In Random Points on a Sphere (Part 1), we learned an interesting fact. You can take the unit sphere in , randomly choose two points on it, and compute their distance. This gives a random variable, whose moments you can calculate.

And now the interesting part: when n = 1, 2 or 4, and seemingly in no other cases, all the even moments are *integers*.

These are the dimensions in which the spheres are *groups*. We can prove that the even moments are integers because they are differences of dimensions of certain representations of these groups. Rogier Brussee and Allen Knutson pointed out that if we want to broaden our line of investigation, we can look at other groups. So that’s what I’ll do today.

If we take a representation of a compact Lie group we get a map from group into a space of square matrices. Since there is a standard metric on any space of square matrices, this lets us define the distance between two points on the group. This is different than the distance defined using the shortest geodesic *in the group*: instead, we’re taking a straight-line path in the larger space of matrices.

If we randomly choose two points on the group, we get a random variable, namely the distance between them. We can compute the moments of this random variable, and today I’ll prove that *the even moments are all integers*.

So, we get a sequence of integers from any representation of any compact Lie group So far we’ve only studied groups that are spheres:

• The defining representation of on the real numbers gives the powers of 2.

• The defining representation of on the complex numbers gives the central binomial coefficients

• The defining representation of on the quaternions gives the Catalan numbers.

It could be fun to work out these sequences for other examples. Our proof that the even moments are integers will give a way to calculate these sequences, not by doing integrals over the group, but by counting certain ‘random walks in the Weyl chamber’ of the group. Unfortunately, we need to count walks in a certain weighted way that makes things a bit tricky for me.

But let’s see why the even moments are integers!

If our group representation is real or quaternionic, we can either turn it into a complex representation or adapt my argument below. So, let’s do the complex case.

Let be a compact Lie group with a unitary representation on This means we have a smooth map

where is the algebra of complex matrices, such that

and

where is the conjugate transpose of the matrix

To define a distance between points on we’ll give its metric

This clearly makes into a -dimensional Euclidean space. But a better way to think about this metric is that it comes from the norm

where is the trace, or sum of the diagonal entries. We have

I want to think about the distance between two randomly chosen points in the group, where ‘randomly chosen’ means with respect to normalized Haar measure: the unique translation-invariant probability Borel measure on the group. But because this measure and also the distance function are translation-invariant, we can equally well think about the distance between the identity 1 and *one* randomly chosen point in the group. So let’s work out this distance!

I really mean the distance between and so let’s compute that. Actually its square will be nicer, which is why we only consider *even* moments. We have

Now, any representation of has a character

defined by

and characters have many nice properties. So, we should rewrite the distance between and the identity using characters. We have our representation whose character can be seen lurking in the formula we saw:

But there’s another representation lurking here, the dual

given by

This is a fairly lowbrow way of defining the dual representation, good only for unitary representations on but it works well for us here, because it lets us instantly see

This is useful because it lets us write our distance squared

in terms of characters:

So, the distance squared is an integral linear combination of characters. (The constant function 1 is the character of the 1-dimensional trivial representation.)

And this does the job: it shows that all the even moments of our distance squared function are integers!

Why? Because of these two facts:

1) If you take an integral linear combination of characters, and raise it to a power, you get another integral linear combination of characters.

2) If you take an integral linear combination of characters, and integrate it over you get an integer.

I feel like explaining these facts a bit further, because they’re part of a very beautiful branch of math, called character theory, which every mathematician should know. So here’s a quick intro to character theory for beginners. It’s not as elegant as I could make it; it’s not as simple as I could make it: I’ll try to strike a balance here.

There’s an abelian group consisting of formal differences of isomorphism classes of representations of , mod the relation

Elements of are called **virtual representations** of Unlike actual representations we can subtract them. We can also add them, and the above formula relates addition in to direct sums of representations.

We can also multiply them, by saying

and decreeing that multiplication distributes over addition and subtraction. This makes into a ring, called the representation ring of

There’s a map

where is the ring of continuous complex-valued functions on This map sends each finite-dimensional representation to its character This map is one-to-one because we know a representation up to isomorphism if we know its character. This map is also a ring homomorphism, since

and

These facts are easy to check directly.

