text:
\section{Appendix}\label{app}
In this Appendix,
we clarify the derivation of the
central limit theorem of Theorems \ref{precise1}
and \ref{rg1complexe} and  Lemma \ref{control}.
We follow
 the ideas
of \cite{bolt}, where only sums of i.i.d entries
$N^{-1}\sum_{i=1}^N  x_i$  were
considered rather than ponderated sums
$N^{-1}\sum_{i=1}^N \l_i x_i$.
 We consider the case of Theorem
 \ref{rg1complexe} which is the most complicated;
\begin{equation}
\label{f11}
I_N(\t, E_N)
= \prod_{i=1}^N\sqrt{\z_i} \, \, e^{N\theta v}
\int \exp\left\{ N\theta{\g_N(v\g_N-\hat \g_N)\over
1+\g_N}\right\} e^{ -{1\over 2}\sum_{i=1}^N
g_i^2}\prod_{i=1}^N dg_i,
\end{equation}
where we recall that $\z_i := (1+2\t v - 2 \t \l_i)$,
$\g_N := \frac 1 N \sum_{i=1}^N \z_i g_i^2 - 1$ and $\hat \g_N =
 \frac 1 N \sum_{i=1}^N \l_i \z_i g_i^2 - v$.
We denote
$$J_N(\t, E_N)=\sqrt{2\pi}^{-N}\int \exp\left\{
N\theta{\g_N(v\g_N-\hat \g_N)\over
1+\g_N}\right\} e^{ -{1\over 2}\sum_{i=1}^N
g_i^2}\prod_{i=1}^N dg_i.$$
The idea is the following :

\begin{itemize}
\item The first step is to derive a large deviation
principle for $(\g_N, \hat \g_N)$ under
the Gibbs measure
$$\mu_N^\theta(dg)=J_N(\t, E_N)^{-1}\exp\left\{ N\theta{\g_N(v\g_N-\hat \g_N)\over
1+\g_N}\right\} \prod_{i=1}^N P(dg_i).$$
As we showed that the unique minimizer is zero, it
entitles us to write
$$J_N(\t, E_N)=(1+\d(\e,\e',N))J_N^{\e,\e'}(\t, E_N)$$
with
$$J_N^{\e,\e'}(\t, E_N)=
\int_{|\g_N|\le \e, |\hat \g_N|\le\e'}
\exp\left\{ N\theta{\g_N(v\g_N-\hat \g_N)\over
1+\g_N}\right\} \prod_{i=1}^N P(dg_i)$$
where $\d(\e,\e',N)$ goes to zero
as $N$ goes to infinity for any $\e,\e'>0$.

\item Let us assume that we can take above
$\e=M/\sqrt{N},\e'= M'/\sqrt{N}$ with
$\d(M\sqrt{N}^{-1},M'\sqrt{N}^{-1} ,N)$
going to zero as $N$  and then $M,M'$ go to infinity. On the set
$\{|\g_N|\le N^{-\frac{1}{2}}M, |\hat \g_N|\le N^{-\frac{1}{2}}M'\}$,
$$f( \sqrt{N} \g_N,\sqrt{N} \hat\g_N)
= N\theta{ \g_N(v \g_N- \hat \g_N)\over
1+\g_N}= N\theta{\g_N(v \g_N-
\hat \g_N)} +
O((M+M')^3 N^{-\frac{1}{2}})$$
and $f( \sqrt{N}\g_N,
\sqrt{N} \hat\g_N)$ is uniformly bounded. Further, the law
of $( N^{\frac{1}{2}} \g_N,
N^{\frac{1}{2}} \hat\g_N)$ converges under $P^{\otimes N}$ towards
a two-dimensionnal complex Gaussian process with covariance matrix $
K'(\theta)$.
Hence, we can apply dominated convergence theorem
to see that
$$\lim_{N\ra\infty}
\int_{|\g_N|\le N^{-\frac{1}{2}}M,
|\hat \g_N|\le N^{-\frac{1}{2}}M'}
\exp\left\{ N\theta{\g_N(v\g_N-\hat \g_N)\over
1+\g_N}\right\} \prod_{i=1}^N P(dg_i)$$
$$
=(2\pi)^{-2}\mbox{det}(K'(\theta))^{-\frac{1}{2}}\int_{|x|\le M,|y|\le M'}
e^{ \theta x(vx-y)-\frac{1}{2} <(x,y), K'(\theta)^{-1} (x,y)> }
dxdy.$$
In the proof of Theorem \ref{precise1}, we established that
the bilinear
form $x,y\ra
\theta x(vx-y)-\frac{1}{2} <(x,y), K'(\theta)^{-1} (x,y)>$
 is strictly negative for $|\theta|$ small enough,
therefore we can now let $M,M'$ going to infinity
to obtain a limit.

\item To see that we can take $\e=M/\sqrt{N},\e'=M'/\sqrt{N} $,
we can simplify the argument by recalling that
the spherical integral does not depend on $\gamma_N$.
Therefore,
$$
(1-P^{\ot N}(\e\ge |\gamma_N|\ge M\sqrt{N}^{-1}))
J^{\e,\e'}_N(\t, E_N)
=J^{MN^{-\frac12},\e'}_N(\t, E_N)$$
But, $\sqrt{N}\gamma_N= G_N^1+iG_N^2$ has, under $P^{\ot N}$,
sub-Gaussian exponential moments since
$$\E[e^{a G_N^j}]=\prod_{i=1}^N [ (1-2a\sqrt{N}^{-1}
\zeta_j(\l_i))^{-\frac{1}{2}}e^{- a\sqrt{N}^{-1}\zeta_j(\l_i)}]
\le e^{c a^2}$$
for some finite
constant $c$ which only depends
on a  uniform bound on the $\zeta_j(\l_i)$,
where we recall that $\zeta_j(\l_i) = \Re \zeta_i$ if $j=1$ and
$\zeta_j(\l_i) = \Im \zeta_i$ if $j=2$.
By Chebychev's inequality, we therefore conclude
that for $M$ big enough,
$$P^{\ot N}(|\gamma_N|\ge M\sqrt{N}^{-1})\le e^{-\frac{c}{8} M^2}.$$
Finally let us consider
$$J_N^{M,M',\e'}= \int_{|\g_N|\le  M\sqrt{N}^{-1},
 M'\sqrt{N}^{-1}\le |\hat \g_N|\le\e'}
\exp\left\{ N\theta{\g_N(v\g_N-\hat \g_N)\over
1+\g_N}\right\}\prod_{i=1}^N P(dg_i).$$
Clearly,  we find a finite constant $C$ (depending on $\theta$
and $\e'$) such that
$$|J_N^{M,M',\e'}|\le e^{CM^2} \int_{|\g_N|\le  M\sqrt{N}^{-1},
 M'\sqrt{N}^{-1}\le |\hat \g_N|\le\e'}
\exp\left\{ C M|\sqrt{N}\hat \g_N|\right\} dP^{\ot N}(g).$$
Again, $\sqrt{N}\hat \g_N$ has sub-Gaussian tail
so that we find $C'>0$ so that
$$|J_N^{M,M',\e'}|\le e^{(C+\frac{C^2}{C'}) M^2-C'(M')^2}.$$
Now, by
the previous point,
we know that
\begin{eqnarray*}
I(\theta,\mu_E)&=&\lim_{M,M'\ra\infty}
\lim_{N\ra\infty}
\int_{|\g_N|\le N^{-\frac{1}{2}}M,
|\hat \g_N|\le N^{-\frac{1}{2}}M'}
e^{\left\{ N\theta{\g_N(v\g_N-\hat \g_N)\over
1+\g_N}\right\}} \prod_{i=1}^N P(dg_i)\\
&=&
(2\pi)^{-2}\mbox{det}(K'(\theta))^{-\frac{1}{2}}
\int
e^{ \theta x(vx-y)-\frac{1}{2} <(x,y), K'(\theta)^{-1} (x,y)> }
dxdy
 \\
\end{eqnarray*}
exists and moreover goes to one
as $\theta$ goes to zero.
Hence, for $|\theta|$
small enough,
this term dominates $J_N^{M,M',\e'}$ for $N,M,M'$ large
enough ( $M'\gg M$) and we conclude
that
$$\lim_{N\ra\infty} J_N(\t, E_N)=\lim_{M,M'\ra\infty}
\lim_{N\ra\infty} J_N^{MN^{-\frac{1}{2}},M'N^{-\frac{1}{2}}}=
I(\theta,\mu_E).$$
Of course, this strategy only
requires non-degeneracy of the minimum
and $I(\theta,\mu_E)\neq 0$.
In the setting of Theorem \ref{precise1},
this is verified on the whole interval
 $2\theta\in \HE([\l_{\rm min},\l_{\rm max}]^c)$.
In Lemma \ref{control}, we can also apply it by
noting that $L_N(\theta,A,B)$ does
not depend on \linebreak
$(||g^{(1)}||, ||g^{(2)}||, ||g^{(3)}||,
||G||)$ to localize these quantities and proceed.

\end{itemize}

config: arxiv


text:
\begin{document}
\maketitle

\begin{abstract}
The present note generalizes Debarre's Bertini-type results for inverse images of Schubert varieties with the extension of formal functions.
\end{abstract}

\section*{Introduction}

The aim of this short note is to develop Bertini-type results related to the extension of formal functions for inverse images of Schubert varieties of grassmannians. We start with some notation and our fundamental definition.

Given a closed subscheme $Y$ of a variety $X$ over a closed field $k$. Let $X_{/Y}$ be the formal completion of $Y$ along $X$. Let $K(Z)$ be the ring of rational functions of a scheme (formal or not) $Z$ (i.e. the global sections of the sheaf of rational functions). Then it is well known that there exists an injective homomorphism $\alpha:K(X)\to K(X_{/Y})$. We now recall the following definition:

\begin{df}

If $\alpha$ is an isomorphism, we say that $Y$ is G3 in $X$.
\end{df}

The first Bertini-type result regarding extension of formal functions is probably the following:

\begin{theorem}(Faltings, \cite{F})
Let $f:X\to \P^n$ a proper morphism. If dim$(f(X))\ge d$, then for any $d-1$ hyperplanes $H_1,...,H_{d-1}\subset\P^n$, $f^{-1}(H_1\cap...\cap H_{d-1})$ is G3 in $X$.
\end{theorem}

The proof in Faltings' paper involves local rings and local cohomology. Recently, Bonacini, Del Padrone and Nesci found in \cite{BDN} an inspiring global proof whose ideas have already been used for products of projective spaces in \cite{B2}. Our aim is to find an analogous for grassmannians. In order to state our main results, we need a definition regarding special subvarieties of grassmannians.

\begin{df}
Let $\G(k,n)$ be the grassmannian of $k$-planes in $\P^n$. Let $a<b\le n$ be two nonnegative integers. Let $\Pi_0,...,\Pi_k$ be subspaces of $\P^n$ such that dim$\Pi_i=d_i$, with $0\le d_0< ...< d_k\le n$. We introduce the following notation:
\[
\Om(\Pi_0,...,\Pi_k)=\{\Lambda\in\G(k,n)\ |\ \mbox{dim}(X\cap\Pi_i)\ge i,\ \forall i=0,...,k\}
\]
Let also $\Om(d_0,...,d_k)$ be its class in the Chow ring of $\G(k,n)$. They are said to be a Schubert variety and a Schubert cycle (respectively). The dimension of $\Om(d_o,...,d_k)$ is
\[\sum(d_i-k)=\sum d_i+\frac{k(k+1)}{2}.\]
\end{df}

Now we can state the main results of this note:

\begin{main1}
{\it Let $X$ be an irreducible complete variety, and let $j\in\{0,...,n\}$. Let $f:X\to \G(k,n)$ be a morphism such that the intersection product $[f(X)]\Om(d_0-1,d_1-1,..,d_j-1,n-k+j+1,...,n)$ in the Chow ring of $\G(k,n)$ is nonzero for some $0<d_0<...<f_j\le n-k+j$. Then for any $\Pi_i\in\G(d_i,n)$, $i=0,...,j$, we have that
\[f^{-1}(\Om(\Pi_0,...,\Pi_j,\P^{n-k+j+1},...,\P^n))\]
is G3 in $X$.}
\end{main1}

The following result is analogous in the case of the grassmannians of lines. It improves the connectedness theorems of Debarre with some complementary conditions (that, in some cases, are strictly milder).

\begin{main2}
{\it Let $X$ be an irreducible complete variety. Let $f:X\to \G(1,n)$ be a morphism such that the intersection product $[f(X)]\Om(a,a+1)$ in the Chow ring of $\G(1,n)$ is nonzero, given $0\le a<b\le n$. Then for any $\tilde\Lambda\in\G(a,n)$, $\tilde\Pi\in\G(b,n)$, we have $Y:=f^{-1}(\Om(\tilde\Lambda,\tilde\Pi))$ is G3 in $X$.}
\end{main2}

These two Main Results (in the case of grassmannians of lines) can be merged in the following way:

\begin{theorem}
Let $X$ be an irreducible complete variety. Let $f:X\to \G(1,n)$ be a morphism such that one the intersection products
\begin{itemize}
\item $[f(X)]\Om(a,a+1)$, or
\item $[f(X)]\Om(a-1,b-1)$ (or $[f(X)]\Om(a-1,n)$, if $b=n$)
\end{itemize}
in the Chow ring of $\G(1,n)$ are nonzero, given $0\le a<b\le n$. Then for any $\tilde\Lambda\in\G(a,n)$, $\tilde\Pi\in\G(b,n)$, we have $Y:=f^{-1}(\Om(\tilde\Lambda,\tilde\Pi))$ is G3 in $X$.
\end{theorem}

The paper is distributed as follows. First section containssome results thart are useful for the proofs in the paper. Second section contains the proofs of the main results of this note.

{\bf Acknowledgements:} This paper was written during a two-year stay the Universit\`a degli Studi di Genova with a postdoctoral Fundaci\'on Ram\'on Areces grant. I would like to thank Lucian B\u adescu for introducing me to the problem and Enrique Arrondo for some useful discussions.

\section{Preliminaries}

This section contains several known results that will be needed for the proof of the main result. Some related to G3 property and some related to connectedness.

