4

The questions is

Show that if $X$ is compact and all fixed points of $X$ are Lefschetz, then $f$ has only finitely many fixed points.

n.b. Let $f: X \rightarrow X$. We say $x$ is a fixed point of $f$ if $f(x) = x$. If $1$ is not an eigenvalue of $df_x: TX_x \rightarrow TX_x$, we say $x$ is a Lefschetz fixed point.

I have proved that $x$ is a Lefschetz fixed point of $f$ if and only if $\mathrm{graph}(f)$ and $\Delta = \mathrm{graph}(\mathrm{identity})$ intersect transversally at $(x,x) \in X \times X$, but not sure how to proceed.

Thank you!

1LiterTears
  • 4,572
  • 6
    Hint: Prove that Lefschetz fixed points are isolated. – Henry T. Horton May 28 '13 at 02:39
  • @HenryT.Horton Thank you Henry. Then I start to wonder why the union of isolated points are closed...Is that because the union contains all the limit points? This would work for the 1/n case, but I've been wondering how to justify the general case. – 1LiterTears May 28 '13 at 02:44
  • 3
    What's true if you have the transverse intersection of a compact submanifold and a closed submanifold of complementary dimension? – Ted Shifrin May 28 '13 at 02:56
  • Hi @TedShifrin, thanks for your hint - the intersection has trivial dimension. - so it must be an isolated point? – 1LiterTears May 28 '13 at 03:21
  • 2
    Elaborate. The transverse intersection of submanifolds is ... And where does the topology come in? – Ted Shifrin May 28 '13 at 03:28
  • @TedShifrin Because graph($f$) transverse $\triangle$, so dim graph($f$) + dim$\triangle$ = dim $V \times V$.

    But dim graph($f$) = dim$V$, dim$\triangle$ = dim $V$, so graph($f) \cap \triangle = \emptyset$.

     – 1LiterTears May 28 '13 at 03:40
  • 2
    NO. This means $\dim \text{graph}(f)\cap\Delta = 0$. This kind of reasoning will be huge throughout diff. top. – Ted Shifrin May 29 '13 at 23:59
  • Hi @TedShifrin, thanks for your generous help again. But what's the difference between $\emptyset$ and dim 0? Thank you! – 1LiterTears May 30 '13 at 00:08
  • Any set of isolated points is a $0$-dimensional manifold. (Each point has a neighborhood that is diffeomorphic to $\mathbb R^0 = {0}$. The empty set has no dimension (or some would say it has every dimension). As you proceed through G&P, you will see that $0$-dimensional submanifolds play a pivotal role ... defining intersection numbers, for starters. – Ted Shifrin May 30 '13 at 00:11
  • @HenryT.Horton Hint: Prove that Lefschetz fixed points are isolated..... why we should prove this? – Idonotknow Oct 29 '18 at 20:04

1 Answers1

2

Some key points for this answer is contributed by @tessellation:

Suppose $x_0$ is a Lefschetz fixed point. Take a chart $(U,\phi)$ around $x_0.$ Then in this coordinate neighborhood (after composing with proper coordinate functions) I can think of $f$ as a map from open ball in $\mathbb{R}^n,\,\,$ (say $B$) to itself with $f(0)=0.$ Now consider we have a function $f:B\rightarrow B$ such that $f(0)=0.$ Consider the function $g=f-id.$ Then $g(0)=0$ and by lefschetz condition $det(dg)(0)\neq 0$ (as 1 is not an eigenvalue of $df$). Hence $g$ is a local diffeomorphism by inverse function theorem and we are done.

If $g$ is a diffeomorphism then $g$ is bijective in that small neighborhood. If $f$ has another fixed point a in that neighborhood then $g(a)=0$. This will contradict the bijcetivity of $g$. Therefore the Lefschetz fixed point is a 0-manifold. In other words, isolated.

Then we consider such an open cover of $X$: the open set around each fixed points, and the complement of the set of fixed poinots. Since $X$ is a compact manifold, so there must be finitely many of subcovers. Therefore, there are finitely many Lefschetz fixed points.

1LiterTears
  • 4,572
  • 1
    You don't need to resort to covers. An infinite subset of a compact space (in $\mathbb R^n$) has a limit point. – Ted Shifrin May 30 '13 at 00:01
  • 2
    Hi @TedShifrin I have thought of that $\forall x_i \in$ Lefschetz fixed point, $$\lim_{i \rightarrow \infty}f(x_i) = \lim_{i \rightarrow \infty} x_i = x,$$ where $x$ is also a Lefschetz fixed point. There seems some reasoning breakage to me here before get the conclusion of finiteness. – 1LiterTears May 30 '13 at 00:27
  • 1
    Not quite. You get a convergent subsequence. But then you need to say $x$ is another fixed point. Contradiction, why? Note that from the transversality viewpoint, $0$-manifolds must consist of isolated points. – Ted Shifrin May 30 '13 at 00:38
  • @TedShifrin Why it has to do with isolated points? – 1LiterTears May 30 '13 at 00:52
  • 1
    You said so yourself above. I'm not sure what you're asking. – Ted Shifrin May 30 '13 at 00:59
  • Hi sorry for the confusion @TedShifrin. You said I don't need to resort to covers, so I am trying to find the other way out. Thanks! – 1LiterTears May 30 '13 at 01:22
  • Why is the complement of a set of fixed points open? It is an intersection of complements to different fixed points, which is not necessarily open – Grisha Taroyan Oct 17 '22 at 08:40