4
$\begingroup$

$\DeclareMathOperator\GL{GL}\DeclareMathOperator\M{M}\DeclareMathOperator\Tr{Tr}$Consider the diagonal action of $\GL(n,\mathbb{C})$ on the variety of $k$-tuples of matrices, $\M_{n\times n}(\mathbb{C})^k$ through conjugation. Then it is known that the ring of invariants $\mathbb{C}[\M_{n\times n}(\mathbb{C})^k]^{\GL(n,\mathbb{C})}$ is generated by the functions obtained by first evaluating a non-commutative polynomial on the tuple of matrices and then applying the trace to the resulting matrix.

So for example if $k=2$, we need to look at functions like $(A,B) \mapsto \Tr(AB)$, $(A,B) \mapsto \Tr((AB)^2 A)$, etc. Details can be found in:

The invariant theory of $n \times n$ matrices, Claudio Procesi, Advances in Mathematics, Volume 19, Issue 3, March 1976, Pages 306-381

Now assuming this, suppose we are looking for the ring of invariants of tuples as before but now for the diagonal action of a reductive group $G$ on the variety of tuples $\mathfrak{g}^k$.

Then will it be true that the ring of invariants in this situation is also generated as above by first evaluating non-commutative polynomials and then taking the trace, but now we do this on the matrices obtained by considering various representations of $\mathfrak{g}$ (or $G$), just like what happens in the case of $k=1$?

$\endgroup$
10
  • $\begingroup$ I found the repeated inline use of \underbrace very hard to read, so edited them to exponents (as well as adding some appropriate uses of \DeclareMathOperator). I hope that this was all right. $\endgroup$
    – LSpice
    Apr 23, 2021 at 13:09
  • $\begingroup$ How does the determinant (for $k = 1$) arise in the framework that you describe? $\endgroup$
    – LSpice
    Apr 23, 2021 at 13:10
  • 1
    $\begingroup$ @LSPice, determinant? determinant of an $n\times n$ matrix $A$ can expressed in terms of $\{Tr(A),\ldots,Tr(A^n)\}$. .. for example for $2 \times 2$, it is if i am not mistaken $Tr(A)^2 - Tr(A^2)$ (off by a sign or a constant could be) .. $\endgroup$
    – skeptic
    Apr 23, 2021 at 13:16
  • 1
    $\begingroup$ @LSpice I thought you could recover all the invariants like that , see for example this related question mathoverflow.net/questions/25439/… $\endgroup$
    – skeptic
    Apr 23, 2021 at 13:22
  • 1
    $\begingroup$ @Malkoun, you are welcome ! :) $\endgroup$
    – skeptic
    Apr 26, 2021 at 16:36

2 Answers 2

3
$\begingroup$

For some Lie groups you will need additional invariants. For example, for the even orthogonals you will need the Pfaffian in addition to the Trace.

See for example:

Invariant theory of special orthogonal groups, Helmer Aslaksen, Eng-Chye Tan, Chen-bo Zhu Pacific J. Math. 168(2): 207-215 (1995).

However, you can already see this with one copy of $\mathfrak{so}(2,\mathbb{F})$ and $\mathrm{SO}(2,\mathbb{F})$ acting by conjugation. For in that case the Lie algebra is $\left\{\left(\begin{array}{cc}0&a\\-a&0\end{array}\right)|\ a\in \mathbb{F}\right\}$ and the conjugation action is trivial. In this case the trace is identically 0 and the Pfaffian is the only useful invariant.

In case you are interested, the Lie group version of this question was addressed by myself and Sikora here:

Varieties of Characters, Sean Lawton & Adam S. Sikora Algebras and Representation Theory volume 20, pages 1133–1141(2017).

Remark 1: In the above answer, I assumed the OP wanted a fixed representation of $G$. As noted in the answers to this MO question, if you vary over all representations, then for $k=1$ the answer is yes. But for $k\geq 2$ the answer appears to me to still be no. See Theorem 1 in Sikora's paper SO(2n,C)-character varieties are not varieties of characters; it is not exactly the same thing, but it seems to imply the result (see the comments for a strategy to fill in the details).

