Subject: Doug Ravenel's second invariant theory question From: "Campbell, Eddy" Date: Tue, 2 Aug 2005 13:59:27 -0230 Hi Doug et al. David Wehlau and I have been having some fun trying to answer Doug Ravenel’s latest question on the Top-List quoted below my signature here. Let G be the symmetric group \Sigma_n. Doug wishes to count the number of times a given G-set occurs in the symmetric algebra of the standard representation of G acting as permutations of n variables, i.e., the multiplicity of a given G-set occurring among the monomials of degree d as d varies 0 to infinity. The G-set consisting of the n variables is here referred to as \rho. First of all, because the action we are examining is on this set of monomials, the stabilizers of such are all Young subgroups, hence the G-sets that arise in the decomposition he seeks (of the symmetric algebra of \rho) are all of the form G/H for H a Young subgroup. Of course, the conjugacy classes of Young subgroups are in one-to-one correspondence with the partitions of n. Let’s call the number of such partitions P(n). Folks working in this area use the “dominance” partial order on partitions (by partial sums of entries, or we could say that one partition is smaller than another if it is a refinement of the other). Under the correspondence, this is the partial order induced by inclusion on the conjugacy classes of Young subgroups. David and I have been fooling around with the sub-ring of the Burnside ring A(G) constructed from these. If we choose such a subgroup H and consider the corresponding G-set G/H, then we want to find the “Poincare” series P_H (t) = \sum_{d=0}^{\infty} (multiplicity of G/H in monomials of degree d) t^d for each H. This series lives in A(G)[[t]] There are also exactly P(n) irreducible representations of G over any sufficiently large field. Take one of these, call it W, and consider the Poincare series Q_W(t) = \sum_{d=0}^{\infty} (multiplicity of W in the span of the monomials of degree d) t^d This is the Poincare series that can be calculated via Molien’s Theorem. That is, Q_W (t) = (1/|G|) \sum_{g \in G} \frac{\chi_W(g)}{det(Id_\rho – tg)} where \chi_W is the character associated to W. This series lives in Green(G)[[t]], where Green(G) denotes the Green ring of G. There is a P(n) by P(n) matrix K whose rows are indexed by the irreducible representations W and whose columns are indexed by the G-sets G/H. The entry in row W and column G/H is the multiplicity of W in the representation space with basis G/H. Let’s denote the entries by K(W,H). In the literature, this matrix is called the Kostka matrix (see “The Symmetric Group” by Sagan, page 85, GTM 203 for example. Our definition is equivalent to the usual definition, see Sagan’s theorem 2.11.2). Notably, if the ordering of both the rows and the columns of K is compatible with the dominance partial order on partitions then the Kostka matrix is upper triangular with 1’s along the diagonal. Other features are that the row corresponding to the trivial representation consists of all 1’s, and that the column corresponding to the “regular” G-set has entries equal to the dimensions of the irreducible representations, in other words, that K(W,trivial group) = dim W. Consequently, if we write P for the column vector of Poincare Series of G-sets indexed by conjugacy classes of Young subgroups and Q for the column vector of Poincare series of irreducible representations, then we have K P = Q. The equation takes place in Green(G)[[t]]. We know Q from Molien’s theorem and we seek P. There are combinatorial methods for computing K(W,H): we found such in Sagan, see above, or in James and Kerber, “Representation of the Symmetric Group” Young’s rule on page 89, Encyclopaedia of Mathematics, V16). Hence, David and I note that we can answer Doug’s question by using this matrix, back substitution and Molien’s theorem. Here is a simple example when n = 3. There are the three G-sets corresponding to Young subgroups. These were called 1, \rho and \sigma by Doug. We have 1 = the G-set coming from H = \Sigma_3, \rho = the G-set coming from H = \Sigma_2 x \Sigma_1, and \sigma = the G-set coming from H = the trivial group = \Sigma_1 x \Sigma_1 x \Sigma_1. P is the transpose of (P_{\Sigma_3}(t), P_{\Sigma_2}(t), P_{trivial group})(t)) There are three irreducible representations over the complex numbers, let’s call them \theta (the trivial representation), \bar \rho (the 2 dimensional representation) and \det (the non-trivial 1-dimensionlal representation). Q is the transpose of (Q_{\theta}(t), Q_{\bar \rho}(t), Q_{\det}(t)). Let us write D = [(1-t)(1-t^2)(1-t^3)]. Then by Molien’s theorem we have: Q_{\theta}(t) = 1/D, the Poincare series of the ring of invariants, Q_{\bar \rho} (t) = [t + t^2]/D Q_{\det}(t) = t^3/D Here is the matrix K (which we computed using character theory): 1 \rho \sigma \theta 1 1 1 \bar \rho 0 1 2 \det 0 0 1 The last row of K shows that Q_{det} (t) = P_1 (t) = P_{trivial group} (t) The middle row of K shows that P_{\Sigma_2}(t) + 2P_1(t) = = Q_{\bar \rho} (t) which when we back substitute gives P_{\Sigma_2}(t) = Q_{\bar \rho}(t) – 2Q_{det} Now the first row shows that P_{\Sigma_3}(t) + P_{\Sigma_2}(t) + P_1(t) = Q_{\theta} which when we back substitute gives P_{\Sigma_3}(t) = Q_{\theta} – Q_{\bar \rho} + Q_{\det} This finishes our calculation of the answer to Doug’s question in this case. That is, P_1(t) = Q_{\det}(t) = t^3/D P_{\Sigma_2}(t) = [t + t^2 - 2t^3]/ D P_{\Sigma_3}(t) = [1 – t - t^2 + t^3]/D David and I did take a look at trying to extend this technique to answer analogues of Doug’s question for permutation representations of G other than \rho. However, these results are particular to \rho. In summary, the method described here is better than the method provided by Webb for answering Doug’s question, but the method provided by Webb is better for any other analogue of Doug’s question. The difficulty with the approach coming from Webb’s theorem is that we are required to examine the full subgroup lattice of G and indeed to perform a double sum over it. Our approach requires the computation of the Kostka matrix of size P(n) x P(n) and back substitution. Regards, Eddy H E A Campbell, Vice-President (Academic) Memorial University of Newfoundland A1C 5S7 709-737-8246 (O) eddy@mun.ca Subject: Another question on invariant theory From: "Douglas C. Ravenel" Date: Thu, 21 Jul 2005 11:39:12 -0400 (EDT) Here is another question about invariant theory. Let the symmetric group $G=\Sigma_{p}$ act on the polynomial ring (over the integers) on $p$ variables in the usual way. Then it acts on the set of monomials of degree $n$ via a permutation representation $\rho_{n}$. This information can be encoded in a Poincar\'e series with coefficients in the Burnside ring $A[G]$, \begin{displaymath} g (t) = \sum_{n \ge 0}\rho_{n}t^{n}. \end{displaymath} For $p=3$, $A[G]$ is generated by two elements, the regular representation $\rho $ of degree 3, and the standard representation $\sigma $ of degree 6. These are subject to the relations $\rho^{2}=\rho +\sigma$, $\rho \sigma =3\sigma $, and $\sigma^{2}=6\sigma$. It is easy to compute $g (t)$ for $p=3$, and the answer is \begin{displaymath} 1/g (t) = (1-t) (1-\alpha t+\beta t^{2}), \end{displaymath} \noindent where $\alpha =\rho -1$ and $\beta =\sigma -2\rho +1$. My question is: Is there a similar formula for general $p$? Is there a conceptual (rather than computational) proof? Doug