Corollary \ref{sc and sx} (ii).
\end{description}
\end{proof}

Here we mention that wild ramification of the restrictions to all curves also determines the characteristic cycle.
In \cite[Definition 4.10]{S},
for a constructible complex $\mathcal{F}$
on a smooth scheme $X$ over a perfect field,
the characteristic cycle $CC\mathcal{F}$ is defined as a cycle on the cotangent bundle $T^*X$.

\begin{Cor}
Let $X$ be a smooth scheme over a perfect field
and $\mathcal{F}$ and $\mathcal{F}'$ be constructible complexes
of $\Lambda$-modules and $\Lambda '$-modules respectively on $X$.
Assume that $\mathcal{F}$ and $\mathcal{F}'$ have universally the same conductors.
Then, $CC\mathcal{F}=CC\mathcal{F}'$.
\end{Cor}

\begin{proof}
By \cite[Proposition 3.4]{SY}, it suffices to prove that
$\mathcal{F}$ and $\mathcal{F}'$ have universally the same Euler-Poincar\'e characteristics.
Hence the assertion follows from Corollary \ref{sc and sx} (ii)$\Rightarrow$(iii).
\end{proof}

\subsection{Curve case}
First, we show Proposition \ref{p-power case} holds if $\dim\overline{X}\leq1$ and if $G$ is of prime-power order:
\begin{Lem}
\label{curve case}
Let $X$ be a regular $S$-curve
and $\mathcal{F}$ and $\mathcal{F}'$ smooth complexes
of $\Lambda$-modules and $\Lambda'$-modules respectively on $X$.
Assume that
$\mathcal{F}$ and $\mathcal{F}'$
are trivialized by a $\mathbb{Z}/p^e\mathbb{Z}$-torsor over $X$
for a prime number $p$ and an integer $e\geq0$
and have universally the same conductors over $S$.
Then they have the same virtual wild ramification over $S$.
\end{Lem}

\begin{proof}
Let $W\to X$ be a $G=\mathbb{Z}/p^e\mathbb{Z}$-torsor trivializing $\mathcal{F}$ and $\mathcal{F}'$
and $X\to\overline{X}$ a canonical regular compactification of $X$ over $S$.

Take a geometric point $x$ of $\overline{X}$.
We show that for every pro-$p_x$ element
$\sigma\in\pi_1(\overline{X}_{(x)}\times_{\overline{X}}X,y)$,
we have $\Gamma^\sigma([\mathcal{F}_y])=\Gamma^\sigma([\mathcal{F}'_y])$,
where $y$ is a geometric point.
If $x$ lies over a point of $X$, this is trivial.
We assume $x$ lies over a point of $\overline{X}\setminus X$.

We note that
$K=\Gamma(\overline{X}_{(x)}\times_{\overline{X}}X,\mathcal{O})$
is a strictly henselian discrete valuation field
and that
$\overline{W}_{(w)}\times_{\overline{W}}W\to\overline{X}_{(x)}\times_{\overline{X}}X$
is an $I_x$-torsor,
where $\overline{W}$ is the normalization of $\overline{X}$ in $W$,
$w$ is a geometric point of $\overline{W}$ which lies over $x$,
and $I_x$ is the inertia group, i.e. the stabilizer of $w$.
The elements $[\mathcal{F}_y]$ and $[\mathcal{F}'_y]$
of $K(\Lambda[\pi_1(\overline{X}_{(x)}\times_{\overline{X}}X,y)])$ come from
elements $M$ and $M'$ of $K(\Lambda[I_x])$.

By the assumption that
$\mathcal{F}$ and $\mathcal{F}'$ have universally the same conductors,
it follows that for every subgroup $H$ of $I_x$ we have
$\Sw(M|_H)=\Sw(M'|_H)$.
In fact, if we put $V=W/H$,
then $\overline{W}_{(w)}\times_{\overline{W}}W\to\overline{V}_{(v)}\times_{\overline{V}}V$
is an $H$-torsor,
where $\overline{V}$ is the normalization of $\overline{X}$ in $V$
and $v$ is the geometric point under $w$.
Then the assertion follows from Lemma \ref{if the same sw then the same fixed parts}.
\end{proof}

\subsection{General case}
We now prove Proposition \ref{p-power case}
for the case where $X$ is general and $G\simeq\mathbb{Z}/p^e\mathbb{Z}$
for some prime number $p$ and some integer $e\geq0$:
\begin{Lem}
\label{p-power cov}
Let $X$ be a regular scheme separated of finite type over $S$.
Let $\mathcal{F}$ and $\mathcal{F}'$ be constructible complexes of
$\Lambda$-modules and $\Lambda'$-modules respectively on $X$
and $W\to X$ a $\mathbb{Z}/p^e\mathbb{Z}$-torsor trivializing $\mathcal{F}$ and $\mathcal{F}'$.
Assume that $\mathcal{F}$ and $\mathcal{F}'$ have universally the same conductors over $S$
and that $X$ has a regular compactification over $S$.
Then $\mathcal{F}$ and $\mathcal{F}'$ have the same virtual wild ramification over $S$.
\end{Lem}
\begin{proof}
Let $\overline{X}$ be a regular compactification of $X$ over $S$.
It suffices to show that
for every geometric point $x$ {\it over a closed point} of $\overline{X}$
and every pro-$p_x$ element $\sigma\in\pi_1(\overline{X}_{(\bar{x})}\times_{\overline{X}}X,y)$,
we have $\Gamma^\sigma([\mathcal{F}_y])=\Gamma^\sigma([\mathcal{F}'_y])$.
By Lemma \ref{taking a good curve},
we can find
an $S$-curve $C$, an $S$-morphism $g:C\to X$, and a geometric point $v$ of $\overline{C}$ over $x$
such that the composite
$\pi_1(\overline{C}_{(v)}\times_{\overline{C}}C,y')\to\pi_1(\overline{X}_{(x)}\times_{\overline{X}}X,y)\to I_x$
is surjective.
By Lemma \ref{curve case},
$g^*\mathcal{F}$ and $g^*\mathcal{F}'$ have the same virtual wild ramification along $\overline{C}\setminus C$.
Therefore the assertion follows.
\end{proof}

