Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
    
Sie scheinen auch ein Nachtvogel zu sein! –  Georges Elencwajg Jul 19 '10 at 0:16
1  
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. –  Victor Protsak Jul 19 '10 at 7:20
    
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. –  darij grinberg Jul 19 '10 at 8:25
1  
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 $\text{char} K=0.$) –  Victor Protsak Jul 19 '10 at 8:55
1  
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. –  Victor Protsak Jul 19 '10 at 9:02

1 Answer 1

up vote 4 down vote accepted

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.

share|improve this answer
    
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... –  darij grinberg Jul 19 '10 at 14:47

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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