# Is this a valid way to show $\chi(SL_n(\mathbb{R}))=0$?

Why does $\chi(SL_n(R))=0$? I'm going about it like this. Let $X:=SL_n(R)$.

Define a map $f:X\to X$ such by $A\mapsto BA$, where $B$ is the identity matrix, except with an extra $1$ in the upper right corner entry. So $\det(BA)=1$, and the map is smooth. I also know $f$ has no fixed points, since $A=BA$ implies $B=I_n$ since all the $A$ are invertible. But $f$ is also homotopic to the identity, say through a family of maps $f_t$ where $f_t$ is left multiplication by the matrix which has $1$ on the diagonal, and $t$ in the upper right entry.

From Lefschetz fixed point theory, I know the Lefschetz number of the identity is equal to the Euler characteristic, and is homotopy invariant. But the Lefschetz number of a map with no fixed points is $0$, so it would follow that $$\chi(X)=L(id)=L(f)=0.$$

However, all my searching seems to show that Lefschetz fixed point theory is for compact oriented manifolds. But $SL_n(R)$ is not compact, although I think it is orientable. Does this idea work without the compactness assumption, or is there a better way to compute this?

The polar decomposition defines a deformation retract of $$SL(n, \mathbb{R}) \to SO(n, \mathbb{R})$$, and since Euler characteristic $$\chi$$ is homotopy invariant, we may just as well compute $$\chi(SO(n, \mathbb{R}))$$. But $$SO(n, \mathbb{R})$$ is compact and orientable (all Lie groups are orientable), so your Lefschetz fixed point argument seems to apply to it. In fact, for any Lie group $$G$$ and any nonidentity element $$g \in G$$, the map $$h \mapsto gh$$ fixes no point, so this argument in fact shows that any compact, connected, nontrivial Lie group has Euler characteristic zero.
For the case of $$SO(n, \mathbb{R})$$ (again, equivalently $$SL(n, \mathbb{R})$$, here's an alternate argument that uses doesn't use a fixed point argument, and is more "natively topological": The Euler characteristic is multiplicative for nice fibrations. More precisely, given a fibration $$E \to M$$ with path-connected base $$M$$ and fiber $$F$$ (and a technical orientability condition that holds here), the Euler characteristics of $$E$$, $$M$$, and $$F$$ are related by $$\chi(E) = \chi(M) \chi(F).$$ For $$n > 1$$, $$SO(n)$$ is the total space for the fibration $$SO(n) \to \mathbb{S}^{n - 1}$$ induced by the standard action of $$SO(n)$$ on $$\mathbb{R}^n$$, and its fiber is $$SO(n - 1)$$, so $$\chi(SO(n)) = \chi(\mathbb{S}^n) \chi(SO(n - 1)).$$ By induction, $$\chi(SO(n)) = \chi(\mathbb{S}^n) \cdots \chi(\mathbb{S}^1) \chi(SO(0)),$$ but $$\chi(\mathbb{S}^1) = 0$$, and so $$\chi(SO(n)) = 0.$$
• Thanks Travis. How does polar decompositon define a deformation retract? If $A=UP$ is the polar decomposition, do you just map $A\mapsto U=AP^{-1}$? – YN Chew Sep 21 '14 at 5:39
• A quick way to compute the Euler characteristic of $SO(n)$ is via the Poincare-Hopf theorem. Every Lie group admits a nowhere-zero vector field (given by pushing a tangent vector around by the multiplication action) and so has Euler characteristic zero. – Kevin Arlin Sep 22 '14 at 3:30
• @KevinCarlson That's a nice observation. At least in the $\chi(X) = 0$ case it seems to be an infinitesimal version of the fixed point theorem approach mentioned in the original question. – Travis Willse Sep 22 '14 at 3:35