We will need to consider the existence of a compactification to prove Proposition \ref{p-power case}.
\begin{Lem}
\label{cptf}
Let $X$ be a normal $S$-scheme separated of finite type
and $V\to X$ be a finite \'etale morphism.
Then, for any normal compactification $\overline{V}$ of $V$ over $S$,
there exists a normal compactification $\overline{X}$ of $X$ over $S$
such that the normalization of $\overline{X}$ in $V$ dominates $\overline{V}$.
\end{Lem}
\begin{proof}
The proof is the same as the latter half of the proof of \cite[Proposition 2.1.1 (ii)]{V}.
For the reader's convenience, we include the proof.
Take an arbitrary compactification $Y$ of $X$
and a compactification $W$ of $V$ dominating $\overline{V}$
with a morphism $W\to Y$ extending $V\to X$.
By \cite[I.5.2.2]{GR}, we can find
a blow-up $Y'\to Y$ along a closed subscheme contained in $Y\setminus X$
such that $W'\to Y'$ is flat,
where $W'\to W$ is the blow-up induced by $Y'\to Y$.
Then, $W'\to Y'$ is finite,
and hence it suffices to take $\overline{X}$ to be the normalization of $Y'$.
\end{proof}
\begin{Lem}
\label{red to the cyclic case}
Let $X$ be a normal $S$-scheme separated of finite type,
$G$ a finite group,
$\mathcal{F}$ and $\mathcal{F}'$ smooth complexes of
$\Lambda$-modules and $\Lambda'$-modules respectively on $X$,
and $W\to X$ a $G$-torsor trivializing $\mathcal{F}$ and $\mathcal{F}'$.
Consider subgroups $H_1,\ldots,H_k$ of $G$ such that
any element $\sigma\in G$ of prime-power order belongs to
$H_i$ for some $i=1,\ldots,k$.
Set $V_i=W/H_i$ and
assume $\mathcal{F}|_{V_i}$ and $\mathcal{F}'|_{V_i}$
have the same virtual wild ramification over $S$.
Then $\mathcal{F}$ and $\mathcal{F}'$
have the same virtual wild ramification over $S$.
\end{Lem}
\begin{proof}
For each $i$, we take a normal compactification $\overline{V_i}$ of $V_i$ over $S$
such that
$\mathcal{F}|_{V_i}$ and $\mathcal{F}'|_{V_i}$
have the same virtual wild ramification along $\overline{V_i}\setminus V_i$.
By Lemma \ref{cptf},
we can take a normal compactification $\overline{X}$ of $X$ over $S$
such that the normalization $\overline{V'_i}$ of $\overline{X}$ in $V_i$ dominates $\overline{V_i}$
for every $i$.
We claim that $\mathcal{F}$ and $\mathcal{F}'$ have the same virtual wild ramification along $\overline{X}\setminus X$.

Let $x$ be a geometric point of $\overline{X}$ and
$\sigma$ be a pro-$p_x$ element of $\pi_1(\overline{X}_{(x)}\times_{\overline{X}}X,y)$,
where $y$ is a geometric point.
Fix a geometric point $w$ of the normalization of $\overline{X}$ in $W$ lying above $x$
and take $i$ such that the image $\bar{\sigma}$ of $\sigma$ in $G$ belongs to $H_i$.
Then $\bar{\sigma}$ belongs to the inertia group $I_v\subset H_i$ at the image $v\to\overline{V'_i}$ of $w$
and hence comes from a pro-$p_v$ element $\sigma'$ of $\pi_1(\overline{V'_i}_{(v)}\times_{\overline{V'_i}}V_i,y')$.
Since $\Gamma^\sigma([\mathcal{F}_y])=\Gamma^{\sigma'}([\mathcal{F}_{y'}])$ and the same for $\mathcal{F}'$,
the assertion follows.
\end{proof}

\begin{proof}[Proof of Proposition \ref{p-power case}]
Take all maximal cyclic subgroups $H_1,\ldots,H_k$ of prime-power order of $G$.
By the assumption, we can apply Lemma \ref{p-power cov} to
$\mathcal{F}|_{W/H_i}$ and $\mathcal{F}'|_{W/H_i}$
and thus $\mathcal{F}|_{W/H_i}$ and $\mathcal{F}'|_{W/H_i}$
have the same virtual wild ramification over $S$.
Then, by Lemma \ref{red to the cyclic case},
$\mathcal{F}$ and $\mathcal{F}'$ have the same virtual wild ramification over $S$.
\end{proof}

\begin{proof}[Proof of Theorem \ref{A to W}]
By virtue of Proposition \ref{W to A}, it suffices to show that
having universally the same conductors implies having the same virtual wild ramification
for $\mathcal{F}$ and $\mathcal{F}'$.
By devissage, we may assume $X$ is regular and $\mathcal{F}$ and $\mathcal{F}'$ are smooth.

Since every regular $S$-surface has a regular compactification over $S$ \cite{L},
the assertion follows from Proposition \ref{p-power case}.
\end{proof}

\subsection{Existence of good curve}
The following lemma will be used to reduce the proof of Proposition \ref{p-power case} to the curve case.
Let $p$ be a prime number.
\begin{Lem}
\label{taking a good curve}
Let
$X$ be a dense open subscheme of
a regular connected excellent noetherian scheme $\overline{X}$.
Let $G$ be a cyclic group of $p$-power order
and $W\to X$ a $G$-torsor.
Then, for any geometric point $x$ over a closed point of $\overline{X}$,
there exists
	a regular scheme $\overline{C}$ of dimension one,
	a finite morphism $\bar{g}:\overline{C}\to \overline{X}$,
	and a geometric point $v$ of $\overline{C}$ over $x$
such that
$C=X\times_{\overline{X}}\overline{C}$ is not empty
and the composite
$\pi_1(\overline{C}_{(v)}\times_{\overline{C}}C,y')
\to\pi_1(\overline{X}_{(x)}\times_{\overline{X}}X,y)\to I_x$
is surjective,
where $I_x$ is an inertia subgroup of $G$ at $x$
and where $y'$ is a geometric point $\overline{C}_{(v)}\times_{\overline{C}}C$
and $y$ is its image.
\end{Lem}

To prove this we use the following lemma:
\begin{Lem}[Lemma 2.4 in \cite{KS}]
\label{KS}
Let $X=\Spec A$ be a spectrum of a normal excellent noetherian local ring $A$.
Let $U$ be a dense open subscheme of $X$
and $V\to U$ an $\mathbb{F}_p$-torsor ramified along a closed subscheme $D\subset X$ of codimension one.
Then, there exists a closed subscheme $C\subset X$ of dimension one
such that the $\mathbb{F}_p$-torsor $V\times_X\widetilde{C}\to U\times_X\widetilde{C}$ is ramified along $D\times_X\widetilde{C}$,
where $\widetilde{C}$ is the normalization of $C$.
\end{Lem}

\begin{proof}[Proof of Lemma \ref{taking a good curve}]
If $I_x=0$, there is nothing to prove.
We assume $I_x\neq0$.
First we reduce the general case to the case where $I_x\simeq\mathbb{F}_p$.
Let $H$ be the subgroup of index $p$ of $I_x$
and $W'$ the quotient $W/H$.
Then, the inertia subgroup $I_{W'/X,x}$ at $x$ of the Galois group $G/H$ of $W'\to X$
	is isomorphic to $\mathbb{F}_p$.
If the $I_x\simeq\mathbb{F}_p$ case is known,
we can find
a regular scheme $\overline{C}$ of dimension one,
a finite morphism $\overline{C}\to\overline{X}$,
and a geometric point $u$ of $\overline{C}$ lying above $x$
such that
$C=\overline{C}\times_{\overline{X}}X$ is not empty
and
the composite
$\pi_1(\overline{C}_{(v)}\times_{\overline{C}}C,y')
\to\pi_1(\overline{X}_{(x)}\times_{\overline{X}}X,y)\to I_{W'/X,x}$
is surjective
and the surjectivity of
$\pi_1(\overline{C}_{(v)}\times_{\overline{C}}C,y')\to I_x$
follows.

