Invariants of Symmetric group - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T07:21:39Z http://mathoverflow.net/feeds/question/91472 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/91472/invariants-of-symmetric-group Invariants of Symmetric group mark 2012-03-17T15:49:20Z 2012-03-18T17:35:42Z <p>The celebrated Chevalleyâ€“Shephardâ€“Todd theorem says that $\mathbb C[V]^{S_n}$ is a polynomial algebra and gives the generators of this algebra, where $V$ is the standard (or natural) representation of the symmetric group $S_n$. I am just curious to know for what other representations of $S_n$ the generators of this algebra is known ? When is this algebra a polynomial algebra ? </p> http://mathoverflow.net/questions/91472/invariants-of-symmetric-group/91477#91477 Answer by Mark Wildon for Invariants of Symmetric group Mark Wildon 2012-03-17T17:56:11Z 2012-03-17T22:15:55Z <p>Let $n \ge 7$. If $V$ is an irreducible representation of $S_n$ such that $\mathbb{C}[V]^{S_n}$ is a polynomial algebra then either $V$ is the trivial representation, the sign representation or the $(n-1)$-dimensional standard representation. </p> <p><i>Outline Proof</i>: Let $\rho : S_n \rightarrow \mathrm{GL}(V)$ be an irreducible representation affording the irreducible character $\chi^\lambda$ where $\lambda$ is a partition of $n$. Suppose that $V$ is $d$-dimensional. By the Chevalley-Shephard-Todd theorem, $\mathbb{C}[V]^{S_n}$ is a polynomial algebra if and only if $\rho(S_n)$ is generated by pseudo-reflections. If $\rho(g)$ is a pseudo-reflection then $\rho(g)$ is similar to a diagonal matrix $\mathrm{diag}(1,1,\ldots,1,\zeta)$ where $\zeta$ is a root of unity. Hence $\chi(g) = d-1 + \zeta$. However, the irreducible characters of symmetric groups are real valued, so if $g \not= 1_{S_n}$ then $g$ is an involution and $\chi^\lambda(g) = d-2$. </p> <p>It therefore suffices to show that if $n \ge 7$ and $g \in S_n$ is an involution such that $\chi^\lambda(g) = \chi^\lambda(1)-2$ then $g$ is a transposition and either $\lambda = (1^n)$ or $\lambda = (n-1,1)$. This follows by induction using the Murnaghan-Nakayama rule. (The details are fiddly but routine.) </p> <p><b>Edit:</b> Geoff Robinson shows in his answer (posted at the same time as mine) that $g$ is a product of at most $3$ transpositions. This leads to a quick inductive proof: suppose that $\lambda$ has two removable boxes, whose removal gives partitions $\mu$ and $\nu$ of $n-1$. If $\mu \not= (n-1)$ and $\nu \not= (n-1)$ then, by induction, we have $\chi^\lambda(g) \le \chi^\lambda(1)-4$. Hence $\lambda = (n-1,1)$. In the remaining case $\lambda$ is a rectangular partition, and provided $n\ge 8$, we can repeat this argument after removing two boxes from $\lambda$ in two different ways.</p> <p>For smaller $n$ there are two exceptional cases, corresponding to the partitions $(2,2)$ of $4$ and $(2,2,2)$ of $6$. The former representation is $2$-dimensional and is obtained from the standard representation of $S_3$ via the quotient map $S_4 \rightarrow S_3$. The latter is $5$-dimensional, and can be obtained by applying the outer automorphism of $S_6$ to the standard $5$-dimensional representation of $S_6$; the relevant character value is $\chi^{(2,2,2)}(g) = 3$ where $g = (12)(34)(56) \in S_6$.</p> http://mathoverflow.net/questions/91472/invariants-of-symmetric-group/91478#91478 Answer by Geoff Robinson for Invariants of Symmetric group Geoff Robinson 2012-03-17T18:04:39Z 2012-03-18T17:35:42Z <p>I only concern myself with faithful representations of $S_n$ and for $n >4$. The only way to get a polynomial algebra of invariants is to represent $S_n$ as a complex reflection group (so generated by pseudo-reflections, that is elements wih a fixed-point space of codimension $1$). A complex relfection group is easily checked to always be a direct product of ireducible complex reflection groups, so from now on I consider only irreducible representations. The question then becomes: is there an irreducible representation of $S_n$ other than the usual $n-1$-dimensional irreducible constituent of the natural permutation character, where $S_n$ is represented as a complex reflection group? Note that tensoring with the sign representation of the stated $n-1$-dimensional representation does not produce a representation of $S_n$ as a complex reflection group. There is still some content to this question. It can probably be easily addressed by the Murnaghan-Nakayama rule, or from a more detailed knowledge of the character table of $S_n,$ but I attempt a direct argument here. Since the characters of $S_n$ are rational-valued, any element represented as a pseudo reflection in the given representation must be represented as a genuine (or "real") reflection, since its trace must be rational. The generating reflections for $S_n$ in the given representation act with determinant $-1,$ so lie outside the derived group $A_n.$ Hence they are odd permutations, and expressible as a product of an odd number of (disjoint) transpositions, since they have order $2$. Since $A_n$ is simple ( as $n \geq 5$,) any single conjugacy class of these reflections generates the whole of $S_n,$ so we assume that all the generating reflections are conjugate. Note that we are not yet entitled to assume that the generating reflections are transpositions. Now we note that the product of two generating reflections has order $1,2,3,4$ or $6.$ For such a product has a fixed point space of codimension $0$ or $2,$ has determinant $1$, and has a rational trace. Now if $k >1$ is odd, a product of $k$ transpositions inverts a $2k+1$-cycle in $S_{2k+1}.$ Hence if $k > 3,$ and $n \geq 2k+1,$ then $S_{n}$ contains a pair of conjugate permutations, each a product of $k$ disjoint transpositions, whose product has order $2k+1.$ hen $n = 2k$ we may express a $2k$-cycle as a product of two permutations, one a product of $k$ disjoint transpositions, the other a product of $k-1$ disjoint transpositions. Since $k > 3,$ if we adjoin a transposition interchanging the fixed points of the second permutation, we get a product of two permutations, each a product of $k$ disjoint transpositions, where the product is a product of two disjoint cycles of lengths (allowing a $1$-cycle) whose sum adds to $n.$ Since $n \geq 10,$ sucha product never has order $1,2,3,4,5$ or $6.$ Hence we can assume that our generating reflections are products of at most three transpositions. Furthermore, if $n \geq 8,$ we can express a $5$-cycle in $S_n$ as a product of permutations, each a product of three disjoint transpositions. For we may express a $5$-cycle in $S_5$ as a product of two permutatons, each a product of two disjoint two-cycles. Affix a transposition commuting with the $5$-cycle to each of them, and this does not change the product. Hence if our reflections are products of $3$ disjoint $2$-cycles, we are left with the case $n =6.$ In fact, this case does occur. "Twist" the natural (non-unimodular) irreducible $5$-dimensional representation of $S_6$ by the exceptional outer automorphism of $S_6,$ and we obtain a reflection representation of $S_6$ in which products of $3$ disjoint $2$-cycles act as reflections. It remains to deal with reflection representations of $S_n$ in which transposition act as reflections. I presume that the Murnaghn-Nakayama rule or the character table of $S_n$ then shows that only the expected representation arises, but I leave that issue open here. (Later Edit: This now seems to be covered by Mark Wildon's answer).</p>