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Prove that for a path connected topological group $G$ the Euler characteristic $\chi (G)$ is zero.

I'm trying to use Lefschetz for this proof. Bare with me.

The Lefschetz Fixed Point Theorem states that:

If $X$ is a finite simplical complex, or more generally a retract of a finite simplicial complex, and $f:X\longrightarrow X$ is a map with $\tau (f) \neq 0$, then $f$ has a fixed point. Here, the Lefschetz Number $\tau (f)$ is defined to be $\tau (f) = \Sigma _n(-1)^ntr(f_{*}:H_n(X)\longrightarrow H_n(X))$.

Here is a blurb explaining the proof: "The Lefschetz fixed point theorem tells us that if an endomorphism on a space has Lefschetz number (the alternating sum of the traces of its induced maps on homology) not equal to zero, then it must have a fixed point. Groups famously have lots of topological automorphisms with no fixed points (left-multiply by any nontrivial element g). One can also prove that such a function is homotopic to the identity whenever the group is path connected (just take any path from 1 to g and appropriate translations thereof performed simultaneously). Hence, since homotopic maps induce the same maps on homology, the Lefschetz number of the identity map must be zero, and the Lefschetz number of the identity map is precisely the Euler characteristic!"

But I'm a little confused. When they say "such a function is homotopic to the identity" are they referring to the function $f: X\rightarrow X$? I feel like they mean homotopic to the constant map. I know $X$ is path-connected, but I thought if $X$ was contractible, then it was homotopic to a constant map.

Moving past that, so if $f$ is homotopic to the constant map, then we would also have that $f_{*}$ is homotopic to a constant map. Which would leave $\tau (f) = 0$. Since $f$ is the identity map on $X$, then $\tau (f) = \chi (X)$. So, $\chi (X) = 0$.

Can someone help clarify this explanation and/or give an alternative way to prove this.

Also, I'm not given that $X$ is a simplicial complex, so I might be in the wrong with even using Lefschetz. All we're given about $X$ is that it is path connected topological group.

Michael
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    Counterexample: $G = \mathbb{R}$. .You need compactness to use Lefschetz. I only know of easier proofs with additional assumptions on $G$, for example, that it's a Lie group. – Jason DeVito - on hiatus Aug 10 '16 at 20:04
  • @JasonDeVito What argument does the Lie group structure give you? – Elle Najt Aug 10 '16 at 20:37
  • @AreaMan: If you have smooth structure, you can talk about vector fields and Lie algebras. Then, for example, every non-trivial connected Lie group has a non-vanishing vector field, and one can then use the Poincare-Hopf theorem on vector fields to learn that $\chi(G) = 0$ in this case. – Jason DeVito - on hiatus Aug 11 '16 at 04:05

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They mean homotopic to the identity. The constant map induces zeros on homology - it is the map that sends everything to a point, and the identity is homotopic to the constant map iff the space is contractible (by definition).

The identity map induces the identity on homology, so the trace gives the dimension of the vector space. (Trace of the identity matrix.)

Hence the Lefschetz number for the identity map is the alternating sums of the dimensions of the homology groups, i.e. the Euler characteristic. But then because the identity is homotopic to a right translation by some $g \not = e$, and the Lefschetz number is invariant under homotopies (because homotopic maps induce the same map on homology), and because $R_g$ has no fixed points, the lefschetz number must be zero.

(As JasonDeVito points out - you need compactness to apply Lefschetz.)

This is a cool argument - thanks for sharing it. Let me know if you have more questions.

I don't know the answer to your question about simplicial complexes, though I wouldn't worry about it too much - I think most spaces you encounter can be given that structure.

Elle Najt
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  • Since we don't have compactness, does that mean we can't use Lefschetz? We have to find another way to prove this? – Michael Aug 10 '16 at 20:48
  • Where did this right translation come from? I don't know what that is. and what is $R_g$? – Michael Aug 10 '16 at 20:49
  • @Michael Well it's not true without compactnes, JasonDeVito gave a counter example of the reals in his comment. The translation is just a nontrivial element in your group, g, and $R_g$ is right multiplication by $g$. So I guess you have to assume that the group is not trivial also - otherwise the euler characteristic is 1. – Elle Najt Aug 10 '16 at 21:05
  • @Michael Right translation is the right multiplication action. Explicitly, it is the map $R_g: h \mapsto hg$. This is a very useful map when you’re trying to construct something without a fixed point. – Nancium Apr 18 '22 at 14:43
  • A similar problem I encountered is that G has a finite CW complex structure, so compactness would be assumed in that context. – Nancium Apr 18 '22 at 17:26