We assume $I_x\simeq\mathbb{F}_p$.
By Zariski-Nagata's purity theorem,
there exists a point $\xi\in\overline{X}$ of codimension one
such that $x$ lies over the closure $\overline{\{\xi\}}$ and $W\to X$ is ramified at $\xi$.
Let $V$ be the quotient $W/I_x$
and $\overline{V}$ the normalization of $\overline{X}$ in $V$.
Take a geometric point $v$ of $\overline{V}$ lying over $x$
and a point $\eta\in\overline{V}$ lying over $\xi$ such that $v$ lies over the closure $\overline{\{\eta\}}$.
Then, $W\to V$ is ramified at $\eta$ and $\eta$ is of codimension one.
By Lemma \ref{KS},
We can find
a regular scheme $\overline{C}$ of dimension one,
a finite morphism $\overline{C}\to\overline{V}$,
and a geometric point $u$ of $\overline{C}$ lying above $v$
such that
$C=\overline{C}\times_{\overline{V}}V$ is not empty
and the $\mathbb{F}_p$-torsor $W\times_VC\to C$ is ramified at $u$.
Since an inertia group $I_{W/V,v}$ at $v$ is isomorphic to $\mathbb{F}_p$,
we have the equality $I_{C\times_VW/C,u}=I_{W/V,v}$ of inertia groups.
Thus $\pi_1(\overline{C}_{(u)}\times_{\overline{C}}C,y')\to I_{W/V,v}=I_x$ is surjective.
\end{proof}

\subsection{Statement of the main theorem}
It is easy to show that
having the same virtual wild ramification implies having universally the same conductors:
\begin{Prop}
\label{W to A}
Let $X$ be a scheme separated of finite type over $S$
and $\mathcal{F}$ and $\mathcal{F}'$ constructible complexes of
$\Lambda$-modules and $\Lambda'$-modules respectively on $X$.
Assume $\mathcal{F}$ and $\mathcal{F}'$ have the same virtual wild ramification over $S$.
Then they have universally the same conductors over $S$.
\end{Prop}
\begin{proof}
By Lemma \ref{pullback} and Lemma \ref{char of sc},
we may assume that
$X$ is a regular connected $S$-curve and $\mathcal{F}$ and $\mathcal{F}'$ are smooth
and it suffices to show $\Sw_x(\mathcal{F})=\Sw_x(\mathcal{F}')$
for every geometric point $x$ of $\overline{X}\setminus X$,
where $X\to\overline{X}$ is a canonical regular compactification of $X$ over $S$.
We may further assume $S=\overline{X}$.
Again by Proposition \ref{pullback}, we may assume
$S$ is a strictly local trait and $X$ is the generic point of $S$.
Then the assertion follows from Lemma \ref{sw}.
\end{proof}

We now state our main theorem;
Conjecture \ref{conj} holds for sheaves on an $S$-surface:
\begin{Thm}
\label{A to W}
Let $X$ be an $S$-surface
and $\mathcal{F}$ and $\mathcal{F}'$ constructible complexes of
$\Lambda$-modules and $\Lambda'$-modules respectively on $X$.
Then the followings are equivalent.
\begin{enumerate}[{\rm (i)}]
\item
\label{thmswr}
$\mathcal{F}$ and $\mathcal{F}'$ have the same virtual wild ramification over $S$.
\item
\label{thmsc}
$\mathcal{F}$ and $\mathcal{F}'$ have universally the same conductors over $S$.
\end{enumerate}
\end{Thm}
We show that,
under the existence of a regular compactification,
Conjecture \ref{conj} holds for smooth complexes on a regular $S$-scheme:
\begin{Prop}
\label{p-power case}
Let $X$ be a regular connected scheme separated of finite type over $S$
and $\mathcal{F}$ and $\mathcal{F}'$ smooth complexes of
$\Lambda$-modules and $\Lambda'$-modules respectively on $X$.
Take a finite group $G$ and a $G$-torsor $W\to X$
which trivializes $\mathcal{F}$ and $\mathcal{F}'$.
We assume that $\mathcal{F}$ and $\mathcal{F}'$ have universally the same conductors over $S$
and that
for every subgroup $H$ of $G$ which is maximal among cyclic subgroups	of prime-power order,
the quotient $W/H$ has a regular compactification over $S$.
Then
$\mathcal{F}$ and $\mathcal{F}'$ have the same virtual wild ramification over $S$.
\end{Prop}

\section{The main theorem}
Let $S$ be an excellent noetherian scheme and
assume that all closed points of $S$ have perfect residue fields.
\input{wildram.rel-state}
\input{wildram.rel-curvecase}
\input{wildram.rel-goodcurve}
\input{wildram.rel-goodcptf}

\section{Preliminaries on the Swan conductor}
\label{swbr}
In this section, we see some elementary properties of the Swan conductor.

We need the following lemma on the Brauer trace.
For the definition of the Brauer trace we refer to \cite[18.1]{Se}
and denote it by $\Trbr$.
Let $p$ be a prime number
and $\Lambda$ a finite field of characteristic different from $p$.
\begin{Lem}[Lemma 4.1 in \cite{SY}]
\label{brtr}
Let
$M$ be a $\Lambda$-vector space of finite dimension
and $\sigma$ an endomorphism of $M$ of order a power of $p$.
Then for a subfield $K$ of the fraction field of the Witt ring $W(\Lambda)$
which is finite degree over $\mathbb{Q}$
and contains $\Trbr(\sigma,M)$,
we have
$$
\frac{1}{[K:\mathbb{Q}]}\Tr_{K/\mathbb{Q}}\Trbr(\sigma,M)=\frac{1}{p-1}(p\cdot\dim M^{\sigma}-\dim M^{\sigma^p}).
$$
\end{Lem}

Let $G$ be a pro-finite group.
For a pro-$p$ subgroup $H$ of $G$,
we define the homomorphism $\Gamma^H:K(\Lambda[G])\to\mathbb{Z}$
by assigning the class of each finitely generated $\Lambda[G]$-module $M$ to $\dim M^H$,
where $M^H$ denotes the fixed part.

Let $K$ be a henselian discrete valuation field with algebraically closed residue field of characteristic $p$
and $L$ be a finite Galois extension of $K$ with Galois group $G$.
The Swan character $sw_G:G\to\mathbb{Z}$ is defined in \cite[19.1]{Se}.
Let $M$ be an element of $K(\Lambda[G])$.
The Swan conductor $\Sw(M)$ of $M$ is defined in \cite[19.3]{Se}.

\begin{Lem}
\label{sw}
We have
$$
\Sw(M)=\frac{1}{|G|}\sum_{\sigma\in P}sw_G(\sigma)\cdot\frac{p\cdot\Gamma^\sigma(M)-\Gamma^{\sigma^p}(M)}{p-1},
$$
where $P$ is the $p$-Sylow subgroup of $G$.
\end{Lem}

\begin{proof}
We may assume $M$ is a $\Lambda$-vector space with action of $G$. We have
$$
\Sw(M)=\frac{1}{|G|}\sum_{\sigma\in P}sw_G(\sigma)\cdot\Trbr(\sigma,M).
$$
\cite[19.3]{Se}.
Since $\Sw(M)$ is an integer, the assertion follows from Lemma \ref{brtr}.
\end{proof}

Let $K'$ be an intermediate extension of $L/K$ with Galois group $H=\Gal(L/K')$
and $M\in K(\Lambda[G])$ a virtual representation of $G$.
We regard the restriction $M|_H$ as a representation of the absolute Galois group of $K'$,
and hence $\Sw(M|_H)$ denotes the Swan conductor with respect to $K'$.

