Let $k$ be a field and $k[{\bf x}] = k[x_{ij}: 1 \leq i, j \leq n]$ be a polynomial algebra that I can think of as the algebra of functions on $n \times n$ matrices that are polynomial in each coordinate. I'd like to understand the polynomials $f$ in $k[{\bf x}]$ that are invariant by left translations by the unipotent group $N$ of upper triangular matrices: that is, those $f \in k[{\bf x}]$ that satisfy $f(ng) = f(g)$ for all $g \in M_n(k)$ and all $n \in N$.

In this blog post on the representations of $GL_n$, I think that David Speyer claims that the algebra of $N$-invariants as described above is generated, as an algebra over $k$, by all the bottom-justified minors: that is, for every subset $\sigma =\{\sigma_1, \ldots, \sigma_r\} \subset \{1, \ldots, n\}$ let $f_{\sigma}$ be the polynomial that corresponds to the $r \times r$ minor given by the bottom $r$ rows and the columns $\sigma_1, \ldots, \sigma_r$. Then the algebra generated by all of the $f_\sigma$ (something like a Plücker algebra) is supposedly exactly the algebra of $N$-invariants.

It's not hard to convince yourself that all the $f_\sigma$s are $N$-invariant, but how can you show that there's nothing else? Speyer tantalizingly suggests that there are various arguments, and that Theorem 14.11 of Miller-Sturmfels provides one, but I can't figure out how it does that. For what it's worth, the theorem says that the $f_\sigma$s form a sagbi basis for the Plücker algebra with any term order where the leading term of each $f_\sigma$ is the diagonal (or antidiagonal) one. You can see the statement here; search inside for "diagonal or antidiagonal". (Keep in mind that Miller-Sturmfels uses top-justified minors, so the Plücker algebra there is invariant by left translation by the lower-triangular unipotents. but I don't think that changes things much.)

Any ideas would be much appreciated! I'd love to understand how Theorem 14.11 might be used to prove that the $f_\sigma$s generate the full algebra of $N$-invariants. I'm also terribly curious about other possible arguments.

  • $\begingroup$ Just wanted to say that I have seen this question, and I now realize that Miller-Sturmfels proves that the plucker coordinates are a SAGBI basis for the ring generated by the Plucker coordinates, not for the ring of N-invariant polynomials, and the latter may a priori be larger. This will take some thought to fix, and I don't have time to give it that thought. But here should be the basic ideas: $\endgroup$ – David E Speyer Nov 1 '11 at 12:07
  • $\begingroup$ Let $R$ be the ring of N-invariant functions. By general SAGBI nonsense, it is enough to show that, for any $f \in R$, the leading monomial of f is divisible by the leading monomial of a Plucker coordinate. Suppose not. Then Hall's marriage theorem imposes strong restrictions on what the leading monomial of f can look like -- strong enough that we should be able to show f is not N invariant. $\endgroup$ – David E Speyer Nov 1 '11 at 12:09
  • $\begingroup$ I'm hoping someone else will finish the work here. In case it helps, I've plucked down a bounty for an elementary answer which proves this without assuming the representation theory of $GL_n$ as known. $\endgroup$ – David E Speyer Nov 1 '11 at 12:11
  • $\begingroup$ @DavidSpeyer: Thank you! -- for confirming that I'm not crazy, for suggesting an argument, and for experimenting with the bounty (oh, and for the original post, of course!). I'm getting a sense for the kind of combinatorial messing around an elementary argument would require. It's turning out that that's enough for me for the moment... $\endgroup$ – sibilant Nov 11 '11 at 5:43

Ok, here is an argument that I think will work. Let $k = \mathbb C$.

In that same blog post, Speyer essentially argues that the algebra of $N$-invariants-by-left-translation -- let's call it $A$ -- decomposes as a direct sum of all the irreducible polynomial representations of $G = GL_n({\mathbb C})$, where the action of $G$ is now translation on the right.

(He actually argues that the left $N$-invariants of $\mathcal O[G] = {\mathbb C}[x_{ij}, \det^{-1}]$ decomposes as the direct sum of all the algebraic representations of $G$. But if you look at how each $V$ gets to sit inside ${\mathcal O}[G]$ -- by its coefficients -- it's clear that each irreducible one is in a homogeneous component and that the polynomial ones are in the polynomial part of ${\mathcal O}[G]$, which is ${\mathbb C}[G]$.

This part of his argument is so lovely; let me just summarize it. By Peter-Weyl, ${\mathcal O}[G] = \bigoplus V^* \otimes V$, as representations of $G \times G$; the direct sum is over all the irreducible algebraic representations of $V$. Incidentally, the map from $V^* \otimes V$ to ${\mathcal O}[G]$ is just $\lambda \otimes v \mapsto \{g \mapsto \lambda(gv)\}$, so to its coefficients. The action of $G \times G$ on ${\mathcal O}[G]$ is by left-right translation: $\big({}^{(g, h)}f\big)(x) = f(g^{-1} x h)$. If you take the $N \times 1$ invariants on each side, you get exactly the left-translation $N$-invariants on the ${\mathcal O}[G]$ side. And on the other side, since $V^*$ is irreducible and has a unique highest-vector line, you get $\mathbb C \otimes V = V$ in each direct summand. So in ${\mathcal O}[G]$, you realize each irreducible $V$ under right translation inside the highest-weight vectors of $V^*$ under left translation! Isn't that great?)

But anyway, back to matter at hand. Let $P \subset \mathbb C[x_{ij}]$ be the Plucker subalgebra as described above. We want to see that $A = \mathbb C[x_{ij}]^N = \bigoplus_{V\ {\rm poly}} V$ is the same as $P$; we see by computation that $P \subset A$. Construct all the irreducible polynomial representations of $G$ as Schur modules over the standard representation. This is done in chapter 8 of Fulton's "Young Tableaux", for example. At the end of section 8.1, Fulton also shows that each $V = V^{\lambda}$ maps to $P$, in a $G$-equivariant manner, where the action on $P$ is right translation, just as we were thinking about it.

So I guess that does it, by Schur's lemma and dimension tracking in each homogeneous component, or maybe also the semisimplicty of nice algebraic representations of $G$: $$A = \bigoplus V \hookrightarrow P \subset A.$$

However, I still don't see how this has to do with sagbi bases or Theorem 14.11 of Miller-Sturmfels. And it's a bit unsatisfying to have to appeal to some other construction of all the algebraic representations of $GL_n$...


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.