As described in the title, what is the (topological) Euler characteristic of the homogeneous space $SL_m(\mathbb{C})/SO_m(\mathbb{C})$?

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    $\begingroup$ This is something very classical. You can replace the groups by their maximal compacts up to homotopy. Hence the space is homotopic to the quotient $SU(n)/SO(n)$. The Euler characteristic is zero unless $n=2$. For compact connected groups $H\subset G$, the Euler characteristic is zero unless both $G,H$ have the same rank, in which case, it is strictly positive. You may see A.Borel's early papers. $\endgroup$ Sep 20, 2019 at 6:51
  • $\begingroup$ Thanks @Venkataramana, that's very helpful, I will look them up in details. $\endgroup$
    – Winnie_XP
    Sep 23, 2019 at 0:26

1 Answer 1


Let $G$ be a connected complex reductive affine algebraic group, and let $H$ be an algebraic subgroup (assume reductive). Then the Euler characteristic of the homogeneous space $G/H$ can be computed in many different ways.

As noted in the comments, $G/H$ is homotopic to $K/J$ where $K\subset G$ and $J\subset H$ are maximal compact subgroups.

By Homogeneous Spaces with Non-Vanishing Euler Characteristics, Hsien-Chung Wang, Annals of Mathematics, Second Series, Vol. 50, No. 4 (Oct., 1949), pp. 925-953:

$K/J$ has Euler characteristic 0 if and only if $\mathrm{Rank}(K)\not=\mathrm{Rank}(J)$. Moreover, if the Euler characteristic is non-vanishing, then $\chi(K/J) = |W_K|/|W_J|$ where $W_K$ and $W_J$ are the Weyl groups of $K,J$ respectively.

A different way to compute the Euler characteristic without appealing to the homotopy between $G/H$ and $K/J$ is to think of $G/H$ as a variety. Then one can count the number of points of this variety over finite fields $\mathbb{F}_q$. This will give a specialization of the mixed Hodge polynomial called the $E$-polynomial. Taking the limit as $q\to 1$ will give the Euler characteristic.

As $G\to G/H$ is principal bundle, the $E$-polynomial of $G/H$ is the quotient of the $E$-polynomial of $G$ with the $E$-polynomial of $H$.

To learn more about this in further generality see:

The virtual Poincare polynomials of homogeneous spaces by Michel Brion, Emmanuel Peyre.

Note that for the groups under consideration here, the mixed Hodge polynomials have been determined by Deligne (and hence the $E$-polynomials as well); see Theorem 9.1.5 in P. Deligne, Théorie de Hodge III. Publ. Math. I.H.E.S. 44 (1974) 5-77.

For fun, let's see this play out in your specific case.

First, it is a pretty easy exercise to determine that the number of $\mathbb{F}_q$-points in $\mathrm{SL}_m$ is $$q^{m(m−1)/2}\prod_{k=2}^m(q^k − 1).$$

In particular, the zero at $q=1$ has order $m-1$.

Now, in Orthogonal Matrices Over Finite Fields by Jessie MacWilliams (The American Mathematical Monthly, Vol. 76, No. 2 (Feb., 1969), pp. 152-164) one can find a nice description of the number of $\mathbb{F}_q$-points in the orthogonal and special orthogonal groups. In particular, the order of the zero of $q=1$ is $(m-1)/2$ if $m$ is odd, and $m/2$ or $m/2-1$ if $m$ is even (depending if $-1$ is a quadratic residue or not).

From this we conclude that the zero of $q=1$ is dominant in the numerator whenever $m$ is odd (of course greater than 1) and if $m$ is even and greater than 2. So we are left with only the case $m=2$. In that case, we explicitly have $$\chi(\mathrm{SL}_2(\mathbb{C})/\mathrm{SO}_2(\mathbb{C}))=\lim_{q\to 1}\frac{q(q^2-1)}{q-1}=\lim_{q\to 1}q(q+1)=2.$$

On the other hand, since the Weyl group of $\mathrm{SL}_2(\mathbb{C})$ has order 2 and the Weyl group of $\mathrm{SO}_2(\mathbb{C})$ is trivial, we again compute the Euler characteristic is 2.

  • 1
    $\begingroup$ Thank you @Sean for the detailed explanation ! That's very helpful ! $\endgroup$
    – Winnie_XP
    Sep 23, 2019 at 0:27

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