We can integrate continuous complex-valued functions on so we get a map

The first non-obvious fact in character theory is that we can compute inner products of characters as follows:

where the expression at right is the dimension of the space of ‘intertwining operators’, or morphisms of representations, between the representation and the representation

What matters most for us now is that this inner product is an *integer*. In particular, if is the character of any representation,

is an integer because we can take to be the trivial representation in the previous formula, giving

Thus, the map

actually takes values in

Now, our distance squared function

is actually the image under of an element of the representation ring, namely

So the same is true for any of its powers—and when we integrate any of these powers we get an integer!

This stuff may seem abstract, but if you’re good at tensoring representations of some group, like you should be able to use it to compute the even moments of the distance function on this group more efficiently than using the brute-force direct approach. Instead of complicated integrals we wind up doing combinatorics.

I would like to know what sequence of integers we get for A much easier, less thrilling but still interesting example is This is the 3-dimensional real projective space which we can think of as embedded in the 9-dimensional space of real matrices. It’s sort of cool that I could now work out the even moments of the distance function on this space by hand! But I haven’t done it yet.

I do not have the stamina to make any attempt at your SU(3) challenge, but reading this post made me think of a question.

We’ve gone away from the spheres to consider distances in the norm of representations of other groups.

But there is a different generalization we could have made from the original problem. Specifically, you said that it’s only n=0,1,3 where there is some Lie group which has topology . But,

everyis a homogeneous space that comes from the quotient of groups, namely .Is there a representation-theoretic approach to computing moments of distances in based on being a group quotient? Is this something that might come from induced or restricted representations?

Yes, I suspect one could compute the even moments of the distances on other spheres using the representation theory of The problem is that I don’t see any interesting thing to do with this approach. In the case of spheres that are groups we had a nice mystery to solve: why are the even moments integers? Representation theory gave an easy way to explain this phenomenon. I don’t see a comparably exciting puzzle in the general case that I can solve using representation theory.

But that’s not to say that such a puzzle won’t appear, or that someone better at representation theory could solve one of the existing ones now.

For example: it seems that infinitely many moments are _non_integral except for spheres in 1, 2 or 4 dimensions. Why? I have no idea how to prove this, and if I had to try I’d start thinking hard about Pomerance’s paper on central binomial coefficients. But maybe someone smarter could spot a connection to representation theory.

I guess duetosymmetry thought you might be interested in checking out whether you obtain the quotients in your and Greg Egan’s table via Group Quotients. Since the Group quotients involve products instead of tensor products this might be though considerably more difficult, I don’t know.

This problem has been solved by Jean-Marie Souriau with Lie Groups Thermodynamics and by defining a Gibbs density that is covariant under the action of the Group:

See MDPI book “Differential Geometrical Theory of Statistics”:

http://www.mdpi.com/books/pdfview/book/313

with papers:

[1] Marle, C.-M. From Tools in Symplectic and Poisson Geometry to J.-M. Souriau’s Theories of Statistical Mechanics and Thermodynamics. Entropy 2016, 18, 370

http://www.mdpi.com/1099-4300/18/10/370/pdf

[2] Barbaresco, F. Geometric Theory of Heat from Souriau Lie Groups Thermodynamics and Koszul Hessian Geometry: Applications in Information Geometry for Exponential Families. Entropy 2016, 18, 386.

http://www.mdpi.com/1099-4300/18/11/386/pdf

[3] de Saxcé, G. Link between Lie Group Statistical Mechanics and Thermodynamics of Continua. Entropy 2016, 18, 254.

http://www.mdpi.com/1099-4300/18/7/254/pdf

Could you point me to a specific equation or theorem number in which this problem is solved?

“This problem” is to prove that the even moments of the distance function on a compact Lie group, thought of as a subvariety of one of its matrix representations, are integers.

See in paper of Marle:

7.3. Examples of Generalized Gibbs States

7.3.1. Action of the Group of Rotations on a Sphere

This paper is very nice, but it does not mention the problem we’re concerned with here, which is to compute the even moments of the distance function on a compact Lie group, thought of as a subvariety of one of its matrix representations—or prove that these moments are integers.

If G is compact connected, we can use character theory as follows to compute the dimension of the invariant space in ; it’s the constant term in the Laurent polynomial

where that product is the Weyl denominator.

For the SU(2) case, we’re looking at the 0th minus 2nd coefficient of , so , the Catalan number.

