This Week's Finds in Mathematical Physics (Week 157)

John Baez

I never write issues of This Week's Finds about topics that people request. I only write about what I happen to be studying at a given moment - nothing else seems to work. But when my friend Minhyong Kim asked me to do an issue on Young diagrams, I decided to break this rule just once. Young diagrams are too cool to ignore.

Physics relies a lot on symmetry to simplify problems, and there are two kinds of diagrams that show up a lot in this context: Dynkin diagrams and Young diagrams.

Dynkin diagrams first show up when you study shapes with lots of reflection symmetries, like crystals and Platonic solids. They wind up being good for all sorts of other stuff, like classifying simple Lie groups and their representations. I talked about them in "week62" - "week65".

But what about Young diagrams? These are also important for studying group representations, but for a more limited class of groups: the "classical" groups.

As with composers of music, there's no precise list of groups that count as "classical". But in general, a classical group should consist of linear transformations that preserve some nice geometrical structure on a vector space. A good example is SU(N), the group of all linear transformations of an N-dimensional complex vector space that preserve an inner product and volume form. In less elevated language, SU(N) is the group of all N x N unitary matrices with determinant 1.

The symmetric group Sn may also be considered an honorary classical group, even though it's defined in terms of a set rather than a vector space. Sn is the group of all permutations of an n-element set.

Rather amazingly, Young diagrams can be used to classify all 3 of these things, which at first seem quite different in flavor:

• conjugacy classes in Sn
• irreducible representations of Sn
• irreducible representations of SU(N)

Let me sketch how this goes, and then say a bit about the other things you can do with Young diagrams.

Say we have any permutation g in Sn, like this:

```1 → 2
2 → 4
3 → 3
4 → 1
5 → 6
6 → 5
7 → 7
```
Note that 1 gets mapped to 2, which gets mapped to 4, which gets mapped back to 1 again. Similarly, 5 gets mapped to 6, which gets mapped back to 5. The number 3 gets mapped to itself right away, as does 7. No matter where we start, we always cycle back eventually. So our permutation consists of a bunch of "cycles":

(1,2,4) (5,6) (3) (7)

and writing down this "cycle decomposition" completely describes the permutation. To simplify life, we always write down these cycles in order of decreasing length. We also write the lowest number in each cycle first.

Now suppose we conjugate our permutation g by some other permutation, say h. This gives the permutation hgh^{-1}. How does the cycle decomposition of this compare with that of g? It looks almost the same! For example, it might look like this:

(2,7,6) (1,3) (4) (5)

There are the same number of cycles, each the same length as before. The only thing that changes are the numbers in each cycle. These get switched around by means of the permutation h.

In short, when we conjugate a permutation, all that remains unchanged is the picture we get by writing down its cycle decomposition and blotting out the specific numbers in each cycle, like this:

```(X,X,X) (X,X) (X) (X)
```
Folks usually write each cycle as a row, like this:
```X X X
X X
X
X
```
This is called a "Young diagram"! But instead of X's, people usually draw boxes. So a Young diagram is just a bunch of rows of boxes, arranged in order of decreasing length.

Okay: so far I've shown how conjugacy classes of permutations in Sn correspond to Young diagrams with a total of n boxes. Now I want to do the same for irreducible representations of Sn.

This is cool for the following reason: for any finite group, the number of irreducible representations is the same as the number of conjugacy classes of group elements. But in general there's no natural way to match up irreducible representations with conjugacy classes. The group Sn just happens to be specially nice in this way.

Here I must turn up the math level slightly... for example, I'll assume you know what "irreducible representations" means! I'll even show off by calling them "irreps" for short. But to be nice, I'll start by reviewing some general facts about representations of finite groups.

Suppose G is a finite group. Then G has only finitely many irreps, all finite-dimensional. Every finite-dimensional representation of G is a direct sum of copies of these irreps.

