Polarizations generate the ring of invariants?

The symmetric group $S_n$ acts on $\mathbb R^n$ by permuting the coordinates and the ring of polynomial invariants is generated by the elementary symmetric polynomials. If we restrict the action to the alternating group $A_n$ then the invariant ring is generated by elementary symmetric polynomials along with the discriminant $\prod_{i <j}(x_i-x_j)$. If we take $m$ copies of $\mathbb R^n$ i.e., $\oplus_m \mathbb R^n$, then $S_n$ acts naturally and a theorem of Hermann Weyl says that the ring of invariants is generated by the polarizations of the elementary symmetric polynomials. Now if we restrict the action to $A_n$ then what is a minimal generating set for the ring of invariants ? Is it generated by the polarizations of the above invariants ?

• you might want to check that this will give you enough invariants for $n=3$ (i.e. for the cyclic group of order 3). It is known that you will need $\binom{m-1}{3}$ degree 3 invariants in any generating set. Dec 31 '14 at 10:41
• more precisely, the degree 3 homogeneous component has at least this dimension, $\binom{m-1}{3}$. Dec 31 '14 at 13:42
• @Ram: Added a tag, plus minor edits. Maybe add a modern reference to Weyl's work? Dec 31 '14 at 13:52

To complete Peter's answer, I would like to give a concrete counterexample: by computing (e.g. using GAP) the Molien series for $S_3$ (resp. for $A_3$) acting naturally on $\mathbb{R}^3\oplus \mathbb{R}^3$, one sees that already in degree 2 the dimension of the space of $A_3$-invariant polynomials is bigger (by 1) than the one of $S_3$.

This extra degree 2 invariant cannot come from polarization of $A_3$-invariants in the natural action, as the difference with $S_3$ only starts occurring in degree 3.

(I also checked for $n=5$ - there the difference starts in degree 6, again too low for polarization to kick in).

• In this case we can say exactly what the invariant rings are: complexify and strip out the 2-dimensional trivial submodule, then pick a basis so that $(1\; 2\; 3)$ sends $(a,b,s,t)\in{\mathbb C}^4$ to $(\zeta a,\zeta^{-1} b,\zeta s,\zeta^{-1}t)$ (where $\zeta=e^{2\pi i/3}$) and $(1\; 3)$ swaps $a$ and $b$, resp. $s$ and $t$. Then the invariant ring for the $S_3$-action is generated by $ab,st,at+bs$ in degree 2 and $a^3+b^3,a^2s+b^2t,as^2+bt^2,s^3+t^3$ in degree 3. Meanwhile, the invariants for the $A_3$-action are generated by $ab,st,bs,at,a^is^{3-i},b^it^{3-i}$ with $0\leq i\leq 3$. Jan 1 '15 at 23:12
• In any way can we conclude at least that the degrees of the invariants in a generating set is $1,2, \cdots, n$ and $n \choose 2$, i.e., same as the degrees of the basic invariants ?
– Ram
Jan 3 '15 at 7:14
• in this toy case, yes, but in general there is no reason for this coincidence. Jan 3 '15 at 8:21

Maybe the following paper helps:

• Mark Losik, Peter W. Michor, Vladimir L. Popov: On Polarizations in Invariant Theory , J. Algebra 301 (2006), 406-424. (pdf)
• there is a comment there saying that it is not so easy to find counterexamples to failure of polarizations to generate the ring of invariants (equation (2) in the text). It seems that noone tried to compute the $Alt(3)$ example explicitly. :-) Jan 3 '15 at 8:26

I think the paper

Gobel, M. (1995). Computing Bases for Permutation-Invariant Polynomials. Journal of Symbolic Computation 19, 285{291 MR 96f:13006

should be helpful for finding a small set of generators. It doesn't deal with the polarization aspect of your question though - just the fact that you have a permutation representation.