# Euler characteristic of matrix manifolds

I'm reading through examples of computing Euler characteristic of manifolds. I know how to compute it for generic manifolds like sphere and torus. But what about matrix manifolds? I'd like to know how to compute the Euler characteristic of a matrix group, say $$SL_3(\mathbb{R})$$, for example.

What I know: The definition of Euler characteristic for a manifold $$M$$, I'm using is $$\chi(M)=L(Id)$$, where $$L$$ is the Lefschetz number of the identity map on $$M$$, which is basically the intersection number of the diagonal of the identity with itself. I also know the Poincare-Hopf theorem.

Any help is appreciated. Thanks!

• math.stackexchange.com/questions/13260/… – Zircht Dec 8 '18 at 2:29
• @Zircht I have no knowledge of Lie theory. Most of what's mentioned there makes no sense to me. I'd like to use the definition I mentioned. There has to be a simpler way. – tangentbundle Dec 8 '18 at 2:36
• The tangent bundle of a Lie group is parallelizable via left invariant vector fields. Hence it is trivial and has Euler number zero. – Charlie Frohman Dec 8 '18 at 3:09
• @Zircht The tools used there are waaaaay too hard for this problem. You need those only if you care about calculating the entire homology groups. – user98602 Dec 9 '18 at 1:06

2) Every connected Lie group has a compact subgroup that it deformation retracts onto. For $$SL_n$$ it is $$SO(n)$$: this is a continuous version of the Gram-Schmidt procedure.
3) Now, and only now, may we apply Lefshcetz: Pick any non-identity elemeny of your connected compact group $$G$$. Left multiplication $$L_g$$ by that element is a continuous map with no fixed points, so $$L(L_g) = 0$$. Picking a path from $$g$$ to the identity $$e$$ gives a homotopy between $$L_g$$ and $$L_e = \text{Id}$$. Because the Lefschetz number is (defined to be!) a homotopy invariant, $$L(L_e) = \chi(G) = 0$$.