Let $X_r\subset Mat_{n\times n}$ denote the matrices of rank at most $r$, and let $S_{\pi}C^n$ denote the irreducible $GL_n$module corresponding to the partition $\pi$. One can check that degree($X_r$)=dim($S_{(nr)^{nr}}C^n)$=$\Pi_{i=0}^{nr1}\frac{(n+i)!i!}{(r+i)!(nr+i)!}$ Does anyone have a geometric (or any) explanation for this equality? Had it been observed previously? Here $(nr)^{nr}=(nr,...,nr)$ has Young diagram the square box.
There are a number of better statements (i.e. yes, this is "wellknown"). To begin with, wouldn't you rather have the character of this representation, instead of just its dimension? You can get that as the class $[X_r] \in H^*_{T^n}(Mat_{n\times n})$. This improves on your statement, because for $X \subseteq V$ a cone in a vector space, $[X] = deg(X) t^{{\rm codim}(X)} \in H^*_{\rm dilation}(V) \cong {\mathbb Z}[t]$, and the character of $t\cdot Id_n$ on $V_\lambda$ is $\dim(V_\lambda) t^{\lambda}$. (Of course, the character is a symmetric polynomial. Correspondingly, we should look at $[X_r] \in H^*_{GL(n)}(Mat_{n\times n}) = H^*_T(Mat_{n\times n})^{S_n}$.) Next, let's go beyond square matrices to general $k\times n$. But most satisfyingly, instead of asking that the whole thing have rank $r$, let $X_{\vec n}$ be the $k \times \infty$ matrices where the left $n_i$ columns has rank $\leq i$, for each $i=0,\ldots,k$. After column $n_k$ there are no conditions, so including those columns has no effect on the degree. The case you're thinking about is $n_{i} = n+max(0,ir)$. The data of $\vec n$ is equivalent to a partition with $\leq k$ rows. You might be most satisfied looking at $$ [X_{\vec n}] \in H^*_{GL(k)}(Mat_{k\times \infty}) = H^*_{GL(k)}(\text{full rank matrices}) = H^*(GL(k) \backslash \text{full rankers}) = H^*(Gr_k({\mathbb C}^\infty)) \cong Rep(End({\mathbb C}^k)) $$ where the last two ringswithbases are well known to be isomorphic (see Why do LittlewoodRichardson coefficients describe the cohomology of the Grassmannian? ). In particular the degree of $X_{\vec n}$ is the dimension of the corresponding irrep of $GL(k)$ (which extends to $End({\mathbb C}^k)$). (Note that to go from all $k\times \infty$ matrices to the full rank ones, the locus removed is $\infty$codimensional, so doesn't affect the cohomology.) (Here's a Gröbner basis computation you might like. Replace each of the determinants defining $X_{\vec n}$ by its diagonalproduct summand. Decompose the resulting monomial scheme into components, which are coordinate spaces inside matrices. We can draw a component on a $k\times \infty$ strip, where $m_{ij}=0$ is drawn as a $j$ at position $(i,j)$. Now move those $j$s Northwest until they get stuck, and you will always get the same shape worth of numbers (determined by $\vec n$), with a semistandard Young tableau in it. From this one can show that the character is given by a sum over SSYT. This stuff is in my paper with Miller and Yong. For a first example, try $k=2$, $n_0 = 1, n_1 = 3, n_2 = 4$. Then the equations are $m_{11} = m_{21} = \underline{m_{13} m_{24}}  m_{14} m_{23} = 0$. The drawings are $$ \begin{pmatrix} 1 & 1 & &\cdots \\ 2 & & &\cdots \end{pmatrix}, \begin{pmatrix} 1 & & & \cdots \\ 2 & &2 &\cdots \end{pmatrix} $$ which march NW to be the two SSYT for the $V_{2+1}$ irrep of $GL(2)$.) 


