Why did people originally like Frobenius algebras? - MathOverflow

Why did people originally like Frobenius algebras?

2010-07-16T15:51:27Z

These days, lots of people are excited by Frobenius algebras because commutative Frobenius algebras are the same thing as 2D topological quantum field theories.

...but this seems like teaching an old dog new tricks. Can anyone sum up (using only diet representation theory :-P), why Frobenius algebras were invented and what was so good about them?

Also, any nice texts and/or papers along this line would be much appreciated. I'm working through the old Nakayama papers now, but perhaps this material exists in a friendlier and more modernised form somewhere?

Answer by Jack Schmidt for Why did people originally like Frobenius algebras?

2010-07-16T18:58:13Z

Frobenius's original turn-of-the-century perspective was the nonvanishing of a determinant. Brauer–Nesbitt–Nakayama studied some equivalent definitions in the late 30s and early 40s. For instance, an equivalence between the left and right regular representations is a rare and beautiful thing; this gives an equivalence between projectivity and injectivity that is explained in modern language in Lam's Lectures on Modules and Rings. This also gives a "perfect duality" studied by Dieudonné in the late 50s. I added the missing early sources to the wikipedia article, including the Brauer–Nesbitt announcement in PNAS which is pretty easy to read.

Answer by Kevin Lin for Why did people originally like Frobenius algebras?

2010-07-16T19:44:48Z

I learned the following from Constantin Teleman, and from these lectures of David Ben-Zvi.

Let $G$ be a finite group. Let $A = \mathbb{C}[G]^G$. This is an algebra under convolution:

$$(\phi \cdot \psi)(g) := \sum_{h \in G}\phi(gh^{-1})\psi(h).$$

We also have a trace $t : A \to \mathbb{C}$ given by $t(\phi) = \phi(1)/|G|$.

This is a Frobenius algebra. (Maybe this is the original example of a Frobenius algebra?)

In fact, this is a semisimple Frobenius algebra: Let $P_i = \frac{\chi_i \operatorname{dim}\chi_i}{|G|}$, where $\chi_i$ are the irreducible characters. Then $P_i$ is a basis for $A$ and $P_i\cdot P_j = \delta_{ij}P_i$.

Note that $1 = \sum_i P_i$. We have $t(1) = t(\sum_i P_i) = \frac{\sum_i (\operatorname{dim} \chi_i)^2}{|G|^2}$. On the other hand we have $t(1) = 1/|G|$. We get the formula $$|G| = \sum_i (\operatorname{dim} \chi_i)^2.$$

Pretty cool, no?

I think the correspondence between Frobenius algebras and 2D TQFTs is more than just "teaching an old dog new tricks". The 2D TQFT corresponding to this Frobenius algebra is the "finite group ('pure') gauge theory" of Dijkgraaf-Witten. From the TQFT perspective, $t(1)$ is the number which is assigned to the sphere $S^2$. It's some kind of "path integral"(?) over all maps $S^2 \to BG = \ast/G$. This number is a certain "weighted count" of $G$-bundles over $S^2$. And of course we have a similar story for all other surfaces. When the surfaces have boundary, we look at $G$-bundles with prescribed holonomy around the boundary circles.

Moreover, the 2D TQFT can be "extended" to manifolds with corners, which is also interesting. Check out Ben-Zvi's lecture notes for more on this stuff.