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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~~~

1LiterTears
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2 Answers2

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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.

Anthony Carapetis
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  • No, I am not familiar with the theory of Lie Groups... – 1LiterTears Aug 19 '13 at 04:38
  • What do you mean by "non-zero vector at the identity", and "translate it everywhere"? Thank you~~ – 1LiterTears Aug 19 '13 at 04:39
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    @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 04:48
  • Thank you @tessellation. What is $v$? – 1LiterTears Aug 19 '13 at 05:41
  • @Jellyfish v is a tangent vector at identity i.e. $v\in T_{I}O(3).$ – tessellation Aug 19 '13 at 07:33
  • Thank you @tessellation - but which direction? Or I just pick an arbitrary direction on the tangent space of each point? – 1LiterTears Aug 19 '13 at 15:24
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    @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
  • Thanks Anthony. Still a bit confused: if I translate, then it would not lie in the tangent plane of some points...? – 1LiterTears Aug 19 '13 at 16:35
  • Oh, or you mean project it down onto the tangent plane? – 1LiterTears Aug 19 '13 at 16:38
  • You're not translating in the affine space $GL(3) \supset O(3)$; you're translating in the group $O(3)$ using the left action of the group on itself. See e.g. the definition of $\xi^L$ on page 9 of http://www.math.toronto.edu/mein/teaching/lie.pdf – Anthony Carapetis Aug 19 '13 at 16:39
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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.

tessellation
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