Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm interested in methods of computing the homology groups and Euler characteristics of the classical Lie groups ($GL(n,\mathbb{R}), SL(n,\mathbb{R})$, etc.). (But I'd be interested in techniques which are more generally applicable to arbitrary Lie groups as well.)

Here's the current state of my knowledge on this issue with some explicit questions interspersed:

  • A compact Lie group of positive dimension admits a nowhere vanishing vector field and hence has Euler characteristic zero.

  • I remember reading in the past that a connected Lie group deformation retracts onto its maximal compact subgroup, which by the foregoing has Euler characteristic zero. Given this, we can say that a connected Lie group is either contractible or has Euler characteristic zero. However, my knowledge of Lie theory is (very) weak at best, so I am not fully comfortable accepting this fact. Is there a simple proof? (I feel that there should be but I'm just missing it.)

  • The Euler characteristic respects fiber bundle structures: if $F \rightarrow E \rightarrow B$ is a fiber bundle with total space $E$, base $B$, and fiber $F$, then $\chi(E) = \chi(F)\cdot\chi(B)$. Some suitable conditions are needed; I think $B$ path-connected suffices, but I'm not 100% certain. I can think of one explicit example of this: Identify $S^3$ with $SU(2)$ and implement the Hopf fibration $S^1 \rightarrow S^3 \rightarrow S^2$. The subgroup $U(1)$ is realized as $S^1$ and the quotient $SU(2)/U(1)$ is realized as $S^2$. So we have a fibration $SU(2) \rightarrow SU(2)/U(1)$ with fiber $U(1)$; hence $\chi(SU(2)) = \chi(SU(2)/U(1)) \cdot \chi(U(1)) = 0$. But this is already known since $SU(2)$ is compact.

  • The most direct approach to computing both the homology and the Euler characteristics would possibly be to find an explicit cellular decomposition of these groups. Again, my knowledge of basic Lie theory is failing me here: I don't see how to find such decompositions of the matrix groups.

  • Surely there are other methods of computing the homology which would exploit the Lie group structure, but I don't know of any.

share|cite|improve this question
The tool that lets you go from a general Lie group to its maximal compact subgroup is the Iwasawa decomposition. – Matt E Dec 6 '10 at 17:47
That is in general, but in the case of $GL$ and $SL$ you can use Gram-Schmidt. – Sean Tilson Dec 6 '10 at 18:58
up vote 9 down vote accepted

So we know that $GL_n(\mathbb{R})$ and $SL_n(\mathbb{R})$ are homotopy equivalent to $O_n$ and $SO_n$ respectively (Similar things are true for working over $\mathbb{C}$). Now you can work inductively using a Serre SS. Look at the fibration coming from the inclusion of one classical group into the next higher one, if you set it up right the homogenous space will be a sphere. This should get most of the work done for you, and it doesn't require lie theory. I don't think it will generalize though to exceptional lie groups though, you need to understand the homogenous spaces in order for this to work.

Here is another way using the Serre SS: Use the fact that we know what $H^*(BG)$ is for $G$ unitary or orthoganl, special or not (with integer or mod 2 coefficients depending on whether or not it is $U$ or $O$). We know these rings because they give us chern and stiefel-whitney classes respectively (I believe the effect of looking a the special groups is just to kill $c_1$ or $w_1$). Also, the dimension of the vector space the group is acting on tells you the index of the top characteristic class. Now use our good old friend the loop path space fibration: $\Omega BG \to PBG \to BG$ and remember that $\Omega BG$ is homotopy equivalent to $G$ and $PBG$ is contractible.

Both of these approaches give you the ring structure.

Then use a Bockstein SS to get the integral cohomology for the stuff with mod two coefficients ($sq^1$ is the differential!). Once you get the cohomology you can compute the euler characteristic by taking the alternating sums of the ranks.

share|cite|improve this answer
You can push your first paragraph to compute stuff for $G = G_2$ or $G = F_4$. The Lie group $G_2$ fits into a sequence $SU(3)\rightarrow G_2\rightarrow S^6$. Likewise, for $F_4$ one gets $Spin(9)\rightarrow F_4\rightarrow \mathbb{Ca}P^2$, where $\mathbb{Ca}P^2$ denotes the Cayley plane (which is not a sphere, but it's topology is still well understood). – Jason DeVito Dec 7 '10 at 0:32
awesome! What about $E_6$, $E_7$, and $E_8$? – Sean Tilson Dec 7 '10 at 2:40
awesome is right.... both answer and comment! – BBischof Dec 7 '10 at 4:06

The explicit cellular decomposition you're looking for is called Bruhat decomposition. Counting the number of cells in each dimension automatically gives you the homology of classical groups over $\mathbb{C}$ (since then there are only cells in even dimensions), but I think over $\mathbb{R}$ you have to work harder (although I think you do get the Euler characteristic, in any case).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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