5
$\begingroup$

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?

$\endgroup$
5
$\begingroup$

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 $$\chi(SO(n)) = 0.$$

$\endgroup$
5
  • $\begingroup$ Thanks Travis. How does polar decompositon define a deformation retract? If $A=UP$ is the polar decomposition, do you just map $A\mapsto U=AP^{-1}$? $\endgroup$ – YN Chew Sep 21 '14 at 5:39
  • $\begingroup$ Yes, that's the correct map, though of course there's a little more to do to verify that it is indeed a deformation retract. $\endgroup$ – Travis Willse Sep 21 '14 at 7:11
  • $\begingroup$ 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. $\endgroup$ – Kevin Arlin Sep 22 '14 at 3:30
  • $\begingroup$ @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. $\endgroup$ – Travis Willse Sep 22 '14 at 3:35
  • $\begingroup$ Yes, that's exactly what it is. $\endgroup$ – Kevin Arlin Sep 22 '14 at 3:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.