To get our hands on these irreps, let C[G] be the space of formal linear combinations of elements of G. This is called the "group algebra" of G, since it becomes an algebra using the product in G. Any representation of the group G becomes a representation of C[G] in an obvious way, and vice versa.

With some work, one can show that C[G] is isomorphic to an algebra of block diagonal matrices. For example, C[S3] is isomorphic to the algebra of matrices like this:

```   * * 0 0
* * 0 0
0 0 * 0
0 0 0 *
```
where the * entries can be any complex number whatsoever. Since matrices act on vectors by matrix multiplication, we can use this to get a bunch of representations of C[G], and thus of G - one representation for each block. And this trick gives us all the irreps of G! For example, S3 has one 2-dimensional irrep, coming from the 2 x 2 block in the above matrix, and two 1-dimensional irreps, coming from the two 1 x 1 blocks.

This wonderful fact does not solve all our problems. If someone hands us a finite group G, we still need to work to find which algebra of block diagonal matrices C[G] is isomorphic to. How do we do this?

The trick is to find elements of C[G] corresponding to matrices that are the identity matrix in one block and zero in the rest, like these:

```            1 0 0 0            0 0 0 0             0 0 0 0
0 1 0 0            0 0 0 0             0 0 0 0
0 0 0 0            0 0 1 0             0 0 0 0
0 0 0 0            0 0 0 0             0 0 0 1

p1                 p2                  p3
```
If we can find these guys, the rest is easy: C[G] is a direct sum of "blocks"

{pi a pi : a in C[G]}

each of which is isomorphic to some algebra of n x n matrices.

How do we find these guys pi in C[G]? It's actually pretty straightforward to characterize them:

• They are idempotent: pi2 = pi.
• They are central: pi x = x pi for all x in C[G].
• They are minimal: if pi is the sum of two central idempotents, one of them must be zero.

So we've reduced the problem of finding the irreps of a finite group G to the problem of finding "minimal central idempotents" in the group algebra C[G].

To go further, we need to know more about our group G. So now I'll take G to be the permutation group Sn and tell you how to get the minimal central idempotents. We'll get one for each Young diagram with n boxes!

Say we have a Young diagram with n boxes, like this:

```X X X
X X
X
X
```
Then we can pack it with numbers from 1 to n like this:
```1 2 3
4 5
6
7
```
There are a bunch of permutations in Sn called "column permutations", that only permute the numbers within each column of our Young diagram. And there are a bunch called "row permutations", that only permute the numbers within each row.

We can form an idempotent p in C[Sn] that antisymmetrizes over all column permutations. We get p by taking the sum of all even column permutations minus the sum of all odd column permutations, and then dividing by the total number of column permutations.

Similarly, we can form an idempotent q in C[Sn] that symmetrizes over all row permutations. We get q by taking the sum of all row permutations divided by the number of row permutations.

Now here's the cool part: pq is a minimal central idempotent in C[Sn], and we get all minimal central idempotents this way! This isn't very obvious, but I went over the proof before writing this, so I know it's true.

Consider n = 3, for example. There are 3 Young diagrams in this case:

```      X X X            X X               X
X                 X
X
```
so S3 has 3 minimal central idempotents and thus 3 irreps, confirming something I already said.

There is a lot more to say about this, but now I want to switch gears and tell you how representations of SU(N) are classified by Young diagrams. Since SU(N) consists of N x N matrices, it has an obvious representation on the vector space C^N, which people call the "fundamental" representation. This is an irrep. If we're trying to cook up irreps of SU(N), this is an obvious place to start.

How can we get a bunch of representations of SU(N) starting from the fundamental representation? One way is to take the fundamental representation and tensor it with itself a bunch of times, say n times:

```           C^N tensor C^N tensor ... tensor C^N tensor C^N
|------------------n copies-------------------|

```
There's no reason in the world this new representation should be irreducible. But we can try to chop it up into irreducible bits. And the easiest way to do this is to look for bits that transform in nice ways when we permute the n copies of C^N. In physics lingo, we have a space of tensors with n indices, and we can look for subspaces consisting of tensors that transform in specified ways when we permute the indices. For example, there will be a subspace consisting of "totally symmetric" tensors that don't change at all when we permute the indices. And a subspace of "totally antisymmetric" tensors that change sign whenever we interchange two indices. And so on....

