TL;DR version.

Given a commutative ring $\mathbf{k}$ and $n+m$ "generic" vectors $\mathbf{x}_1, \mathbf{x}_2, \ldots, \mathbf{x}_n, \mathbf{y}_1, \mathbf{y}_2, \ldots, \mathbf{y}_m$ in $\mathbf{k}^k$ (really over a polynomial ring, not over $\mathbf{k}$), how can we describe the polynomial relations that hold between the $nm$ scalars $\left(\mathbf{x}_i\right)^T \mathbf{y}_j$ ? Is the ideal of these relations spanned by the $\left(k+1\right)\times\left(k+1\right)$-determinants that one would expect?

Long version.

Here is a classical result.

Let $\mathbf{k}$ be a commutative ring with $1$. Let $n$ and $m$ be two nonnegative integers. Let $A$ be the polynomial ring $\mathbf{k}\left[z_{i,j} \mid 1\leq i\leq n, \ 1 \leq j \leq m\right]$, and let $B$ be the polynomial ring $\mathbf{k}\left[x_1,x_2,\ldots,x_n, y_1,y_2,\ldots,y_m\right]$. We define a $k$-algebra homomorphism $\phi : A \to B$ by sending each $z_{i, j}$ to $x_i y_j$. Then, $\operatorname{Ker}\phi$ is the ideal of $A$ generated by all elements of the form $z_{a, b} z_{c, d} - z_{a, d} z_{c, b}$ with $1 \leq a \leq n$, $1 \leq b \leq m$, $1 \leq c \leq n$ and $1 \leq d \leq m$.

A nice proof that Bjorn Poonen suggested to me in 2011 can be found on m.se (see also the original AoPS thread).

Now, let me attempt to generalize this, inspired by a suggestion from Alexander Chervov on a rather different problem in MO question #102874.

Let $\mathbf{k}$ be a commutative ring with $1$. Let $k$, $n$ and $m$ be three nonnegative integers. Let $A$ be the polynomial ring $\mathbf{k}\left[z_{i,j} \mid 1\leq i\leq n, \ 1 \leq j \leq m\right]$ (this is the same $A$ as before), and let $B_k$ be the polynomial ring $\mathbf{k}\left[\left(x_{i, \ell} \mid 1 \leq i \leq n,\ 1 \leq \ell \leq k \right) \cup \left(y_{j, \ell} \mid 1 \leq j \leq m, \ 1 \leq \ell \leq k\right)\right]$ (this is a polynomial ring in $nk+mk$ variables). For every $1 \leq i \leq n$, we let $\mathbf{x}_i$ be the vector $\left(x_{i,1}, x_{i,2}, \ldots, x_{i,k}\right)^T \in \left(B_k\right)^k$. For every $1 \leq j \leq m$, we let $\mathbf{y}_j$ be the vector $\left(y_{j,1}, y_{j,2}, \ldots, y_{j,k}\right)^T \in \left(B_k\right)^k$. We define a $k$-algebra homomorphism $\phi_k : A \to B_k$ by sending each $z_{i, j}$ to $\left(\mathbf{x}_i\right)^T \mathbf{y}_j$. The question is what $\operatorname{Ker}\left(\phi_k\right)$ is.

I have a guess: Let $W_k$ be the ideal of $A$ formed by all determinants of the form

$\det\left( \left( \left( \mathbf{x}_{i_a} \right)^T \mathbf{y}_{j_b} \right) _{1 \leq a \leq k+1, \ 1 \leq b \leq k+1} \right)$

ranging over all pairs $\left(\left(i_1,i_2,\ldots,i_{k+1}\right), \left(j_1,j_2,\ldots,j_{k+1}\right)\right) \in \left\{ 1, 2, \ldots, n \right\}^k \times \left\{ 1, 2, \ldots, m \right\}^k$. (Of course, the ideal does not change if we restrict these pairs by requiring $i_1 < i_2 < \cdots < i_{k+1}$ and $j_1 < j_2 < \cdots < j_{k+1}$.)

