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Let $M$ be a 4-dimensional closed simply connected manifold. Show that every continuous map $f:M\longrightarrow M$ which is homotopic to the identity has a fixed point.

We have the Lefschetz Fixed Point Theorem that says: If $X$ is a finite simplicial complex, or more generally a retract of a finite simplicial complex, and $f: X\rightarrow X$ is a map with $\tau (f) \neq 0$, then $f$ has a fixed point.

Hatcher says (p.179) that since $f$ is homotopic to the identity, we have $\tau (f) = \chi (X)$. (The Lefschetz number of $f$ equals the Euler characteristic of $M$.)

I also found in a paper that: "When $X$ is simply connected manifold, the Lefschetz number provides a complete invariant in that the map $f$ is deformable to a fixed point free map if and only if $\tau (f) = 0$."

(So, we've used simply connectedness of $M$ and that $f$ is homotopic to the identity. All that's left to use from our assumptions is the fact that $M$ is a 4-dimensional closed manifold. )

So, if I can say that $\chi (M)\neq 0$, then I'm done.

However, I can't seem to find what the Euler characteristic of $M$ is.

Any suggestions?

Michael
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  • You haven't actually used simply connectedness. If $\chi(M)\neq 0$, then you're already done, without using the result you quoted. All that result would give you that you couldn't do already is a sort of converse (if $\tau(f)=0$, then $f$ is homotopic to a map with no fixed points). – Eric Wofsey Aug 15 '16 at 03:44

1 Answers1

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We have $\chi(M) = h_0 - h_1 + h_2 - h_3 + h_4$, where $h_i$ is the rank (dimension) of $H_i(M;\mathbb{F})$. Since $M$ is simply connected, $h_1 = 0$. Also, using Poincar\'{e} duality and the universal coefficient theorem, $h_3 = 0$ too. So assuming $M$ is nonempty, we have $\chi(M) > 0$.

JHF
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  • What about $h_0,h_2,h_4$? – Michael Aug 15 '16 at 04:11
  • +1, very nice. @Michael: since $M$ is nonempty, $h_{0}$ is nonzero (it counts the number of connected components of $M$.) Hence, $\chi(M) = h_{0} + h_{2} + h_{4} > h_{0} > 0$. – Alex Wertheim Aug 15 '16 at 04:20
  • Okay, got it thanks! – Michael Aug 15 '16 at 04:23
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    For Poincare duality don't you need $M$ to be an n-dimensional oriented closed manifold? We don't have that $M$ is orientable. – Michael Aug 15 '16 at 04:34
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    @Michael You can either use $\Bbb Z/2$ coefficients (whose alternating sum also gives the Euler characteristic) or note that simply connected manifolds are orientable. –  Aug 15 '16 at 06:03
  • Did you use Poincare Duality to find that $H^1(M)\cong H_3(M)$? And then how did you use the Universal Coefficient theorem to find that $h_3 = 0$? – Michael Aug 17 '16 at 14:49
  • Yes for Poincare duality, and then UCT says $H^i(M; \mathbb{F}) \to \operatorname{Hom}(H_i(M; \mathbb{F}), \mathbb{F})$ is an isomorphism, so the ranks of $H^i(M; \mathbb{F})$ and $H_i(M; \mathbb{F})$ are equal. – JHF Aug 17 '16 at 15:08