But to make the "and so on" precise, we need Young diagrams. After all, these describe all the representations of the permutation group.

Here's how it works. The space

```      V =  C^N tensor C^N tensor ... tensor C^N tensor C^N
|------------------n copies-------------------|

```
is not only a representation of SU(N); it's also a representation of Sn. And the actions of these two groups commute! This means that we can chop up V into subspaces using the minimal central idempotents in Sn, and each of these subspaces will be a representation of SU(N).

This much is obvious. The really cool part is that all these subspaces are irreducible representations of SU(N). Even better, we get all the irreps of SU(N) by this process, as we let n vary.

In other words, any Young diagram gives us an irrep of SU(N) consisting of tensors that transform in a certain way under permutation of indices, and we get all irreps this way.

If you think about it, some of these irreps will be a bit silly. If we have a Young diagram with more than N rows, we'll be antisymmetrizing over more than N indices, which gives a zero-dimensional representation of SU(N). We can ignore these.

Also, if we have a Young diagram that has just one column and exactly N rows, we'll get the space of completely antisymmetric tensors with N indices. This is a 1-dimensional space. Applying a matrix in SU(N) to a tensor of this sort just multiplies it by the determinant of that matrix, which is 1 by the definition of SU(N). So this Young diagram gives the trivial representation of SU(N). That's not too silly - the trivial representation is important, in its own trivial sort of way. But notice: the trivial representation is already described by the Young diagram with no boxes! So it's redundant to also consider the Young diagram with one column and N rows.

By the same logic, we can remove any column with exactly N rows from a Young diagram without changing the rep of SU(N) that we get.

So here's the bottom line: irreps of SU(N) correspond in a 1-1 way with Young diagrams having fewer than N rows.

Okay, I've shown you how Young diagrams classify conjugacy classes of Sn, irreps of Sn, and irreps of SU(N). But this is really just the tip of the iceberg!

First of all, we can use Young diagrams packed with numbers, called "Young tableaux", to do all sorts of calculations involving irreps of Sn and SU(N). Say we tensor two irreps and want to decompose it as a direct sum of irreps: how do we do it? Well, we play a little game with Young tableaux and out pops the answer. One relevant buzzword is "Littlewood-Richardson rules". Or say we have an irrep of Sn and want to know how it decomposes into irreps when we restrict it to a subgroup like Sn-1. Or the same for SU(N) and SU(N-1). How do we do this? More messing with Young tableaux. Here one relevant buzzword is "branching rules".

I'll warn you right now: there is an enormous literature on this stuff. The combinatorics of Young diagrams is one of those things that everyone has worked on, from hardnosed chemists to starry-eyed category theorists. It takes a lifetime to master this material, and I certainly have not. But learning even a little is fun, so don't be too scared.

Second of all, Young diagrams are also good for studying the representations of other classical groups, notably GL(N), SL(N), O(N), SO(N), U(N) and Sp(N). All these groups have an obvious "fundamental representation", and we can cook up lots of reps by taking the nth tensor power of the fundamental representation and hitting it with minimal central idempotents in C[Sn]. The story I just told you for SU(N) can be repeated with slight or not-so-slight variations for all these other groups.

Third, we can "q-deform" the whole story, replacing any one of these classical groups by the associated "quantum group", and replacing C[Sn] by the corresponding "Hecke algebra". This is really important in topological quantum field theory and the theory of type II subfactors.

Fourth, there are nice relationships between Young diagrams and algebraic geometry, like the "Schubert calculus" for the cohomology ring of a Grassmanian.

And there's a lot more, but I have to stop somewhere.