\begin{Lem}
\label{if the same sw then the same fixed parts}
Assume that the extension $L/K$ is cyclic of degree $p^e$ for an integer $e\geq0$.
For virtual representations $M\in K(\Lambda[G])$ and $M'\in K(\Lambda'[G])$,
if $\rk(M)=\rk(M')$ and if $\Sw(M|_H)=\Sw(M'|_H)$ for every subgroup $H$ of $G$,
then we have $\Gamma^\sigma(M)=\Gamma^\sigma(M')$ for every $\sigma\in G$.
\end{Lem}

\begin{proof}
We prove the assertion by induction on $e$.
The $e=0$ case is clear.
Suppose $e\geq1$.
Consider the subgroup $G^p$ of $G$ of index $p$.
Then we can apply the induction hypothesis to $M|_{G^p}$ and $M'|_{G^p}$,
and hence $\Gamma^g(M)=\Gamma^g(M')$ for every $g\in G^p$.
Let $\sigma$ be a generator of the cyclic subgroup $G$.
By Lemma \ref{sw},
$$
\Sw(M)=\frac{1}{|G|}\sum_{g\in G}sw_G(g)\cdot\frac{p\cdot\Gamma^g(M)-\Gamma^{g^p}(M)}{p-1},
$$
and similarly for $\Sw(M')$.
Here, by the assumption, $\Sw(M)=\Sw(M')$.
Since $\Gamma^g(M)=\Gamma^g(M')$ for every $g\in G^p$,
we have
$$
\left(\sum_{g\in G\setminus G^p}sw_G(g)\right)\cdot\Gamma^\sigma(M)=
\left(\sum_{g\in G\setminus G^p}sw_G(g)\right)\cdot\Gamma^\sigma(M).
$$
The extension $L/K$ is of degree a power of $p$ and hence $sw_G(g)>0$ for every $g\in G\setminus\{1\}$.
Thus we have $\Gamma^\sigma(M)=\Gamma^\sigma(M')$ and the assertion follows.
\end{proof}

We denote the absolute Galois group of a field $F$ by $G_F$.
In Section 5, we will use the induction formula for the Swan conductor:
\begin{Lem}
\label{conductor discriminant}
Let $K'$ be a finite extension of $K$
and $M'$ a continuous representation of $G_{K'}$ with respect to the discrete topology of $M'$.
Then we have
$$\Sw (\Ind_{G_{K'}}^{G_K}M')=\Sw M'+\dim M'\cdot(d+[K'_s:K]-1),$$
where
$\Ind_{G_{K'}}^{G_K}M'$ denotes the induced representation,
$K'_s$ the maximal separable extension in $K'/K$,
and $d$ the length of the discriminant of the extension $K'_s/K$.

\end{Lem}
\begin{proof}
If $K'/K$ is separable then the assertion follows from  \cite[Proposition 1.(c)]{R}.
Hence we may assume $K'/K$ is purely inseparable.
Here, $\Ind_{G_{K'}}^{G_K}M'$ of $G_K$ is
the representation induced by $M'$ and by the isomorphism $G_{K'}\to G_K$.
Hence, by Lemma \ref{p insep} below, we have $sw_{L/K}=sw_{L'/K'}$ for finite Galois extension $L$ of $K$
and for the composite $L'$ of $L$ and $K'$. Thus the assertion follows.
\end{proof}

\begin{Lem}
\label{p insep}
Let $K'$ be a finite purely inseparable extension of $K$ of degree $p^s$.
Let $L$ be a finite Galois extension of $K$ with Galois group $G$
and $L'$ a composite of $L$ and $K'$.
Take a uniformizer $\pi$ of $L$.
Then $\pi^{1/p^s}$ belongs to $L'$ and is a uniformizer of $L'$
and we have $v(\sigma\pi-\pi)=v'(\sigma\pi^{1/p^s}-\pi^{1/p^s})$
for any element $\sigma\neq1$ of $G$,
where $v$ and $v'$ are the valuation of $L$ and $L'$ respectively.
\end{Lem}
\begin{proof}
We may assume $K$ is complete with respect to its valuation.
Since every complete discrete valuation ring is excellent we can apply Lemma \ref{p insep ext of cdvf} below
and hence $\pi^{1/p^s}$ belongs to $L'$ and is a uniformizer of $L'$.
The $p^s$-th power map $L'\to L$ induces an isomorphism
	between their rings of integers
	and hence we have $v(x^{p^s})=v'(x)$ for every $x\in L'$.
Thus the assertion follows.
\end{proof}
\begin{Lem}
\label{p insep ext of cdvf}
Let $K$ be a discrete valuation field of characteristic $p>0$ with perfect residue field
and $\pi$ a uniformizer of $K$.
Assume that the ring of integer $\mathcal{O}_K$ is excellent.
Then any finite purely inseparable extension of $K$
is of the form $K(\pi^{1/p^s})$ for some integer $s\geq0$.
\end{Lem}
\begin{proof}
We note that $K^p$ is the field of fraction of the excellent discrete valuation ring $\mathcal{O}_K^p$
	and that $\mathcal{O}_K$ is the integral closure of $\mathcal{O}_K^p$ in $K$,
	which is finite over $\mathcal{O}_K^p$.
Since the ramification index of the extension $K/K^p$ is $p$
and the extension of the residue fields is trivial,
we have $[K:K^p]=p$ and hence $[K^{1/p^s}:K]=p^s$ for every $s\geq0$.
Since every finite purely inseparable extension of $K$ of degree $p^s$
	is contained in $K^{1/p^s}$, the assertion follows.
\end{proof}

\section{The Swan conductors of cohomology groups}
Let $S$ be an excellent trait with perfect residue field of characteristic $p>0$
and generic point $\eta=\Spec K$.
For a scheme $X$ separated of finite type over $K$
and a constructible complex $\mathcal{F}$ of $\Lambda$-modules on $X$, we denote by $\Sw(X,\mathcal{F})$
the alternating sum $\sum (-1)^i\Sw H_c^i(X_{\bar{K}},\mathcal{F})$.

The following proposition is a consequence of the result of Vidal (i.e, Proposition \ref{l indep fixed pt}):
\begin{Prop}
\label{svwr to sgc}
Let $X$ be a separated scheme of finite type over $K$
and $\mathcal{F}$ and $\mathcal{F}'$ constructible complexes
of $\Lambda$-modules and $\Lambda'$-modules respectively on $X$.
If $\mathcal{F}$ and $\mathcal{F}'$ have the same virtual wild ramification over $S$
then we have $\Sw(X,\mathcal{F})=\Sw(X,\mathcal{F}')$.
\end{Prop}
\begin{Prop}
\label{l indep fixed pt}
Assume $S$ is strictly local.
Let $X$ be a normal separated scheme of finite type over $\eta=\Spec K$.
We denote the structure morphism by $f$.
Let $\mathcal{F}$ be a smooth complex of $\Lambda$-modules on $X$
and $W\to X$ a $G$-torsor, for a finite group $G$, trivializing $\mathcal{F}$.
We denote by $M$ the virtual representation of $G$ corresponding to $\mathcal{F}$
and take a finite Galois extension $K'$ of $K$ with Galois group $I$
trivializing the smooth complex $Rf_!\mathcal{F}$.
We put $H=G\times I$.
Then $H$ acts on $H_c^*(W_{\bar{\eta}},\mathbb{Q}_l)$ in a natural way
and the followings hold, where $\bar{\eta}$ is a geometric point over $\eta$
and $l$ the characteristic of the finite field $\Lambda$.
\begin{enumerate}[{\rm (i)}]
\item {\rm \cite[Corollary 6.4]{V}}
For $\sigma\in I$, we have
$$
\Tr(\sigma,H_c^*(X_{\bar{\eta}},\mathcal{F}))
=\frac{1}{|G|}\sum_{g\in G}
\Tr ((g,\sigma),H_c^*(W_{\bar{\eta}},\mathbb{Q}_l))\cdot
\Trbr(g,M).
$$

\item {\rm\cite[Proposition 4.2]{V}
}
For $h\in H$,
the trace $\Tr (h,H_c^*(W_{\bar{\eta}},\mathbb{Q}_l))$ is independent of $l\neq p$.