Remark 2: As noted already by Professor Procesi, his work cited by the OP implies the answer is yes for the Lie groups $\mathrm{GL}_n$, $\mathrm{O}_n$, and $\mathrm{Sp}_{2n}$ (by restricting $k$-tuples of generic matrices to the subvariety $\mathfrak{g}^k$). From this one can also deduce that the answer is yes for the Lie groups $\mathrm{SL}_n$ and $\mathrm{SO}_{2n+1}$.

Remark 3: As I said in the comments, I believe the work of G. Schwarz probably implies the answer is also yes for a representation of $G_2$ (he also addresses some Spin groups). The question for the other exceptional groups is open as far as I know. If I were to guess, I would say it is probably true for all of them except $E_6$ which has additional symmetry as in the case of $\mathrm{SO}_{2n}$ (where the answer is no as I already indicated).

Remark 4: Once one knows the answer for a given $G$, then one also knows it for finite central quotients of $G$ (which is how one goes from the orthogonal case to the special orthogonal case for odd $n$). Also, if one knows the answer for $G$ and $H$, then one also knows it for $G\times H$. Putting these observations together with the known cases reduces the entire problem down to the (simply connected forms of the) exceptional groups and the spin groups.

$\endgroup$
8
  • 1
    $\begingroup$ Yes. You are comparing the $G$-invariants of $\mathrm{Hom}(F_k,G)\cong G^k$ versus $\mathfrak{g}^k$. The invariant rings are "morally'' the same. $\endgroup$ Apr 26, 2021 at 16:21
  • 1
    $\begingroup$ I see .. thanks .. this is nice .. if I may ask .. is it very obvious how to pass from $G$ to $\mathfrak{g}$ (or the other way) ? $\endgroup$
    – skeptic
    Apr 26, 2021 at 16:34
  • 1
    $\begingroup$ There is something to do here. First understand the simple case of $\mathrm{SL}(n,\mathbb{C})$. In that case the invariants you care about are those of $nxn$ traceless matrices. So there are two steps to related the two invariant rings. Step 1: add in the traces of single generic matrices; and Step 2: quotient by $det-1$ for each generic matrix. In general, the relationship is in one direction you take the tangent space of the identity in the $G$-character variety of $F_k$ to obtain $\mathfrak{g}^k//G$ (and invariants of the former give invariants of the latter). $\endgroup$ Apr 26, 2021 at 16:44
  • 1
    $\begingroup$ To go the other way around, you need to use a "variation function" $F:G\to \mathfrak{g}$ which is conjugation invariant. This will turn $G$-invariants of $\mathfrak{g}^k$ into $G$-invariants of $G^k$. This is essentially "integration". Again, a simple example is $\mathrm{SL}(n,\mathbb{C})$ which is $X\mapsto X-(tr(X)/n)I$. In general, you take an orthogonal structure $B$ on $\mathfrak{g}\times \mathfrak{g}$ which always exists for reductive $G$ (killing form if semisimple) and solve: $B(F(A),X)=(d/dt)|_{t=0}f(AexptX)$ for any $G$-invariant $f:G\to\mathbb{C}$. $\endgroup$ Apr 26, 2021 at 16:50
  • 1
    $\begingroup$ @ Sean Lawton, thanks once again $\endgroup$
    – skeptic
    Apr 26, 2021 at 17:06
4
$\begingroup$

The orthogonal and symplectic case are treated in my old paper as well, there is a paper by Gerry Schwarz on G_2, always in characteristic 0, the theory in positive characteristic is presented in
The Invariant Theory of Matrices – 30 dicembre 2017 di Corrado De Concini (Autore), Claudio Procesi (Autore)

the case of 2\times 2 matrices is very special and can be described in full detail see

Rings With Polynomial Identities and Finite Dimensional Representations of Algebras – 30 dicembre 2020 Eli Aljadeff (Autore), Antonio Giambruno (Autore), Claudio Procesi (Autore), Amitai Regev (Autore)

$\endgroup$
1
  • $\begingroup$ Many thanks! Prof Procesi $\endgroup$
    – skeptic
    Apr 27, 2021 at 2:05

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.