For the SU(3), I’m getting the coefficients

1, 6, 38, 252, 1722, 11986, 84347, 597078, 4235854, 30021888, 211934817, 1485161772, 10288303886, 70055772234, 464923042633, 2965117407662, 17696545167474, 92953547867476, 347577444678438, -472718240879256, …

so not all positive. Of course my program could be wrong. (No, this is not in the OEIS.)

Cool!

The average of any power of the distance must be nonnegative, so I don’t think that negative number can be right. This makes me wonder how many of the entries in your sequence I can trust!

But it’s cool that you have a way to get at these things pretty efficiently. I need to re-learn the Weyl character formula, and maybe this time it will stick.

One way to understand WCF is as the Euler characteristic of the BGG resolution. The th term in the BGG resolution has, as character, where (so, only dominant if ) and is the Weyl denominator from above. Hence, if you multiply by and look at multiplicities of dominant weights only, you see the -multiplicities.

In the SO(3) case, do you get 1, 3, 10, 35, 126, 462, 1716, 6435, 24310, 92378, … ?

I was expressing the simple roots wrongly in terms of the fundamental weights. Now I’m getting

1,

6,

38,

250,

1692,

11706,

82426,

588846,

4258010,

31110384,

229347792,

1704100272,

12750305736,

95996382930,

726836098674,

5531550989062,

42296249058042,

324821371944692,

2504614899752820,

19385423092317708,

150573060108385344,

1173463240919004144,

9174138207471038448,

71939453721618678480,

565737960449151026032,

4461251634562667480256,

35273121820700443391936,

279598150264510560049728,

2221721114362793423931600,

17695991343217778226366402

which again, is not in the OEIS.

I’m not sure when I’m gonna have time to study this calculation, but thanks for doing it! This should be really fun to think about. It looks it grows like asymptotically. I wonder what is!

The ratio starts at 6 and seems to keep increasing, at least out to n=40, where it’s gotten to about 8.1.

Hmm, maybe a logarithmic correction?

You might already know this, but this sequence satisfies the recurrence relation:

In the large-n limit, this is satisfied by:

Actually, a limit of 8 will also satisfy the quadratic for the ratio between successive terms, but empirically it’s easy to check that the actual ratio approaches 9. At n=10000 it’s 8.9964.

Where did you get that recurrence relation from, Greg? Dd you spot the pattern in Allen’s data, or did you prove something from scratch about moments of distances between randomly chosen points in SO(3), perhaps using the representation theory?

Either way, it’s impressive. I don’t see where the 9 in

is coming from, but I feel it must ultimately be something simple.

I don’t really follow how Allen got these numbers (mainly because I’m far too lazy right now to go off and learn new things about Lie algebras), so I found the recurrence formula from his raw data. But whatever he did to calculate these numbers, he can probably get the recurrence formula fairly easily just by inspecting the algorithm he used.

The asymptotic ratio comes from setting:

in the recurrence formula, and then taking the limit as to get a quadratic in :

which has roots .

8 is the dimension of SU(3), and 9 is the

complexdimension of the algebra of matrices it’s a subgroup of, but those could be coincidences.What can we say about the large-n asymptotics of the nth moments of a compactly supported probability distribution? There could be some easy very general results.

What’s the maximum distance between two points of SU(3)? I have a feeling that may play a role. If we have a probability distribution on [0,a] that vanishes only at the endpoints, a very high moment

should grow sort of like though not quite that fast. I think it should at least grow faster than for any

In our problem, the maximum distance between points of SU(3) should play the role of the number

I’m too lazy to calculate anything rigorous, but some quick numerical calculations very strongly suggest that the maximum squared distance between elements of SU(3) is 9.

Excellent!

This is the sequence I get for (the defining representation of) SU(3), too.

While this sequence isn’t in the OEIS, a little playing around shows that it agrees with

where is the inverse binomial transform. A151366 is “Number of walks within (the first quadrant of ) starting and ending at (0,0) and consisting of n steps taken from {(-1, 0), (-1, 1), (0, -1), (0, 1), (1, -1), (1, 0)}” (at least to all the terms recorded here), which can be interpreted as something like the number of closed walks of length n in a Weyl chanmber for A2 and based at the origin. This can be reinterpreted (as I think was mentioned for another group elsewhere in the comments) as the sequence of the dimensions of the invariant subspaces (i.e., the number copies of the trivial representation) of the nth tensor powers of the defining representation of SU(3).