\item {\rm\cite[Corollary 6.5]{V}
}
Let $\overline{X}$ be a normal compactification of $X$ over $S$
and $\overline{W}$ the normalization of $W$ in $\overline{X}$.
Then for $h=(g,\sigma)\in H$,
if $\Tr (h,H_c^*(W_{\bar{\eta}},\mathbb{Q}_l))\neq0$,
then $h$ is of order a power of $p$
and the induced map $g:\overline{W}\to\overline{W}$ fixes some closed point of the closed fiber of $\overline{W}$.
\end{enumerate}
\end{Prop}

\begin{proof}[Proof of Proposition \ref{svwr to sgc}]
We have
$$
\Sw(X,\mathcal{F})=\frac{1}{|I|}\sum_{\sigma\in I}
sw_I(\sigma)\cdot
\Trbr(\sigma,(Rf_!\mathcal{F})_{\bar{\eta}}).
$$
By Proposition \ref{l indep fixed pt} (i), we have
$$
\Sw(X,\mathcal{F})=\frac{1}{|H|}\sum_{h=(g,\sigma)\in H}
sw_I(\sigma)\cdot
\Tr (h,H_c^*(W_{\bar{\eta}},\mathbb{Q}_l))\cdot
\Trbr(g,M).
$$
Since $\Sw(X,\mathcal{F})$ is an integer,
by Lemma \ref{brtr},
$$
\Sw(X,\mathcal{F})=\frac{1}{|H|}\sum_{\substack{h=(g,\sigma)\in H\\\text{ of $p$-power order}}}
sw_I(\sigma)\cdot
\Tr (h,H_c^*(W_{\bar{\eta}},\mathbb{Q}_l))\cdot
\frac{p\cdot\Gamma^g(M)-\Gamma^{g^p}(M)}{p-1}.
$$
Then, the assertion follows from Proposition \ref{l indep fixed pt} (ii) and (iii).
\end{proof}

Now we prove the following lemma, which is needed
to reduce Theorem \ref{sc to sgc} to the curve case (i.e. the $S$-surface case).
\begin{Lem}
\label{red2}
Let $f:X\to Y$ be a morphism of separated schemes of finite type over $K$
and $\mathcal{F}$ and $\mathcal{F}'$ constructible complexes
of $\Lambda$-modules and $\Lambda'$-modules respectively on $X$.
Assume that
$\mathcal{F}$ and $\mathcal{F}'$ have universally the same conductors over $S$
and that
for any regular $S$-curve $\xi$ and any morphism $g:\xi\to Y$,
we have $\Sw(X\times_Y\xi,\mathcal{F})=\Sw(X\times_Y\xi,\mathcal{F}')$.
Then, $Rf_!\mathcal{F}$ and $Rf_!\mathcal{F}'$ have universally the same conductors over $S$.
\end{Lem}

\begin{Lem}
\label{x over a local field}
Let $Z$ be a separated scheme of finite type over $K$
and $\mathcal{F}$ and $\mathcal{F}'$ constructible complexes
of $\Lambda$-modules and $\Lambda'$-modules respectively on $Z$.
If $\mathcal{F}$ and $\mathcal{F}'$ have universally the same conductors over $S$,
then they have universally the same conductors over $\bar{\eta}$,
where $\bar{\eta}$ is the spectrum of an algebraic closure of $K$.
\end{Lem}
\begin{proof}
By Proposition \ref{W to A}, it suffices to show that
for every smooth curve $C$ over $\bar{\eta}$
and every morphism $g:C\to Z\times_\eta\bar{\eta}$ over $\bar{\eta}$,
$g^*p^*\mathcal{F}$ and $g^*p^*\mathcal{F}'$ have the same virtual wild ramification over $\bar{\eta}$,
where $p$ is the projection $Z\times_\eta\bar{\eta}\to Z$.

Take a smooth curve $C$ and a morphism $g:C\to Z\times_\eta\bar{\eta}$ over $\bar{\eta}$.
We can find
a finite extension $\eta'$ of $\eta$,
a  smooth curve $C'$ over $\eta'$,
and a morphism $g':C'\to Z\times_\eta\eta'$ over $\eta'$
such that $g'$ induces $g$ by base extension.
Then by the assumption,
$g'^*p'^*\mathcal{F}$ and $g'^*p'^*\mathcal{F}'$ have universally the same conductors over $S'$,
where $p'$ is the projection $Z\times_\eta\eta'$.
Since $C'$ is an $S$-surface, we can apply Theorem \ref{A to W}.
Thus $g'^*p'^*\mathcal{F}$ and $g'^*p'^*\mathcal{F}'$ have the same virtual wild ramification over $S'$.
Then the assertion follows from Lemma \ref{pullback}.
\end{proof}

\begin{proof}[Proof of Lemma \ref{red2}]
We may assume $S$ is strictly local.
By Lemma \ref{char of sc}, it suffices to show
\begin{enumerate}[{\rm (i)}]
\item
For every geometric point $y$ of $Y$, we have $\rk Rf_!\mathcal{F}_y=\rk Rf_!\mathcal{F}'_y$.
\item
For every finite extension $\xi$ of $\eta$
and every morphism $\xi\to Y$ over $\eta$,
we have $\Sw Rf_!\mathcal{F}|_\xi=\Sw Rf_!\mathcal{F}'|_\xi$.
\end{enumerate}

To show (i), we may assume $y$ is the spectrum of an algebraically closed field.
Then the assertion follows from Lemma \ref{x over a local field} and Theorem \ref{A to X}.

