2
$\begingroup$

a) Prove that nondegenerate zeros are isolated.

b) Furthermore, show that at a nondegenrate zero $x$, $ind_x (\vec v)=+1$ if the isomorphism $d(\vec v_x )$ preserve orientation, and $ind_x (\vec v)=-1$ if the isomorphism $d(\vec v_x )$ reserve orientation

For part a)

I know that a zero $x$ is nondegenerate if $d(\vec v_x )$ is isomorphism. So this guantee on part b), $\vec v_x:X\to X$ is diffeomorphism.

Let $f_t:X\to X$ be defined as $f_t(x)=\pi(x+t\vec v_x)$. From one of my theorem, at a zero $x$ of $\vec v$, $d(f_t)=I-td(\vec v_x)$ as a linear map $T_x (X)$ to itself

At a nondegenerate zero, $d(\vec v_x ):T_x (X)\to T_x (X)$ is isomorphism. I'm trying to find the connection between this and the above statement, but I can't see any. How can I show that in a very small neighbor hood $U$ of $x$, I can't find any other zero?

From one of my previous exercise, I have shown that for $f:X \to Y$ is diffeomorphism where $X$ is compact manifold and $Y$ is connected manifold, then $deg(f)=+1$ if $f$ preserve the orientation and $deg(f)=-1$ if $f$ reserve the orientation. Can I use any of this result for part b)?

$\endgroup$
2
  • $\begingroup$ I believe Guillemin & Pollack has a typo, where "$d(f_t)_x=I-t \, d \vec v_x$" should be "$d(f_t)_x=I + t \, d \vec v_x$". $\endgroup$
    – Kyle
    Dec 2, 2014 at 21:17
  • $\begingroup$ @TedShifrin The above comment may be an addition to your list of errata (this exercise is on p.139). $\endgroup$
    – user557
    Apr 9, 2018 at 23:44

1 Answer 1

3
+50
$\begingroup$

Please note: It sounds like you're reading Guillemin and Pollack ("G&P"), so I'll reference it when convenient. However, since this is such a standard text, so I'm not going to write out all the details in the proofs. There's a lot to be learned by struggling with these exercises and forcing yourself to identify and overcome the parts where you get stuck.

Background: A zero $x$ of a vector field $\vec v$ on a smooth manifold $X \subset \mathbb{R}^n$ is called nondegenerate if $d\vec v_x : T_x X \to T_x X$ is an isomorphism. By the $\epsilon$-Neighborhood Theorem, there's a neighborhood $X^\epsilon$ of $X$ in $\mathbb{R}^n$ and a "nearest point" projection $\pi: X^\epsilon \to X$. As you wrote, for sufficiently small $t$, we can define maps $f_t: X\to X$ by $f_t(x)=\pi(x + t \vec v(x))$.

Lemma (G&P 3.5.4). At a zero $x$ of $\vec v$, we have \begin{equation}d(f_t)_x=I + t \, d\vec v_x.\tag{$*$}\end{equation}

Proof. Consider a curve $\gamma: \mathbb{R} \to X$ satisfying $\gamma(0)=x$. Then \begin{align} d(f_t)_x( \gamma'(0))&= d\pi_{x+t \cdot \vec v(x)} \left(\frac{d}{ds} \big( \gamma(s)+ t \vec v(\gamma(s)) \big)\bigg|_{s=0}\right) = d\pi_x \left( \gamma'(0) +t \, d\vec v_x (\gamma'(0)) \right). \end{align} Since $x$ is a zero of $\vec v$, the derivative $d \vec v_x: T_x X \to \mathbb{R}^n$ sends $T_X X$ to itself; see 3.5.3 in G&P if this is unclear. It follows that $\gamma'(0)+t \, d \vec v_x (\gamma'(0))$ lies in $T_x X$, which is fixed by $d\pi_x$ because $X$ is fixed by $\pi$. The claim follows. $\blacksquare$

Proof that nondegenerate zeroes are isolated: Lefschetz fixed points of diffeomorphisms $X \to X$ are isolated, so it suffices to show that nondegenerate zeroes of $\vec v$ are Lefschetz fixed points of $f_t$; this is the hint provided in G&P. If $x$ is a nondegenerate zero of $\vec v$, then $d \vec v_x (\xi)\neq 0$ for all $\xi \neq 0\in T_x X$. Thus, using ($*$), we see that $d(f_t)_x(\xi)\neq \xi \, \text{ for } t \neq 0.$ It follows that $x$ is a Lefschetz fixed point of $f_t$ (for nonzero $t$). $\blacksquare$

Note: The above argument easily extends to show that the nondegenerate zeroes of $\vec v$ are precisely the Lefschetz fixed points of $f_t$ (for $t \neq 0$).

Proof that $\boldsymbol{\operatorname{ind}_x(\vec v)=+1}$ (resp., $\boldsymbol{-1}$) if $\boldsymbol{d\vec v_x}$ preserves (resp., reverses) orientation: We know that nondegenerate zeroes are Lefschetz fixed points of $f_t$ (for $t\neq 0$), so we have $L_x(f_t)=\operatorname{ind}_x (\vec v)$, where $L_x(f_t)$ is the local Lefschetz number of $f_t$ at $x$; see the Proposition in $\S$3.5 of G&P. We know that $L_x(f_t)=\pm 1$ depending on whether $d(f_t)_x-I$ (an isomorphism because $x$ is a Lefschetz fixed point) preserves ($+1$) or reverses ($-1$) the orientation of $T_x X$; see the Proposition on p121 $\S$3.4 of G&P. Since ($*$) implies that $d(f_t)_x - I= d\vec v_x$, the claim follows. $\blacksquare$

$\endgroup$
1
  • $\begingroup$ Regarding the very last line: the lemma says that $d(f_t)_x-I=tdv_x$, why do you say that $d(f_t)_x-I=dv_x$? $\endgroup$
    – user557
    Apr 9, 2018 at 22:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .