Show that the Euler characteristic of $O[3]$ is zero.

Question

Show that the Euler characteristic of $O[3]$ is zero.

Consider a non zero vector $v$ at the tangent space of identity matrix. Denote the corresponding matrix multiplication by $\phi_A$. Define the vector field $F$ by $F(A)=(\phi_A)_*(v)$. Where $\phi_*$ is the derivative of $\phi$, and $v$ is a tangent vector at identity with a fixed direction.

So $$A = \begin{pmatrix} \cos (\pi/2) & -\sin (\pi/2) & 0 \\ \sin(\pi/2) & \cos(\pi/2) & 0\\ 0&0&1 \end{pmatrix}$$ is homotopic to identity map.

Then how shall I proceed....?

Thank you~~~

Answer 1

Are you familiar with the theory of Lie Groups? You can just take any non-zero vector at the identity and translate it everywhere, generating a non-vanishing smooth vector field on $O(3)$. From here it's easy with Poincare-Hopf.

Answer 2

The following approach is topological but very useful (although not very easy to prove).

Theorem: If $p: E\rightarrow B$ is a fibration with fiber $F$, with the base $B$ (path-connected), and the fibration is orientable , then $$\chi(E)=\chi(F)\chi(B) . $$

Now you have the following fibration $$O(n)\rightarrow O(n+1)\rightarrow S^n.$$ Use Induction.