I suspect that $\operatorname{Ker}\left(\phi_k\right) = W_k$. At least $W_k \subseteq \operatorname{Ker}\left(\phi_k\right)$ is easy to prove, so the guess looks reasonable.

This seems to be related to semistandard Young tableaux -- after all, there is a bitableau basis of $A$, of which I know nothing beyound the definition; and there is a straightening rule that reduces a monomial in $A$ modulo the ideal $W_k$, giving a sum of monomials which contain no products $z_{i_1, j_1} z_{i_2, j_2} \cdots z_{i_k, j_k}$ with $i_1 < i_2 < \cdots < i_k$ and $j_1 < j_2 < \cdots < j_k$ (a condition that seems to make more sense in terms of the RSK tableau corresponding to the monomials). But I am not sure if this straightening rule is confluent and, generally, whether this is a good way to move forward. The whole thing could also be the second fundamental theorem for some algebraic group, whence the invariant-theory tag.

  • $\begingroup$ I think you want the second fundamental theorem of classical invariant theory for vectors and covectors, the group being $GL_n$. I don't have it in front of me now but I think Weyl's book on classical groups has that. You might also try the recent textbooks by Procesi or Nolan and Wallach. $\endgroup$ – Abdelmalek Abdesselam Apr 4 '15 at 22:43
  • $\begingroup$ @AbdelmalekAbdesselam: Good reference! Procesi's "Lie Groups", on page 536, says that "the double standard tableaux in these elements $\overline{x}_{ij}$ with at most $m$ columns are a basis of the ring $A_m$", which at least sounds close to what I am looking for. I'll need to make sure that it means what I think it means... $\endgroup$ – darij grinberg Apr 4 '15 at 22:56
  • $\begingroup$ Ah, even better, what I want is the Theorem in §8.1 of his Chapter 13. Care to post this as an answer? $\endgroup$ – darij grinberg Apr 4 '15 at 22:57
  • $\begingroup$ That's fine. If you got the answer you were looking for, this is all that matters. $\endgroup$ – Abdelmalek Abdesselam Apr 4 '15 at 23:34
  • $\begingroup$ Surely I would not mind something self-contained and elementary, but what you said is perfectly a good answer. $\endgroup$ – darij grinberg Apr 4 '15 at 23:42

This is the second fundamental theorem of classical invariant theory for $GL$ acting on vectors and covectors. This is Theorem 8.1 from Ch. 13 of the book Lie Groups: An Approach Through Invariants and Representations by C. Procesi (I still don't have it in front of me but I trust Darij on this).

Alternatively, it is Theorem 5.2.15 (SFT for $GL(n)$) in the book Representations and Invariants of the Classical Groups by Goodman and Wallach. I have that one on front of me, the old 1998 edition.

The problem is about finding equations for the variety of $n\times m$ matrices $Z$ which factor as $XY$ where $X$ is $n\times k$ and $Y$ is $k\times m$. The condition is that the rank is at most $k$. By elementary linear algebra the $(k+1)\times(k+1)$ minor determinants are set-theoretic equations, however Darij's question is about showing these are also ideal-theoretic equations and that needs a bit more work.

  • $\begingroup$ The appropriate reference in my edition of Goodman and Wallach (2009) seems to be Theorem 12.2.12. It actually does the ideal-theoretic equations, not just the zero locus. However, it being in a section on multiplicity-free spaces seems to suggest that the proof uses characteristic $0$ in a nontrivial way. Then again, this is a book that proves PBW using Ado's theorem, so I am not surprised by some roundaboutness :) $\endgroup$ – darij grinberg Apr 5 '15 at 20:24
  • $\begingroup$ Good to point that out. The two editions differ a lot as the authors added a lot of material in the new one. $\endgroup$ – Abdelmalek Abdesselam Apr 5 '15 at 20:31

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.