Trying this sort of thing out for a sampling of other small compact groups G and small representations gives that this behavior holds for those groups, too: The sequence is something like , with negatives appearing in some places, where are the dimensions of the invariant subspaces of the nth tensor powers of (surely this is already known in some guise to people who know more representation theory than I do). The sequences are recorded in the OEIS (sometimes in other guises) for at least several small G.

Playing around with some other small examples reveals a few curiosities. To name one, the sequences for the defining representations of and agree to high order, at least for small . For both sequences start 1, 4, 17, 76, but the next terms for and are, respectively, 355 and 356. I don’t immediately see a good explanation for this phenomenon.

That’s incredibly cool—thanks, Travis! More mysteries.

What’s Do you mean or maybe

There should be some fairly easy way to work out the even moments for the product of two groups (in given representations) if we know them for each one separately, thanks to the Pythagorean theorem.

By D2 I mean SU(2) x SU(2).

One can compute algorithmically the moments of a product from the moments of the factors using above characterization in terms of the binomial transform and the fact that the dimensions of invariant subspaces of a tensor product are given by are

I agree, though, that there should be a better way to do this, but I can’t see it yet. (It’s a nice exercise featuring some interesting sequences to verify this for G = H = SU(2) and taking both representations to be the defining representation, which in particular verifies which incarnation of D2 was meant.)

It turns out there’s a simple explanation, by the way, for the observed similarity of sequences for the defining representations of C_n and D_n. For large n, the sequences of dimensions of invariant subspaces of the tensor powers of the defining representations V of B_n, C_n, D_n start 1, 0, 1, 0, 3, 0, 15, 0, 105, … (the double factorials of the odd numbers alternating with zero). Here, the second 1 corresponds to the subspace spanned by the symmetric 2-form (for B_n and D_n) or the skew 2-form (C_n) stabilized by the group. Now, the defining representations of C_n and D_n both have dimension 2n, and the even moments of both begin with the subsequences .

For any n this pattern breaks at the (2n)th term, at least in part because for D_n there is a trivial representation in spanned by a volume form for the bilinear form, but there is no such invariant for smaller tensor powers. For n = 2 the sequences are 1, 0, 1, 0, 3, … and 1, 0, 1, 0, 4, … , and so applying the binomial transform gives that precisely the first four even moments of C_2 and D_2 all agree, but the 8th moments disagree, exactly as observed.

Something similar happens with G2 and B3; as you know G2 preserves an inner product on its 7-dimensional irrep V, giving an embedding . G2 preserves (in fact is characterized by stabilizing) a single certain 3-form up to scale, but B3 preserves no 3-form, so applying gives that the 0th, 2nd, and 4th moments of G2 and B3 agree (1, 7, 50), but the 6th moments do not (363 and 364, resp.).

I think we almost answered the question about the maximum distance between elements of SU(3) before, when we were looking at the hypocycloids from the trace of elements of SU(n).

In this comment:

https://johncarlosbaez.wordpress.com/2013/12/03/rolling-hypocycloids/#comment-34248

I wrote:

“if an element of has n equal eigenvalues (which must be nth roots of unity), its trace will achieve the maximum possible absolute value of n.”

The squared distance between 1 and g in SU(3) is:

If g has 3 identical eigenvalues of , we have , both of which give squared distances of 9.

That doesn’t rigorously prove the result, but if you look at the deltoid obtained from the trace, that these two cusps maximise the distance is obvious.

Great! For with even, it’s somehow obvious that the point farthest from the identity is the matrix Since 3 is odd, Last night I guessed that the farthest points from the identity might be the matrices with a cube root of 1 down the diagonal. But it wasn’t obvious, and I was too tired to compute the distance.

Now that I think about it, it’s sort of bizarre that there are

twopoints in that are maximally distant from the identity… and thus, two points maximally distant from each point. What kind of weird shape works like that? Why would a group be so ‘bumpy’? Can’t you get a point that’s even farther, somewhere between the two supposedly farthest points?Of course the group is ‘bumpy’ in exactly the same way: it’s triangular. And it’s a subgroup of , namely the center. But I never thought of as looking sort of ‘triangular’.

Maybe I should have, given our work on hypocycloids.

I guess the next really fun case is There are 5 cube roots of unity but I guess the farthest points from the identity in are the diagonal matrix with all diagonal entries and the diagonal matrix with all diagonal entries

SU(3)/conjugation is an equilateral triangle, and the distance along the group (not matrix distance) is given by distance in that triangle. More generally a compact simply-connected simple group, mod conjugation, is modelable by the Weyl alcove. (So you should, indeed, think of SU(3) as triangular.)

