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$\DeclareMathOperator\GL{GL}\DeclareMathOperator\M{M}\DeclareMathOperator\Tr{Tr}$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  Claudio Procesi - The invariant theory of $n \times n$ matrices.

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 $k=1$?

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

$\DeclareMathOperator\GL{GL}\DeclareMathOperator\M{M}\DeclareMathOperator\Tr{Tr}$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. …. (The details can be found in Claudio Procesi - The invariant theory of $n \times n$ matrices.)

Now asssuming 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 noncommutative polynomials and then taking 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 $k=1$?

$\DeclareMathOperator\GL{GL}\DeclareMathOperator\M{M}\DeclareMathOperator\Tr{Tr}$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 Claudio Procesi - The invariant theory of $n \times n$ matrices.

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 $k=1$?

$\DeclareMathOperator\GL{GL}\DeclareMathOperator\M{M}\DeclareMathOperator\Tr{Tr}$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. …. (the details can be found in the following paper by Claudio Processi http://online.kottakkalfarookcollege.edu.in:8001/jspui/bitstream/123456789/974/1/1-s2.0-000187087690027X-main.pdf)

Now asssuming 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 noncommutative polynomials and then taking 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 $k=1$?

$\DeclareMathOperator\GL{GL}\DeclareMathOperator\M{M}\DeclareMathOperator\Tr{Tr}$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. …. (The details can be found in Claudio Procesi - The invariant theory of $n \times n$ matrices.)

Now asssuming 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 noncommutative polynomials and then taking 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 $k=1$?