Skip to main content
2 of 2
edited title

Why does the degree of the variety of rank at most $r$ $n\times n$ matrices equal dim$S_{(n-r)^{n-r}}C^n$?

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($S_{(n-r)^{n-r}}C^n)$=$\Pi_{i=0}^{n-r-1}\frac{(n+i)!i!}{(r+i)!(n-r+i)!}$ Does anyone have a geometric (or any) explanation for this equality? Had it been observed previously? Here $(n-r)^{n-r}=(n-r,...,n-r)$ has Young diagram the square box.