# Representation of $\mathfrak{S}_p$ over $\mathbb{F}_p$

I'm learning modular representation theory from the 3rd part of Serre's book. Through the process called "reduction mod. $$\mathfrak{m}$$", we obtain representations in positive characteristic from representations in characteristic $$0$$. Howerer, the reduction of irreducible representations may not be irreducible any longer.

I want to test myself, so I try to compute the representation of the symmetric group $$\mathfrak{S}_p$$ in $$\mathbb{F}_p$$ (or its algebraic closure). By Brauer--Nesbitt theorem, the number of irreducible $$\mathbb{F}_p$$-representations is equal to the number of $$p$$-regular conjugacy classes in $$\mathfrak{S}_p$$, so there is exactly one representation "missing" since the only $$p$$-singular conjugacy class is the one containing $$(12\cdots p)$$. Yet I still have no ideas about how to find them exactly.

Serre gives some examples in his book, say $$\mathfrak{S}_4$$ and $$\mathfrak{A}_5$$, but his computations are done by case-by-case analysis. Generally, we have many approaches to construct the representations of a symmetric group over $$\mathbb{C}$$ or $$\mathbb{Q}$$, say the Young symmetrizer and the Specht module. Can we generalize some constructions to positive characteristic?

• Finding the irreducible representations of $\mathfrak{S}_n$ over $\mathbb{F}_p$, or even just their dimension, is a major problem in representation theory, and we know extremely little about them. So it's possible that you will be disappointed in what is known on that topic. Commented May 5, 2022 at 7:30
• Thank you. I know the general case is complicated, Kleshchev wrote a book on this topic. But what about this special case? Commented May 5, 2022 at 8:13

Here is one way to answer your question. I'll begin with the general story, then explain how it simplifies in your example. For any $$n$$ and $$p$$, the reduction modulo $$p$$ of the Specht modules $$S^\lambda$$ carries a symmetric bilinear form with the property that if it is non-zero, the quotient by its radical is irreducible; in addition, the non-zero quotients that arise this way are a complete set of non-isomorphic simple $$F_p S_n$$-modules. Moreover, the set of partitions for which the form is non-zero is the set of $$p$$-restricted partitions of $$n$$: those for which the difference between consecutive parts is at most $$p-1$$. Additionally, the irreducibles corresponding to $$p$$-restricted partitions belong to the same block of $$F_p S_n$$ if and only if the $$p$$-cores of the partitions are equal.
In your case $$n=p$$ this implies that the corresponding quotients of the Specht modules for all partitions of $$n$$ other than the trivial partition $$(n)$$ give a complete set of non-isomorphic irreducible representations, and that moreover there is a unique non-semisimple block in the category of $$F_p S_p$$-modules, containing the hooks. Every other Specht module is actually simple upon reduction modulo $$p$$. I believe it should be true that moreover there is an exact sequence
$$0 \to S^{(1^n)} \to S^{(2,1^{n-2})} \to \cdots \to S^{(n-2,1,1)} \to S^{(n-1,1)} \to S^{(n)} \to 0$$
with the property that the images are the desired unique irreducible quotients (and whose existence is closely related to the fact that we are in the case here of a block of defect 1). You can work out the dimensions of the interesting irreducible $$F_p S_p$$ modules from this.
• Thank you, sir. But I have a question: What do you mean by "the corresponding quotients of the Specht modules for all partitions of $n$ other than the trivial partition $(n)$"? I think the trivial partition corresponds to the trivial representation, which is irreducible under reduction. Commented May 6, 2022 at 2:24
• @Estwald I mean the quotients of those Specht modules whose bilinear form is non-zero---depending on the conventions you use, it often happens for the trivial representation that its bilinear form is zero (as you point out it is nonetheless irreducible---the point is that it may occur as a quotient of another Specht module in certain cases such as yours, and we use the partition indexing that other Specht module instead of $(n)$ as part of our indexing set for the isoclasses of irreps mod p). Commented May 6, 2022 at 14:18