One of the main tools we can use to prove extension of formal functions is the following theorem:

\begin{theorem}\label{HiroMatsu}(Hironaka-Matsumura, \cite{HM}, see also e.g. \cite[Corollary 9.13]{B1})
Let $f:X'\to X$ be a proper surjective morphism of irreducible varieties over an algebraically closed field $k$. Suppose $Y$ is G3 in $X$. Then $f^{-1}(Y)$ is G3 in $X'$.
\end{theorem}

The following is an useful consequence of Theorem \ref{HiroMatsu}

\begin{theorem}\label{Badescu}(see e.g. \cite[Proposition 9.23]{B1})
Let $f:X'\to X$ be a proper surjective morphism of irreducible varieties over an algebraically closed field $k$, and let $Y\subset X$ and $Y'\subset X'$ be closed subvarieties such that  $f(Y')\subset Y$. Assume that the rings $K(X_{/Y})$ and $K(X'_{/f^{-1}(Y)})$ are both fields. If $Y'$ is G3 in $X'$, then $Y$ is G3 in $X$.
\end{theorem}

In order to use the previous result, we first set a framework where the ring of formal rational functions is a field:

\begin{theorem}\label{connectfield}(see e.g. \cite[Corollary 9.10]{B1})
Let $X$ be an algebraic variety over an algebraically closed field $k$, and let $Y$ be a closed subvariety $X$. Let $u:X'\to X$ be the (birational) normalization of $X$. Then $K(X_{/Y})$ is a field if and only if $u^{-1}(Y)$ is connected.
\end{theorem}

The following result, due to Debarre, is the startup and the inspiration for the two main Bertini-type results, Main Results 1 and 2. It is also used for the proof of the first one (the second having hypotheses milder than this one).

\begin{theorem}(Debarre, \cite[Th\'eor\`eme 8.1 (b)]{D})\label{Debarre}
Let $X$ be a projective variety, $f:X\to \G(k,n)$ a morphism. We consider $0\le d_0<...<d_k\le n$ (we also define $d_{k+1}=n+1$) and fix $\Pi_i\simeq\P^{d_i}\subset\Pi_{i+1}\simeq\P^{d_{i+1}}\subset\P^n$, for all $i=0,...,k-1$. If all the intersection products:
\begin{itemize}
\item $[f(X)]\Om(d_i-i-1,...,d_{i}-2,d_i-1,d_{i+1},...,d_k)$, for all $i$ such that $d_i>i$ and $d_{i+1}\ne d_i+1$.
\item $[f(X)]\Om(0,...,i,i+1,d_{i+2},...,d_k)$, for all $i$ such that $d_i=i$ and $d_{i+1}\ne i+1$.
\end{itemize}
in the Chow ring of $\G(1,n)$ are different from zero, then $f^{-1}(\Om(\Pi_0,...,\Pi_k))$ is connected.
\end{theorem}

Again to make Theorems \ref{HiroMatsu} and \ref{Badescu} useful, we need an starting point, which is given by the following result:

\begin{theorem}\label{G3P}(Hironaca-Matsumura \cite{HM} for $\P^n$, Babakarian-Hironaka \cite{BH} for general case)
Any positive-dimensional variety in a grassmannian (not necessarily of lines) is G3.
\end{theorem}

\section{Proofs of the results}

In this section we prove the two main results of this note, just after sketching the general method.

\begin{remark}\label{metodo}
We introduce in this remark the general method we use to prove Bertini-type results regarding extension of formal functions, as extracted from \cite{BDN} and \cite{B2}.

We start with a closed subscheme $Y$ of a scheme $X$. In order to prove that $Y$ is G3 in $X$, we consider an incidence variety $Z$ and the incidence diagram:
\[\begin{array}{ccccl}
& &Z&\subset &X\times P\\
&\swarrow_p & & \searrow^q\\
X& & & & P
\end{array}\]
with $P$ a sufficiently good space (in our cases a projective space and a grassmannian).

We sketch the steps we follow for the proofs of the main results:
\begin{enumerate}
\item prove that $Z$ is irreducible.
\item prove that both $p$ and $q$ are proper and surjective.
\item find a subscheme $L\subset P$ satisfying:
\begin{itemize}
\item $L$ is G3 in $P$.
\item $Y':=q^{-1}(L)\subset p^{-1}(Y)$.
\end{itemize}
\item when the previous steps are completed, we automatically get that $Y'$ is G3 in $Z$ by Theorem \ref{HiroMatsu} to the projection $q:Z\to P$.
\item prove that both $K(X_{/Y})$ and $K(Z_{/Y'})$ are fields.
\item to finish, by Theorem \ref{Badescu} applied to the projection $p:Z\to X$, we get that $Y$ is G3 in $X$.
\end{enumerate}
\end{remark}

We finish the section with the proofs of the main results.

\begin{pf} \emph{(of Main Result 1)}
We will write the proof for the case $j=k$, the other being similar.

Using Theorem \ref{HiroMatsu}, we can suppose $X\subset \G(k,n)$ and $f$ to be the inclusion. Let us define, as in Remark \ref{metodo}, the incidence variety:
\[\begin{array}{ccccl}
&&Z& := & \left\{ (\Lambda,[s_{d_0},...,s_n])\ \left|\
{\scriptsize\begin{array}{l}
s_{d_k|\Lambda}=...=s_{n|\Lambda}=0\\ rk(s_{d_i|l},...,s_{d_{i+1}-1|l})\le 1,\mbox{ when } d_i\ne d_{i+1}-1
\end{array}}\right. \right\} \\
& \swarrow_p && \searrow^q\\
X& & & & P=\P(H^0(\P^n,\O_{\P^n}(1)))
\end{array}\]
Obviously, when $d_i=d_{i+1}-1$, the rank 1 condition means nothing (moreover, in that case, $\Pi_i$ does not impose condition for $\Lambda$). In the case $j<n$ we would just omit the vanishing condition.

First step is to prove that $Z$ is irreducible. We first need to describe $Z$ better. The fiber of $p$ on a point $\Lambda\in X$ consists of all $(n-d_0+1)$-tuples of linear sections such that, when restricted to the line $l$, the last $n-d_k+1$ sections vanish and any pack $s_{d_i},...,s_{d_{i+1}-1}$ have rank at most one, all after restriciton to $\Lambda$. This makes $p$ a bundle over $X$. To study the fiber, we put $x_0,...,x_n$ as coordinates in $\P^n$ such that the space $\Lambda_0:=V(x_{k+1},...,x_n)$ lies in $X$. The fiber of $p$ in $\Lambda_0$ consists on packs of $n-d_0+1$ sections that are linear combinations of the variables $x_0,...,x_n$. The coefficients of $x_k+1,...,x_n$ are completely free, but the coefficients of $x_0,...,x_k$:
\begin{itemize}
\item vanish for $s_{d_k},...,s_n$.
\item form a rank 1 matrix for any pack $s_{d_i},...,s_{d_{i+1}-1}$.
\end{itemize}
This means that the ideal of $p^{-1}(\Lambda_0)$ in the graded ring of $P$ coincides with the ideal of
\[\CalC(\P^k\times\P^{d_2-d_1-1})\times...\times\CalC(\P^k\times\P^{d_k-d_{k-1}-1})\times\P^{(n-k)(n-d_k+1)-1}\]
(where $\CalC(Y)$ is the cone of a variety $Y$) in
\[\P^{(d_2-d_1)(k+1)-1}\times...\times\P^{(d_k-d_{k-1})(k+1)-1}\times\P^{(n-d_k+1)(k+1)-1}\]
(which has the same ring of $P$, but is differently graded). Such ideal is irreducible. Then $Z$ is a bundle with irreducible fibers and then irreducible.

To complete second step we must show that both $p$ and $q$ are surjective. The first one defines a bundle on $X$, so it is obviously surjective. The fiber of $q$ at a general point $[t_{d_0},...,t_n]$, $t_i\in H^0(\O_{\P^n}(1))$, is clearly the set of points in $X$ representing a $k$-plane that is contained in $V(t_{d_k+1},...,t_n)\simeq\P^{d_k-1}$ and intersects all $V(t_{d_i+1},...,t_n)\simeq\P^{d_i-1}$ in a $\P^i$. This fiber is not empty due to the condition: $[f(X)]\Om(d_0-1,...,d_k-1)\ne0$.

Step 3 consists of defining the subscheme $L$ in $P$ with its two properties. Let $t_{a+1},...,t_n\in H^0(\P^n,\O_{\P^n}(1))$ satisfy $\Pi_i=V(t_{d_i+1},...,t_n)$ for all $i=0,...,k$. Then we consider the line $L=<(0:t_{d_0+1}:...:t_n),(t_{d_0+1}:...:t_n:0)>$. For any $(s_{d_0}:...:s_n)\in N$, we have that $V(s_{d_i},...,s_n)\subset\tilde\Pi_i$, so $q^{-1}(N)\subset p^{-1}(Y)$. On the other side:
\[q^{-1}(t_{a+1}:...:t_n:0)=\{(t_{a+1}:...:t_n:0)\}\times Y\]
since $V(t_{d_{i}+1},...,t_n,0)=\Pi_i$. Therefore $p(q^{-1}(N))= Y$. Since, by Theorem \ref{G3P}, $N$ is G3 in a projective space, this proves the connectedness of $Y$.

Now, we directly get fourth step, for $Y':=q^{-1}(N)$ is G3 in $Z$ by Theorem \ref{HiroMatsu}.

We continue proving that both $K(Z_{/p^{-1}(Y)})$ and $K(X_{/Y})$ are fields. Let $h:\tilde Z\to Z$ and $g:\tilde X\to X$ birational normalizaions of $Z$ and $X$ respectively. Then, by Theorem \ref{Debarre}, both $h^{-1}p^{-1}(Y)$ and $g^{-1}(Y)$ are connected, since they are preimages of $\Om(\Pi_0,...,\Pi_k)$. Now, by Theorem \ref{connectfield} $K(Z_{/p^{-1}(Y)})$ and $K(X_{/Y})$ are fields

Last step is again automatic, since by Theorem \ref{Badescu}, $Y'$ being G3 in $Z$ implies that $Y$ is G3 in $X$.
\end{pf}

\medskip

We prove now Main Result 2.

\begin{pf} \emph{(of Main Result 2)}
First of all, we can take a resolution of the singularities of $X$ and compose with $f$. Applying Theorem \ref{HiroMatsu}, we can suppose $X$ to be smooth. We again follow the program in Remark \ref{metodo} and define the incidence variety
$$\begin{array}{ccccl}
&&Z& := & \{(x,\Gamma)\ |\ f(x)=L\subset\Gamma\} \\
& \swarrow_p && \searrow^q\\
X& & & & P=\G(a+1,n)
\end{array}$$

To begin the program we show that $Z$ is irreducible. The fiber of $p$ on $L$ is the set of all $\Gamma$ that contain the line $L=f(x)$ which is isomorphic to a $\G(a,n-2)$. Therefore $Z$ is a bundle over $X$ with irreducible and smooth fiber. We so get that $Z$ is both irreducible and smooth.

We have seen moreover that $p$ is surjective. On the other side, the inverse image of a given $\Gamma$ by $q$ is the preimage of the set of all lines $L\in f(X)$ contained in $\Gamma$. Since $[f(X)]$ intersects $\Om(a,a+1)$, it is nonempty and hence $q$ is also surjective. This finishes the second step.

For the third step of the program, we consider the subspace $N\subset\G(a+1,n)$ consisting of all $\Gamma$ such that $\tilde\Lambda \subset \Gamma \subset \tilde\Pi$. Clearly, $N\simeq\P^{b-a-1}$, so $N$ is G3 in $\G(a+1,n)$ by Theorem \ref{G3P}.

To finish this step, we need to prove that $q^{-1}(N)\subset p^{-1}(Y)$. For any subspace $\Gamma\in L$ we have that its fiber by $q$ consists on all points $x\in X$ such that $f(x)=L\subset\Gamma$. We know that $\Gamma\subset\tilde\Pi$ (by definition of $N$). Moreover, since $\tilde\Lambda$ is a hyperplane of $\Gamma$ (also by definition of $N$), we have that every line of $\Gamma$ intersects $\tilde\Lambda$, so all $x\in p(q^{-1}(\{\Gamma\}))$ satisfy $L=f(x)\subset\tilde\Pi$ and $L\cap\tilde\Lambda\ne\emptyset$, so $p(q^{-1}(\{\Gamma\}))\subset Y$ as we wanted.

Fourth step is automatic, $q^{-1}(N)$ is G3 in $Z$ by Theorem \ref{HiroMatsu} and the fact that $N$ is G3 in $\G(a+1,n)$.

Next step is the proof of both $K(Z_{/p^{-1}(Y)})$ and $K(X_{/Y})$ being fields. Since both $X$ and $Z$ are smooth (and then normal), Theorem \ref{connectfield} reduces this question to showing that both $Y$ and $p^{-1}(Y)$ are connected. Moreover, $p$ is a bundle, so proving that $Y$ is connected is enough: For all $x\in Y$, $l=f(x)$ satisfies $L\cap\tilde\Lambda\ne\emptyset$, so $\Gamma:=<L,\tilde\Lambda>$ has dimension $b-1$ and therefore $L\in p(q^{-1}(\{\Gamma\}))$. This proves that $Y\subset p(q^{-1}(N))$. Since the other content was proved in the third step, we have $Y=p(q^{-1}(N)=Y')$. $Y'$ is G3 in $Z$ as shown before, so it is conected. Therefore $Y$ is the image of a connected subset by a continuous function, so it is also connected.

Last step is automatic once more, since by Theorem \ref{Badescu},  $Y'$ is G3 in $Z$ provides the fact that $Y$ is G3 in $X$.
\end{pf}

\begin{remark}
The generalization of Main Result 2 to general grassmannians requires better knowledge on Flag manifolds than that of the author. So an unique statement for general grassmannians is out of reach now.
\end{remark}


text:
\begin{document}
\maketitle

\begin{abstract}
Let $k$ be an algebraically closed field of characteristic $0$, and let $f:X \lra Y$ be a morphism of smooth projective varieties over the ring $k((t))$ of formal Laurent series. We prove that if a general geometric fiber of $f$ is rationally connected, then the Berkovich analytifications of $X$ and $Y$ are homotopy equivalent. Two important consequences of this result are that the homotopy type of the Berkovich analytification of any smooth projective variety $X$ over $k((t))$ is a birational invariant of $X$, and that the Berkovich analytification of a rationally connected smooth projective variety over $k((t))$ is contractible.
\end{abstract}

\vskip 1.25cm

\section{Introduction}

Unless otherwise specified, we work over a field $k$ which is algebraically closed of characteristic $0$. Given a $k$-scheme $S$, we say that $X\lra S$ is a variety over $S$ if $X$ is an integral, finite type scheme over $S$. We use the term \emph{generic point} to mean the point of the variety $X$ corresponding to its field of rational functions, while a \emph{general point} refers to any closed point chosen inside some open set. If $T$ is a second $S$-scheme, then a \emph{general $T$-point} of $X$ refers to any $T$-valued point $T\lra X$ whose image is a general point.

\vskip .5cm

\subsection{A suggestive dictionary}\

Let $K$ be a topological field, topologically complete with respect to a  fixed norm $|-|:K\lra\mathbb{R}_{\ge0}$. Assume that this norm satisfies the ultrametric triangle inequality $|a+b|\le\mathrm{max}\{|a|,|b|\}$ for all $a,b\in K$. We call a field $K$ complete with respect to such a norm a {\em non-archimedean field}.

