\documentstyle{article}
\input macros.tex
\begin{document}
\title{Transversality}
\author{Kevin Iga}
\date{October 1996}
\maketitle
Let $(X,g)$ be a compact Riemannian manifold, with $f$ a Morse
function. We wish to study Morse flows:
\begin{equation}
\frac{dx}{dt}=-\nabla_g f(x(t))\label{eqn:flow}
\end{equation}
As $t\to\pm\infty$, $x(t)$ converges to critical points. For the
remainder of the discussion, we will examine the space of flows between
fixed critical points $a$ and $b$, as it lies in the space of paths
from $a$ to $b$, under the $C^1$ topology.
Let $P$ be the space of smooth paths from $a$ to $b$ in $X$ under the
$C^1$ topology, so that $P=C^1(\re\to X;a,b)$. We define a vector
bundle $E\to P$ so that at each point $p\in P$, the fiber is the
Banach space of $C^0$ vector fields in $TX$ along the path $p(t)$;
i.e. if we consider $p$ as a function $p:\re\to X$, the pullback
$p^*(TX)$ defines a vector bundle over $\re$, and $E$ is the Banach
space of $C^0$ sections of $p^*(TX)$.
We consider the section of $E$:
\begin{equation}
F_g(x)=\frac{dx}{dt}+\nabla_g f(x(t)).
\end{equation}
Let $W\subset P$ be the subspace of paths satisfying the flow
equations (\ref{eqn:flow}), so that $W$ is the preimage of the zero vector
field under $F$. If $x\in W$, the tangent space $TW$ at $x$ is given
by the kernel of $dF_g(x)$. This can be described as follows: The
differential of the projection $E\to P$ is a vector bundle $TE\to TP$ over
$TP$. Then $dF_g(x)$ is the following section of $TE$:
\begin{equation}
dF_g(x)(\xi)=\frac{d\xi}{dt}+H_f(x(t))(\xi)
\end{equation}
where $H_f$ is the Hessian of $f$, considered as a linear operator instead
of as a bilinear form, through the metric.
Next, we define $B$ to be the space of $C^1$ metrics, under the $C^1$
topology.
\begin{theorem}
For a generic choice of $C^1$ metric $g$, the operator
\[F_g(x):P\to E\]
is a submersion on $W$, [unless $x$ is the constant map].
\end{theorem}
Proof:
We first note that (in coordinates) the function $F$ for a given metric
$g$ are:
equations for are:
\begin{equation}
F_g^\mu(x)=\frac{dx^\mu}{dt}+g^{\mu\nu}\partial_\nu f(x(t))=0.
\end{equation}
If we fix a metric $g_0$, and use it to turn $g$ into a linear transformation
$g^*$, we get:
\begin{equation}
F_g(x)=\frac{dx}{dt}+g^*\nabla_{g_0} f(x(t))=0.
\end{equation}
The set of metrics $g$ is thus in bijective correspondence with the
subset of sections $\Gamma(Ad(TX))$ consisting of ($g_0-$)
positive-definite, symmetric linear transformations. This is an open
set of the Banach space of sections consisting of symmetric linear
transformations, so that this Banach space is a good model for the tangent
space to $B$.
We first show that $F:B\times P\to E$ is a submersion on $W$. We first
fix $x(t)\in P$, and fix a target $\eta\in TE$, and show that it is in the
image of $dF_g(x):TB\times TP\to TE$.
\begin{equation}
dF_g(x)(\alpha,\xi)=\frac{d\xi}{dt}+\alpha^*\nabla_{g_0} f(x(t))+g^*
H_f(x)\xi + dg^*(\xi)\nabla_{g_0}f(x(t))=\eta.
\label{eqn:surjeta}
\end{equation}
We first identify a ball around each critical point, inside which the
Hessian is nondegenerate: by the Morse condition, the Hessian is
nondegenerate at the critical points, and by the continuity of the
Hessian and the determinant function, there is a neighborhood of the
critical points for which the Hessian is nondegnerate. We furthermore
require that the balls for two different critical points are disjoint.
Fix a radius $r$ smaller than the necessary radii of all such balls.
Since $x(t)$ converges to $a$ and $b$, for $t\to-\infty$ and $t\to+\infty$,
respectively, there is a $T$ for which $x(t)\in B(a,r)$ for $t<-T$, and
$x(t)\in B(b,r)$ for $t>T$. Then it follows from the above that
along $x(t)$, the Hessian $H$ is nondegenerate for $|t|>T$.
