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*](http://www.math.unibas.ch/~kraft/Papers/KP-Primer.pdf).


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"](http://web.mit.edu/~darij/www/algebra/KP-errata-web.pdf) for a proof). I am still wondering whether it holds for arbitrary $K$ and has a more elementary or combinatorial proof.



Left U_n-invariants of SL_n - an exercise in Kraft-Procesi
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.