In close analogy with the operation of complex analytification, Berkovich showed how to associate a well behaved topological space $X^{\mathrm{an}}$ to each finite type $K$-scheme $X$ \cite[\S 3.4]{Berk}. The space $X^{\mathrm{an}}$ is called the {\em Berkovich analytification} of $X$. Berkovich observed that there exists a partial dictionary between algebro-geometric properties of $X$ and topological properties of $X^{\mathrm{an}}$. Namely:

\vskip .25cm

{\bf Theorem  \cite[Theorem 3.4.8]{Berk}.} Let $X$ be a scheme locally of finite type over $K$. Then:
\begin{itemize}
\item[{\bf (i)}] $X$ is separated $\iff$ $X^{\mathrm{an}}$ is Hausdorff.
\item[{\bf (ii)}] $X$ is proper $\iff$ $X^{\mathrm{an}}$ is Hausdorff and compact.
\item[{\bf (iii)}] $X$ is connected $\iff$ $X^{\mathrm{an}}$ is pathwise connected.
\item[{\bf (iv)}] The algebraic dimension of $X$ $=$ the topological dimension of $X^{\mathrm{an}}$.
\vskip .25cm
\end{itemize}

\vskip .25cm

\noindent
Naturally, one would like to continue Berkovich's dictionary, and so we ask:

\vskip .25cm

\begin{question}\label{giant question} Which other algebro-geometric properties of $X$ correspond to topological properties of $X^{\mathrm{an}}$?
\end{question}

One fundamental invariant of a topological space is its homotopy type. Our main theorem in the present article provides a tool for relating the birational geometry of $K$-varieties to the homotopy types of their Berkovich analytifications, in the special case that $K=k((t))$, the field of formal Laurent series over an algebraically closed field $k$ of characteristic $0$:

\vskip .3cm

\begin{theorem}\label{main}
Let $k$ be an algebraically closed field of characteristic $0$, and let $f: X \lra Y$ be a surjective map of smooth projective varieties over the nonarchimedean field $K= k((t))$. Suppose that for a general geometric point $q$ of  $Y$ the fiber $X_q$ is rationally connected. Then the Berkovich spaces $X^{\mathrm{an}}$ and $Y^\mathrm{{an}}$ are homotopy equivalent.
\end{theorem}

\vskip .4cm

As immediate corollaries, we have

\vskip .4cm

\begin{corollary}\label{pb}
If $Y$ is a smooth $k((t))$-variety and $f:X\lra Y$ is a $\mathbb{P}^{n}$-bundle over $Y$, then the induced map $f^{\mathrm{an}}:X^{\mathrm{an}}\lra Y^{\mathrm{an}}$ is a homotopy equivalence.
\end{corollary}

\vskip .4cm

\begin{corollary}\label{rc}
Let $X$ be a smooth projective $k((t))$-variety. If $X$ is rationally connected, then its Berkovich analytification $X^{\mathrm{an}}$ has the homotopy type of a point, i.e., is contractible.
\end{corollary}

\vskip .4cm

Berkovich showed that for projective space $\mathbb{P}_{\!K}^{n}$ of any dimension $n$, over any non-archimedean field $K$, the analytification $(\mathbb{P}_{\!K}^{n})^{\mathrm{an}}$ is contractible \cite[Theorem 6.1.5]{Berk}. The simplest example of a rationally connected variety is projective $n$-space, so Corollary \ref{rc} can be seen as a broad extension of Berkovich's contractibility result in the special case that $K=k((t))$, and Corollary \ref{pb} can be seen as an extension of Berkovich's result to the relative setting.

Finally, given $X$ and $Y$ smooth, projective and birational, we can choose a common resolution of singularities $Z$ of both $X$ and $Y$, so that Theorem \ref{main} also implies

\vskip .4cm

\begin{corollary}\label{ht}
Let $X$ and $Y$ be smooth projective $k((t))$-varieties. If $X$ and $Y$ are birationally equivalent, then their Berkovich analytifications $X^{\mathrm{an}}$ and $Y^{\mathrm{an}}$ are homotopy equivalent.
\end{corollary}

\vskip .4cm

Note that being rationally connected is a birational invariant. Recent work of Musta{\c{t}}{\u{a}} and  Nicaise  \cite{MN} and Nicaise and Xu \cite{NX} explores connections between the minimal model program (MMP) and the geometry of Berkovich spaces. We adopt many of their techniques in the proof of Theorem \ref{main}. We also use recent results by de Fernex, Koll\'ar, and Xu on the homotopy type of the dual complex of a resolution of singularities \cite{dFKX}. It is our use of these results that forces us to work over the non-archimedean field $k((t))$, with $k$ algebraically closed of characteristic $0$. It remains an open question  whether or not a counterpart to Theorem \ref{main}, or any of its Corollaries \ref{pb} through \ref{ht}, holds over more general non-archimedean fields.

\vskip .25cm

\vskip .5cm

\subsection{Organization of the paper}\

In Section \ref{skeleta}, we briefly review Berkovich analytifications and their skeleta. We use results of \cite{MN} and \cite{NX} to construct a skeleton $\mathrm{Sk}(\mathscr{X})\subset\an{X}$, and we reduce the proof of Theorem \ref{main} to a verification that the map $f:X\lra Y$ appearing in the statement of the theorem induces a homotopy equivalence $\mathrm{Sk}(\mathscr{X})\simeq\mathrm{Sk}(\mathscr{Y})$

In Section \ref{rcvarieties}, we review rational connectivity of algebraic varieties in both the absolute and relative settings.

The main technical contribution of the paper occurs in Section \ref{spreading out section}. Here we show that if $\mathscr{X}$ is an snc $k[[t]]$-model of a rationally connected variety, then we can construct a degenerating family of rationally connected varieties over an algebraic curve, such that the dual complex of the special fiber of this degenerating family is homeomorphic to our skeleton $\mathrm{Sk}(\mathscr{X})$. We actually need a relative version of this statement in order to prove Theorem \ref{main}, and it is the relative version that we prove in Section \ref{spreading out section}. This puts us in a situation where the results of \cite{dFKX} become applicable.

In Section \ref{applying dFKX}, we establish a relative version of \cite[Theorem 41]{dFKX}, and use it to complete the proof of Theorem \ref{main}.

\vskip .4cm

\subsection{Thanks}
The authors would like to thank Matt Baker, Mattias Jonsson, Eric Katz, Mircea Musta{\c{t}}{\u{a}}, James McKernan, Sam Payne, Joe Rabinoff, and David Zurieck-Brown for invaluable discussions over the course of writing this article. Several of these discussions occurred at the BIRS workshop ``Specialization of Linear Series for Algebraic and Tropical Curves'', at which the first author was a participant.  We would like to thank Sam Payne in particular for remarks on an earlier draft of this paper. His remarks lead to a significant strengthening of our results.

Both authors are partially supported by NSF RTG grant DMS-0943832.

\vskip .75cm

\section{Skeleta in Berkovich spaces}\label{skeleta}

\subsection{The Berkovich analytification}\

Let $K$ be a topological field, complete with respect to a continuous, non-archimedean norm $|-|:K\lra\mathbb{R}_{\ge0}$. Assume that this norm is {\em discrete}, in the sense that its value group $|K^{\times}|$ is a discrete subgroup of $\mathbb{R}_{\ge0}$. Let $R=\{a\in K:|a|\le1\}$ denote the ring of integers in $K$, with unique maximal ideal $\mathfrak{m}=\{a\in K:|a|<1\}$, and denote its residue field $k\overset{{}_{\mathrm{def}}}{=}R/\mathfrak{m}$.

Let $X$ be a proper $K$-variety, and let $X^{\mathrm{an}}$ denote its Berkovich analytification. A variety for us will always be separated, and hence its Berkovich analytification will always be a Hausdorff topological space. In this article, we consider $X^{\mathrm{an}}$ solely as a topological space, completely ignoring the analytic structure sheaf on $X^{\mathrm{an}}$.

Let us briefly recall that if $U=\mathrm{Spec}_{\ \!}A$ is an affine local chart on $X$, then $U^{\mathrm{an}}$ is an open subset of $X^{\mathrm{an}}$ given by
$$
U^{\mathrm{an}}
\ =\
\left\{
\begin{array}{c}
\mbox{multiplicative seminorms }
\\
|-|_{x}:A\!\lra\!\mathbb{R}_{\ge0}\mbox{\ that restrict}
\\
\mbox{to the norm }|-|\mbox{ on }K\!\subset\!A
\end{array}
\right\}
$$
Each function $f\in A$ determines a map $\mathrm{ev}_{f}:U^{\mathrm{an}}\lra\mathbb{R}_{\ge0}$ that takes $|-|_{x}\mapsto |f|_{x}$. The topology on $U^{\mathrm{an}}$ is the coarsest topology that renders each of these maps continuous. We glue $X^{\mathrm{an}}$ from these topological spaces $U^{\mathrm{an}}$, and the resulting topological space is independent of the particular affine cover we use.

\vskip .4cm

\subsection{Skeleta}\

The Berkovich analytification tends to be large and complicated. For example, the point set underlying the Berkovich analytification of a curve has a natural structure of an infinite graph. One powerful technique for understanding a given Berkovich analytification is to produce a smaller space inside $X^{\mathrm{an}}$, called a {\em skeleton}, which reflects much of the geometry of the ambient space.  Specifically, this skeleton is supposed to be a finite simplicial complex onto which $X^{\mathrm{an}}$ deformation retracts. In the case of curves, the skeleton turns out to be a finite graph.

A general technique for producing a skeleton is to choose a model $\mathscr{X}$ of $X$ over the local ring $R$ of integers in $K$, such that the special fiber $\mathscr{X}_{k}$ is a reduced simple normal crossing divisor in $\mathscr{X}$. Each divisor corresponds to a point in $\an{X}$, and the full dual complex  of $\mathscr{X}_{k}$ embeds as a closed subspace of $\an{X}$. Note though that because this construction requires the choice of a model $\mathscr{X}$, it is not canonical.

It is often desirable to produce a skeleton that is canonical in the sense that it can be recovered uniquely from $X$. One construction which produces such a skeleton is due to Musta{\c{t}}{\u{a}} and Nicaise \cite{MN}, following work of Kontsevich and Soibelman \cite{KS} on mirror symmetry. Their construction uses weight functions coming from pluricanonical forms on $X$, and Nicaise and Xu \cite{NX} have used this construction to study degenerations of Calabi-Yau varieties.  This construction is in many ways analogous to the theory of minimal models. If $X$ is of general type, and if the canonical divisor $K_X$ is nef, one expects to be able to produce a minimal model $\mathscr{X}$ over $R$, with $K_{\mathscr{X}}$ nef. This model will be unique up to simple birational modifications called flops. In the case where $X$ is a Calabi-Yau variety, Nicaise and Xu \cite{NX} show that this skeleton is given by the dual complex of a good minimal dlt model of $X$, if one exists.

On the other hand, a rationally connected variety has no pluricanonical forms. When one runs the minimal model program on a rationally connected variety $X$, the end result is a Mori fiber space. In contrast to the general type case, there may be  many different ways of expressing $X$ as birational to a Mori fiber space. Thus from the perspective of minimal model theory, one should not expect there to be a distinguished way of building a skeleton for a rationally connected $X$.

\vskip .5cm

\subsection{Construction of a non-canonical skeleton}\

Let $X$ be a proper $K$-variety. An {\em $R$-model} of $X$ is any separated flat $R$-scheme $\mathscr{X}$ with generic fiber $\mathscr{X}_{K}\cong X$. A {\em proper snc-model} of $X$ is any proper $R$-model $\mathscr{X}$ of $X$ such that:
\vskip .2cm
\begin{itemize}
\item[{\bf (i)}]
$\mathscr{X}$ is normal;
\vskip .2cm
\item[{\bf (ii)}] The special fiber $\mathscr{X}_{k}$ is a strict normal crossing divisor in $\mathscr{X}$.
\end{itemize}
\vskip .2cm

Assume that our proper $K$-variety $X$ is regular, connected, and that it has a proper snc-model $\mathscr{X}$. As Musta\c{t}\u{a} and Nicaise explain in \cite[\S 3]{MN}, even when the special fiber $\mathscr{X}_{k}$ is not reduced, so that the $\mathfrak{m}$-adic completion $\widehat{\mathscr{X}}$ is not pluristable, we can still construct a cell complex $D(\mathscr{X}_{k})\subset\mathscr{X}^{\mathrm{an}}$ from $\mathscr{X}_{k}$ as follows.

Points in $D(\mathscr{X}_{k})$ are pairs $(\xi,x)$ where $\xi$ is any generic point of the intersection of finitely many irreducible components of $\mathscr{X}_{k}$, and where $x=(x_{1},\dots,x_{m})$ is any point in the topological space $D^{\circ}(\xi)\subset\mathbb{R}^{m}_{>0}$ cut out by the equation
$$
\underset{\mathrm{containing\ }\xi}{\underset{V_{i}\mathrm{\ of\ }\mathscr{X}_{k}}{\underset{\mathrm{components}}{\sum_{\mathrm{irreducible}}}}}\!\!\!x_{i}\ \ =\ 1
$$
Here $m$ is the number of irreducible components of $\mathscr{X}_{k}$ containing $\xi$.
We can attach these spaces $D^{\circ}(\xi)$ to one another via boundary relations induced by the specialization relations between the generic points $\xi$ (see \cite[\S 3]{Nic} for details). Note that the resulting cell complex $D(\mathscr{X}_{k})$ coincides with the cell complex $\Delta\big(\mathrm{supp}(\mathscr{X}_{k})\big)$ that de Fernex, Koll{\'a}r, and Xu construct in \cite[\S 2]{dFKX}.

At each point $(\xi,x)$ in $D(\mathscr{X}_{k})$, with $x=(x_{1},\dots,x_{m})$, let $V_{1},\dots,V_{m}$ be the set of irreducible components of $\mathscr{X}_{k,\mathrm{red}}$ containing the generic point $\xi$. For each $1\le i\le m$, let $N_{i}$ denote the multiplicity of the component $V_{i}$ in the special fiber $\mathscr{X}_{k}$. Then the coordinatewise rescaled $m$-tuple
$$
v\ =\ \Big(\frac{x_{V_{1}}}{N_{1}},\dots,\frac{x_{V_{m}}}{N_{m}}\Big)
\ \ \ \ \ \ \mbox{in}\ \ \ \ \ \
\mathbb{R}^{m}_{>0}
$$
determines a multiplicative seminorm
\begin{equation}\label{divisorial seminorm}
|-|_{v}:K(X)\lra\mathbb{R}_{\ge0}
\end{equation}
defined as follows. Each rational function $f\in\widehat{\mathscr{O}}_{\mathscr{X}\!,\xi}$ admits an {\em admissible decomposition}, meaning a decomposition of the form
\begin{equation}\label{admiss decomp}
f=\sum_{u\in\mathbb{Z}^{m}_{\ge0}}a_{u}T^{u}
\ \ \ \ \ \ \mbox{with}\ \ \ \ \ \
a_{u}=\left\{
\begin{array}{l}
\mbox{a unit in }\widehat{\mathscr{O}}_{\mathscr{X}\!,\xi}
\\
\mbox{or }0\mbox{ otherwise}
\end{array}
\right.
\end{equation}
(see \cite[Proposition 2.4.6]{MN} for details). When $f$ is in $\mathscr{O}_{\mathscr{X}\!,\xi}$ with an admissible decomposition (\ref{admiss decomp}), the seminorm $|-|_{v}$ returns
$$
|f|_{v}
\ \ =\ \
\underset{a_{u}\ne0}{\underset{u\in\mathbb{Z}^{m}_{\ge0}}{\mathrm{max}_{\ }}}\ \frac{1}{e^{\langle u,v\rangle}}
$$
Here $\langle-,-\rangle:\mathbb{Z}^{m}_{\ge0}\times\mathbb{R}^{m}_{>0}\lra\mathbb{R}_{\ge0}$ denotes the standard pairing. One checks that this gives a well-defined map $|-|_{v}:\mathscr{O}_{\mathscr{X},\xi}\lra\mathbb{R}_{\ge0}$ that extends to a multiplicative seminorm on $K(X)$, and we take this to be our seminorm (\ref{divisorial seminorm}) (see \cite[\S's 2.4 \& 3.1]{MN} for details). The resulting assignment $v\mapsto|-|_{v}$ induces a closed embedding
$$
\xymatrix{\mathrm{Sk}:D(\mathscr{X}_{k})\ar@{^{(}->}[r]&X^{\mathrm{an}}}
$$
(see \cite[\S 3.1.3 \& Proposition 3.1.4]{MN}). We refer to its image as the {\em skeleton} of $X^{\mathrm{an}}$ associated to $\mathscr{X}$, and denote it $\mathrm{Sk}(\mathscr{X})$.