By Lemma \ref{conductor discriminant}, we have
$\Sw(X\times_Y\xi,\mathcal{F})=\Sw(Rf_!\mathcal{F}|_{\xi})+\rk(Rf_!\mathcal{F}|_{\xi})\cdot(d+n-1)$
and the same equality for $\mathcal{F}'$.
Hence the assertion follows from (i) and the assumption.
\end{proof}

\begin{proof}[Proof of Theorem \ref{sc to sgc}]
We may assume $S$ is strictly local.
We prove the assertion by induction on $\dim X$.
If $\dim X\leq1$, by Theorem \ref{A to W},
$\mathcal{F}$ and $\mathcal{F}'$ have the same virtual wild ramification over $S$. Then the assertion follows from Proposition \ref{W to A}.
Assume $\dim X\geq2$.
By the induction hypothesis, it suffices to prove the assertion
after replacing $X$ by a dense open subscheme.
Then, we can take a flat morphism $f:X\to Y$ to a curve $Y$ over $K$
by shrinking $X$ if necessary.
Then for every finite extension $\xi$ of $\eta$
and every morphism $\xi\to Y$, we have
$\Sw(X\times_Y\xi,\mathcal{F})=\Sw(X\times_Y\xi,\mathcal{F}')$.
In fact, since we have $\dim X\times_Y\xi<\dim X$, we can apply the induction hypothesis to
$g^*\mathcal{F}$ and $g^*\mathcal{F}'$, where $g$ is the canonical morphism $X\times_Y\xi\to X$.
Then, by Lemma \ref{red2}, $Rf_!\mathcal{F}$ and $Rf_!\mathcal{F}'$ have universally the same conductors over $S$.
By the $\dim X=1$ case, we get $\Sw(Y,Rf_!\mathcal{F})=\Sw(Y,Rf_!\mathcal{F}')$ and the assertion follows.
\end{proof}

\begin{Def}
\label{def of X}
\upshape
Let $X$ be a separated scheme of finite type over $K$.
For constructible complexes $\mathcal{F}$ and $\mathcal{F}'$
of $\Lambda$-modules and $\Lambda'$-modules respectively on $X$,
we say $\mathcal{F}$ and $\mathcal{F}'$ have {\it universally the same global conductors} over $S$
if for any scheme $Z$ separated of finite type over $K$
and any morphism $g:Z\to X$,
we have
$\chi_c(Z_{\bar{K}},g^*\mathcal{F})=\chi_c(Z_{\bar{K}},g^*\mathcal{F}')$
and $\Sw(Z,g^*\mathcal{F})=\Sw(Z,g^*\mathcal{F}')$.
\end{Def}

\begin{Cor}
\label{sc and sgc}
Let $X$ be a separated scheme of finite type over $K$
and $\mathcal{F}$ and $\mathcal{F}'$ constructible complexes
of $\Lambda$-modules and $\Lambda'$-modules respectively on $X$.
Consider the following three conditions.
\begin{enumerate}[{\rm (i)}]
\item\label{svw s}
$\mathcal{F}$ and $\mathcal{F}'$ have the same virtual wild ramification over $S$.
\item\label{sc s}
$\mathcal{F}$ and $\mathcal{F}'$ have universally the same conductors over $S$.
\item\label{sgc s}
$\mathcal{F}$ and $\mathcal{F}'$ have universally the same global conductors over $S$.
\end{enumerate}

Then, the implications
{\rm (\ref{svw s})}$\Rightarrow${\rm (\ref{sc s})}$\Leftrightarrow${\rm (\ref{sgc s})} hold.
Moreover, if $\dim X\leq1$, then these three conditions are equivalent.
\end{Cor}
\begin{proof}
(\ref{svw s})$\Rightarrow$(\ref{sc s}):
This is nothing but Proposition \ref{W to A}.

(\ref{sc s})$\Rightarrow$(\ref{sgc s}):
Since the property having universally the same conductors is preserved by pullbacks,
this follows from Theorem \ref{sc to sgc}.

(\ref{sgc s})$\Rightarrow$(\ref{sc s}):
We may assume $X$ is connected and finite over $\eta=\Spec K$
and it suffices to show $\rk\mathcal{F}=\rk\mathcal{F}'$ and $\Sw(\mathcal{F})=\Sw(\mathcal{F}')$.
The first equality follows from the equality $\rk f_*\mathcal{F}=[X:\eta]\cdot\rk\mathcal{F}$.
The second equality follows from Lemma \ref{conductor discriminant}.

Finally when $\dim X\leq1$, then the implication (\ref{sc s})$\Rightarrow$(\ref{svw s})
is nothing but Theorem \ref{A to W}.

\end{proof}

\begin{Cor}
\label{functoriality over a local field}
Let $f:X\to Y$ be a morphism between schemes separated of finite type over $K$
and $\mathcal{F}$ and $\mathcal{F}'$ constructible complexes
of $\Lambda$-modules and $\Lambda'$-modules respectively on $X$.
Then,
\begin{enumerate}[{\rm (i)}]
\item
If $\mathcal{F}$ and $\mathcal{F}'$ have universally the same global conductors over $S$,
then so do $Rf_!\mathcal{F}$ and $Rf_!\mathcal{F}'$.
\item
If $\mathcal{F}$ and $\mathcal{F}'$ have universally the same conductors over $S$,
then so do $Rf_!\mathcal{F}$ and $Rf_!\mathcal{F}'$.
\item
Assume $\dim Y\leq1$.
Then, if $\mathcal{F}$ and $\mathcal{F}'$ have the same virtual wild ramification over $S$,
then so do $Rf_!\mathcal{F}$ and $Rf_!\mathcal{F}'$.
\end{enumerate}
\end{Cor}
\begin{proof}
\begin{enumerate}[{\rm (i)}]
\item
For a scheme $Z$ separated of finite type over $K$ and a morphism $g:Z\to Y$,
by the proper base change theorem,
$\chi_c(Z_{\bar{K}},g^*Rf_!\mathcal{F})=\chi_c((Z\times_YX)_{\bar{K}},g'^*\mathcal{F})$
and $\Sw(Z,g^*Rf_!\mathcal{F})=\Sw(Z\times_YX,g'^*\mathcal{F})$,
where $g'$ is the canonical morphism $Z\times_YX\to X$.

\item
This follows from (i) and Corollary \ref{sc and sgc} (ii)$\Leftrightarrow$(iii).

\item
By (i)$\Rightarrow$(iii) of Corollary \ref{sc and sgc},
$\mathcal{F}$ and $\mathcal{F}'$ have universally the same global conductors over $S$.
Then by (i), $Rf_!\mathcal{F}$ and $Rf_!\mathcal{F}'$ have universally the same global conductors over $S$
and the assertion follows from (iii)$\Rightarrow$(i) of Corollary \ref{sc and sgc}.
\end{enumerate}
\end{proof}

redpajama_set_name: RedPajamaArXiv


All industrial wind projects built in Vermont in recent years, including those in Georgia, Lowell and Sheffield, have generated opposition from neighbors — making for protracted legal and regulatory battles. In some cases, developers have bought the homes of opponents. In others, the project has lowered neighboring property assessments.


