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

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JM Landsberg
JM Landsberg
  • 835
  • 5
  • 11

Why does the degree of the vareity of rank at most $r$ $\times n$ 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.