\vskip .5cm

\subsection{Retracting to the skeleton over characteristic-0 Laurent series}\label{skeleton building}\

Let $k$ be an algebraically closed field of characteristic $0$, and equip the field $K=k((t))$ of formal Laurent series with the norm $|-|:k((t))\lra\mathbb{R}_{\ge0}$ defined by
$$
\left|\sum^{\infty}_{n=m}a_{n}t^{n}\right|=\frac{1}{e^{m}}
\ \ \ \ \ \ \mbox{for}\ \ \ \ \ \ a_{m}\ne0
$$
This norm is non-archimedean, and $k((t))$ is complete with respect to the topology that it induces. The resulting ring of integers in $k((t))$ is the ring $R=k[[t]]$ of formal power series, with maximal ideal $\mathfrak{m}=(t)$ and residue field $k=k[[t]]/(t)$.

Let $X$ be an irreducible, smooth projective variety over $k((t))$. For each choice of closed embedding $X\!\xymatrix{{}\ar@{^{(}->}[r]&{}}\!\mathbb{P}^{d}_{k((t))}$, we obtain a $k[[t]]$-model $\mathscr{X}$ of $X$ as the closure
$$
\mathscr{X}\ \overset{\mathrm{def}}{=}\ \overline{X}
$$
inside $\mathbb{P}^{d}_{k[[t]]}$. This $k[[t]]$-scheme $\mathscr{X}$ is flat. Thus it is a bona fide proper $k[[t]]$-model of $X$. Moreover, $\mathscr{X}$ is reduced whenever $X$ is, so we can assume without loss of generality that $\mathscr{X}$ is normal.

We have no assurance, a priori, that the proper $k[[t]]$-model $\mathscr{X}$ is an snc-model. However, because we work over a residue field $k$ of characteristic $0$, and because $X$ is smooth, we can use Hironaka's resolution of singularities to replace $\mathscr{X}$ with such a model if need be.
After making this replacement, the construction of the previous section provides us with a skeleton
$$
\mathrm{Sk}(\mathscr{X})\ \subset\ X^{\mathrm{an}}
$$
Nicaise and Xu prove that, at least over $k((t))$, this skeleton is a strong deformation retract of $X^{\mathrm{an}}$  \cite[Theorem 3.1.3]{NX}. Thus we immediately have the following:

\vskip .3cm

 \begin{lemma}\label{homotopyofdual}
 \begin{proof}[\!\!\!\!{\color{white} .}]
 With $X$ and $\mathscr{X}$ as above, the Berkovich analytification $X^{\mathrm{an}}$ has the same homotopy type as the dual complex $D(\mathscr{X}_{k})$.
 \end{proof}
 \end{lemma}

\vskip .3cm

\begin{remark}
From Lemma \ref{homotopyofdual}, it follows immediately that to prove that the analytification $f^{\mathrm{an}}:\an{X}\lra\an{Y}$ of a given map $f:X\lra Y$ of $k((t))$-varieties, it suffices to show that $f$ induces a homotopy equivalence $D(\mathscr{X}_{k})\simeq D(\mathscr{Y}_{k})$ between the dual complexes of the special fibers in snc $k[[t]]$-models $\mathscr{X}$ and $\mathscr{Y}$ of $X$ and $Y$, respectively.

One would like to establish homotopy equivalence of $D(\mathscr{X}_{k})$ and $D(\mathscr{Y}_{k})$  by using techniques from MMP directly on a $k[[t]]$-model $\mathscr{X}\lra\mathscr{Y}$ of the map $f:X\lra Y$. Unfortunately, the necessary MMP techniques are not established over rings, like $k[[t]]$, which are not of finite type over a field of characteristic $0$. We will have to use a ``spreading out" argument in order to transfer our task into a setting where the needed MMP techniques are known.
\end{remark}

\vskip .75cm

\section{Rationally Connected Varieties}\label{rcvarieties}

\subsection{Definitions and first properties}\

The details in this section are well known to experts; we refer to \cite{Deb} and \cite{Kol}.

One way to distinguish birational classes of varieties is by considering the sets of rational curves they contain. If $X$ is a smooth variety over an algebraically closed field, we say that $X$ is rationally connected if given any two closed points $p$ and $q$ of $X$, there is a map $f: \mathbb{P}^1 \lra X$ such that both $p$ and $q$ are in the image of $f$.  Projective $n$-space is rationally connected for instance, since any two closed points of $\mathbb{P}^n$ are contained in a line. Many other familiar varieties are also rationally connected, including Grassmanians, toric varieties, and smooth Fano varieties.

Because we will be working over $k((t))$, which is not algebraically closed, we will need a slightly more general definition than the one above.

\begin{definition}\label{rationally connected def} Assume that $F$ is a field of characteristic $0$. A proper scheme $X$ over $F$ is {\em rationally connected} if there exists a variety $M$ over $F$, and a morphism $u: M \times\mathbb{P}^1 \lra X$ such that the induced map $u^{(2)}: M \times\mathbb{P}^1\times\mathbb{P}^1 \lra X \times_F X$ is dominant.
\end{definition}

This definition follows \cite[Definition 4.3]{Deb}, and corresponds to the definition of a \emph{separably rationally connected} scheme in \cite[Definition IV.3.2]{Kol}. The two notions are equivalent in characteristic $0$. Note that in Definition \ref{rationally connected def}, it is equivalent to take $u$ to be a rational map instead of a morphism. When the field $F$ is algebraically closed and $X$ is smooth, this definition is equivalent to the condition that every pair of closed points of $X$ be connected by a rational curve.

Definition \ref{rationally connected def} behaves well under birational modification: if $X'$ is birational to $X$, then $X'$ is rationally connected if and only if $X$ is. Thus all rational varieties are rationally connected. Rationally connected varieties also enjoy the additional property that the image of a rationally connected variety is rationally connected:

\begin{proposition}
Let $X$ be rationally connected, and let $f: X \lra Z$ be a surjective morphism. Then $Z$ is rationally connected.

\end{proposition}
\begin{proof}
We have $u: M \times \mathbb{P}^1 \lra X$, with $u^{(2)}$ dominant. Compose to get $f\circ u: M\times \mathbb{P}^1 \lra Z$, and observe that surjectivity of $f$ implies that the induced map $M \times \mathbb{P}^1 \times \mathbb{P}^1 \lra Z$ is dominant.
\end{proof}

\vskip .4cm

\subsection{Rational connectivity in families}\

Rational connectivity behaves well in families. If $\pi:Y\lra S$ is a proper, smooth map of varieties over an algebraically closed field $F$ of characteristic $0$, then the locus of points $s\in S$ with rationally connected fiber $Y_{s}$ is both open in $S$ and is a countable union of closed subsets of $S$ \cite[Corollary IV.3.5.2  \& Theorem IV.3.11]{Kol}. In particular, if $S$ is connected, then every fiber $Y_{s}$ is rationally connected as soon as one of these fibers is.

When $F$ is algebraically closed of characteristic $0$, we say that a morphism $\pi:Y\lra S$ of connected $F$-varieties is a {\em rationally connected fibration} if $\pi$ is proper, and if the fiber of $\pi$ over a general point of $S$ is rationally connected. This implies that every fiber contained in the smooth locus of $\pi$ is rationally connected.
\vskip .4cm

\begin{proposition}\label{rcfib}
Let $S$ be a variety with generic point $\nu$, and let $\pi:Y\lra S$ be a proper map. If the fiber $Y_\nu$ is rationally connected, then $\pi$ is a rationally connected fibration.
\end{proposition}

\begin{proof}
Let $\nu$ be the generic point of $S$. We have a variety $M$ defined over $\nu$ with a map $u:M\times\mathbb{P}^1\lra Y_\nu$ such that the induced map $u^{(2)}: M\times\mathbb{P}^1\times\mathbb{P}^1\lra Y_{\nu}\times_{\nu}Y_{\nu}$ is dominant. The variety $M$ and the map $u$ are defined over some affine open in $S$, so there is an open set $V$ in $S$ such that we have a variety $M_V$ and a rational map $u_V: M_V \times\mathbb{P}^1 \lra Y_V$ such that the induced map $u_V^{(2)}$ is dominant. We can assume $M_V$ is quasiprojective, and then the image of $M_V\times\mathbb{P}^1\times\mathbb{P}^1$ is a constructible set in $Y_V$.

Thus for a general point $p$ in $V$ the image is dense in the fiber $Y_p$, and therefore $Y_p$ is rationally connected.
\end{proof}

\vskip .4cm

\subsection{Rational connectivity and the Minimal Model program}\

Loosely speaking, the prevalence of rational curves on a variety measures the failure of its canonical divisor to be positive, and so being rationally connected is a strict condition on the birational geometry of a variety $X$.  For example, if $X$ is a smooth Fano variety ($-K_X$ is ample) then we expect to find many rational curves on $X$, and in fact such an $X$ is always rationally connected \cite{Cam}. While the converse is not true, in a certain sense rationally connected varieties can be built out of Fano varieties using the minimal model program.

The minimal model program is a program for classifying algebraic varieties up to birational equivalence \cite{KM}. Given a smooth projective variety $X$, the main goal of the MMP is to produce a variety $\widetilde{X}$ birational to $X$, with at worst mild singularities, such that the canonical divisor $K_{\!\widetilde{X}}$ is nef (that is, nonnegative on every curve). One calls the variety $\widetilde{X}$ a \emph{minimal model} of $X$.

The strategy for producing a minimal model of $X$ is to systematically eliminate curves $C$ in $X$ satisfying $K_X\cdot C <0$. This is accomplished by either contracting a divisor $D$ in $X$ covered by such curves, or by applying a birational operation called a flip. Assuming that the procedure of repeatedly applying these two operations terminates after finitely many steps, the end result is either a minimal model or a map $\widetilde{X} \lra Z$ of relative Picard number $1$, such that $-K_{\!\widetilde{X}}$ is relatively ample. The map $\widetilde{X} \lra Z$ is called a \emph{Mori fiber space}. In general, termination of flips is a difficult open problem in birational geometry. However, Birkar, Cascini, Hacon, and McKernan \cite{BCHM} establish the existence of minimal models by showing termination of flips for a special MMP called the {\em MMP with scaling}.

The operations constituting the MMP are closely related to the existence of rational curves on the variety $X$: if there exists a curve $C$ in $X$ such that $K_X \cdot C<0$, then there is always a rational curve with this same property. Whenever we contract a divisor in running the MMP, we do so by contracting a family of rational curves that cover the divisor.

Having a covering family of rational curves on a smooth variety $X$ precludes the existence of a section of $nK_X$ for any positive $n$. In fact slightly more is true:

\begin{proposition}\label{notpsef}
If $X$ is a smooth and rationally connected projective variety over an algebraically closed field, then $K_X$ is not pseudoeffective.
\end{proposition}
Note that this is true under the weaker hypothesis that $X$ is uniruled.
\begin{proof}
By \cite[Ex 4.7, Cor 4.11]{Deb}, there is a family of rational curves on $X$ such that
\begin{enumerate}
\item $-K_X\cdot C \leq -2 $ for any curve $C$ in the family.

\item For a general point $p \in X$, there is a curve $C$ in the family passing through $p$.

Choose any ample divisor $A$ on $X$. Then for sufficiently small $\lambda >0$, $\lambda A \cdot C <2$. Thus $(K_X+\lambda A)\cdot C<0$. Since given any closed subset of $X$ we can choose $C$ to not be contained in that set, we must have that no multiple of $(K_X+\lambda A) $ has a section.
\end{enumerate}
\vskip -.6cm
\end{proof}

\vskip .2cm

As a consequence, by \cite[Cor 1.3.3]{BCHM}, running any MMP with scaling on a smooth rationally connected $X$ produces a Mori fiber space $\tilde{X} \to Z$. The base $Z$ is the image of a rationally connected variety, so it too is rationally connected. If we repeatedly run an MMP with scaling starting with a rationally connected variety, we produce a long sequence of divisorial contractions, flips, and Mori fibrations terminating at a single point.

\vskip .75cm

\section{Spreading out over a curve}\label{spreading out section}

\subsection{Rationally connected spreading out}\

Throughout the remainder of this article, we fix an algebraically closed field $k$ of characteristic $0$, and let $K=k((t))$ with ring of integers $R=k[[t]]$.

The present section is devoted to a proof that if $f:X\lra Y$ is a surjective morphism of smooth projective $k((t))$-varieties, then we can spread any snc $k[[t]]$-model $\mathscr{X}\lra\mathscr{Y}$ of this morphism out to a morphism over a smooth curve. Moreover, we can do so in a way that preserves the morphism's general relative smoothness and relative rational connectivity when these properties hold over general points of $Y$.

We begin by establishing a non-relative version of spreading out that replaces $k[[t]]$ with a finite type $k$-variety $S$.

\vskip .4cm

\begin{proposition} \label{spreadout}
Let $X$ be a projective, rationally connected $k((t))$-variety, and let $\mathscr{X}$ be a projective $k[[t]]$-model of $X$.  Then there exists a finite type $k$-variety $S$, a dominant morphism $\mathrm{Spec}_{\ \!}k[[t]]\lra S$, and a projective variety $V$ over $S$, such that
\[
\mathscr{X}\ \cong\ \mathrm{Spec}_{\ \!}k[[t]]\underset{S}{\times}V
\]
and such that the fiber $V_q$ over a general point $q$ in $S$ is rationally connected.

