I am sorry for spamming MO with questions I have not thought about for more than 3 hours, but currently I am quite busy with preparing a talk on representations of $S_n$, and I don't want these to get lost. I hope this one is not quite as vague as the last one.

This here is an attempt to generalize Exercise I.21 in Kraft-Procesi, Classical Invariant Theory.

Let $K$ be a field - say, infinite, since we are going to do classical invariant theory. Let $K\left[\mathrm{SL}_n K\right]$ denote the $K$-algebra of polynomial functions on $\mathrm{SL}_n K$, which I define either as

$\left\lbrace f\mid_{\mathrm{SL}_n K} \ \mid \ f\in K\left[\mathrm{M}_n K\right]\right\rbrace$

or as $K\left[\mathrm{M}_n K\right]\diagup \left(\det-1\right)$ (proving the equivalence of these two definitions is not the matter, it's rather easy - even easier than Kraft and Procesi try to make one believe).

Now, the group $\mathrm{U}_n K$ of unipotent upper triangular matrices acts on $\mathrm{SL}_n K$ from the left. What is the invariant ring? It is easily seen that

$\det\left(\text{the submatrix formed by the intersection of the rows }i,i+1,...,n\text{ with the columns }j,i+1,i+2,...,n\right)$

is an invariant for any $i\geq j$. These generate the fraction field of the invariants, but do they also generate the ring of the invariants itself?

(The above-mentioned exercise is the above for $n=2$.)

Arguments using Victorian age methods (as opposed to Zariski-topological or other algebro-geometrical) would be particularly preferred.

EDIT: As Allen Knutson has pointed out, my question has a negative answer. However, the (larger) collection of determinants of the form

$\det\left(\text{the submatrix formed by the intersection of the rows }i,i+1,...,n\text{ with the columns }j_1, j_2, ..., j_{n-i+1}\right)$

for $1 < i \leq n$ and $1 \leq j_1 < j_2 < ... < j_{n-i+1} \leq n$ does generate the ring of invariants. When $K$ has characteristic $0$, this can be proven using the standard theory of highest-weight modules and multiplicity-free algebras explained in Kraft-Procesi (see my errata, "Page 9, Exercise 21" for a proof). I am still wondering whether it holds for arbitrary $K$ and has a more elementary or combinatorial proof.

  • $\begingroup$ Sie scheinen auch ein Nachtvogel zu sein! $\endgroup$ – Georges Elencwajg Jul 19 '10 at 0:16
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    $\begingroup$ Darij, I see no problem with many motivated questions on the same topic, but can you, please, reveal whether they are crucial for your upcoming talk? If not, perhaps, you can simply record them and think about them later? The algebra that you are asking about is the homogeneous coordinate ring of the flag variety of $SL_n,$ which is a rational variety, but not the projective space for $n\geq 3,$ hence the answer is negative. I recommend looking at Roger Howe's Schur lectures, Sec 5.6.4-5.6.5. Victorian age stipulation is met by the use of Lewis Carrol identity, if that's indeed his real one. $\endgroup$ – Victor Protsak Jul 19 '10 at 7:20
  • $\begingroup$ Thanks a lot. No, this is not relevant for my talk (at least not more than everything in invariant theory and representation theory is interrelated). Thanks a lot for the answer, although I am not quite sure I understand it. Is it really the flag variety of $SL_n$? My $U_n$ are the unipotent upper triangular matrices, whereas the flag variety of $SL_n$ should be something like $SL_n$ modulo left multiplication by all upper triangular matrices, am I right? Though probably the diagonal factors shouldn't matter too much. $\endgroup$ – darij grinberg Jul 19 '10 at 8:25
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    $\begingroup$ The flag variety itself is a projective variety $G/B$ (in your case, the group acts on the left, but it's easily seen to be equivalent). However, being a projective variety, it admits only constant functions. Instead, you need to consider the $\textit{homogeneous}$ coordinate ring, which is $A=K[G/U],$ with multigrading given by the action of $B/U=T$ ($A_{\lambda}$ is the unique copy of the simple highest weight $G$-module with highest weight $\lambda$, at least if $G=SL_n$ or $\text{char} K=0.$) $\endgroup$ – Victor Protsak Jul 19 '10 at 8:55
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    $\begingroup$ The set of invariants that you've exhibited has the right cardinality $\text{dim}G/U$ (in particular, they are algebraically independent) and the homogeneous coordinate ring of a projective $X$ is a free polynomial algebra if and only if $X$ is (biregularly) isomorphic to a projective space. $\endgroup$ – Victor Protsak Jul 19 '10 at 9:02

No, they don't. $U_n(K)$ is performing upward row operations, so any $m\times m$ minor that uses the last $m$ rows will be $U_n(K)$-invariant, e.g. any single bottom entry. You won't be able to generate those linear functions using your higher-degree functions. (Victor's disproof is nice too!)

What is true is that the invariant ring is generated by those $2^n-1$ many minors (corresponding to nonempty subsets of columns). One nice place to read about them is [Miller-Sturmfels], chapter 14, where they show e.g. that you can degenerate this invariant ring by replacing each minor by the product of its diagonal entries, obtaining the semigroup algebra of the cone of Gel'fand-Cetlin patterns.

  • $\begingroup$ Oh, thanks. I must have been totally blind because I knew of these $m\times m$ minors. Somehow I thought the straigthening laws were linear, or something like that... $\endgroup$ – darij grinberg Jul 19 '10 at 14:47

Your question (in char 0) has already been answered by Knutson and Victor Protsak. I just wanted to say that this holds in greater generality (char 0) but the method is not Victorian.

So, consider the algebraically closed field $K$ of char 0, and $G=SL_n$. Given an irreducible $V_{\lambda}$ with highest weight $\lambda$ (relative to the standard upper triangular subgroup $B=TU$where $T$ is the group of diagonals), we have the decomposition for the action of $G\times G$ on the coordinate ring $k[G]$

$$k[G]=\bigoplus V_{\lambda }^* \otimes V_{\lambda}.$$ Taking $U$ invariants on the left, we have

$$k[U\backslash G]=\bigoplus V_{\lambda},$$ i.e. every irreducible representation $V_{\lambda}$ occurs exactly once and is generated as a $G$ module by the vector $v_{\lambda}$ which is invariant under the group $V$ of $lower \quad triangular$ unipotent matrices. If $\lambda_1, \cdots, \lambda_{n-1}$ are the highest weights of the fundamental representations, and $\lambda =\sum a_i\lambda _i$ with $a_i\geq 0$ then clearly (by multiplicity one) $v_{\lambda}=\prod v_i^{a_i}$ is a product of powers of $v_{\lambda_1}, \cdots ,v_{\lambda _{n-1}}$. Hence the ring of invariants is generated by the functions spanning $V_{\lambda_1},\cdots V_{\lambda _{n-1}}$. Thus the ring of invariants is finitely generated.

The above proof works for any connected reductive group.

  • $\begingroup$ +1 but this looks a lot like the proof I have now added to my errata. Do you ever use the algebraic closedness of $K$, or is it just something you reflexively write? :) $\endgroup$ – darij grinberg Feb 2 '15 at 23:51
  • $\begingroup$ You are right: for $SL_n$ all representations are defined over the rationals and for other $G$, if you take the Chevalley form of $G$, then again all reps are defined over $\mathbb Q$.So, $K$ being alg closed is not necessary, only that $char K=0$. $\endgroup$ – Venkataramana Feb 3 '15 at 0:45
  • $\begingroup$ I should say $G$ is assumed to be a simply connected semi-simple Chevalley group $\endgroup$ – Venkataramana Feb 3 '15 at 16:48

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