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The following exercise is from Guillemin and Pollack, Differential Topology.

Show that the Euler characteristic of the orthogonal group (or any compact Lie group for that matter) is zero. Hint: Consider the left multiplication by an element $A\neq I$ in $O(n)$.

As it is stated, the first clue I had was to apply this proposition:

Let $X$ be a smooth compact orientable manifold. If $X$ admits a smooth map $f:X\to X$ that is homotopic to the identity and has no fixed points then the Euler characteristic of $X$ is zero.

Of course, as the hint suggests we define $f:O(n)\to O(n)$ by $f(B)=AB$ for every $B\in O(n)$, and $A\in O(n)\setminus\{I\}$ is fixed. Clearly $f$ has no fixed points.

So, my question now is if $f$ is homotopic to the identity map? How can we show this?

Thank you very much.

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  • $\begingroup$ Well, I mean, this is not literally true. They need to demand the Lie group is positive-dimensional. In that case, pick an element in the path component of the identity to translate with. $\endgroup$
    – user98602
    May 13, 2016 at 19:01
  • $\begingroup$ Oh, I see. That sounds good. Thank you. But how do you know the path component of the identity is not a singleton? @MikeMiller $\endgroup$
    – Talexius
    May 13, 2016 at 19:03
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    $\begingroup$ Use the positive-dimensional assumption. There's something to do here, so I'll let you think about it. $\endgroup$
    – user98602
    May 13, 2016 at 19:13
  • $\begingroup$ Oh, sorry. I don't know what I was thinking. Every Lie group is locally euclidean so of course path components are not trivial. @MikeMiller $\endgroup$
    – Talexius
    May 13, 2016 at 19:36
  • $\begingroup$ You should write up an answer with all the details; I'll upvote it. $\endgroup$
    – user98602
    May 13, 2016 at 19:36

1 Answer 1

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It is suffice to show some non-trivial left-translation is homotopic to the identity map.

Let $G$ be a Lie group positive-dimensional. For every $x\in G$ we can find an open neighborhood of $x$ homeomorphic to an open ball of $\mathbb{R}^n$ ($\dim G= n>0$), which is path connected. Hence the path component of $x$ is always non-trivial. (Another way to see this is that as $G$ is locally path connected, its path components are open, then they are non-trivial.)

Let $e\in G$ be the identity element. From above, we can find $x\neq e$ and a continuous $\gamma:[0,1]\to G$ such that $\gamma (0)=e$ and $\gamma (1)=x$. Define the left-translation by $L_x(g)=xg$ for every $g\in G$. The function $H:G\times [0,1]\to G$ defined by $H(g,t)=\gamma(t)g$ is an homotopy between $L_x$ and the identity map.

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