Furthermore, if $k[t]]\lra S'$ is a dominant map into any finite type $k$-variety, not necessarily the variety $S$ provided by the previous paragraph, and if $V'$ is a projective $S'$-variety with $\mathscr{X}\cong\mathrm{Spec}_{\ \!}k[[t]]\times_{S'}V'$, then the fiber $V'_{q}$ over a general point $q$ in $S'$ is again rationally connected.
\end{proposition}

\begin{proof}
We first show that we can construct varieties $S$ and $V$ with the required properties.

As in the proof of \cite[Proposition 5.1.2]{MN}, we use the spreading out techniques from \cite[\S 8 \& Th\'eor\`eme 11.6.1]{Groth} to obtain:
\vskip .2cm
\begin{itemize}
\item[$\bullet$] An integrally closed, finite-type $k$-subalgebra $A\subset k[[t]]$;
\vskip .2cm
\item[$\bullet$] An affine open neighborhood $\mathrm{Spec}_{\ \!}A_{0}\subset\mathrm{Spec}_{\ \!}A$ containing the image of the closed point under the map $\mathrm{Spec}_{\ \!}k[[t]]\lra\mathrm{Spec}_{\ \!}A$;
\vskip .2cm
\item[$\bullet$] A flat projective family $V\lra\mathrm{Spec}_{\ \!}A_{0}$ such that
$\mathscr{X}\ \cong\ \mathrm{Spec}_{\ \!}k[[t]]\times_{A_{0}}V$.
\end{itemize}
\vskip .2cm

Our assumption that $X$ is rationally connected means that there exists a $k((t))$-variety $M$ and a map $f:M \times \mathbb{P}^1 \lra X$ such that the induced map
$$
f^{(2)}\ \!:\ M\times\mathbb{P}^{1}\times\mathbb{P}^{1}\xrightarrow{\ \ \ \ \ }X\underset{\!\!\!k((t))\!\!\!}{\times}X
$$
is dominant.  Choose some flat $k[[t]]$ model $\mathscr{M}$ of $M$. We can run the spreading out for $\mathscr{X}$ and $\mathscr{M}$ simultaneously, by taking a simultaneous extension of the relevant subrings inside $k[[t]]$. Thus, changing our choice of $A\subset k[[t]]$ and $A_{0}$ if need be, we also obtain:
\vskip .2cm
\begin{itemize}
\item[$\bullet$] A flat family $N\lra\mathrm{Spec}_{\ \!}A_{0}$ such that
$\mathscr{M}\ \cong\ \mathrm{Spec}_{\ \!}k[[t]]\times_{A_{0}}N$.
\end{itemize}\vskip .2cm

Finally, let $\Gamma$ be the graph of $f$ in $(M\times\mathbb{P}^{1})\times_{k((t))}X$, and let $\mathscr{G} \subset (\mathscr{M}\times \mathbb{P}^1) \times_{k[[t]]} \mathscr{X}$ be a flat $k[[t]]$-model for $\Gamma$. By \cite[Lemme 8.5.2.1, Th\'eor\`eme 8.10.5, \& Th\'eor\`eme 11.6.1]{Groth}, combined with the arguments of the previous paragraph, we can run spreading out for $\mathscr{X}$, $\mathscr{M}$, and $\mathscr{G}$ simultaneously. Changing our choice of $A\subset k[[t]]$ and $A_{0}$ once again if need be, we obtain:
\vskip .2cm
\begin{itemize}
\item[$\bullet$] A closed subscheme $G\subset(N\times\mathbb{P}^{1})\times_{A_{0}}V$, flat over $A_{0}$, satisfying $\mathscr{G}\cong\mathrm{Spec}_{\ \!}k[[t]]\times_{A_{0}}G$.
\end{itemize}
\vskip .2cm

Let $S\subset\mathrm{Spec}_{\ \!}A$ denote the closure of the image of the map $\mathrm{Spec}_{\ \!}k[[t]]\lra\mathrm{Spec}_{\ \!}A_{0}$, so that the map $\mathrm{Spec}_{\ \!}k[[t]]\lra S$ is dominant. We wish to show that over the generic point $\eta\in S$, the fiber $G_{\eta}$ is the graph of a rational map
\begin{equation}\label{phi}
\phi:N_{\eta}\times\mathbb{P}^{1}\!\xymatrix{{}\ar@{-->}[r]&{}}\!V_{\eta}
\end{equation}
Thus, we must show that the map $\mathscr{G}_\eta\lra N_{\eta}\times\mathbb{P}^{1}$ is birational. To this end, note that we have a commutative diagram of function fields:
$$
\xymatrix{ k(G_\eta)\ar[r] & k(\Gamma)
\\ k(N_\eta\times\mathbb{P}^1)\ar[r]\ar[u] & k(M\times\mathbb{P}^1) \ar[u]
\\ k(S) \ar[r] \ar[u] & k((t)) \ar[u]
}
$$
The bottom square is co-Cartesian, as is the large square. Hence the top square is also co-Cartesian. Since $k(\Gamma)$ has degree $1$ over $k(M\times\mathbb{P}^1)$, this implies that $k(\mathscr{G}_\eta)$ has degree $1$ over $k(N_\eta\times\mathbb{P}^1)$. This gives us an open subset $U \subset N_{\eta}\times\mathbb{P}^1$ equipped with a morphism $\phi:U\lra V_{\eta}$. Note that $\phi$ induces a morphism
$$
\phi^{(2)}\ \!:\ U\underset{N_{\eta}}{\times}U\xrightarrow{\ \ \ \ \ \ }V_{\eta}\underset{\!k(S)\!}{\times}V_{\eta}
$$
Because both $f^{(2)}:M\times\mathbb{P}^{1}\times\mathbb{P}^{1}\lra X\times_{k((t))}X$ and $\mathrm{Spec}_{\ \!}k[[t]]\lra S$ are dominant, this morphism $\phi^{(2)}$ is also dominant.

By resolving singularities of $N_\eta$, we can choose $N'_\eta$ so that the rational map
$$
\phi'\ \!:\ N'_\eta \times\mathbb{P}^{1}\!\xymatrix{{}\ar@{-->}[r]&{}}\!V_{\eta}
$$
induced by $\phi$ is defined away from a codimension-$2$ subset.  Let $N''_\eta$ be the complement of the image of this set in $N'_\eta$.  Thus we have a map $\phi':N''_\eta\times\mathbb{P}^{1}\lra V_{\eta}$ such that the induced map
$$
\phi'^{\ \!(2)}\ \!:\ N''_\eta \times\mathbb{P}^{1}\times\mathbb{P}^{1}
\xrightarrow{\ \ \ \ \ \ \ }
V_{\eta}\underset{\!k(S)\!}{\times}V_{\eta}
$$
is dominant, since this map agrees with $\phi^{(2)}$ on an open set. Hence $V_{\eta}$ is rationally connected, and thus by Proposition \ref{rcfib}, the map $V\lra S$ is a rationally connected fibration.

Now suppose we have a finite type $k$-variety $V'$ over a finite type $k$-variety $S'$, and suppose given a dominant map $\mathrm{Spec}_{\ \!}k[[t]] \lra S'$ such that
\[
\mathscr{X}\ \cong\ \mathrm{Spec}_{\ \!}k[[t]]\underset{S'}{\times}V'
\]
We may assume that $S'$ is affine, so that $S'=\mathrm{Spec}_{\ \!}A'$, where $A' \subset k[[t]]$. We can them run the above construction, choosing $A$ so that it is an extension of $A'$ inside $k[[t]]$. We obtain $V$ and $S$ satisfying the proposition, along with a morphism $S \lra S'$ that commutes with the morphisms from $\mathrm{Spec}_{\ \!}k[[t]]$.

After replacing $A$ with another extension inside $k[[t]]$ if necessary, we can assume that $V' \times_S S'$ is isomorphic to $V$ over $S'$, by \cite[Cor 8.8.2.5]{Groth}, since $V' \times_S S'$ and $V$ become isomorphic over $k[[t]]$. Because the map $\mathrm{Spec}_{\ \!}k[[t]] \lra S'$ is dominant, so is $S \lra S'$. Thus the fiber of $V'$ over a general point in $S'$ is the image of the fiber over some general point in $S$. Since the image of a rationally connected variety is connected, the proposition follows.
\end{proof}

\vskip .5cm

\begin{remark}
The previous Proposition \ref{spreadout} allows us to spread a rationally connected snc $k[[t]]$-model to a model over a finite type $k$-variety. In order to apply the techniques of \cite{dFKX}, we need to spread our $k[[t]]$-model not over just any finite type $k$-variety, but over a smooth curve. Theorem \ref{curvemodel} accomplishes this for not only a rationally connected snc $k[[t]]$-model, but for a rationally connected fibration of snc $k[[t]]$-models.
\end{remark}

\vskip .5cm

\begin{theorem}\label{curvemodel}
Let $f:X\lra Y$ be a surjective morphism of smooth projective varieties over $k((t))$. Assume that for a general point $q$ of $Y$ such that the fiber $X_q$ is smooth and rationally connected. If $f_{k[[t]]}:\mathscr{X}\lra\mathscr{Y}$ is a map of snc $k[[t]]$-models that extends $f$, then there exists a pointed affine curve $C\ni p$ over $k$, along with a map $g_C:V_{C}\lra W_{C}$ of smooth $C$-varieties satisfying:

\begin{enumerate}
\item The special fibers $V_p$ and $W_p$ are snc inside $V_C$ and $W_C$.
\item The dual complexes of the special fibers $V_p$ and $W_p$ are identified with those of $\mathscr{X}$ and $\mathscr{Y}$, respectively.
\item The fiber of $g_C$ over a general point $q$ of $W_p$ is rationally connected.
\end{enumerate}
\end{theorem}

\begin{proof}

As above, we can choose an affine $k$-variety $S$, along with varieties $V$, $W$, and a $\Theta \subset V \times_{S} W$ such that $V$, $W$, and $\Theta$ pull back to $\mathscr{X}$, $\mathscr{Y}$, and the graph $\Gamma$, respectively. We may assume $\Theta$ is isomorphic to $V$ over $S$, so that $\Theta$ gives the graph of a morphism $g:V \lra W$. Let $A$ denote the coordinate ring of $S$.

The construction of the curve $C$ and flat families $V_{C}\lra C$ and $W_{C}\lra C$ proceeds exactly as in the proof of \cite[Proposition 5.1.2]{MN}: Given a positive integer $n$, we use Greenberg approximation \cite{Green} to find a morphism
\begin{equation}\label{morphism to Henselization}
A\lra\mathscr{O}^{\ \!\mathrm{h}}_{\!\mathbb{A}^{\!1}_{k},(t)}
\end{equation}
to the Henselization of the stalk at the origin in $\mathbb{A}^{\!1}_{k}$, such that the composite of (\ref{morphism to Henselization}) with the projection
\[
\xymatrix{
\widehat{\mathscr{O}^{\ \!\mathrm{h}}_{\!\mathbb{A}^{\!1}_{k},(t)}}\ \cong\ k[[t]]
\ar@{->>}[r]
&
k[t]/(t^{n})
}
\]
factors through a morphism $A\lra k[[t]]/(t^{n})$. Because the Hensilization is the stalk for the \'etale topology, we can find a smooth curve $C$ and a morphism
\begin{equation}\label{curve morphism}
C\lra\mathrm{Spec}_{\ \!}A
\end{equation}
through which the morphism dual to (\ref{morphism to Henselization}) factors. If we define $V_{C}$ and $W_{C}$ to be the pullbacks of $V$ and $W$ along this morphism (\ref{curve morphism}), then it follows immediately from Proposition \ref{spreadout} that
\[
\mathrm{Spec}_{\ \!}k[t]/(t^n)\underset{C}{\times}V_{C}
\ \ \cong\ \
\mathrm{Spec}_{\ \!}k[t]/(t^n)\underset{\!\!k[[t]]\!\!}{\times}\mathscr{X}
\ \ \cong\ \
\mathrm{Spec}_{\ \!}k[t]/(t^n)\underset{S}{\times}V
\]
and similarly for the triple $W_{C}$, $\mathscr{Y}$, and $W$. Since the dual complexes only depend on the special fibers, the second statement of the proposition is immediate.

Now, consider the flat locus of $g: V \lra W$. This is an open set, so its image in $S$ is open. Let $T \subset S$ be the complement. Now, because the map $\mathrm{Spec}_{\ \!}k[[t]] \lra S$ is dominant, it does not factor through $T$. Thus there is some $n \geq 2$ such that the induced map from $\mathrm{Spec}_{\ \!}k[t]/(t^n)$ does not factor through $T$. Thus we may assume that  $C$ is not contained in $T$.

The requirement that $\mathscr{X}$ and $\mathscr{Y}$ be snc over $k[[t]]$ includes the requirement that they are both regular. Because the Zariski tangent space at points in the special fiber depends only on the fiber over $\mathrm{Spec}_{\ \!}k[t]/(t^2)$, this implies that $V_{C}$ and $W_C$ are both smooth in a neighborhood of their respective special fibers.  Localizing if necessary, we may therefore choose $C$ so that $V_C$ and $W_C$ are smooth \cite[Proposition 5.1.2.(6)]{MN}.  In particular, this implies that a general fiber of the morphism from $V \lra W$ is smooth.

Let $\overline{K}$ be an algebraic closure of $K=k((t))$, and consider the closed points of $Y_{\overline{K}}$. By hypothesis, we know that there is an open set of $Y_{\overline{K}}$ where the fiber of $f_{\overline{K}}: X_{\overline{K}} \lra Y_{\overline{K}}$ is smooth and therefore also rationally connected. Now, since the $\overline{K}$ points are dense in $Y_{\overline{K}}$, their images are dense in $W$. Each $\overline{K}$ point $\overline{q}$ is defined over a finite extension of $k((t))$ of the form $k((t^{1/n}))$. We can therefore apply Proposition \ref{spreadout} to see that the fiber of $V \lra W$ over the image of $\overline{q}$ is rationally connected. Applying Proposition \ref{rcfib}, we have that a dense set of closed points inside the closure of the image of $\overline{q}$ is rationally connected, so we must have that there is no proper closed set of $W$ containing all the closed points with rationally connected fibers. Hence these points are dense.

Note that irreducibility of $Y$ implies that $W$ is irreducible.

