# Ring of invariants for $n$-tuples of Lie algebras

$$\DeclareMathOperator\GL{GL}\DeclareMathOperator\M{M}\DeclareMathOperator\Tr{Tr}$$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 $$k=1$$?

• 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. Apr 23, 2021 at 13:09
• How does the determinant (for $k = 1$) arise in the framework that you describe? Apr 23, 2021 at 13:10
• @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) .. Apr 23, 2021 at 13:16
• @LSpice I thought you could recover all the invariants like that , see for example this related question mathoverflow.net/questions/25439/… Apr 23, 2021 at 13:22
• @Malkoun, you are welcome ! :) Apr 26, 2021 at 16:36

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.

• 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. Apr 26, 2021 at 16:21
• 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) ? Apr 26, 2021 at 16:34
• 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). Apr 26, 2021 at 16:44
• 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 $f:G\to\mathbb{C}$. Apr 26, 2021 at 16:50
• @ Sean Lawton, thanks once again Apr 26, 2021 at 17:06

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)

• Many thanks! Prof Procesi Apr 27, 2021 at 2:05