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Problem of the Week #55 - June 17th, 2013

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Chris L T521

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Jan 26, 2012
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Here's this week's problem.

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Problem: Let $G$ be a non-trivial, connected, compact Lie group. Show that $\chi(G)=0$. (i.e. it's Euler characteristic is zero)

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Hint:
Use the Lefschetz fixed point theorem.

Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

Well-known member
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Jan 26, 2012
995
No one answered this week's question. You can find my solution below.

Proof: Let $G$ be a nontrivial compact, connected Lie Group. Since any Lie group can be given a simplicial complex structure, it follows that $G$ is also a polyhedron. Furthermore, since $G$ is compact, it is a space with finitely generated homology. Let $m_e:G\rightarrow G$ denote the left multiplcation map by $e$ (i.e. the identity of $G$). The map $m_e$ is clearly the identity map; furthermore, by one of the definitions of the Euler characteristic, we have that the Lefschetz number of $m_e$ is $\chi(G)$. Now, let $g\in G$ be any non-identity element. Define the map $m_g:G\rightarrow G$ by $g\mapsto gx$ for $x\in G$ (i.e. left multiplcation by $g$). Then $m_g$ has no fixed points in $G$.

Claim: The Lefschetz number of $m_g=0$.
Proof of Claim: Suppose the Lefschetz number of $m_g\neq 0$. Then, since $G$ is a compact polyhedron, we must have a fixed point by the Lefschetz fixed point theorem, which contradicts the fact that $m_g$ has no fixed points. Hence $m_g$ has a Lefschetz number of zero.$\hspace{.25in}\blacksquare$

We now show that $m_g$ and $m_e$ have the same Lefschetz number. Since $G$ is connected and a manifold, it is path-connected. Now, let $\gamma:[0,1]\rightarrow G$ be a path with $\gamma(0)=e$ and $\gamma(1)=g$. Consider the mapping $M:G\times[0,1] \rightarrow G$ given by $M(x,t)=\gamma(t)x$. Then $M(x,0)=m_e(x)$ and $M(x,1)=m_g(x)$. Furthermore, $M$ is continuous. Hence, $m_e$ and $m_g$ are homotopic maps and since Lefschetz numbers are homotopy invariant, they have the same Lefschetz numbers, i.e. $\chi(G)=0$.$\hspace{.25in}\blacksquare$
 
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