How do I know if an irreducible representation is a permutation representation?

I have a vague question, a less vague question and a lot of vaguer questions about permutation representations of a finite group $G$.

• Vague question. Recall that if $G$ acts on a finite set $X$, we get a permutation representation $$G \to GL_{\lvert X \vert}(\mathbb C).$$ (Unless $X$ is very small,) this representation is never irreducible, for $\mathbb C (1, 1, \ldots, 1)^T$ splits off as a subrepresentation. What's left is the permutation representation $V_X$ associated to $G$. A classical fact is that $V_X$ is irreducible if and only if the original action was $2$-transitive. My question is: conversely, how do I know if some irreducible representation $V$ (whose character I know) is obtained by this construction? Clearly, a necessary condition is that $\chi_V$ should only take integer values $\geq -1$, but is it sufficient? If not, do we have another criterion?

• Less vague question. (A special case of the first question). If I have not blundered, $\mathfrak S(6)$ has a degree-nine representation whose character only takes the values $-1$, $0$, $1$, $3$ and $9$. If you believe that $\mathfrak A(6) \simeq PSL_2(\mathbb F_9)$, you could get it by first considering the permutation representation $V_{\mathbb P^1(\mathbb F_9)}$ of $\mathfrak A(6)$, and inducing it to $\mathfrak S(6)$. This splits off as $$\mathrm{Ind}_{\mathfrak A(6)}^{\mathfrak S(6)} V_{\mathbb P^1(\mathbb F_9)} = V \oplus (V \otimes \epsilon),$$ where $\epsilon$ is the sign morphism. Is this $V$ a permutation representation? (Note that $\mathfrak S(6)$ is not isomorphic to $PGL_2(\mathbb F_9)$, so the action on the projective line does not extend to $\mathfrak S(6)$ in a trivial way.) I believe this representation is not permutation, but I have no proof and not much confidence in my intuition (a brute-force examination of all index-nine subgroups of $\mathfrak S(6)$ would work, but I'd rather avoid hard work and I don't know any conceptual arguments or how to use modern computer tools to figure this out).

• Vaguer questions. The permutation representation machinery gives a morphism $$\mathbf{Burn}(G) \to \mathbf{R}(G)$$ from the Burnside ring of $G$ to its representation ring. Better still, it gives a morphism to all representation rings (whatever the field). What can we say about this morphism, beyond Wikipedia's example showing it can be noninjective and nonsurjective? Is there any textbook where this morphism is considered quite thoroughly? Do you know of any work on representation theory where the $\mathbf{Burn}(G)$-algebra structure on $\mathbf R(G)$ has been usefully exploited?

I realize a lot of my questions are very imprecise and I apologise unreservedly for that. I would nonetheless appreciate any comment, answer or reference recommendation about this topic.

(Comment: I asked this question on MSE a few days ago, but got no answer, hence this question.)

• "projective plane" - it must be "projective line" – Dima Pasechnik Nov 18 '17 at 18:33
• $S_6$ has $10$ Sylow $3$-subgroups, corresponding to the $10$ ways of dividing a $6$-element set into two sets with $3$ elements each. This yields a $2$-transitive action on $10$ points. – Frieder Ladisch Nov 18 '17 at 19:42

First of all, note that $S_6$ has a doubly transitive action on 10 points, obtained from the action of $PSL_2(9)=A_6$ on the projective line over $\mathbb{F}_9$ by adding the Galois automorphism of $\mathbb{F}_9$, and this is the only faithful permutation action of $S_6$ of degree 10. $S_6$ has two irreducible characters of degree 9, one obtained from the other by tensoring with the alternating degree 1 charater. Only one of them, say $\chi$, can come up in the doubly transitive permutation character, as all the values $\chi$ are bounded from below by $-1$.

With GAP, it is very easy to answer everything on $S_6$. (One can replace this by an more human argument; you can also look up the character table in question in the literature, e.g. in the "Atlas of Finite Simple Groups".) In particular you can compute the irreducible characters of degree 9:

[ 9, -3, 1, -3, 0, 0, 0,  1, 1, -1, 0 ]
[ 9,  3, 1,  3, 0, 0, 0, -1, 1, -1, 0 ]

Thus $\chi$ is the latter one, and it does occur in the doubly transitive degree 9 permutation character of $S_6$ (add 1 to each entry).

To give you an example that having entries at least $-1$ is not enough to get a constituent of a doubly transitive character, note that $M_{11}$ (the smallest Mathieu sporadic simple group) has degrees 44 and 55 $\mathbb{Z}$-valued irreducible characters with values at least $-1$, but no degree 45 and 56 doubly transitive permutation representations (for case of 56, the group order is not divisible by 56).

The map from the Burnside ring to the representation ring is well studied. Andreas Dress wrote many papers on this in the 1970's. See also work of Tammo tom Dieck, who was applying things to equivariant topology.

There is a detecting character ring for each of these. Characters for elements of the Burnside ring are integer valued functions from conjugacy classes of subgroups of $G$, and an evident map between the two character rings. A classic character $\chi$ that is integer valued is in the image of this map exactly when $\chi(g) = \chi(h)$ if $g$ and $h$ generate conjugate cyclic groups.

In a 1987 paper in Comm. Math. Helv., Peter Webb gives an explicit basis for the kernel of the map from the Burnside ring to the Representation ring, after inverting the order of $G$.

This should get you going, if you want to learn about this.

• is the terminology in that paper different? "Green ring"? – Dima Pasechnik Nov 18 '17 at 23:30
• I think I am missing something basic here: Let $G$ be the quaternion $8$ group and let $\chi$ be the character of the $2$-dimensional irreducible representation. This is integer valued and obeys the condition that $\chi(g) = \chi(h)$ if $g$ and $h$ generate conjugate cyclic groups. But, for every subgroup $H$ of $G$, if $\psi$ is the permutation representation on $G/H$, then $\psi(1) \equiv \psi(-1) \bmod 8$, whereas $\chi(1) = 2$ and $\chi(-1)=-2$. So it seems to me that $\chi$ should not be in the image of the Burnside ring. What am I missing? – David E Speyer Jul 23 '18 at 16:19
• I’m interested in @DavidESpeyer’s comment and the question of whether there is a simple criterion for determining if a linear representation is a permutation representation. – Sam Hopkins Oct 11 at 17:02