So, how does one start learning this stuff?

If you have a certain amount of patience for old-fashioned terminology, I might recommend going back to the classic text on classical groups:

1) Hermann Weyl, The Classical Groups, Their Invariants and Representations, Princeton U. Press, Princeton, 1997.

Weyl coined the term "classical groups" for the purposes of this book, which was first published in 1939. His prose is beautiful, but I warn you, this book is not the way to learn Young diagrams in a hurry.

For a user-friendly approach that's aimed at physicists, but still includes proofs of all the key results, you can't beat this:

2) Irene Verona Schensted, A Course on the Applications of Group Theory to Quantum Mechanics, NEO Press, Box 32, Peaks Island, Maine.

A girlfriend of mine gave me a copy when I was a college student, but only much later did I realize how great a book it is. Unfortunately it's out of print! Someone should reprint this gem.

Here's another book that covers Young diagrams together with applications to physics:

3) Shlomo Sternberg, Group Theory and Physics, Cambridge U. Press, Cambridge, 1994.

Both these books, but especially the latter, describe applications of Young diagrams to particle physics, like Gell-Mann's famous "eight-fold way", which was based on positing an SU(3) symmetry between the up, down and strange quarks.

Then there are more advanced texts, for when your addiction to Young diagrams becomes more serious. For the combinatorial side of things, these are good:

4) Gordon Douglas James and Adalbert Kerber, The Representation Theory of the Symmetric Group, Addison-Wesley, Reading, Massachusetts, 1981.

5) Bruce Eli Sagan, The Symmetric Group: Representations, Combinatorial Algorithms, and Symmetric Functions, Wadsworth and Brooks, Pacific Grove, California, 1191.

For a more conceptual approach to representation theory that puts Young diagrams in a bigger context, try this:

6) Roe Goodman and Nolan R. Wallach, Representations and Invariants of the Classical Groups, Cambridge University Press, Cambridge, 1998.

It's sort of an updated version of Weyl's book. Finally, here's a mathematically sophisticated book that really gives you a Young diagram workout:

7) William Fulton, Young Tableaux: With Applications to Representation Theory and Geometry, Cambridge U. Press, Cambridge, 1997.

Now, my friend Allen Knutson is a real Young diagram fiend. Together with Terry Tao, he helped prove something called "Horn's conjecture", which had been bugging people for decades, and has implications for a huge number of questions. I have a feeling Allen is going to send me a nasty email saying that I didn't actually say anything interesting about Young diagrams. In an attempt to pacify him, I'll direct you to Fulton's excellent review article on this subject:

8) William Fulton, Eigenvalues, invariant factors, highest weights, and Schubert calculus, Bull. Amer. Math. Soc. 37 (2000), 209-249, also available as math.AG/9908012.

as well as Allen and Terry's papers on the subject:

9) Allen Knutson and Terence Tao, The honeycomb model of GL(n) tensor products I: the saturation conjecture, preprint available as math.RT/9807160

10) Allen Knutson, The symplectic and algebraic geometry of Horn's problem, preprint available as math.LA/9911088.

11) Allen Knutson and Terence Tao, Honeycombs and sums of Hermitian matrices, preprint available as math.RT/0009048

But I should also mention the question that Horn's conjecture settles!

There are many ways to phrase it; here's the easiest one. If you know the eigenvalues of two n x n Hermitian matrices A and B, what are the possible eigenvalues of their sum? There are a bunch of linear inequalities that must hold; find a necessary and sufficient set.

This may not seem related to Young diagrams, but it is....

Devin had been studying this region for ten years, poking his way through a place not much larger than the town in which he lived, and had still not deciphered half its routes. This hugeness inside of smallness creates a matrix of intersections, precious and incalculable channels one after the next. It is a fractal landscape like the surface of a leaf, veins within veins, or the arborescent feathers of ice forming barbs within barbs across the surface of a pond. - Craig Childs, Soul of Nowhere