text:
Scientists at work
Today's young people don't play outside as much as their predecessors and are heavy users of electronics. (Shutterstock)
Too much screen time linked to an epidemic of myopia among young people
February 12, 2019 5.48pm EST
Langis Michaud, Université de Montréal
Langis Michaud
Professeur Titulaire. École d'optométrie. Expertise en santé oculaire et usage des lentilles cornéennes spécialisées, Université de Montréal
Langis Michaud has received funding from: Johnson and Johnson, Bausch and Lomb, Cooper Vision and is a member of l'Association des Optométristes du Québec.
Université de Montréal provides funding as a founding partner of The Conversation CA-FR.
AUF (Agence Universitaire de la Francophonie) provides funding as a member of The Conversation FR.
Ask anyone about the benefits of technology and it probably won't take them long to rhyme off a list of examples: it helps broaden your knowledge, connect with friends, both new and old, and allows you to see things you've never seen before.
But what about the drawbacks? Here's a major one: increased screen time is hard on your eyes.
In fact, more and more young people are wearing glasses to correct their nearsightedness. This trend has prompted optical health experts to determine if the use of electronic devices such as computers, tablets and smart phones is leading to a deterioration in sight that is reaching epidemic proportions.
Let's take the example of Pauline — although she could just as easily be a David or a Jason. She is 10 years old and a studious child, doing well in school. Like any young person her age, she loves to use her tablet computer to study or have fun. She's on it two hours a day, plus weekends.
Pauline may be in for a change, however. She has recently been diagnosed with nearsightedness and her optometrist has strongly recommended that she — and her parents — limit her use of electronic devices and spend more time playing outside.
Pauline isn't happy. She thinks it's not fair.
But her case raises the question: what is the link between the use of electronic devices and the appearance of myopia?
A public health issue
Forty per cent of North Americans are affected by myopia. The number of cases doubled between 1972 and 2004 and continues to grow at a rate that qualifies the phenomenon as an epidemic.
In Europe, myopia is prevalent among 42.2 per cent of adults aged 25 to 29 years, almost twice that of adults aged 55 to 59 years.
This suggests there is a real public health issue facing us — and not just the common problem of an error in the refraction of the eye which shows up as blurred vision when looking into the distance.
In fact, high amounts of myopia significantly increases the risk of major ocular health disorders such as retinal tearing (21 times greater), glaucoma (40 times) or cataracts (six times).
An eye that becomes short-sighted becomes longer. The stretching is proportional to the increase in myopia. The more the eye stretches, the more the retina, which lines the inside of the eye, becomes thinner. Symptoms such as cracks, abnormal development of subretinal blood vessels and bleeding may appear.
Ultimately, the very nearsighted patient has a more than 50 per cent risk of spending the rest of his or her life legally blind — in other words living with vision reduced by 60 per cent. This means that the length of their eye exceeds 28 mm (the normal length is 23 mm) or that the myopia levels exceeds six diopters. (Diopters is a unit of measurement used to calculate eyesight — the further away from zero indicates a worsening in vision.) It is important to intervene before this happens to avoid these levels.
(In Canada, legal blindness is defined by a visual acuity of less than 20/200 in the best eye with the help of glasses or contact lenses. A legally blind person therefore sees, at best, 20 times worse than a person with normal acuity.)
What's behind the epidemic?
There are many causes of onset myopia. Genetics plays an important role but epigenetics — the environment in which the child evolves — is a more important factor.
And what has changed in our environment to explain the recent myopia epidemic? The impact of technology, which has seen a boom in recent years, is being closely examined.
For example, a rapid increase in visual problems has been noted since the introduction of the smartphone in 2007. While the device itself does not emit harmful radiation, it requires the user to read its screen at a distance of 20 cm rather than the normal distance of 45 cm to 50 cm. It has been suggested that this close distance boosts the risk of developing myopia by eight times, especially if both parents are myopic.
Ambient lighting also plays a role because the use of fluorescents, such as in classrooms, also promotes myopia. When a tablet is used in such an environment, the effect is multiplied tenfold.
Unlike paper books, tablet and computer screens emit blue light. The short wavelengths from blue light are perceived in front of the others by the eye, which generates a myopia stimulus. Shutterstock
Unlike books printed on paper, tablet and computer screens are optically associated with so-called chromatic aberrations. The shortest wavelengths (blue light) are perceived in front of the others by the eye, which generates a myopia stimulus. There is a dose and response effect here, suggesting that prolonged use leads to more negative effects.
Today's young people don't play outside as much as their predecessors and are heavy users of electronics. However, exposure to daylight has a protective effect against myopia. These beneficial effects are decreased, but not negligible, during less sunny months or when myopia is already present.
Possible solutions
The purpose of myopia treatment is to reduce the rate of progression of the dioptric, but above all to slow down the elongation of the eye. If, obviously, we cannot play with genetics, it is imperative to influence epigenetics and therefore the environment in which the child evolves.
Therefore, using any electronic media before the age of two should be avoided, even if it's only for a few minutes. A limit of one hour per day should be the rule for those between two and five years of age and the emphasis should be put on educational sites or applications that promote interactions between the parent and child.
While school requests should be taken into consideration, the use of electronic media should continue to be limited as the child gets older. A maximum of one hour, besides school work, must be observed for good eye health. A two-minute break after every 30 minutes of device use will also lessen harmful effects and exposure to a device should be avoided at least one hour before sleep.
Children should also get a minimum of 45 minutes of daylight per day. They can get this by walking to school or participating in such activities as regular outdoor sports.
The next step is to ensure the co-ordination of both eyes, from far away but especially close up. An optometrist will need to perform an orthopic check-up and correct, by exercise or optical means, any anomaly that may have been detected.
Finally, optical correction must be chosen with the goal of slowing down the progression. In rare cases, glasses with anti-myopia lenses may be prescribed. They provide a deceleration rate of about 30 per cent . Specialized contact lenses are often preferred and offer control ranging from 50 per cent to 80 per cent. The contact lens treatment is available from the age of seven, and will be periodically reassessed according to the child's changing needs.
In short, myopia is not just a commonplace vision defect. It is a significant risk factor for serious eye disease. We must therefore do everything possible to slow its progress and protect our children's vision — and that means also reviewing their relationship with electronic devices.
This article was originally published in French
Near-sightedness
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Labor laws fill volumes and are quite complex. Plus, labor laws vary from state to state! It gets worse—the government is ramping up efforts to check up on organizations and crack down. The Obama administration has allotted $25 million for the sole purpose of investigating those misclassified as "independent contractors," hoping to reclaim lost tax revenue.. Be smart. Remember that stiff penalties and lawyer fees can decimate your organization. Are your workers properly classified?
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The global life sciences deal market sector is 'renormalizing' and almost half of industry leaders plan to actively pursue deals in the coming 12 months.
A report from EY shows that, although M&A in the sector has eased recently, it is "far from over" and 58 per cent of those surveyed said that acquisitions and divestitures are topping their boardroom agenda.
Jeff Greene, EY's Global Life Sciences Transaction Advisory Services Leader commented that, following the inversion-focused acquisitions of recent years, specialty pharma firms will be assessing their acquisitions and portfolios.
"Not surprisingly, pharmaceutical M&A deals will be more subdued this year as big pharma becomes more selective in its dealmaking. Yet big pharma still has relatively strong M&A firepower and its need for filling growth gaps may spawn some hostile bids for deal targets, such as those in the surging biotech subsector," he said.
The wives of three human rights lawyers detained by Chinese authorities, have released a video calling for their release.
Hundreds of activists and lawyers were arrested in 2015 and relatives were told to follow rules including not hiring their own lawyers or talking to the media.
However, some including Wang Qialing whose husband Li Heping has been detained, have chosen to speak out. "Facing these illegal and groundless requirements from the government, we as family members didn't give in," she told CNN.
The women say that the Chinese authorities said they must speak on camera to urge their husbands to plead guilty but they have refused.
A class action lawsuit against one US presidential hopeful Donald Trump's former businesses has escalated after a US judge ordered the release of certain documents.
The lawsuit alleges that the defunct Trump University made claims to students about the credibility of its tutors which are disputed; and the promise that the courses would ensure students would be successful in real estate are also challenged.
Donald Trump's lawyers have argued that business documents should not be released as they contain information that may be used in the future if the business is relaunched.
The International Business Times reports that US federal judge Gonzalo Curiel dismissed the appeal against releasing the papers prompting Trump to tell supporters at a rally that "he is a hater of Donald Trump." He also highlighted the judge's Hispanic heritage.


