John Carlos Baez on Nostr: HARDCORE MATH POST Any action of a finite group G on a finite set x gives a linear ...

HARDCORE MATH POST

Any action of a finite group G on a finite set x gives a linear representation of G on the vector space with basis X. This is called a 'permutation represention'. And this raises a natural question: how many representations of finite groups are permutation representations?

Most representation are *not* permutation representations, since every permutation representation has a vector fixed by all elements of G, namely the vector that's the sum of all elements of X. In other words, every permutation representation has a 1-dimensional trivial rep sitting inside it.

But what if we could 'subtract off' this trivial representation?

There are different levels of subtlety with which we can do this. For example, we can decategorify, and let:

• the 'Burnside ring' of G be the ring A(G) of formal differences of isomorphism classes of actions of G on finite sets;

• the 'representation ring' of G be the ring R(G) of formal differences of isomorphism classes of finite-dimensional representations of G

In either of these rings, we can *subtract*.

There's an obvious map β: A(G) → R(G) since any action of G on a finite set gives a permutation representation of G on the vector space with basis X

So I asked on MathOverflow is this map typically *surjective*, or typically *not* surjective?

In fact everything depends on what field we're using for our vector spaces! If we use the rational numbers, it seems β is surjective for the vast majority of finite groups. If we use the complex numbers, it seems β is almost always *not* surjective.

I explain why here:

https://golem.ph.utexas.edu/category/2025/03/how_good_are_permutation_repre.html

If we use the rationals, I believe the smallest group where β is not surjective has 24 elements.