Week 11.5: Semidirect Products of Cyclic Groups (continued)

Welcome back! In my previous post, I laid most of the groundwork to attack one specific case of these semidirect products: C_p⋊C_n, where p is a prime that gives a remainder of 1 when divided by n. Today, my goal is to finish this proof. To do this, I’ll need to take a new approach.

Solving the C_p⋊C_n Case

A new approach

First, I’ll discuss the change I’m making to my approach. Initially, I was following Naidu and Rowell’s arguments: to show that the category of bimodules is pointed, we find enough simple objects to guarantee that all simple objects are invertible. This essentially “counts” total Frobenius-Perron dimension.

However, a complexity in this approach is that we need to be able to exactly count how many bimodule structures each object has. To get around this, we can instead model our approach after the work in [1]. Specifically, we chase the loftier goal of directly classifying simple bimodules. I expect that this is usually the harder approach, and that may even be true in this case; regardless, I usually like more theoretical arguments so I’m more comfortable with this approach.

The first step is to classify simple bimodules which are made of higher-dimensional representations. First, we need to understand more about their characters.

Higher-dimensional representations in the character table

In my previous post, I showed the following:

For some χ, the one-dimensional representations are given by the powers of χ. Furthermore, for all higher-dimensional representations V, χ⊗V = V.

I also explained why I’d expect that all other representations have dimension n. I didn’t actually prove that, because I hadn’t yet figured it out and I didn’t think it was important. As I’ve worked more on this problem, however, I’ve realized that it is necessary. Fortunately, I’ve also found a beautiful proof. To understand this, we first need an idea from linear algebra.

Eigenvalues of similar matrices

Two matrices A and B are called similar if we can write B = PAP^-1 for some invertible matrix P. Then, a key fact is that any two similar matrices have the same set of eigenvalues. In fact, these matrices are each similar to their Jordan Canonical Form. I won’t discuss this in too much detail, but it’s a matrix which has their eigenvalues along its diagonal.

This fact is important in its own right, but another corollary we’ll use relates to traces. Recall that the trace of a matrix is the sum of its diagonal elements, and that any two similar matrices have the same trace. The trace of a Jordan Canonical Form is, by construction, the sum of its eigenvalues. Therefore, in general, the trace of a matrix is the sum of its eigenvalues.

If we use this fact along with the argument I’ll soon make, we can even fully write the character table. However, I don’t actually need the entire table, so all I’ll use for now is the fact that two similar matrices have the same eigenvalues.

Eigenvalues of the a terms

Let the representation map a to the matrix A. Then it maps a^p to A^p. Because a^p is the group identity, A^p must be the identity matrix.

First, I’ll briefly discuss why A has an eigenvalue other than 1. If A‘s eigenvalues were all 1, then A would be the identity matrix itself. Then, the resulting representation would just be a representation of C_n, trivially extended to our semidirect product. Because C_n is a commutative group, its irreducible representations are only one-dimensional, so the representation we’re considering can’t be irreducible.

Now, let λ be an eigenvalue of A other than 1. Then λ^p is an eigenvalue of the identity matrix, so λ^p = 1. Therefore, λ is a p-th root of unity.

We also know that λ² is an eigenvalue of A², and generally, λ^k is an eigenvalue of A^k. Therefore, each power of λ (other than 1) is the eigenvalue of some power of A (other than the identity matrix).

The elegant finish

Because λ is a p-th root of unity, these p-1 powers of it are all distinct. Furthermore, they are split among the conjugacy classes of the powers of a. As I showed last post, there are (p-1)/n such classes.

Each class has one set of eigenvalues, and the number of eigenvalues is the representation’s dimension. Therefore, for (p-1)/n classes to total at least p-1 eigenvalues, the representation must have dimension at least n.

Finally, another calculation I discussed last post is that among the remaining (p-1)/n representations, the sum of the squares of their dimensions is (p-1)•n. Combining this with the fact that each dimension is at least n, we can conclude that each dimension is exactly n, as desired.

Classifying the simple bimodules

Let A be the unique semisimple algebra structure on 1⊕χ⊕…⊕χ^n-1.

Let M be any simple bimodule which contains a copy of some higher-dimensional representation V. By Lemma 6.2 of [1], M must exactly be V.

Now, suppose that some simple bimodule M is composed only of the χⁱ representations. Note that χⁱM is also a simple bimodule for all exponents i.

Now, let’s look at the bimodule M⊕χM⊕χ²M⊕…⊕χ^n-1M. We just showed one way to decompose it in terms of simple bimodules. Now, let’s find another.

By the distributive property, we can factor this as A⊗M. Now, recall that M is composed of powers of χ. Let’s apply the distributive property again for this decomposition. The key idea is that for any exponent i, A⊗χⁱ is still A. Therefore, A⊗M is just some tensor power of A.

In a semisimple category, the decomposition of an object into the sum of simple objects must be unique (just as how prime factorizations are unique). Therefore, we must have M=A.

The finish

Now, we just need to show that these simple objects are invertible in the bimodule category. We do this by looking at their Frobenius-Perron dimension.

Notice that the simple bimodules are all either higher-dimensional representations, or A itself. Importantly, the first thing we proved today is that all higher-dimensional representations have dimension n. Therefore, all the simple bimodules have the same dimension in our original fusion category, and this matches the dimension of A.

When we switch from our original fusion category to the category of bimodules, the Frobenius-Perron dimensions are proportional. Since A must have dimension 1 in its bimodule category, so must everything else. Therefore, all simple objects have Frobenius-Perron dimension 1 in the bimodule category, so they are invertible. Consequently, this category is pointed, so our original category is group-theoretical.

Conclusion

Today, I finished the proof that I started in the previous post. This is a quite general result, and it seems pretty important. Last week, I mentioned the case where we replace C_p with any Abelian group, or where we replace n with any positive integer that shares a factor with p-1. I’ve worked more on each of these cases, and they don’t seem like something I can tackle at the moment. I have made some progress on the case where we replace p with any prime power, but this is still quite messy and I don’t expect it to work out cleanly.

[1] P. Etingof, S. Gelaki, V. Ostrik. Classification of fusion categories of dimension pq, International Mathematics Research Notices (2004)