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Psyrassa oaxacae is a species of beetle in the family Cerambycidae.

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Health care vote puts pressure on dozens of vulnerable GOP reps
Heidi M Przybyla
WASHINGTON — Days after President Trump's inauguration, Rep. Mimi Walters of California vowed to protect patients with pre-existing conditions in any plan to replace the Affordable Care Act.
On Wednesday she cast a vote for a plan allows insurance companies in some cases to charge higher premiums to people with cancer, diabetes and other common preexisting conditions — even pregnancy.
It's a moment that could have significant consequences for the Republican Party, particularly House lawmakers like Walters. Her Orange County district is among a couple dozen that have just become more competitive for Democrats due to a vote on a Republican health care plan passed without a single Democratic vote, according to the nonpartisan Cook Political Report.
There's no way GOP leaders could have gotten the bill approved without the support of vulnerable lawmakers like Walters, who sits in a district that Democrat Hillary Clinton won by six points in November. Republicans say the plan will eventually lower health care premiums, even if it results in fewer Americans being insured than under the current health system.
"It's called spending your political capital," said David Wasserman, House editor of the nonpartisan Cook Political Report. "Elections have consequences and majorities tend to use their majority to do things that are unpopular," said Wasserman, who told USA TODAY he is changing ratings on close to two dozen races to reflect the increased political liability.
"It gives Democrats a really big vote to attack," he said, citing members like Walters as well as fellow Californians Ed Royce and Dana Rohrabacher. Even before the vote, there were regular demonstrations outside all of their district offices.
In January, Walters tweeted that she is "committed to protecting patients w/ pre-existing conditions."
In a statement following the vote, she didn't mention pre-existing conditions. "I gave the people of Orange County my word that I would work to ensure all Americans have access to an affordable, patient-centered healthcare system," she said.
Protecting sick Americans
The imperative of protecting sick Americans from the whims of a more capitalist, free market health care system was a key talking point for Democrats as well as Trump throughout the 2016 election. As recently as a few days ago, Trump even insisted such protections were part of his plan to replace the Affordable Care Act.
The current law does not and the original GOP bill would not allow insurers to set premiums based on health status. But an amendment would allow it for those who do not maintain continuous coverage, defined as a lapse of 63 days or more over the previous 12 months. Such policyholders could be charged higher premiums for pre-existing conditions for one year. After that, provided there wasn't another 63-day gap, the policyholder would get a new, less expensive premium that was not based on health status.
The GOP vow to repeal and replace Obamacare was so central — both to Trump's campaign and throughout the past seven years of Republican control of the House — that Republicans made a bet that doing nothing could be more costly if their base voters rebelled. Republicans say the legislation would bring down the price of premiums and stimulate more competition among insurers who are currently withdrawing from certain markets.
"Donald Trump was elected president on the promise to repeal and replace Obamacare," said Rep. Chris Collins, a Republican representing New York's 27th district. "If we weren't able to repeal and replace Obamacare, it would have been a bad midterm for us. I think we will at least hold our own, if not pick up seats," he said.
Indeed, conservative groups known for spending against Republicans who fail to back their agenda, including the Club for Growth, lauded the bill. "Republicans owe a debt of gratitude to conservatives for greatly improving the AHCA," said Club for Growth president David McIntosh.
The legislation is unlikely to go anywhere in the Senate in its current form, with critical moderate Republicans, including Nevada's Dean Heller, registering their disapproval within minutes after the vote. Yet members including Walters are now on the record supporting it going into the 2018 election cycle. It "is a serious problem for Republicans who come from Clinton districts who voted for this," said Wasserman.
In all, there are 23 Republicans representing districts Democratic nominee Hillary Clinton won, and 14 of them supported the bill. Democrats need 24 seats to retake the House.
Yet the danger isn't limited to these districts. According to an analysis by the New York Times' Upshot, voters in districts that voted for Trump have the most to lose if the legislation were to become law. Those eligible for at least $5,000 less in tax credits supported Trump by a margin of 59% to 36%, according to the analysis based on the previous version of the bill.
At the same time, taxpayers in counties that backed Trump would see less of a gain in tax relief (about $6.6 billion) compared to counties that backed Clinton (about $21.9 billion), according to a Bloomberg analysis based on the initial bill.
Moments after the vote, the Democrats' official campaign organization blasted out an email announcing a digital ad campaign in 30 Republican-held districts, including those of some of the most vulnerable Republicans who were initially skeptical of the bill but supported it after a new amendment was added, including Reps. David Valadao and Jeff Denham of California. The amendment worth $8 billion was intended to offset some of the costs of high-risk patients.
The big question is whether the GOP health plan will be as big a liability for the party as Obamacare was for Democrats in 2010, when they lost control of the House. According to a Kaiser Family Foundation tracking poll, the number of Americans viewing Obamacare favorably was up to 48% by late April.
Democrats say they are targeting even vulnerable Republicans like Virginia's Barbara Comstock who voted against the bill. "Every single House Republican owns this," said Tyler Law, a Democratic Congressional Campaign Committee spokesman.
The House bill would not fully repeal the 2010 Affordable Care Act, but it would kill major elements of that Democratic law. It would prune taxes, subsidies and mandates under the law and trim Medicaid. Current tax credits, which are based on income and the cost of health insurance in their local market, would be replaced with less generous tax credits based on age. It would create pools for high-risk patients but also allow insurance companies to opt out of "essential benefits" including maternity and emergency care. Prior to the Affordable Care Act, the majority of states did not have such coverage protections.
Democrats say allowing insurance companies to discriminate will inevitably hurt the sickest, oldest and most vulnerable. They cite opposition by major groups representing doctors, nurses, hospitals and seniors, including the AARP and the American Medical Association. The bill effectively creates an "age tax" on old Americans, Democrats say, and it will actually raise premiums in short term, according to CBO.
Republicans including Tim Walberg of Michigan cited a lot of "myths that are being spun," citing a number of markets where insurers are withdrawing from Obamacare's individual exchange markets and arguing that more competition among insurance companies will bring down premiums. "We tried the Obamacare way. It is failing remarkably and the American people are demanding a change," said Rep. Kevin McCarthy of California, the House majority leader.
Yet Republicans did not receive a score for the legislation from the nonpartisan Congressional Budget Office after a final amendment was added to draw more moderate Republican support. The amendment threw an extra $8 billion intended to offset the costs of high-risk patients.
The prior version would have resulted in 24 million losing insurance, according to CBO. The Center for American Progress estimated on Tuesday that the version the House voted on underfunds high-risk pools by $200 billion over 10 years.
Town hall pressure
Walters is among the Republicans already under significant pressure from her constituents, a number of whom have created a group called the California 45th Town Hall Group to pressure her to meet face-to-face with constituents.
According to the group, it was formed because constituents have made "repeated, formal" requests for an open format town hall. She has declined and earlier told a radio station that the "whole goal is to try to get me to say something that they could use against me."
That pressure is only likely to intensify over a congressional break that starts this weekend.
"You have every provision of this bill tattooed on your forehead. You will glow in the dark," said Rep. Nancy Pelosi, the top Democrat in the House.
Experts: Pre-existing coverage in House GOP bill would fall far short
In major victory for Republicans, House passes Obamacare repeal
Trump claims victory on repealing Obamacare, but he still has a long way to go