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

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~~~

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.

• No, I am not familiar with the theory of Lie Groups... – 1LiterTears Aug 19 '13 at 4:38
• What do you mean by "non-zero vector at the identity", and "translate it everywhere"? Thank you~~ – 1LiterTears Aug 19 '13 at 4:39
• @jellyfish Consider a non zero vector $v$ at the tangent space of identity matrix. Now matrix multiplication is smooth. So take any matrix $A$. 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 $\phi.$ – tessellation Aug 19 '13 at 4:48
• Thank you @tessellation. What is $v$? – 1LiterTears Aug 19 '13 at 5:41
• @Jellyfish: choose an arbitrary direction at the origin, then define the vector field elsewhere by the differential of left translation (as tesselation described in his comment above). This gives a smooth vector field. – Anthony Carapetis Aug 19 '13 at 15:33

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.