In this post, I want to discuss to what extent a group’s character table determines it up to isomorphism.

First, let’s do the easy case of Abelian groups. The set of characters of an Abelian group G is itself a group called given by pointwise multiplication. In fact, G is isomorphic to (though not canonically) as follows: we can write where the are prime powers. Fix generators for each cyclic summand. For each j, the function given by sending the generator of to gives an element of order in , and the map given by is injective. Since the number of characters of G is equal to the order of G, we conclude that it is an isomorphism. To do this canonically (without picking the direct sum decomposition), we can just do it twice (picking the decomposition twice ends up cancelling the fact that we made a choice), and we get an isomorphism by sending x to the character of defined by evaluation: . This is a special instance of Pontryagin duality, which holds more generally for any locally compact Abelian group. So we know that the characters determine the group up to isomorphism in this case.

Now let’s look at the noncommutative case. The first value of n for which there exist two nonisomorphic noncommutative groups of order n is 8, in which case we have the dihedral group which is the symmetries of the square, and the quaternion group .

The group has the presentation . For a character, we need to pick a fourth root of unity for a, along with a sign for b, and the relation becomes . Hence we see the four characters: assign to a and b in all possible ways. The other representation can be defined by sending a and b to the matrices and .

The conjugacy classes are , so we get the following character table:

For the quaternions, we use the presentation . Comparing with the usual definition as , we can take a=i and b=j. This presentation shows that images of (a,b) under the 4 characters must be . For the 2-dimensional representation, we can use the standard matrix representation of the quaternions: .

The conjugacy classes are . This looks familiar, and in fact, the character table is exactly the same as that of .

So there is no hope of recovering the group from its character table, but in fact, one can reconstruct the group from its representations. More precisely, we need to look at the category of its representations Rep(G). So far it still looks bad because the dimensions of the hom spaces are given by inner product of characters, and the tensor product is given by multiplication, so there must be some additional information in there somewhere. In fact there is, as Tannaka pointed out. We consider not just Rep(G), but also its forgetful functor F to Vect, the category of finite-dimensional complex vector spaces. First we note that an element x of G gives a natural transformation from F to itself by defining to be multiplication by x whenever V is a representation. This association realizes G as a subgroup of the monoid of natural transformations from F to itself. Is there a way to intrinsically characterize this subgroup?

We notice two more properties of . First, it’s the identity map on the trivial representation of G, and it preserves tensor products in the sense that as maps (we’ll call both of these properties tensor-preserving). It is also self-conjugate: Given a vector space V, we define a conjugate space by and for . Then if V is a representation, then so is by saying that where as a matrix is g with its entries conjugated. We can conjugate a natural transformation u by defining for V a representation and x in V. In fact, these two properties are enough: Tannaka’s theorem says that all tensor-preserving self-conjugate natural transformations from F to itself are of the form for some x in G. Just as in the case of Pontryagin duality, Tannaka’s theorem holds more generally for arbitrary compact groups (one can define a topology on the set of endomorphisms of F).

So where’s the Krein and where’s the duality? Given the above information, one should think of Rep(G) as a sort of dual to G, and Krein classified the categories which are of the form Rep(G). Given a category C of vector spaces with a tensor product and an involution (the conjugation above), then C is dual to a compact group G if and only if the following three properties hold:

- (Identity axoim) There exists I, which is unique up to isomorphism, such that for all A.
- (Krull–Schmidt axoim) Every V can has a minimal direct sum decomposition (the summands are not isomorphic to nontrivial direct sums).
- (Schur’s lemma axiom) If A and B are minimal (with respect to direct sum), then Hom(A,B) is 1-dimensional if A and B are isomorphic, and is 0 otherwise.

Then C = Rep(G) where G is the tensor-preserving self-conjugate endomorphisms of the forgetful functor F.

Things don’t end here of course, there’s extensions to quantum groups and algebraic groups (Grothendieck’s Galois theory), and other things. If I ever learn this stuff, I’ll try to write a sequel to this post.

-Steven

I’m trying to ascertain exactly where the difference between the two groups and manifests itself in the above. Does this result reduce to saying that a finite group is determined by its group algebra considered as a Hopf algebra? (Is this last question I asked even true?)

By:

Peter McNamaraon May 20, 2009at 12:45 AM

Hi Peter,

I like your question, it made me do a little bit of hunting. Here is the summary:

Of course, by Wedderburn’s theorem for semisimple algebras we know that . The last statement you made is true though: if we have a monoid algebra , then the group-like elements (those such that ) is a monoid isomorphic to M.

The self-conjugate, tensor-preserving property looks similar to the group-like property for the Hopf algebra, but I don’t see how to conclude Tannaka’s theorem for finite groups from this fact.

By:

Steven Samon May 20, 2009at 2:36 AM

What can one conclude if the group algebras over Z are isomorphic?

By:

random-o-sauron May 22, 2009at 6:03 AM

random-o-saur:

I just did a search of the literature, and I have a positive statement and a negative statement:

First, the positive statement. If we have two groups G and H such that , then their Abelianizations are isomorphic. In particular if G and H are Abelian, then if and only if . This is proved using group homology, see

Guram Donadze and Manuel Ladra, On the groups with isomorphic integral group rings,

Int. J. Algebravol. 3 (2009), no. 11, 525–529.The original result about Abelian groups and their integral group rings was known for a long time, I think at least by Higman, though I can’t find the precise source.

For the negative answer, I’ll point you to

Martin Hertweck, A counterexample to the isomorphism problem for integral group rings,

Annals of Math.,154(2001), 115–138.There he constructs a finite solvable group X with order whose integral group ring contains a group of units Y such that but Y and X are nonisomorphic. This paper is also cohomological in nature.

By:

Steven Samon May 22, 2009at 4:31 PM

Thanks!

By:

random-o-sauron May 23, 2009at 12:24 AM

There is also Tannaka formalism for Lie algebra g.

Given a Semisimple Lie algebra. We can use category of finite dimensional representations of g and fiber functor to reconstruct this Lie algebra.

Moreover, using the same formalism we can associate a Lie algebra to locally compact Lie group G which is isomorphic to Lie(G) and we can from a semisimple Lie algebra g to construct a Lie group G such that Lie(G)=g

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