Because rational connectedness is both open and closed in smooth families, we have that for points in the flat locus of $g$, the fiber of $g$ is rationally connected whenever it is smooth. Since both $V_C$ and $W_C$ are smooth, and $C$ is not contained in $T$, this means that over a general point of $W_C$, the fiber of $g_C$ is rationally connected.
\end{proof}

\vskip .75cm

\section{Homotopy type of the dual complex}\label{applying dFKX}

\subsection{Singularities of the MMP}\

De Fernex, Koll\'{a}r, and Xu have shown recently that running the MMP on certain varieties with snc divisors preserves the homotopy type of the dual complex  \cite{dFKX}. As a result they are able to show the following
\begin{theorem}\cite[Thm 1]{dFKX}
Let $X$ be an isolated log terminal singularity. Then for any log resolution, the dual complex is contractible.
\end{theorem}
\noindent
Applying their techniques to the case of a degeneration of rational varieties, they are able to show

\begin{theorem}\cite[Thm 4]{dFKX}\label{degen}
Let $X\lra C$ be a degeneration of rationally connected varieties over a curve, with special fiber $X_0$. Assume $X$ is smooth and that $X_0$ is snc. Then the dual complex of $X_0$ is contractible.
\end{theorem}

Log terminal singularities are a natural class of singularities arising in the MMP. Examples of log terminal singularities are provided by quotient singularities and Gorenstein rational singularities. We will need to introduce some related notions for pairs $(X,\Delta)$ so that we may apply the techniques of \cite{dFKX}. The standard reference is \cite{KM}.

Let $X$ be a normal variety over an algebraically closed field of characteristic $0$, and let $\Delta$ be an effective $\mathbb{Q}$-divisor on $X$ such that $K_X+\Delta$ is Cartier. Let $f:Y \lra X$ be a {\em log resolution} of $(X,\Delta)$. This means that:
\vskip .2cm
\begin{itemize}
\item[{\bf (i)}] The variety $Y$ is smooth;
\item[{\bf (ii)}] The map $f$ is proper and birational;
\item[{\bf (iii)}] If we let $\{E_i\}$ be the set whose elements are all components of the exceptional divisors of $f$ and all components of the strict transform of $\Delta$, then the $E_i$ form a simple normal crossing divisor.
\end{itemize}
\vskip .2cm

We can compare the divisor $K_X+\Delta$ on $X$ with its counterpart $K_Y+\sum E_i$ in $Y$. These must agree away from the $E_i$, and so we have
$$
K_Y+\sum E_i\ =\ f^*(K_X+\Delta)+\sum a_i E_i
$$
where the pullback of $K_X+\Delta$ is a well defined $\mathbb{Q}$-divisor, and where the $a_i$ are rational numbers, called the {\em log discrepancies}. If every one of the log discrepancies satisfies $a_i \geq 0$, then we say that $(X, \Delta)$ is \emph{log canonical}, or \emph{lc}. If each of the log discrepancies satisfies $a_i >0$, then we say that $(X, \Delta)$ is \emph{Kawamata log terminal}, or \emph{klt}. We say that $X$ itself is \emph{log terminal} if the pair $(X,0)$ is klt.

A subvariety $V \subset X$ is called a {\em log canonical center} if it is of the form $f(E_i)$ for some $E_i$ with vanishing log discrepancy: $a_i=0$.  The pair $(X,\Delta)$ is called \emph{divisorial log terminal}, or \emph{dlt}, if for every log canonical center $V$ of $(X,\Delta)$, there is a neighborhood of the generic point of $V$ in which $(X,\Delta)$ becomes snc. The related notion of \emph{quotient-dlt}, or \emph{qdlt}, is less well known, so we refer to \cite[Section 5]{dFKX}. The pair $(X,\Delta)$ is qdlt if every codimension-$d$, log canonical center of $(X,\Delta)$ is the intersection of $d$ components of $\Delta$, each with coefficient $1$. The name ``quotient dlt'' comes from the fact that a qdlt singularity is characterized by being locally a quotient of a dlt singularity by an abelian group \cite[Prop 34]{dFKX}.

These singularity classes obey the following logical dependencies:
$$
\text{klt}  \Rightarrow \text{dlt} \Rightarrow \text{qdlt} \Rightarrow \text{lc}
$$

\begin{example}
Let $X=\mathbb{A}^2$ and let $\Delta$ be the coordinate axes both with coefficient $1$. Then $(X,\Delta)$ is dlt, qdlt, and lc, but not klt, and the log canonical centers are $V(x)$, $V(y)$, and $V(x,y)$. Let $X'$ be the affine quadric cone and $\Delta'$ the union of two lines on this cone. Then $(X',\Delta')$ is the quotient of $(X,\Delta)$ by a $\mathbb{Z}/2$ action which preserves the components of $\Delta$, so $(X',\Delta')$ is qdlt and lc. However, $(X',\Delta')$ is not dlt (nor klt) because it is not snc in a neighborhood of the vertex of the cone, which is a log canonical center.
\end{example}

The first important property of these types of singularities is that each class is preserved under running a $K_X+\Delta$ MMP. More precisely, if $g: X \dashrightarrow X'$ is a divisorial contraction or a flip for the pair $(X,\Delta)$, which is klt/dlt/qdlt/lc, then the pair $(X', g_* \Delta)$ is also klt/dlt/qdlt/lc.

\vskip .5cm

\subsection{Proof of the Main Theorem}\

Before giving a proof of Theorem \ref{main}, we need to establish an analogous result for a morphism of varieties over a curve.

\vskip .3cm

\begin{theorem}\label{mapdegen}
Let $C$ be an affine curve over an algebraically closed field $k$ of characteristic $0$, and let $p$ be a $k$-valued point of $C$. Suppose given projective $C$-varieties
$$
\pi_X: X \lra C
\ \ \ \ \ \ \mbox{and}\ \ \ \ \ \
\pi_Y: Y \lra C
$$
and let $\Delta_X$ and $\Delta_Y$ be the reduced fibers of $X$ and $Y$ over $p$. Assume that $(X,\Delta_X)$ and $(Y,\Delta_Y)$ are both snc pairs, and let $f:X \lra Y$ be a surjective morphism over $C$, such that for a general point $q \in Y$, the fiber $X_q=f^{-1}(q)$ is rationally connected. Since $X$ is smooth, we may assume that $X_q$ is smooth.

Then the dual complex of $(X,\Delta_X)$ is homotopy equivalent to the dual complex of $(Y,\Delta_Y)$.
\end{theorem}

The proof of this theorem is adapted directly from \cite{dFKX}. The key point is that under these hypotheses, every birational step of the MMP induces a specific type of homotopy equivalence, called a collapse, on the dual complex. It then suffices to check that Mori fiber spaces also preserve the dual complex.

\begin{proof}
Choose a small $\varepsilon>0$, and consider the pair $\big(\ \!X,\ \!\Delta_X- \varepsilon \pi^{-1}(p)\ \!\big)$. This is a klt pair, so by \cite{BCHM} we can run MMP with scaling for this pair. Because $\varepsilon \pi^{-1}(p)$ is numerically trivial, the MMP with scaling is also a $K_X+\Delta_X$ MMP, and it preserves the property of being dlt.

Since $X_q$ is smooth and rationally connected, $K_{X/Y}$ cannot be pseudoeffective, as otherwise for a general fiber $X_q$, the canonical divisor $K_{X_q}$ would be pseudeoffective, contradicting Proposition \ref{notpsef}. Thus running a relative MMP with scaling over $Y$ eventually terminates in a Mori fiber space $X' \lra Z$ over $Y$ \cite[Cor 1.3.3]{BCHM}.  Let $\Delta'_{X'}$ be the reduced special fiber of $X'$, which is the strict transform of $\Delta_X$. Then the pair $(X',\Delta'_{X'})$ is also dlt, since it is the result of running a $K_X+\Delta_X$ MMP. By \cite[Cor 22]{dFKX}, the dual complex of $(X,\Delta_X)$ collapses to the dual complex of $(X',\Delta'_{X'})$. These complexes are thus homotopy equivalent.

Let $\Delta_Z$ be the reduced special fiber of $\pi_Z: Z \lra C$. The pair $(Z,\Delta_Z)$ may no longer be dlt, but it is qdlt by \cite[Prop 40]{dFKX}. Because the morphism $X' \lra Z$ has relative Picard number $1$, we can identify the dual complex of $(X',\Delta'_{X'})$ with that of $(Z,\Delta_Z)$.

Now, let $\phi: Z'\lra Z$ be a log resolution of singularities of $Z$.  This map induces maps
$$
f_{Z'}:Z' \lra Y
\ \ \ \ \ \ \mbox{and}\ \ \ \ \ \
\pi_{Z'}:Z' \lra C
$$
By \cite[Theorem 3]{dFKX}, the dual complex of the special fiber $(Z', \Delta_Z')$ is homotopy equivalent to the dual complex of $(Z,\Delta_Z)$. Moreover, if we consider a general point $q$ of $Y$, then there is a dominant rational map on the fibers $X_q \dashrightarrow Z'_q$. This implies that $Z'_q$ is rationally connected.

The variety $Z'$ satisfies the hypotheses of the theorem. If the dimension of a general fiber is $0$, then since $Z'\lra Y$ is a resolution of singularities, and $(Y, \Delta_Y)$ is dlt, by  \cite[Theorem 3]{dFKX} the dual complexes of $(Z', \Delta_Z')$ and $(Y, \Delta_Y)$ are homotopy equivalent. Otherwise we may proceed by induction, since $\dim Z' < \dim X$.
\end{proof}

\vskip .5cm

\begin{proof} [Proof of Thm \ref{main}]
Let $R$ be the spectrum of the power series ring $k[[t]]$. Using resolution of singularities, extend the map $f: X \lra Y$ to a map on snc-models $f_R: \mathscr{X} \lra \mathscr{Y}$ over $R$. By Cor \ref{homotopyofdual}, it suffices to show that the dual complexes $D(\mathscr{X}_{k})$ and $D(\mathscr{Y}_k)$ are homotopy equivalent.

If $f _R$, $\mathscr{X}$, and $\mathscr{Y}$ are pulled back from a curve, then the theorem follows immediately by Theorem \ref{mapdegen}. Otherwise, by Proposition \ref{curvemodel} we can produce a pointed $k$-curve $C\ni p$, along with a map of smooth varieties $g_C: V_C \lra W_C$ over $C$, such that:
\begin{itemize}
\item[$\bullet$]
The fiber over a general point of $W$ is smooth and rationally connected;
\item[$\bullet$]
The special fiber $W_p$ is snc inside $W$ and is isomorphic to $\mathscr{X}_k$;
\item[$\bullet$]
The special fiber $V_p$ is snc inside $V$ and is isomorphic to $\mathscr{Y}_k$.
\end{itemize}
Then by Theorem \ref{mapdegen}, the dual complexes $D(V_p)$ and $D(W_p)$ are homotopy equivalent. Thus $D(\mathscr{X}_{k})$ and $D(\mathscr{Y}_{k})$ are homotopy equivalent and the theorem follows.
\end{proof}

\vskip 1cm

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\end{document}


TITLE: Is There a Mayer-Vietoris Spectral Sequence of Motivic Cohomology for Closed Coverings?
QUESTION [2 upvotes]: For etale cohomology, there is a spectral sequence of the following form ("Mayer-Vietories spectral sequence for closed covers"):
$E_{1}^{p,q}=\oplus_{i_{0}< \cdots < i_{p}} H_{ Y_{i_{0} \cdots i_{p} }}^{q} (X, F) \Longrightarrow H_{Y}^{q-p}(X, F).$
Here, $X$ is a scheme, $Y\to X$ is a closed subscheme, and $Y_{i}$'s are a closed covering of $Y$.
$Y_{i_{0} \cdots i_{p}}$ denotes $Y_{i_{0}} \cap \cdots \cap Y_{i_{p}}.$
$F$ is an etale sheaf on $X$, and $H_{Z}^{\ast}$ denotes the cohomology with supports in a closed subscheme $Z$ on $X$.
My questions are: is there any spectral sequence of the similar form for motivic cohomology (or higher Chow group)?
If yes, then how one can prove it?
If no, then why is it?
Please give me any advice.

REPLY [2 votes]: I believe that cycle groups do not behave nicely for singular varieties. On the other hand, one could define certain reasonable motivic cohomology via Voevodsky's motives. In any case, look at (section 2 of) the paper http://www.math.uiuc.edu/K-theory/0529/glr.pdf


TITLE: A family of countable sets
QUESTION [3 upvotes]: Let $(A_\alpha)_{\alpha \in I}$ be a family of countable sets such that for any $\alpha, \beta \in I$, either $A_\alpha \subset A_\beta$ or $A_\beta \subset A_\alpha$. Is it true that $\bigcup A_\alpha$ countable?

REPLY [3 votes]: This is not true. The least uncountable ordinal, $\omega_1$ is the union of all the smaller ordinals, all of which are countable, and all of which make a linearly ordered family as in your question.
However, you can prove that under the requirements you pose, the only counterexamples are those of cardinality $\aleph_1$. So without additional hypothesis (specifically the continuum hypothesis) you cannot use the real numbers as a counterexample.
Also, your attempt to use the rational numbers is futile, since the set of rational numbers is countable.

REPLY [1 votes]: Sure, $I$ doesn't have to be countable. But in general, your result does hold only if $I$ is countable. Take for example $I = \omega_{1}$ and for each $\alpha \in I$ let $A_{\alpha} = \alpha$. Then for $\alpha \le \beta < \omega_{1}$ we have $A_{\alpha} \subseteq A_{\beta}$, each $A_{\alpha}$ is countable, but $\omega_{1} = \bigcup_{\alpha \in I} A_{\alpha}$ is uncountable.


TITLE: Which one is correct for $\sqrt{-16} \times \sqrt{-1}$? $4$ or $-4$?
QUESTION [3 upvotes]: As we can find in order to evaluate $\sqrt{-16} \times \sqrt{-1}$, we can do it in two ways.
FIRST
\begin{align*}
     \sqrt{-16} \times \sqrt{-1} &= \sqrt{(-16) \times (-1)}\\
     &= \sqrt{16}\\
     &=4
\end{align*}
SECOND
\begin{align*}
     \sqrt{-16} \times \sqrt{-1} &= \sqrt{16i^2} \times \sqrt{i^2}\\
     &= 4i \times i\\
     &=4i^2\\
     &=-4
\end{align*}
Incidentally if the above is input in  complex mode of Casio scientific calculator, the result comes out as $-4$.
Which of the above solutions is correct?

REPLY [1 votes]: The answer to each sqrt after removing i is +-, so for a simple explanation you have either (+)(+) or (-)(+) or (-)*(-), depending on the domain or context of the particular problem.


TITLE: Accuracy of LMM $y_{n+2}-\frac{4}{3}y_{n+1}+\frac{1}{3}y_n=\frac{2}{3}kf_{n+2}$
QUESTION [1 upvotes]: I am trying to determine how to find the accuracy of a LMM. Specifically the BDF2 method, $$y_{n+2}-\frac{4}{3}y_{n+1}+\frac{1}{3}y_n=\frac{2}{3}kf_{n+2}$$ for solving the problem $y'=f(y,x),y(0)=n$ where f is Lipshitz continuous.
I was able to show it was consistent using the characteristic polynomials, but I am not sure how to find the order of accuracy in general for LMM.
Any help on this would be much appreciated. I know how to do this for 1 step methods, but I am struggling on this two step one.

REPLY [1 votes]: Let $a_i$ be the coefficient of $y_{n+i}$ and $b_i$ be the coefficient of $f_{n+i}$, where $i \in \{0,\dots,k\} = \{0,1,2\}$. Then, define $d_j = \sum_{i=0}^k \frac{i^j}{j!}a_i - \frac{i^{j-1}}{(j-1)!}b_i$, $d_0 = \sum_{i=0}^k a_i$. Then, the least $j$ such that $d_j\neq 0$ is the order of the method.

