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 $r$?

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).

  • $\begingroup$ Can you explain that $M_{2,3,1}$ is nonorientable ? $\endgroup$
    – HK Lee
    May 29, 2016 at 16:38
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    $\begingroup$ @HK Sure, I'll add a brief sketch in the question. $\endgroup$
    – Ivo Terek
    May 29, 2016 at 16:41
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    $\begingroup$ I do not have an answer but a suggestion. There are two obvious maps from $M_{m,n,r}$ to the Grassmannians of $r$-dimensional subspaces in $R^m$ and $R^n$ respectively (sending the matrix to its row-span and column spans). The thing to do is to investigate these maps, e.g. are they fiber bundles, etc., and relate orientability of the constant rank matrix spaces to orientability of Grassmannians. $\endgroup$ May 30, 2016 at 22:28
  • $\begingroup$ A little question: what do you mean when you say "These $\alpha_i$ are good parametrizations"? $\endgroup$ Jun 14, 2016 at 13:23

2 Answers 2


$M_{m,n,r}$ 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.

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    $\begingroup$ Very nice! And based on the results below together with Ivo's calculations for $M_{3,3,2}$ your claim that orientability depends only on $m+n$ seems very likely to be true. Certainly it matches the behavior of orientability of Grassmannians, where orientability depends only on the ambient vector space and not on the dimension of the subspace you're picking out. $\endgroup$
    – dvitek
    Jun 14, 2016 at 19:20

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.

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.


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