In the theory of quantum groups Hopf algebras arise via the Fourier transform:

A third point of view is that Hopf algebras are the next simplest category after Abelian groups admitting Fourier transform.

At least for nice functions, a Fourier series is just a Laurent series on a circle (which means just substituting $z = re^{i\theta}$ into a Laurent series), so I can see Fourier analysis on Abelian groups as generalizing this simple example.

How do I see (in an easy way) that Hopf Algebras are the natural generalization of Fourier theory to algebras, in a way that motivates what a Hopf algebra is, so that I have some feel for quantum groups?

  • 2
    $\begingroup$ The keyword to look for here is Tannaka-Krein duality (e.g. on nLab), a natural extension of Pontryagin duality. Perhaps someone else here is capable of giving some simple explanation of it, but I don't think I can do better than what's already on nLab or Wikipedia. $\endgroup$ – Logan M Oct 17 '14 at 3:37

I've just released an early version of a paper that has to do with this so I'm updating my answer rather completely.

The discrete Fourier transform in one dimension is appropriate to situations where a function depends on a set of n points equally spaced on a circle. This is an Abelian symmetry. That is, the system is symmetric under rotations that move the points around, and when you take two consecutive rotations, say one by 15 degrees and then another by 30 degrees, the result is 45 degrees no matter which order you take the two rotations. The question is how do we generalize the discrete Fourier transform to a symmetry that is non Abelian? A "group algebra over the complex numbers" is a simple example of a Hopf algebra, and we will show that a non Abelian generalization of the discrete Fourier transform is a transformation of this type.

An Abelian symmetry appropriate to a rotation can be generated by a single element. (Here I'm considering proper rotations only and not symmetries that reverse the order of the points.) Such a symmetry group has a character table with just as many columns as rows, and the elements of the character table are the coefficients of the discrete Fourier transform, up to some normalization constants that depend on the size of the group and for each irrep, the character of the identity. For a half dozen examples, see the character table for the cyclic point group symmetries C_n on the web here: https://en.wikipedia.org/wiki/List_of_character_tables_for_chemically_important_3D_point_groups#Cyclic_groups_(Cn)

This suggests that a discrete Fourier transform for a non Abelian group G will involve the character table of the group. Non Abelian groups have character tables that are not square and so cannot directly be transformation matrices. The reason Abelian groups do have square character tables is that each of the classes of an Abelian group has only one element and the number of irreducible representations is equal to the number of classes. So while character tables are square, the problem is that they have too few columns and too few rows to define a transformation matrix on the group. It's easy to fix the "too few columns" problem label the columns with the group elements instead of the classes. Then any two elements that are in the same class will have identical columns but we can fix that problem when we also increase the number of rows. The irreducible representations (rows) of a character table are orthogonal and they will still be orthogonal when we write the columns as group elements instead of classes. The problem is that now the columns are not orthogonal, but then again, the matrix isn't square so they couldn't be.

From the point of view of diagonalizing the group algebra, each line defines the projection operator for a jxj square block on the diagonal where j is the character of the group identity. For an easy to read explanation of this, see Hammermesh's inexpensive Dover book "Group Theory and its Applications to Physical Problems", chapter 3, section 17 titled "The General Theorems; Group Algebra."

The projection operator for a jxj square block is simply a matrix that happens to have 1s on the diagonal of the jxj block and is zero everywhere else. That tells us exactly what the missing elements of the Fourier transform are. They are the traceless jxj square matrices, i.e. su(j).

This means we can fill out a character table for a non Abelian symmetry group to be a generalization of the discrete Fourier transformation by adding lines for the su(j) degrees of freedom in its jxj blocks on its diagaonalization (for j>1, as the 1x1 blocks have no traceless degrees of freedom so they're taken care of by the corresponding irrep).

This means that for a non Abelian symmetry group, the equivalent of the discrete Fourier transform is of the form su(j) x su(k) x ... x su(m). This is sufficiently reminiscent of the Standard Model symmetry that I looked around to see if I could find one that worked.

For an example of how this works, consider the permutation group on 3 elements. This 6 element group has three classes and therefore three irreps. The character table is:

$\begin{array}{ccc} 1&1&1\\ 1&1&-1\\ 2&-1&0\end{array}$.

The three classes have size 1, 2 and 3. Replacing the classes with their group elements gives a widened character table:

$\begin{array}{cccccc} 1& 1& 1& 1& 1& 1\\ 1& 1& 1&-1&-1&-1\\ 2&-1&-1& 0& 0& 0\end{array}$.

Note that each row is still orthogonal. The missing lines all belong in the the third irrep. So, as elements of the group algebra, they will be annihilated by the first two irreps and will be unchanged by the third. Unlike the other lines, there is a gauge freedom in choosing these last three lines. The third irrep defines a 2x2 block on the diagonal so the traceless parts of that block have an su(2) symmetry. And we have an su(2) gauge freedom in choosing them.

As far as choosing these Pauli spin matrices, note that the 2nd and 3rd columns of the widened character table are identical so to get an orthogonal expanded table we will have to distinguish them with something that is either (0,0) or a multiple of (1,-1) in those two columns. Accordingly, we choose $\sigma_x$ to be an appropriate multiple of (1,-1) in those columns. To get the multiple correct, note that $\sigma_x^2 = 1$, so we choose our normalization so that the group element squares to give the irrep for that 2x2 block. This will make those two columns of the newly expanded table orthogonal, but there are still three identical columns at the right. Take the first two of those and put (1,-1) in them and normalize this to give $\sigma_y$. Then define $\sigma_z$ by $i\sigma_x\sigma_y$ and we have a complete "discrete Fourier transform" on the non Abelian permutation group with 3 elements. I obtained:

$\begin{array}{c|cccccc} &(\;)&(123)&(132)&(12)&(13)&(23)\\ A &1/6& 1/6& 1/6& 1/6& 1/6& 1/6\\ B &1/6& 1/6& 1/6&-1/6&-1/6&-1/6\\ C &2/3&-1/3&-1/3& 0& 0& 0\\ C_x&0&i/\sqrt{3}&-i/\sqrt{3}&0&0&0\\ C_y&0&0&0&1/\sqrt{3}&-1/\sqrt{3}&0\\ C_z&0&0&0&1/3&1/3&-2/3 \end{array}$,

where I've also normalized the rows.

As an example of the normalization, let's compute $C_x^2=C$, the equivalent of $\sigma_x^2 = 1$ for the Pauli spin matrices. That is, we're going to square the 4th row $C_x$ of the above table and see if we get the 3rd row $C$. We have

$[i(123)/\sqrt{3}-i(132)/\sqrt{3}]^2 =-[(123)-(132)]^2/3\\ =[-(123)^2-(132)^2 +(123)(132)+(132)(123)]/3\\ =[-(132)-(123)+(\;)+(\;)]/3 \\ = 2(\;)/3-(123)/3-(132)/3 = C$

as required.

| cite | improve this answer | |
  • $\begingroup$ The calculation was taken from an unpublished paper that is available (at least temporarily) on the web here: vixra.org/abs/2004.0354 $\endgroup$ – Carl Brannen Apr 22 at 4:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.