We wish to find $\xi$ for $|t|>T$ so that
\begin{equation}
\frac{d\xi}{dt}+g^* H_f(x)\xi=\eta.\label{eqn:surj2eta}
\end{equation}
We do this in the next section, and indeed we define $\xi$ for $|t|>T$
which satisfies the equation (\ref{eqn:surj2eta}) and converges to 0
as $t\to \pm\infty$.
We then define $\alpha=-dg^*(\xi)$ for $t<-T$, and note that
$(\alpha,\xi)$ solves (\ref{eqn:surjeta}) for $t<-T$.
We apply the same procedure for $t\to+\infty$ at the critical point there,
in that neighborhood. In this way, we have $(\alpha,\xi)$ defined for
$|t|>T$ which satisfies (\ref{eqn:surjeta}) for $|t|>T$. We can extend
$\xi$ arbitrarily (using, say, a cutoff function) over all $t$.
The idea next is to extend $\alpha$ to satisfy (\ref{eqn:surjeta}) for
$|t|\le T$. Intuitively, since $\nabla f(x(t))$ is never zero, we
simply choose any $g_0$-symmetric endomorphism that sends $\nabla f(x(t))$
to
\[\eta-\frac{d\xi}{dt}-g^*H_f(x)\xi - dg^*(\xi)\nabla_{g_0}f(x(t)).\]
To show we can do this globally, we first need a general lemma:
\begin{lemma}
Let $B$ be a manifold, and $E\to B$ be a fiber bundle where fibers are
affine spaces, and transition functions are affine maps. Then $E$ has
a section.
\end{lemma}
Proof:
We fix a trivialization $\{U_\alpha,E_\alpha,\phi_{\alpha\beta}\}$. Over
each neighborhood $U_\alpha$ we take a constant section $s_\alpha$. We
then take a partition of unity $\Phi_\alpha$ over $\{U_\alpha\}$ and the
section we take is
\[s=\sum_\alpha \Phi_\alpha s_\alpha.\]
Since $\sum_\alpha \Phi_\alpha(x)=1$, this sum remains in the affine space.
So $s$ is a section of $E$.
\qed
(end of Lemma)
\begin{corollary}
If $(B,g_0)$ is a Riemannian manifold (not necessarily compact) and
$v$ a nonvanishing vector field, and $w$ any vector field, there
exists a section $A$ of $\Hom(TB,TB)$ which is $g_0$-symmetric so that
$Av=w$.
\end{corollary}
Proof:
Consider the vector bundle of $g_0$-symmetric bilinear forms in $\Hom(TB,TB)$.
At each point, the space of such that assigns $w$ to $v$ forms an affine
space, nonempty and of the correct dimension if $v$ is nonzero. As a
sub-fiber bundle, this is an affine bundle as above. Applying the lemma
yields the result.
\qed
Now $\nabla f(x)$ is nonzero in a neighborhood of the curve $x(t)$ for
$|t|0$ so that for each $n$ there
exists a $x_n$ so that either
\[|\Phi(x_n)(f_n(x_n)-f(x_n))|\ge 2\epsilon\]
or
\[|D^k(\Phi(x_n)f_n(x_n))-D^k(\Phi(x_n)f(x_n))|\ge 2\epsilon.\]
Because $f_n$ is Cauchy, for all $\epsilon>0$, we can find an $N$ so that
$||f_n-f_m||_{C^k_{\alpha,\beta}}<\epsilon$ whenver $n, m>N$.
So in particular, when $n$ and $m$ are chosen as such,
\[|\Phi(x)f_n(x)-\Phi(x)f_m(x)|<\epsilon\]
and
\[|D^k(\Phi(x)f_n(x))-D^k(\Phi(x)f_m(x))|<\epsilon\]
for all $x$. If we fix $n=N+1$, then putting these four inequalities together
we get that
\[|\Phi(x_n)f_m(x_n)-\Phi(x_n)f(x_n)|\ge\epsilon\]
or
\[|D^k(\Phi(x_n)f_m(x_n))-D^k(\Phi(x_n)f(x_n))|\ge \epsilon\]
for all $m\ge n$. So in a compact neigborhood of $x_n$, $f_n$ does not
converge to $f$ in $C^k$ norm. But this contradicts the defining property
of $f$. So $f_n$ converges to $f$ in $C^k_{\alpha,\beta}$ norm.
