Orientability of $m\times n$ matrices with rank $r$

Question

I know that $$M_{m,n,r} = \{ A \in {\rm Mat}(m \times n,\Bbb R) \mid {\rm rank}(A)= r\}$$is a submanifold of $\Bbb R^{mn}$ of codimension $(m-r)(n-r)$. For example, we have that $M_{2, 3, 1}$ is non-orientable, while some others are, such as $M_{3,3,1}$ and $M_{3, 3,2}$.

Is there a way to decide in general if $M_{m,n,r}$ is orientable, in terms of $m,n$ and $r$?

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For example, to see that $M_{2,3,1}$ is non-orientable, we parametrize it using two maps. Call $V_i$ the open set in $M_{2,3,1}$ on which the $i-$th row is a multiple of the other non-zero one. We have $M_{2,3,1} = V_0 \cup V_1$. Put $$\alpha_1:(\Bbb R^3 \setminus\{{\bf 0}\})\times \Bbb R\to V_1, \quad \alpha_1(v,t) = (tv,v),$$and similarly for $\alpha_2$, where we look at these pairs as rows of the matrix. These $\alpha_i$ are good parametrizations that cover $M_{2,3,1}$. One then checks that $\alpha_1^{-1}(V_1\cap V_2)$ has two connected components, namely, $(\Bbb R^3\setminus\{{\bf 0}\}) \times \Bbb R_{>0}$ and $(\Bbb R^3\setminus\{{\bf 0}\}) \times \Bbb R_{<0}$. And $\det D(\alpha_2^{-1}\circ \alpha_1)(v,t)$ changes sign there, so $M_{2,3,1}$ is non-orientable.

The strategy for proving that $M_{m,n,r}$ is a submanifold is different and writes it locally as an inverse image of regular value. This is an early exercise in Guillemin & Pollack's Differential Topology book.

It seems difficult to attack the general case using parametrizations this way (computing inverses is hard).

Answer 1

$M_{m,n,r}$ is an homogenous space under the natural action of the (orientable) Lie group $G=Gl^+(m)\times Gl^+(n)$, through the action $(P,Q) M= PMQ^{-1}$. The isotropy group of the matrix $I_{m,n,r}= \left( \begin{array}{cc} I_r & 0_{n-r,r} \\ 0_{r,m-r} & 0_{m-r,n-r} \end{array} \right)$ is the set of matrices $(P,Q)$ with $P= \left( \begin{array}{cc} R & B \\ 0 & D \end{array} \right) , Q= \left( \begin{array}{cc} R & 0 \\ C & H \end{array} \right)$ with $R,D,H$ invertible of rank $r, m-r, n-r$, $B,C$ arbitrary.

The orientability can be checked by looking at the action of this subgroup $H$ on the tangent space at the identity in $G/H$, and see if it preserves or not an orientation.

To see this let us note that by continuity, on the connected component of the identity in $H$ the determinant must remains positive, and we just have to check what happens on the other components.

Note that, unless $r=m$ or $r=n$, the group $h$ has exactly two components : the identity component and the component of the pair $\epsilon=(P,Q)$ of matrices $\alpha= \left( \begin{array}{cc} S_r & O \\ 0 & S_{m-r} \end{array} \right) , \beta= \left( \begin{array}{cc} S_r & 0 \\ 0 & S_{n-r} \end{array} \right)$ where $S_r$ is the diagonal matrix with one eigenvalue equal to $-1$, the other equal to 1.

The tangent space of $G$ at the identity $\mathcal G$ is the product $M_m\times M_n$, it contains the tangent space $\mathcal H$ of $H$. In order to check that the action of $\epsilon$ on $\mathcal G/\mathcal H$ preserve the orientation or not, it is enough to check if its action on $\mathcal H$ does.

Identify $\cal H$ with the set of matrices $M=(\left( \begin{array}{cc} X & Y \\ 0 & Z \end{array} \right),\left( \begin{array}{cc} X & U\\ 0 & V \end{array} \right)) $, and compute the action of $\epsilon$, one find $X\to S_rXS_r^{-1}$, $Y\to S_r Y, Z\to S_{m-r}Z; U\to US_{n-r}^{-1}, V\to VS_{n-r}^{-1}$. Its determinant is $(-1)^{2r+r+m-r+r+n-r}= (-1)^{m+n}$

So (unless mistakes on calculations), the orientability depends on the parity of $m+n$. The cas $r=m$ or $r=n$ can be treated in a similar way.

Answer 2

I think you can do the $r = 1$ case as follows. Pick nonzero vectors $c \in \mathbb{R}^m-\{0\}$, $v \in \mathbb{R}^n-\{0\}$. Then form the outer product of $c$ and $v$; this is an $m \times n$ matrix with rank 1, and all rank-1 matrices can be written in this form. So we get a surjective map $$\left(\mathbb{R}^m-\{0\}\right) \times \left(\mathbb{R}^n-\{0\}\right) \twoheadrightarrow M_{m,n,1}$$ with kernel $\mathbb{R}^* = \mathbb{R}-\{0\} = \mathbb{R}^+ \times \mathbb{Z}_2$, because $(c,v)$ and $(xc,v/x)$ clearly give the same matrix for $x \in \mathbb{R}^*$.

This tells us that $M_{m,n,1} = \mathbb{R}^+ \times (S^{m-1} \times S^{n-1})/ \mathbb{Z}_2$. For $m, n \ge 2$, the diagonal action by $-1$ has degree $(-1)^{m+n}$, and the quotient is orientable iff $(-1)^{m+n} = 1$, i.e. if $m + n$ is even.

For $m = 1$ or $n = 1$, $S^0$ behaves very differently from $S^1, S^2, \cdots$ since it's disconnected, so the analysis is different. In particular $M_{k,1,1} = M_{1,k,1} = \mathbb{R}^k-\{0\}$ is always orientable.

Hence $M_{m,n,1}$ is orientable if $m = 1$, $n = 1$, or $m + n$ is even. This agrees with your calculation for $M_{2,3,1}$. I'd be curious to see a direct argument that $M_{2,2,1}$ is orientable.

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I also want to record some useful observations for the general problem, but I've not thought much about the $r \ge 2$ case:

• $M_{m,n,r} \cong M_{n,m,r}$ via the transpose, since row rank is equal to column rank. Hence we can assume $m \ge n \ge r \ge 1$.
• $M_{m,r,r}$ is easy to write as a homogeneous space, and so you should be able to do the orientability calculations without too much trouble.

I'd love to see your proof that $M_{3,3,2}$ is orientable.

My suspicion is that there's an elegant decomposition of $M_{m,n,r}$ as a product of oriented Grassmannians (and possibly a $GL^+$ or two) modulo some kernel as above - the spheres are simply the oriented Grassmannians $\tilde{\mathrm{Gr}}_1(\mathbb{R}^n)$, after all. But I haven't found a way to see this in the quick thinking I've done.