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



Euler Characteristic of $SL_m(\mathbb{C})/SO_m(\mathbb{C})$
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](https://arxiv.org/abs/math/0102052)* 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.