\qed
\begin{lemma}
Suppose we have $\xi\in C^1$, $\eta\in C^0$ so that
\[\left(\frac{d}{dt}+H(t)\right)\xi=\eta\]
and so that $\xi$ and $\eta$ converge to zero as $t\to-\infty$.
Then there exists $\tilde{\xi}\in C^1_{\alpha,\beta,(-\infty,0]}$ and
$\tilde{\eta}\in C^0_{\alpha,\beta,(-\infty,0]}$
so that
\[\left(\frac{d}{dt}+H(t)+\alpha I\right)\tilde{\xi}=\tilde{\eta}.\]
\end{lemma}
Proof:
Let $\tilde{\xi}=e^{-\alpha t}\xi$, and $\tilde{\eta}=e^{-\alpha t}\eta$.
Now $\Phi(t)\tilde{\xi}(t)=\xi(t)$, so $\tilde{\xi}$ has bounded
$C^1_{\alpha,\beta,(-\infty,0]}$ norm. Similarly, $\tilde{\eta}$ has
bounded $C^0_{\alpha,\beta,(-\infty,0]}$ norm. In fact,
\[||\tilde{\xi}||_{C^1_{\alpha,\beta,(-\infty,T]}}=
||\xi||_{C^1_{(-\infty,T]}}\]
and
\[||\tilde{\eta}||_{C^0_{\alpha,\beta,(-\infty,T]}}=
||\eta||_{C^0_{(-\infty,T]}}.\]
The fact that the
equation is solved can be shown directly.
\qed
\section{Transversality Theorem}
\begin{theorem}
Let $X$, $E$, and $B$ be Banach manifolds where $E\to B$ is a vector bundle.
Suppose $f:X\times B\to E$ is a fiberwise section, i.e., that for each value
$x\in X$, $f_x:B\to E$ ($f_x(b)=f(b,x)$) is a section of $E$. Suppose
$f$ is transverse to the zero section. Then for a dense set of $x\in X$,
$f_x$ is transverse to the zero section.
\end{theorem}
Proof:
We first take a trivialization of $E$ over an open covering of $B$, so that
without loss of generality, $E$ is a trivial bundle $F\times B$. Then
we will write $f:X\times B\to F$ and $f_x:B\to F$.
We will study the projection map
\[\pi_0:X\times B \to X\]
and restrict to the subset $S=f^{-1}(0)$ so we have
\[\pi:S\to X\]
and note that by the Sard-Smale theorem, the set of regular values of $\pi$
forms a dense subset of $X$.
We now show that $x$ being a regular value for $\pi$ is equivalent to
$f_x$ being transverse to zero under the hypothesis that
$f$ is transverse to zero.
Suppose $x$ is a regular value. This means that for every $(b,x)\in
\pi^{-1}(x)$
we have $d\pi:T_{(x,b)}S\to T_x X$ surjective. The hypothesis that $f$ be
transverse to the zero section means that if $(x,b)\in S$,
then $df:T_xX\times T_bB\to T_0F$ is surjective.
To show that $f_x$ is transverse to zero, we suppose $(x,b)\in S$, and
show that $df_x:T_bB\to T_0F$ is surjective. In order to do this, we
let $v\in T_0F$ be arbitrary, and note that by the surjectivity of
$df$ we can write $v=df(u+w)$ where $u\in T_xX$ and $w\in T_bB$. By
the surjectivity of $d\pi$, we write $u=d\pi(u')$ where $u'\in
T_{(x,b)}S\subset T_xX\times T_bB$. In that splitting, $u'=u+w'$
where $u\in T_xX$ is as before and $w'\in T_bB$.
Putting it all together we get $v=df(u'-w'+w)$. Now $u'\in T_{(x,b)}S$,
so $df(u')=0$. Then $v=df(w-w')$. Since $w-w'\in T_bB$, $df(w-w')=
df_x(w-w')$ so that $v$ is in the image of $df_x$.
So $x$ being a regular value for $\pi$ implies $f_x$ is transverse to $0$.
We don't need the other direction, but I include it here for completeness.
Suppose $f_x$ is transverse to $0$. This means that $df_x:T_bB\to
T_0F$ is surjective whenever $(x,b)\in S$. We wish to show that $d\pi$
is surjective whenever $(x,b)\in S$. Suppose $u\in T_xX$. We wish to
find $w\in T_bB$ so that $u+w\in T_{(x,b)}S$.
This is equivalent to $df_{(x,b)}(u+w)=0$, i.e. $df_x w=-df_{(x,b)}u$.
We can find such a $w$ by the surjectivity of $df_x$.
\qed
\end{document}