# invariant polynomials on 3 by 3 matrices

Hi there:

A freshman level question here. A polynomial p on the entries of n by n matrices is said to be invariant if p(A)=p(sAs^{-1}) for every invertible matrix s. For example, for 3 by 3 matrices, tr(A^2) and (trA)^2 are two linearly independent invariant polynomials of degree 2. But are there other degree 2 invariant polynomials on 3 by 3 matrices? Any reference is very appreciated.

Ron

-
No. At least not if you ware working over a field of characteristic $\neq 2$. math.unibas.ch/~kraft/Papers/KP-Primer.pdf §2.3 Proposition. –  darij grinberg Mar 12 '11 at 17:01
Thanks! It is very helpful. Ron –  Ron Yang Mar 15 '11 at 0:03
I guess my comment is worth expanding into an answer. Over an arbitrary field $k$, an invariant polynomial on $\mathcal{M}_n(k)$ extends to an invariant polynomial on $\mathcal{M}_n(\bar{k})$. Since the diagonalizable matrices are Zariski dense in $\mathcal{M}_n(\bar{k})$, such a polynomial is determined by what it does to diagonal matrices, and it must be a symmetric polynomial of the entries of any diagonal matrix. Conversely, over an arbitrary field (in fact over an arbitrary commutative ring $R$) the elementary symmetric polynomials (all of which are coefficients of the characteristic polynomial, hence all of which really do come from invariant polynomials) generate the ring of symmetric polynomials in $n$ variables.
Hence the invariants of degree $d$ are precisely the symmetric polynomials of degree $d$ over $k$. A basis of the symmetric polynomials of degree $2$ is always given by $\{ e_1^2, e_2 \}$ where $e_i$ is the $i^{th}$ elementary symmetric polynomial. When $\text{char}(k) \neq 2$ we can instead use $\{ p_1^2, p_2 \}$ where $p_i$ is the $i^{th}$ power sum, since $p_1 = e_1$ and $p_2 = \frac{e_1^2 - e_2}{2}$, but if $\text{char}(k) = 2$ this change of coordinates is not well-defined.
Wikipedia should have proofs of the statements I made above about symmetric polynomials; alternately, see for example Chapter 7 of Stanley's Enumerative Combinatorics Vol. II. Note that on the one hand this result constrains what the possible characters of a "nice" representation of $\text{GL}_n$ can look like, and on the other hand suggests that the characters of the "nice" irreducible representations of $\text{GL}_n$ form a distinguished basis for the symmetric polynomials. These are precisely the Schur polynomials.
This raises an interesting question of what happens over $\mathbb Z_2$. –  Jim Conant Mar 13 '11 at 3:03