Isolatedness of Lefschetz map - almost there

Question

$f$ is a Lefschetz map on a compact manifold X. And I need to show the Lefschetz fixed point is isolated.

I proved that the graph of f is transversal to the diagonal inside $X \times X$, then I don't know how to proceed from here. Thank you very much for your help.

Please give much more information. The more you explain about your problem, the easier it will be for people to help - so give as much information as possible! What is $f$? What is $V$? What, exactly, is it that you are trying to prove about them? — Zev Chonoles, May 28 '13 at 03:42
@ZevChonoles Thanks - I am adding much more information... — 1LiterTears, May 28 '13 at 03:49

tessellation · Accepted Answer · 2013-05-29T05:16:01.727

5

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.
Just out of curiosity are you reading the book differential topology by Guillemin and Pollack.

edited May 29 '13 at 05:16

answered May 28 '13 at 05:37

tessellation

1,554

Yes I am reading the book differential topology by Guillemin and Pollack. – 1LiterTears May 28 '13 at 23:22
Hi, given 1 is not an eigenvalue of $df$, why we get det$(f-id) \neq 0$? I thought it should be det$(df-id) \neq 0$? – 1LiterTears May 29 '13 at 00:37
1

@MathSnail Sorry for the typo. It should be $det(df-id).$ – tessellation May 29 '13 at 05:15
Hi @tessellation, thank you for the clarification. But I could follow you at the very last step - why we are done after showing $g$ is a local diffeomorphism? Thank you! – 1LiterTears May 29 '13 at 06:14
1

If g is a diffeo then g is bijective in that small nbd. If f has another fixed point a in that nbd then g(a)=0. This will contradict the bijcetivity of g. – tessellation May 29 '13 at 06:52
Thank you so much @tessellation! – 1LiterTears May 29 '13 at 07:05

Krzysztof C · Answer 2 · 2013-09-12T09:52:25.610

An alternative solution, directly continuing Jellyfish's beginning of a solution to G&P's Problem 5-10, Chapter 1.

Since $df_x$ has no eigenvalues equal $1$, by Problem 5-9 of G&P

$$\text{graph}(df_x) \pitchfork \Delta(T_x(X)\times T_x(X)).$$

It's an easy exercise to check that $\Delta(T_x(X)\times T_x(X)) = T_{x,x}(\Delta(X\times X)),$ so we now have $$\text{graph}(f) \pitchfork \Delta(X\times X).$$ The overlap of these two sets is $W := \{(x,f(x))\colon x\in X, x=f(x)\},$ in bijective (in fact, diffeomorphic) correspondence with the set of fixed points of $f$. Now, by the transversality theorem from G&P, $W$ is a submanifold of $X\times X$. Furthermore, $\text{codim}(W) = \text{codim}(\text{graph}(f)) + \text{codim}(\Delta(X\times X)) = 2n.$ But this shows that $W$ is in fact a manifold of zero dimension, that is a set of isolated points. Thus the set of fixed points of $f$ is also an isolated set. As a subset of compact $X$ it must thus be finite.

how did you say that we now have graph(f)⋔Δ(X×X) ? Is it general theorem ? — bytrz, Apr 07 '16 at 19:20

Isolatedness of Lefschetz map - almost there

2 Answers2