The irreducible representations of the Symmetric group $S_5$ are classified by the partitions of $5$. For the standard representation which corresponds to the partition (4,1) the ring of invariants is generated by the elementary symmetric polynomials and hence is a polynomial ring. For other irreducible representations except for the trivial and sign representation, $S_5$ does not represent as a pseudo-reflection group, hence for these representations the ring of invariants is not a polynomial ring any more. I also learnt that a minimal generating set for these rings is not possible (or hard) to compute using the programs available till date. Do we have any combinatorial technique to compute them explicitly ?

For example if we take the tensor product of standard and sign representation (which corresponds to the partition 2+1+1+1) is there any way to get a minimal generating set for the ring of invariants ?

Further it would be a great help if someone could provide me an example of a representation of $S_5$ (not necessarily irreducible and except standard, natural, sign and trivial) whose ring of invariants in terms of generators and relations is known.

  • $\begingroup$ Can you give some hints as to what you know already and what you tried already? Invariant theory is a vast topic, so providing some context to your question would be very important in order to get a meaningful answer. $\endgroup$ Jan 18, 2016 at 21:49
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    $\begingroup$ Somehow I know the dimensions of the symmetric power invariants of the tensor product of standard and sign representation. They are quite big for higher symmetric powers and difficult to extract a minimal generating set. $\endgroup$
    – Karthik
    Jan 19, 2016 at 10:45
  • $\begingroup$ Molien series (dimensions of graded pieces) is of course easily computed, but the structure of the ring can be harder to get. $\endgroup$ Jan 27, 2016 at 12:02

1 Answer 1


In fact at least in the case you seem most interested in, the (4-dimensional) standard representation tensored by the sign representation, it is possible to compute the invariants by computer, using the standard tools in MAGMA. The result is a minimal system of generating invariants of degrees 2, 4, 6, 8, 10, 13, and 15. When printerd out, the invariants look quite terrible. The invariant ring is not Gorenstein. Here's the MAGMA session:

> G:=MatrixGroup<4,Rationals() | [0,-1,0,0, -1,0,0,0, 0,0,-1,0, 0,0,0,-1],
> [0,1,0,0, 0,0,1,0, 0,0,0,1, -1,-1,-1,-1]>;
> R:=InvariantRing(G);
> time prim:=PrimaryInvariants(R);
Time: 0.030
> time fund:=FundamentalInvariants(R);
Time: 0.430
> [TotalDegree(f): f in fund];
[ 2, 4, 6, 8, 10, 13, 15 ]
> H<t>:=HilbertSeries(R);
> H * &*[1-t^TotalDegree(f): f in PrimaryInvariants(R)];
t^15 + t^13 + t^8 + 1

I don't have the matrices for the other irreducible representations (of degrees 5,6, and 6). If you give them to me, there should be a fair chance of being able to compute the invariant rings. On the other hand, in view of the ugliness of the above example, it may not be worthwhile.

Let me add that computations become much harder for sums of irreducible representations. Unlike representation theory, invariant theory does not reduce to the irreducible case. For example, invariants of direct sums of copies of the same representation, known as "vector invariants", are notoriously ugly.

  • $\begingroup$ Thank you Prof. Kemper for the effort and the generous answer. Is there someway to find the relations between the invariants, especially one containing the degree 13 invariant. $\endgroup$
    – Karthik
    Jan 27, 2016 at 12:40
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    $\begingroup$ Yes, the relations can also be computed. You just say "rel:=Relations(R);" in the above MAGMA session. It took five minutes to come back with six (minmal) relations. Here are their degrees (extracted by "WeightedDegree"): 16, 21, 23, 26, 28, and 30. Even the first relation is rather long ans has giant coefficients. $\endgroup$ Jan 28, 2016 at 12:23

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