We can do this more generally for any determinantal variety, so just consider $n \times m$ matrices where we don't need to assume $n=m$. Naively, we could calculate the degree from a free resolution of the determinantal variety, but unfortunately that's a complicated procedure. For just the purposes of getting the degree, it suffices to use any rank $1$ module supported in the determinantal variety. It turns out there are such modules which are both CohenMacaulay and have a linear free resolution $F_\bullet$ (without the rank assumption, these are known as Ulrich, or maximally generated maximal CohenMacaulay (MGMCM), modules in the literature). This is good because then ${\rm rank}(F_i) = {\rm rank}(F_0) \cdot \binom{c}{i}$ where $c = (nr)(mr)$ is the codimension of the determinantal variety. But then an easy computation with Hilbert series shows that the degree of the determinantal variety must be ${\rm rank}(F_0)$. The complex is described in Exercise 6.34 of Weyman, Cohomology of Vector Bundles and Syzygies. Here's a brief construction: let $k^n \to k^m$ be the generic matrix thought of as a 2term chain complex with $k^m$ in homological degree $0$. Apply the Schur functor $S_{(mr) \times (nr)}$ (here $(mr) \times (nr)$ means the rectangular partition $(nr, nr, \dots, nr)$ repeated $mr$ times). Then we get a complex (Schur functors can be used in any symmetric monoidal Abelian category, here our category is chain complexes) which has homology only in degree $0$ (and this is the rank $1$ module we were looking for) and the $0$th term is the representation $S_{(mr) \times (nr)}(k^m)$, which specializes to your example. The homology vanishing requires some work, but it can be proven using standard commutative algebra techniques (the BuchsbaumEisenbud acyclicity criterion) and some properties of Schur complexes. A bonus is that all of what I said is independent of characteristic (unlike the resolution of the coordinate ring of the determinantal variety!). Now, to advertise my own work, this has been generalized from determinantal varieties to matrix Schubert varieties: http://arxiv.org/abs/1006.5514 using the Schubert functors of Kraskiewicz and Pragacz: http://dx.doi.org/10.1016/j.ejc.2003.09.016 Finally, let me address the comment asking about (skew)symmetric matrices. The same technique works here, and is partially discussed in Exercises 6.35 annd 6.36 of Weyman's book. The only difference is that we use staircase partitions instead of rectangular partitions, and the module won't have rank $1$, but some power of $2$, which can be calculated. First consider the case of symmetric $n \times n$ matrices of rank at most $r$. This has codimension $\binom{nr+1}{2}$. The staircase partition you want is $(nr, nr1, \dots, 1)$ (the size of the partition is the codimension). We apply the corresponding Schur functor to the generic symmetric matrix $(k^n)^* \to k^n$. Again it has no homology in positive degrees and its $0$th term is $S_{(nr, nr1, \dots, 1)}(k^n)$. The rank of its $0$th homology is $2^d$ where $d=\binom{nr}{2}$. So the degree is $2^{d} \dim(S_{(nr,nr1,\dots,1)}(k^n))$. Sanity check: when $r=n1$, the partition is $(1)$, so the complex is just the generic symmetric matrix, and the determinant has degree $n$. Now we consider skewsymmetric $n \times n$ matrices of rank at most $r$. Now $r$ is even. This has codimension $\binom{nr}{2}$. The staircase partition you want is $(nr1, nr2, \dots, 1)$. Do the same thing as above. The rank of the $0$th homology is $2^d$ where $d=\binom{nr}{2}$. So the degree is $2^{d} \dim(S_{nr1,nr2, \dots, 1}(k^n))$. Sanity check: when $r=n2$, the partition is $(1)$ so the complex is just the generic skewsymmetric matrix, and the Pfaffian has degree $n/2$. The formulas for (skew)symmetric matrices can also be found in a paper of Harris and Tu: http://dx.doi.org/10.1016/00409383(84)900260 though it's stated in a different, but closely related, form. There are also some formulas in Enright and Hunziker (section 6) which don't just give the degree, but the numerator of the Hilbert series in terms of dimensions of some representations: http://dx.doi.org/10.1016/S00218693(03)001595 but I don't see how to specialize those formulas to get the ones discussed here. 

