Why is every $N$-invariant polynomial function on $n\times n$ matrices in the Plücker algebra? 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. 
 A: 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 $GL_n$... 