Edit: To specify, if $k$ is the step size, then the error is $\mathcal{O}(k^j)$ where $j$ is the smallest number such that $d_j\neq 0$, as above. Some consider this to mean $j$ is the order of the method, while others consider $j-1$ to be the order. Consult your text/professor for clarification.


TITLE: smallest distance to place the capital?
QUESTION [0 upvotes]: A country has four cities where should the capital be built. it is desirable that the sum of distances of the cities from the capital be as small as possible?
how do i solve it ?

REPLY [0 votes]: The location of the capital is called the geometric median of the points.  The wikipedia article mentions specific results for four coplanar points.
Update: If the alternative interpretation of @DavidG.Stork and @YNK is correct, the solution is trivial: for each of the four cities, calculate the sum of distances to the other three cities, and the minimum of these four sums indicates which city should be the capital.


TITLE: The size of monochromatic submatrix
QUESTION [3 upvotes]: We say a matrix $(a_{ij})$ is 0-1 matrix if $a_{ij}\in \{0,1\}$ for all $i,j$. We say a matrix $(a_{ij})$ is monochromatic if for some $a$, $a_{ij} = a$ for all $i,j$.

Question: Let $c\geq 1/2$ be a constant and $n$ be very large. Given a $n\times n$ 0-1 matrix $M$, must there be a $c\log_2 n\times c\log_2 n$ submatrix $M'$ of $M$ that is monochromatic?

When $c<1/2$ such submatrix $M'$ clearly exists (see below) but we wonder the case of $c\geq 1/2$.
The case $c<1/2$: It suffices to construct a $2c \log_2 n\times 2c \log_2 n$ matrix $\hat{M}$ so that each row of $\hat{M}$ is monochromatic (but different row may have different "color"). The row of $\hat{M}$ is simply the first $2c \log_2 n$ rows of $M$. To select columns of $\hat{M}$, inductively screen out columns of $M$. The first row screen out a set $K$ of columns of $M$ with $|K|\leq n/2$ where $a_{0j} = a$ for all $j\in K$ and $a_{0j} = 1-a$ for all $j\notin K$. The second row screen out a set $K'\subseteq K$ of columns with $|K'|\leq |K|/2$ where $a_{1j} = a'$ for all $j\in K'$ and $a_{1j} = 1-a'$ for all $j\in  K\setminus K'$. Keep doing this for each rows of $\hat{M}$ (which is $2c\log_2 n$ many times) and there will be at least $n^{1-2c}$ many columns remained.

REPLY [2 votes]: Even $c=1-\varepsilon$ works when $n$ is large enough. This question is about bounding the diagonal bipartite Ramsey, see the recent achievement of David Conlon here
http://people.maths.ox.ac.uk/~conlond/Bipartite.pdf


TITLE: Why are topological phases described by modular tensor categories?
QUESTION [9 upvotes]: After some reading, I have an inuitive idea what topological phases of matter are. But where is the connection to modular tensor categories? Is there fundamental literature where this is covered?
Edit: A topological phase is characterized by a TQFT as low-energy effective theory. Furthermore, every modular tensor category leads to a TQFT, as shown by Turaev. However, according to Wang, "Topological Quantum Computation" (CBMS, Vol. 112, 2010), the converse is only a conjecture. Is it already proven that a strict fusion category of a TQFT can be extended uniquely to a modular tensor category compatible with the TQFT? Even if it is: Is there a more illustrative explanation why modular tensor categories are studied as mathematical models for topological phases?

REPLY [6 votes]: Modular tensor categories only describe the non-abelian statistics of the point-like topological excitations in 2+1D bosonic topologically ordered phases. So every topologically ordered phase gives rise to a modular tensor category. But the inverse is not true.  Every modular tensor category correspond to infinite many 2+1D bosonic topologically ordered phases, and those phases differ by E8 states.


TITLE: Show $\int_{0}^{\frac{\pi}{2}}\frac{x^{2}}{x^{2}+\ln^{2}(2\cos(x))}dx=\frac{\pi}{8}\left(1-\gamma+\ln(2\pi)\right)$
QUESTION [40 upvotes]: Here is an interesting, albeit tough, integral I ran across. It has an interesting solution which leads me to think it is doable. But, what would be a good strategy?.
$$\int_{0}^{\frac{\pi}{2}}\frac{x^{2}}{x^{2}+\ln^{2}(2\cos(x))}dx=\frac{\pi}{8}\left(1-\gamma+\ln(2\pi)\right)$$
This looks rough. What would be a good start?.  I tried various subs in order to get it into some sort of shape to use series, LaPlace, something, but made no real progress.
I even tried writing a geometric series. But that didn't really result in anything encouraging.
$$\int_{0}^{\frac{\pi}{2}}\sum_{n=0}^{\infty}(-1)^{k}\left(\frac{\ln(2\cos(x))}{x}\right)^{2k}$$
Thanks all.

REPLY [26 votes]: In addition to the nice set of references by Raymond Manzoni, here is my proof of the identity. Frankly, I have not seen these references yet, thus I am not sure if this already appears in one of them.
Here I refer to the following identity
$$ \binom{\alpha}{\omega} = \frac{1}{2\pi} \int_{-\pi}^{\pi} \left(1 + e^{i\theta}\right)^{\alpha} e^{-i\omega \theta} \; d\theta, \ \cdots \ (1) $$
whose proof can be found in my blog post.
Now let $x$ be a real number such that $|x| < \frac{\pi}{2}$. Then simple calculation shows that
$$ \log\left(1+e^{2ix}\right) = \log(2\cos x) + ix \quad \Longleftrightarrow \quad \Im \left( \frac{-x}{\log\left(1+e^{2ix}\right)} \right) = \frac{x^2}{x^2 + \log^2(2\cos x)},$$
hence we have
$$ \begin{align*}I
&:= \int_{0}^{\frac{\pi}{2}} \frac{x^2}{x^2 + \log^2(2\cos x)} \; dx
= -\int_{0}^{\frac{\pi}{2}} \Im \left( \frac{x}{\log\left(1+e^{2ix}\right)} \right) \; dx \\
&= -\frac{1}{8}\int_{-\pi}^{\pi} \Im \left( \frac{\theta}{\log\left(1+e^{i\theta}\right)} \right) \; d\theta
= \frac{1}{8}\Re \left(  \int_{-\pi}^{\pi} \frac{i\theta}{\log\left(1+e^{i\theta}\right)} \; d\theta \right).
\end{align*}$$
Differentiating both sides of $(1)$ with respect to $\omega$ and plugging $\omega = 1$, we have
$$ \frac{1}{2\pi} \int_{-\pi}^{\pi} (-i\theta) \left(1 + e^{i\theta}\right)^{\alpha} e^{-i\theta} \; d\theta = \alpha \left(\psi_0(\alpha) - \psi_0(2)\right). $$
Now integrating both sides with respect to $\alpha$ on $[0, 1]$,
$$ \begin{align*}
-\frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{i\theta}{\log \left(1 + e^{i\theta}\right)} \; d\theta
&= \int_{0}^{1} \alpha \left(\psi_0(\alpha) - \psi_0(2)\right) \; d\alpha \\
&= \left[ \alpha \log \Gamma (\alpha) \right]_{0}^{1} - \int_{0}^{1} \log \Gamma (\alpha) \; d\alpha - \frac{1}{2}\psi_0(2) \\
&= -\frac{1}{2}\left( 1 - \gamma + \log (2\pi) \right),
\end{align*}$$
where we have used the fact that
$$ \psi_0 (1+n) = -\gamma + H_n, \quad n \in \mathbb{N}$$
and
$$ \begin{align*}
\int_{0}^{1} \log \Gamma (\alpha) \; d\alpha
& = \frac{1}{2} \int_{0}^{1} \log \left[ \Gamma (\alpha) \Gamma (1-\alpha) \right] \; d\alpha \\
&= \frac{1}{2} \int_{0}^{1} \log \left( \frac{\pi}{\sin \pi \alpha} \right) \; d\alpha \\
&= \frac{1}{2} \left( \log \pi - \int_{0}^{1}  \log \sin \pi \alpha \; d\alpha \right) \\
&= \frac{1}{2} \log (2\pi).
\end{align*} $$
Therefore we have the desired result.


text:
\section{Identity Element for Addition of Cuts}
Tags: Cuts, Addition

\begin{theorem}
Let $\alpha$ be a [[Definition:Cut (Analysis)|cut]].
Let $0^*$ be the [[Definition:Rational Cut|rational cut]] associated with the [[Definition:Zero (Number)|(rational) number $0$]]:
:$0^* = \set {r \in \Q: r < 0}$
Then:
:$\alpha + 0^* = \alpha$
where $+$ denotes the operation of [[Definition:Addition of Cuts|addition of cuts]].
\end{theorem}

\begin{proof}
Let $r \in \alpha + 0^*$.
$\alpha + 0^*$ is the [[Definition:Set|set]] of all [[Definition:Rational Number|rational numbers]] of the form $p + q$ such that $p \in \alpha$, $q \in 0^*$, that is, $q < 0$.
It follows that:
:$p + q < p$
and so:
:$p + q \in \alpha$
that is:
:$r \in \alpha$
Let $r \in \alpha$.
Let $s \in \Q$ be a [[Definition:Rational Number|rational number]] such that $s > r$ and $s \in \alpha$.
This is possible because $r$ is not the [[Definition:Greatest Element|greatest element]] of $\alpha$ by definition of a [[Definition:Cut (Analysis)|cut]].
Let $q = r - s$.
Then $q < 0$ and so $q \in 0^*$.
Thus we have:
:$r = s + q$
and so:
:$r \in \alpha + 0^*$
We have that:
:$r \in \alpha \implies r \in \alpha + 0^*$
and:
:$r \in \alpha + 0^* \implies r \in \alpha$
and so by definition of [[Definition:Set Equality|set equality]]:
:$\alpha + 0^* = \alpha$
{{qed}}
\end{proof}

config: wiki


TITLE: What is the conditional distribution of this product?
QUESTION [0 upvotes]: Let $D$ be a Bernoulli random variable. When $D=1$, $Pr(D=1)=0.5$. When $D=0$, $Pr(D=0)=0.5$. Let $Z$ be a standard normally distributed random variable such that $Z \sim N(0,1)$.
We let $Y=D\times Z$. Is there any way to know the conditional distribution of $Y$ given $Z$? conditional expectation $\mathbf{E}(Y|Z)$ ? conditional variance $\mathbf{var}(Y|Z)$?
Intuitively, I would think the conditional distribution is still Bernoulli.

REPLY [1 votes]: It looks to me that irrespective of the distribution of Z, any particular value of Z, say z leads to a random variable zY which is two valued $(\pm z)$, so $E(Y|Z)=\frac{z}{2}$ and $var(Y|Z)=\frac{z^2}{4}$.
The distribution function is obviously a little messy, Let $U=Y\times Z$, then $P(U)=0=0.5$ and for the other half, it has the same distribution as Z (multiplied by 0.5).


TITLE: Recursive sequences and solutions
QUESTION [0 upvotes]: Let $s_0$, $s_1$, $s_2$, . . . be a recursive sequence defined by
$s_0 = 4$,$s_1 =3$, $s_n$ =$−6s_{n−1}$ − $9s_{n−2}$ for all integers $n\ge2$
Find an expression for $s_n$ in terms of $n$ that holds for all integers $n \ge 0$.
So far I've got:
The characteristic equation of the recurrence is $x^2 + 6x + 9$
Factorised: $(x+3)^2 = 0$
double root $x=-3$
and this is where I'm stuck

REPLY [0 votes]: $a_n = c_1(r^n)+c_2n^2(r^n) + ... + c_mn^{m-1}(r^n)$ is the general solution when the characteristic equation has m repeated real roots.


TITLE: Why $\{Z \le z\} = \bigcap_{m = 1}^\infty \bigcup_{n=1}^\infty \bigcap_{k=n}^\infty \{ Z_k \le z + 1/m \}$ if $Z=\lim_nZ_n$?
QUESTION [3 upvotes]: I am following  A first look at rigorous probability theory by Rosenthal, and I am having troubles with limits of random variables.
Specifically proposition 3.1.5. (iii) states that if $Z_1,Z_2...$ are random variables such that $\lim_{n \to \infty}Z_n(w)$ exists for each $w \in \Omega $, and $Z(w)=\lim_{n \to \infty}Z_n(w)$, then Z is also a random variable.
The proof starts out with: For $z \in R$ $$\{Z \le z\} = \bigcap_{m = 1}^\infty \bigcup_{n=1}^\infty \bigcap_{k=n}^\infty \{ Z_k \le z + 1/m \}$$
Why is this true? the book hints to the definition of limit $Z(w)=\lim_{n \to \infty}Z_n(w)$.
Thanks in advance for your explanations

REPLY [2 votes]: Suppose $w\in\{Z\leq z\}$ then for each $m$, there is $n(m)$ such that for all $k\geq n$, we have $Z_k(w)\leq z+\frac{1}{m}$, which follows because $Z(w)=\lim Z_n(w)$. This means
$$
w\in\bigcap_{k=n(m)}^\infty\{Z_k\leq z+1/m\}\subset\bigcup_{n=1}^\infty\bigcap_{k=n}^\infty\{Z_k\leq z+1/m\}.
$$
So the rightmost expression contains $w$ for each $m\geq 1$. So we must have
$$
w\in\bigcap_{m=1}^\infty\bigcup_{n=1}^\infty\bigcap_{k=n}^\infty\{Z_k\leq z+1/m\}.\tag{*}
$$
This proves one direction.
For the other direction, suppose $w\in A$ where $A$ is the right hand side of (*). Suppose that $Z(w)>z$ then there is some integer $M$ such that $Z(w)-\frac{2}{M}>z$. But this means for $k$ sufficiently large, $Z_k(\omega)\geq z+\frac{2}{M}>z+\frac{1}{M}$. Consequently,
$$
w\not\in\bigcup_{n=1}^\infty\bigcap_{k=n}^\infty\{Z_k\leq z+1/M\}.
$$
In particular, $w$ cannot be in $A$, giving us the desired contradiction.


text:
\section{Proofs}
Note that from \eqref{eq:derivative} (with $\zeta$ replaced by $\zeta_r$), we know the optimal $\lambda_{\opt}$ should satisfies that
\begin{eqnarray}
    \frac{\tilde{\sigma}^2+\gamma \bbE\frac{g(V)h(V)}{(\zeta_r+1)^2}}{1-\gamma \bbE\frac{\zeta_r^2}{(1+\zeta_r)^2}} \ = \ \frac{\bbE \frac{g(V)h(V)\zeta_r}{(1+\zeta_r)^3}}{\bbE \frac{\zeta_r^2}{(1+\zeta_r)^3}}.
\end{eqnarray}

Hence, we have
\[
R(r,\lambda_{\opt})\ = \ \frac{\bbE \frac{g(V)h(V)\zeta_r}{(1+\zeta_r)^3}}{\bbE \frac{\zeta_r^2}{(1+\zeta_r)^3}}.
\]

Hence, with Jensen's inequality, we can upper bound the denominator of the risk function
\[
1-\gamma\bbE\frac{\zeta^2}{(1+\zeta)^2} \leq 1- \gamma \left(\bbE\frac{\zeta}{1+\zeta}\right)^2 = 1-\frac{1}{\gamma}\left(1-\lambda m_r(-\lambda)\right)^2
\]
On the other hand, when $g(V)h(V)$ and $r(V)$ have different monotonicity of $V$, we can lower bound the numerator of the risk function:
\[
\tilde{\sigma}^2+\gamma \bbE\frac{g(V)h(V)}{(1+\zeta)^2} \geq \tilde{\sigma}^2 +\gamma \bbE g(v)h(V) \bbE\frac{1}{(1+\zeta)^2}\geq
\tilde{\sigma}^2 +\frac{1}{\gamma} \bbE g(v)h(V)\left(\gamma-1+\lambda m(-\lambda)\right)^2
\]

WLOG, we assume that $h\in\cH_I^{+}$ and $g\in\cH_I$. Then our first step is to show that the optimal $r\in \cH_I^{+}\bigcup\cH_D^{+}$ should belong to $\cH_I^{+}$.
........

