## Subfactors and Planar Algebras: Galois Theory March 17, 2008

Posted by Noah Snyder in planar algebras, subfactors.

Suppose you were to run across two fields K < L. In a blatant attempt at foreshadowing instead of a field extension, we’ll refer to this pair as a “subfield.” Further let’s suppose that this “subfield” has “finite index”, namely that [L:K] is finite. As a representation theorist how might you go about studying this “subfield”?

A representation theorist studies a ring by studying the modules over it. So a representation theorist should study a pair of rings by studying the bimodules you can make from them. However, if we study all L-K bimodules than we’ve forgotten the inclusion structure. In particular if we had another inclusion of K into L we would still have the same category of bimodules. So instead let’s study the bimodules that we can make starting with the L-K bimodule ${}_L L_K$ which we get from the inclusion of K into L and the K-L bimodule ${}_K L_L$.

What sort of operations can we perform on this bimodules? Well, we have a tensor product! So let’s consider the “tensor category” we get generated by ${}_L L_K$ and ${}_K L_L$. (I put “tensor” in scare quotes here because we have two tensor products, tensor over K and tensor over L. So it isn’t a monoidal category, instead it’s what I like to call a bi-oidal category. More on this later.) What does this category C(L/K) look like?

First let’s decompose ${}_K L_L \otimes_L {}_L L_K$. This is isomorphic (using the multiplication map) to ${}_K L_K$ (where both actions come from the inclusion K<L). Clearly this breaks up into [L:K] copies of the trivial bimodule ${}_K K_K$. Further tensoring these with anything else won’t do anything interesting.

So we’re halfway there, now let’s try decomposing ${}_L L_K \otimes _K {}_K L_L$. This is a bit trickier, let’s try an example. Let K be the real numbers and L the complex numbers. A basis over K for ${}_L L_K \otimes {}_K L_L$ is given by $1 \otimes 1$, $1 \otimes i$, $i \otimes 1$, $i \otimes i$. We want to decompose this as an L-L bimodule.

A little computation yields the two following simple submodules:

• $v = 1 \otimes 1 - i \otimes i$ and $w = i \otimes 1 + 1 \otimes i$. Notice that $iv = w = vi$ so this is isomorphic to ${}_L L_L$
• $v = 1 \otimes 1 + i \otimes i$ and $w = i \otimes 1 - 1 \otimes i$. Notice that $iv = w = -vi$ so this is not isomorphic to ${}_L L_L$. We denote it ${}_LL^\sigma_L.$

Okay, what can we do with these new modules? Tensoring with ${}_L L_L$ will never do anything interesting. The only new computation is that ${}_L L^\sigma_L \otimes_L {}_L L^\sigma_L \cong {}_L L_L.$

It is not difficult to see how this generalizes to an arbitrary Galois extension:

• The only K-K bimodule that occurs is ${}_K K_K$. In any tensor product that lands in the K-K part of the category you can easily compute how many times this bimodule occurs by computing the K-dimension.
• The only L-K bimodule that occurs is ${}_L L_K$. Again counting its multiplicity in tensor products is easy.
• The only K-L bimodule that occurs is ${}_K L_L$. Again counting its multiplicity in tensor products is easy.
• The L-L bimodules are interesting. For every element $\sigma \in \text{Gal}(L/K)$ we have the bimodule ${}_L L^\sigma_L$. Furthemore the tensor product decompositions are given by ${}_L L_K \otimes_K {}_K L_L \cong \bigoplus_\sigma {}_L L^\sigma_L$ and ${}_L L^\sigma_L \otimes_L {}_L L^\rho_L \cong {}_L L^{\sigma \cdot \rho}_L.$

In summary, the Galois group appears naturally as the set of irreducible L-L bimodules that appear in C(L/K)!

Bonus questions: What if L/K is separable but non-Galois? What if L/K is totally inseparable?

(I’ve done examples in both cases, but they’re fun to look at yourself. I’ll answer questions in comments if people want hints or details.)

1. Scott Carnahan - March 17, 2008

Hi Noah,

In the list near the bottom, I think you want L instead of K to be a L-K bimodule (or a K-L bimodule).

2. Noah Snyder - March 17, 2008

Fixed!

3. Greg Kuperberg - March 17, 2008

I wrote a paper on this topic. The result is that if you have a finite collection of such double field extensions which is closed under tensor product and duality, and if the extensions are all separable and finite, then all of the fields listed are finite extensions of a common subfield F. Note that if L is a field with two finite-index subfields K_1 and K_2, then in general K_1 cap K_2 does not have finite index in L. Closure under tensor product is a necessary condition. Although the theorem could have been proven ages ago and possibly even was, it isn’t trivial.

4. Noah Snyder - March 17, 2008

Here’s the link for Greg’s paper. The result he mentions is Corollary 4.2. I hadn’t read this paper before, but it has a lot of stuff related to this post so if you liked this post check out his paper for a meatier take.

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