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For any matrices $A$,$B$ in $SL(2,\mathbb{C})$, let $M(A,B)$ be the 4 by 4 matrix whose columns are matrices $I$,$A$,$B$,$AB$. Then it is not hard to verify that $det M(A,B)= 2- tr[A,B]$. Is there an analogous identity for matrices in $SL(3,\mathbb{C})$? I suppose if there is one, then $M(A,B)$ would be $9$ by $9$ matrix whose columns are matrices $I$,$A$,$A^{2}$,$B$,$AB$,$A^{2}B$, $B^{2}$, $AB^{2}$, $A^{2}B^{2}$.

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The identity in $SL(2;\mathbb C)$ is false: $M(I_2,B)$ has two equal columns and therefore its determinant is $0$, whereas the formula $2-Tr[I_2,B]$ gives $2$. – Denis Serre Feb 18 '11 at 14:34
The trace of a commutator is $0$ in general... something is wrong here. – darij grinberg Feb 18 '11 at 14:50
And the formula fails for homogeneity reasons too. – darij grinberg Feb 18 '11 at 14:51

Given $X_1,\dots,X_{n^2} \in M_n(\mathbb{C})$, one can form the $n^2 \times n^2$ matrix whose $i$th column is the entries of $X_i$ in some fixed order. The determinant of this matrix (defined up to sign) is called the discriminant of $X_i, \dots , X_{n^2}$ and is denoted $\mathcal{D}(X_i)$. It is a matrix invariant, which means that $\mathcal{D}(X_i) = \mathcal{D}(UX_iU^{-1})$ for any $U \in GL(n ,\mathbb{C})$. The first fundamental theorem of matrix invariants says that any matrix invariant can be expressed in terms of traces. The expression is not unique, since there are "trace identities" which are identically zero on $M_n(\mathbb{C})$. An explicit formula for $\mathcal{D}$ in terms of traces is given on p.46 of my book "The Polynomial Identities and Invariants of $n\times n$ Matrices".

It happens that for $2 \times 2$ matrices, $\mathcal{D}(I,A,B,AB)I = \pm(AB - BA)^2$. In other words, $\mathcal{D}(I,A,B,AB)$ is the value of a central polynomial for $2 \times 2$ matrices. As far as I know, it is not known if $\mathcal{D}(A^iB^j \mid 0 \leq i,j \leq n-1)$ is the value of a central polynomial for $n \geq 3$.

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