Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

share|improve this question
    
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

1 Answer 1

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.

share|improve this answer
    
This raises an interesting question of what happens over $\mathbb Z_2$. –  Jim Conant Mar 13 '11 at 3:03

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.