# Do we know this homogeneous space by another name?

Consider the homogeneous space $GL(3)/GL(2) = GL(3,\mathbb{R})/GL(2,\mathbb{R})$ where $GL(2)$ fixes the first coordinate axis (so can be identified with the subgroup of $2\times 2$ blocks sitting in the 'bottom right' corner of matrices in $GL(3)$).

Is there any 'explicitly known' other homogeneous space which is isomorphic to this one? Say by using orthogonal groups? Is it some sort of Grasmannian/Stiefel manifold?

Am I correct in guessing that $GL(3)/GL(2)$ has dimension 5?

-

As a bundle it fibers over $\mathbb RP^2$ with fiber $\mathbb R^3 \setminus \mathbb R^2$.

I think the most natural way to describe this space would be to consider two bundles over $\mathbb RP^2$: (1) $\mathbb RP^2 \times \mathbb R^3$ and (2) $\{(L,v) : L \in \mathbb RP^2, v \in \mathbb R^3, v \perp L \}$ for this purpose $\mathbb RP^2$ is considered the space of lines in $\mathbb R^3$.

So the total space of bundle (2) is a subspace of the total space of bundle (1). The bundle you're interested in, is the complement of (2) in (1).

So another way to say what this space is, is it's all pairs $(L, v)$ such that $L$ is in $\mathbb RP^2$ and $v$ projects to a non-zero vector in $L$ (using orthogonal projection).

So another way to describe it would be the tangent bundle of $\mathbb RP^2$ fiber-product with a certain bundle over $\mathbb RP^2$ -- as a bundle over $\mathbb RP^2$ this "certain bundle" is the map $\mathbb R \times S^2 \to \mathbb RP^2$ where $(p,q) \longmapsto \pi(q)$ where $\pi : S^2 \to \mathbb RP^2$ is the 2:1 covering map.

Even simpler, as a space it's diffeomorphic to $S^2 \times \mathbb R^3$. You can make this equivarient with respect to the left $GL(3)$ action as well -- $GL(3)$ has its standard (projectivized) action on $S^2$, similarly it acts on $\mathbb R^3$.

-
So it's just $S^2\times \mathbb{R}^3$? That's good. – David Roberts Mar 11 '11 at 5:44

More of a comment than an answer (I know you tagged this as "differential-geometry" and not "homotopy-theory"):

By using simultaneous continuous Gram-Schmidt orthonormalization, this deformation retracts onto $O(3)/(GL(2)\cap O(3))=O(3)/O(2)$. Note that $O(3)/(O(2)\times O(1))=:Gr_2(\mathbb{R}^3)\cong Gr_1(\mathbb{R}^3)=\mathbb{RP}^2$, so taking an $O(1)$-quotient yields $\mathbb{RP}^2$. As $O(3)/O(2)$ is connected (the coset of any $A\in O(3)$ contains an element of $SO(3)$), it must be that $O(3)/O(2)\cong S^2$. (In fact, the fiber bundle $O(2)\rightarrow O(3)\rightarrow S^2$ is just the circle bundle $S(TS^2)$, i.e. the sphere bundle of $TS^2$.) So, your space is homotopy equivalent to $S^2$.

The fact that you're quotienting by an embedded Lie subgroup means that the dimensions do indeed subtract: $\dim(GL(3)/GL(2))=3^2-2^2=5$. So this isn't just $TS^2$ or anything like that. I'd nevertheless be willing to bet that it's some sort of fiber bundle over $S^2$, but I couldn't prove it...

-