Hence, we just need to focus on all possible $r\in\cH_{I}^{+}$.

\begin{eqnarray}
R_{\alpha}(\lambda)\ \triangleq \ \frac{m'_{\alpha}(-\lambda)}{m^2_{\alpha}(-\lambda)}\left(\tilde{\sigma}^2+\gamma \bbE\frac{g(V)h(V)}{(r_{\alpha}(V)m_{\alpha}(-\lambda)+1)^2}\right),\n
\end{eqnarray}

Note that when $\alpha=0$, we have $R_0(\lambda)$ being the risk function for $\vSigma_w=\tilde{\vSigma}_w$ and when $\alpha=1$, we have $R_1(\lambda)$ being the risk function for $\vSigma_w=\vSigma_{\beta}^{-1}$\footnote{For any $\alpha\in(0,1)$, the risk function $R(\alpha)$ is well defined but it may not be the limit of the expected risk for any choice of $\vSigma_w$ unless $r\in\cH_{I}^{+}$}.
Our goal is to show that $\alpha=1$ is a first order stationary point for optimizing the optimal weighted ridge regression i.e., $\min_{\alpha}\min_{\lambda}R_{\alpha}(\lambda)$. Simiarly as

Then, the risk function can be equivalently expressed by the following formula:
\begin{eqnarray}
R_{\alpha}(\lambda_{\opt}) &=& \frac{\tilde{\sigma}^2+\gamma \bbE\frac{g(V)h(V)}{(\zeta_{\alpha}+1)^2}}{1-\gamma \bbE\frac{\zeta_{\alpha}^2}{(1+\zeta_{\alpha})^2}}\label{eq:R_alpha}
\end{eqnarray}

Hence, we have
\[
R_{\alpha}(\lambda_{\opt})\ = \ \frac{\bbE \frac{g(V)h(V)\zeta_{\alpha}}{(1+\zeta_{\alpha})^3}}{\bbE \frac{\zeta_{\alpha}^2}{(1+\zeta_{\alpha})^3}}.
\]
With \eqref{eq:m'alpha}, we have
\begin{eqnarray}
\frac{\dif R_{\alpha}(\lambda_{\opt})}{\dif \alpha}&\stackrel{(i)}{\propto_+}&\left(\bbE\frac{\zeta_{\alpha}}{(1+\zeta_{\alpha})^2}\bbE\frac{\psi_{\alpha}(1-2\zeta_{\alpha})(\psi_{\alpha}-\zeta_{\alpha})}{(1+\zeta_{\alpha})^4}-\bbE\frac{\psi_{\alpha}\zeta_{\alpha}(1-2\zeta_{\alpha})}{(1+\zeta_{\alpha})^4}\bbE\frac{\psi_{\alpha}-\zeta_{\alpha}}{(1+\zeta_{\alpha})^2}\right)\bbE\frac{\zeta_{\alpha}^2}{(1+\zeta_{\alpha})^3}\n\\
&&-\left(\bbE\frac{\zeta_{\alpha}}{(1+\zeta_{\alpha})^2}\bbE\frac{\zeta_{\alpha}(2-\zeta_{\alpha})(\psi_{\alpha}-\zeta_{\alpha})}{(1+\zeta_{\alpha})^4}-\bbE\frac{\zeta_{\alpha}^2(2-\zeta_{\alpha})}{(1+\zeta_{\alpha})^4}\bbE\frac{\psi_{\alpha}-\zeta_{\alpha}}{(1+\zeta_{\alpha})^2}\right)\bbE\frac{\psi_{\alpha}\zeta_{\alpha}}{(1+\zeta_{\alpha})^3}.\n\\
\end{eqnarray}
where in equation (i) we have omitted the following positive quantity
\[
\left((1-\alpha)m_{\alpha}(-\lambda)\left(\bbE \frac{\zeta_{\alpha}^2}{(1+\zeta_{\alpha})^3}\right)^2\bbE\frac{\zeta_{\alpha}}{(1+\zeta_{\alpha})^2}\right)^{-1}
\]

\begin{itemize}
    \item When $h(v)g(v)\in \cH_I$, then the optimal choice of $r\in \cH_{I}\bigcup\cH_D$ belongs to $\cH_I$.
    \item When $h(v)g(v)\in \cH_D$, then the optimal choice of $r\in \cH_{I}\bigcup\cH_D$ belongs to $\cH_D$.
\end{itemize}
Further, $r(v)\stackrel{\text{a.s.}}{=} h(v)g(v)$ with respect to $V$ is a first order stationary point

In general, the sufficient condition to have negative optimal regularization is the following:
\[
\bbE \frac{g(V)\zeta^2}{(1+\zeta)^3}\bbE\frac{\zeta}{(1+\zeta)^2}>\bbE g(V)\left(\frac{\zeta}{(1+\zeta)^2}+\frac{\zeta}{\gamma\xi}\right)\bbE\frac{\zeta^2}{(1+\zeta)^3}.
\]

Further, for special case when $\tilde{\sigma}^2=0$ and $\vSigma_x=\vSigma_{\beta}^{-1}$(i.e., $h(v)g(v)\equiv 1$), we can confirm that the minimal risk is $\gamma$ for the regime $\theta\gamma<1$ while the risk achieved at $\theta=1$
at full model is $\gamma-1$ while the optimal under-parameterized regime is $\gamma$.

Note that from \eqref{eq:smrelation}, we have
\[
    0<\gamma s(-\lambda) = m(-\lambda)-\frac{1-\gamma}{\lambda} = \frac{\gamma \bbE\frac{1}{1+\zeta}}{\lambda},
\]
where the last equality holds due to \eqref{eq:relation_m}. Hence, we have
\begin{eqnarray}
\bbE \frac{1}{1+h(V)m(-\lambda)} < 0, \quad \forall \lambda<0.
\label{eq:under_eq_1}
\end{eqnarray}
Note that the support of $V$ is a single point or interval and $h$ is a continuous function, which means $\bbE \frac{1}{1+h(V)m(-\lambda)}$ is only valid when either $1+h(V)m(-\lambda)<0$ or $1+h(V)m(-\lambda)>0$ holds almost surely with respect to $V$. Hence, due to \eqref{eq:under_eq_1}, we have $1+h(V)m(-\lambda)<0$ holds almost surely if $\lambda<0$ and $m(-\lambda)<0$. Therefore, for the case when $\lambda<0$ and $m(-\lambda)<0$, we have
\[
\bbE\frac{h^2(V)}{(1+h(V)m(-\lambda))^3} < 0,
\]
and \eqref{eq:under_goal_1} holds.


TITLE: Probability density from standard domain - I
QUESTION [3 upvotes]: Pick $x+iy$ at random with respect to hyperbolic measure from $\{z:|z|\geq1,|\mathcal R(z)|\leq\frac12\}$. What does the probability distribution function of $\frac1{\sqrt y}$ look like?

REPLY [2 votes]: $\newcommand{\ii}[1]{\operatorname{\mathbf I}\{#1\}}$
The density (with respect to the Lebesgue measure) of the hyperbolic measure in your region is
\begin{equation}
 f(x,y)=\frac3{\pi y^2}\,\ii{y>0,\ x^2+y^2>1,\ |x|<1/2}
\end{equation}
for $(x,y)\in\mathbb R^2$, where $\ii\cdot$ denotes the indicator, and has been normalized for the region.
Introducing the new variables $(u,v):=(x,1/\sqrt y)$, so that $(x,y)=(u,v^{-2})$, we have the joint density of $(u,v)$:
\begin{equation}
 g(u,v)=f(u,v^{-2})\left|\frac{\partial(u,v^{-2})}{\partial(u,v)}\right|
 =\frac{6v}{\pi} \ii{v>0,\ u^2+v^{-4}>1,\ |u|<1/2}.
\end{equation}
Hence, the density of $v=1/\sqrt y$ is
\begin{equation}
h(v):= \int_{-\infty}^\infty g(u,v)\,du=
 \begin{cases}
 6v/\pi &\text{ if } 0<v<1, \\
 6v\left(1-2 \sqrt{1-v^{-4}}\right)/\pi &\text{ if } 1\leq v<\sqrt[4]{4/3}, \\
 0&\text{ if }v\ge \sqrt[4]{4/3}.
\end{cases}
\end{equation}


TITLE: Let $S=\{ \mathcal{X_s} \;\;|\;\;\ s\in\mathbb{R} \}$. It is possible that $ Span(S) = \mathbb{R}^{\mathbb{R}}$
QUESTION [1 upvotes]: Let $S=\{ \mathcal{X_s} \;\;|\;\;\ s\in\mathbb{R} \}$.
It is possible that $ Span(S)=\langle S\rangle = \mathbb{R}^{\mathbb{R}}=\{f:\mathbb{R} \rightarrow \mathbb{R} \; |\;\; f\;\;\ fuction  \}$?
I do not think this is possible, but I can not prove it. Can anyone give me a hint on how to prove this or a counter example. I also wanted a hint on how to describe the subspace generated $Span(S)=\langle S\rangle.$

REPLY [1 votes]: I suppose that $\chi_s(x)$ is $0$, unless $x=s$, in which case it is equal to $1$. If that's so, you are right, since every element of $\operatorname{span}(S)$ is $0$ outside a finite set.


TITLE: Prove: $\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}\geq\sqrt{x}+\sqrt{y}$
QUESTION [0 upvotes]: $$\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}\geq\sqrt{x}+\sqrt{y}$$ where $x,y>0$

How should I approach this?

REPLY [2 votes]: Fleshing out a comment I left below the OP, I find it convenient to recast problems like this in a form that gets rid of the square root symbol.  In this case that's easily done by setting $\sqrt x=u$ and $\sqrt y=v$, which turns the inequality into
$${u^2\over v}+{v^2\over u}\ge u+v$$
An important thing to note is that we have $u,v\gt0$, since each variable stands for a square root (and appears in a denominator).  That means we can argue as follows:
$$\begin{align}
{u^2\over v}+{v^2\over u}\ge u+v&\iff u^3+v^3\ge u^2v+v^2u\\
&\iff u^2(u-v)+v^2(v-u)\ge 0\\
&\iff (u^2-v^2)(u-v)\ge0\\
&\iff (u+v)(u-v)^2\ge0
\end{align}$$
with the final line being clearly true since $u,v\gt0$ makes $u+v\ge0$, while $(u-v)^2\ge0$ even if the variables are allowed to be negative.
The basic argument here is no different, really, than in some of the other answers.  I just find it easier to see what's going on, and avoid making mistakes, if I don't have a bunch of square root symbols to keep track of.


TITLE: What is the motivation behind free groups?
QUESTION [1 upvotes]: I have started studying free groups but cannot get motivation behind it. I know that a free group $F_A$ over the set $A$ consisting of all combinations or expressions that can be built from the members of $A$.
But why would one want to do such a construction? What is the motivation behind it

REPLY [3 votes]: Well, off the top of my head, there are two big things that free groups do for us.
First of all, they give us a nice generalization of the integers under addition--also known as the free group generated by one element. (One could also argue that the trivial group is the free group generated by zero elements.)
Second of all, every group is the quotient of a free group! Have you ever seen something like the following before?
$$D_n=\left\langle x,y\mid x^n,y^2,xyxy\right\rangle$$
This is shorthand that says $D_n$ can be obtained by starting with the free group $F$ generated by two elements--$x$ and $y,$ say--letting $N$ be the smallest normal subgroup of $F$ containing each of $x^n,y^2,$ and $xyxy,$ and then forming the quotient group $F/N.$


TITLE: How do you graph an inequality on Real/Imaginary plane?
QUESTION [0 upvotes]: Suppose we have $z$ as a complex number, $z \in C$, how do you graph an inequality which has $z$ in it?
This kinds of inequalities arise when we need to graph the shape of stability region of a given numerical method:
For example, for RK2 method we can derive the stability inequality:
$|1+z+\frac{z^2}{2}| < 1$
What I could do so far was to set the inequality to zero and solve for $z$:
$|1+z+\frac{z^2}{2}| = 1$ so $z = 0, -2$. Therefore $z$ must be between $0$ and $-2$.
I also know the existence of circle formula: $(x-h)^2 + (y-k)^2 = r^2$ which sometimes helps to graph an inequality by transforming the inequality into the circle formula at origin $(h, k)$ with radius $r$.
But I can't transform the above RK2 inequality to a circle formula, and also unable to get the imaginary part properly, if I take the 'set to equality' method (though I know $-2< Re{\{z\}} < 0$).

REPLY [2 votes]: Ok the answers provided by Eric Stucky and Morgan Rodgers both pointing at replacing $z$ with $x+iy$ is prefect. The problem is  that in the exams there is no computer to plot the inequality derived in terms of $x$ and $y$.
Below is the approach I learned from http://www.math.ubc.ca/~peirce/M405_607E_Lecture%2018.pdf which I found the most suitable approach when you are in exam room:
First set the inequality to the function G such that:
$G(z) = 1 + z + \frac{z^2}{2}$
For stability we require $|G(z)| < 1$.
Solve $z$ in terms of $G$:
$z = -1 \pm \sqrt{2G-1}$
Now we can start graphing the $z$. First we assume $G$ is a circle of radius 1 at the origin of real/imaginary plane. Then we scale $G$ by two and we get $2G$, then we shift the circle to the left and we get $2G-1$ and so on unil we get $-1 \pm \sqrt{2G-1}$.
The process has been  shown with step by step conformal map in http://www.math.ubc.ca/~peirce/M405_607E_Lecture%2018.pdf
I consider this as the best answer, specially in exam rooms. I will wait for further confirmation by the community and then will accept this as answer after few days.