A quotient of the general linear group

Question

I am very interested in finding the homology of the following space, call it $M$. We take the quotient of $GL(3,\mathbb R)$ by the equivalence relation that identifies two $A,B\in GL(3,\mathbb R)$ by saying that for any $i=1,2,3$ there is a non-zero scalar $\lambda_i \in \mathbb R$ such that the $i$th row of $B$ is a $\lambda$ multiple of the $i$th row of $A.$

This quotient space is very similar to the space $PGL(3,\mathbb R)$ which is the quotient of $GL(3,\mathbb R)$ by the equivalence relation that identifies two matrices that are nonzero scalar multiple, $M$ additionally identifies scaling of the rows.

If $M$ is a well-known space can you tell me what it is? I know the homology of $GL(3,\mathbb{R}),$ how can I calculate the homology of $M$?

Answer 1

Claim. 1. $M$ is homotopy-equivalent to the quotient of $S^3$ by free (linear) action of the dihedral group $D_4$ (of order $8$).

2. Consequently, $H_0(M)\cong H_3(M)\cong {\mathbb Z}$, $H_1(M)\cong {\mathbb Z}_2\times {\mathbb Z}_2$ and $H_2(M)=0$.

In particular, the space $M$ does not have homology groups isomorphic to that of $GL(3, {\mathbb R})$ (apart from the fact that $H_2$ vanishes in both cases).

Here is a sketch of the proof.

i. Using the QR decomposition, we obtain $GL(3, {\mathbb R})= U\cdot O(3)$, where $U$ is the subgroup of upper-triangular matrices with positive diagonal entries. As Lee Mosher noted in his answer, the quotient space $M$ is nothing but $GL(3, {\mathbb R})/T$, where $T$ is a maximal torus in $GL(3, {\mathbb R})$, the subgroup of diagonal matrices. The group $T$ deformation-retracts to its maximal compact subgroup $K_T=T\cap O(3)$ consisting of diagonal matrices with entries in $\{\pm 1\}$. From this, it follows that $M$ is homotopy-equivalent to $O(3)/K_T$.

ii. The group $O(3)$ has two components, one of which is $SO(3)$. From this, it follows that $O(3)/K_T$ is diffeomorphic to $Y=SO(3)/(K_T\cap SO(3))$, where $\Gamma=K_T\cap SO(3)\cong {\mathbb Z}_2\times {\mathbb Z}_2$. Thus, we have to analyze the quotient of $SO(3)$ by that subgroup of order 4. The universal covering group of $SO(3)$ is $SU(2)$, which is diffeomorphic to $S^3$. One verifies that lifting $\Gamma$ to the covering space $S^3$ results in the group $\Gamma^*$ which is the nontrivial central extension of $\Gamma$ by the center of $SU(2)$, i.e. $\Gamma^*$ is isomorphic to the dihedral group $D_4$. In particular, the abelianization of $\Gamma^*$ is isomorphic to $\Gamma\cong {\mathbb Z}_2\times {\mathbb Z}_2$.

iii. Now, I will compute homology of the quotient manifold $Y=S^3/\Gamma^*$. By the Hurewicz theorem, $H_1(Y)$ is isomorphic to the abelianization of $\pi_1(Y)\cong \Gamma^*$. Hence, $H_1(Y)\cong {\mathbb Z}_2\times {\mathbb Z}_2$. The manifold $Y$ is orientable (since $\Gamma^*$ is contained in $SU(2)$) and, obviously, compact and connected. This gives us $H_0(M)\cong H_3(M)\cong {\mathbb Z}$, $H_1(M)\cong {\mathbb Z}_2\times {\mathbb Z}_2$. To compute $H_2$, we use the Poincare Duality: $$ H_2(Y)\cong H^1(Y). $$ The 1st cohomology group one computes using the Universal Coefficients Theorem for Cohomology. The Ext-term ($Ext(H_0(Y), {\mathbb Z})= Ext({\mathbb Z}, {\mathbb Z})$) vanishes and one is left with $$ H^1(Y)\cong Hom(H_1(Y), {\mathbb Z})=0. $$ Thus, $H_2(Y)=0$.

Answer 2

Here's a partial answer.

Your space falls into a well known class of spaces, namely homogeneous spaces of Lie groups, also known as coset spaces of a Lie group modulo a Lie subgroup. In any Lie group $G$, for any Lie subgroup $H < G$, there is an associated homogeneous space, namely the quotient space of right cosets $$H \setminus G = \{Hg \mid g \in G\} $$ In your case $$H = \begin{pmatrix} \lambda_1 & 0 & 0 \\ 0 & \lambda_2 & 0 \\ 0 & 0 & \lambda_3 \end{pmatrix} \qquad G = GL(3,\mathbb R) $$ It's called a homogeneous space for $G$ because the right action of $G$ on itself by multiplication descends to a transitive action of $G$ on $H \setminus G$.

There is always fibration of $G$ over $H \setminus G$ with fiber $H$.

In your case, the fiber $H$ is $(\mathbb R - \{0\})^3$ which has 8 components, each of which is homeomorphic to $(0,\infty)^3$ and is hence contractible. This is simple enough that one should be able to exactly relate the homology groups of $G$ to those of $H$.

I suspect, however, that this is not really a shortcut to computing the homology groups of $H \setminus G$.

I do know of methods for computing homology groups of certain homogeneous spaces by finding reasonably simple cell decompositions. There are certain homogeneous spaces of $GL(3,\mathbb R)$ called Grassmanians for which there are well studied decompositions into Schubert cells. These decompositions allow one to go on and do homological calculations for Grassmanians.

Your homogeneous spacs is not a Grassmanian, but perhaps similar techniques work.