Neat! I had a mental image of compact Lie groups as all being ’round’, which made me wrongly feel that each point had a unique maximally distant point.

So we could look at the matrix distance between points in SU(3)/conjugation, and it would give a different metric on the triangle. What’s that other metric like?

These images might help with visualisation of SU(3).

They’re explained in detail in these G+ posts:

https://plus.google.com/113086553300459368002/posts/BuWJ9eR9Qnw

https://plus.google.com/113086553300459368002/posts/M9oYhoApTxR

John Baez wrote:

“So we could look at the matrix distance between points in SU(3)/conjugation, and it would give a different metric on the triangle. What’s that other metric like?”

If we parameterise diagonal matrices by two of their phases, , then the squared distance between two such matrices is:

Thanks!

I’m feeling a bit confused… that metric doesn’t give the triangle the geometry of a deltoid, does it? I see no reason it

should. In fact I don’t even see why it should be a Riemannian metric! And that makes me realize I don’t understand non-Riemannian metrics on a manifold, or even a solid triangle.If we take the usual Riemannian metric on SU(3) and use that to get a metric on SU(3)/conjugation, does

thatgive this space the geometry of a deltoid? I don’t think it should: the deltoid has sharp cuspy corners, while I think the metric I’m talking about should give corners with 60° angles, because right near a corner we might as well be working with Cartan algebra mod the Weyl group, i.e. the plane mod the symmetry group of an equilateral triangle.By “the geometry of a deltoid” do you mean that distances are found by taking the trace of the matrix (which lies in a deltoid-shaped subset of the complex plane) and then using the standard Euclidean metric on the complex plane?

Distances measured that way wouldn’t agree with distances in the space of 3×3 complex matrices as defined in your post. I don’t

thinkthey’d agree with “the usual Riemannian metric on SU(3)” … though to be honest I’m not really sure what that is! Is it somehow connected to the Haar measure, or is it the induced metric you get from the embedding of SU(3) in 18-dimensional Euclidean space? (Unlike SU(2), it’s not obvious to me that these would be equivalent notions.)Greg wrote:

Yes, or more generally: there’s the metric you wrote down, and there’s the deltoid which is a subset of the flat plane, and one could ask if there’s

anydistance-preserving bijection between them (allowing rescaling to increase one’s chance of success). The bijection you’re describing is the ‘reasonable’ one.Any compact simple Lie group has a Riemannian metric, unique up to a scale factor, that’s invariant under both left and right translations. In general you can get ahold of it using the “Killing form”, which is the unique-up-to-scale invariant metric on the Lie algebra: you can use

where is any nontrivial representation of the Lie algebra. But thanks to uniqueness it’s also the induced metric you get from the usual embedding of SU(3) in 3×3 matrices.

In the case of SU(2), where we know the moments exactly, the maximum squared distance between elements of the group is 4, and the asymptotic form we get from the Catalan numbers turns out to be:

With a bit of numerical tinkering, I conjecture that:

This is not a great fit at the highest value Allen posted above, n=30, but by n=200,000 it’s within half a percent.

I forgot that Allen’s sequence 1,6,38,… starts at n=0, not n=1, so I was out by one position. The correct asymptotic form is:

Also, we’re measuring norms here a bit differently on than when we were treating it as a unit sphere, because:

With that norm, the even moments have extra factors of times the Catalan numbers, and the maximum squared distance between elements of the group is 8. So the asymptotic form is:

This is what it looks like if you take the diagonal subgroup of SU(3), a 2-torus, embed (multiple copies) in the plane using the flat metric that comes from its Killing form, and then take the quotient under the where you permute the diagonal entries, giving the equilateral triangles here that are SU(3)/conjugation.

The colours here encode the trace.

I think since we’re looking at a 2 dimensional space, since all 2-metrics are conformally flat, we can perform a conformal transformation to put this in a standard form of the metric on Euclidean 2-space. So, the angles at the corners will be the same, but somehow the shapes of the triangles will get deformed.

Leo, when you write “we can perform a conformal transformation to put this in a standard form of the metric on Euclidean 2-space”, by “this” do you mean the matrix-based metric where the distance squared between diagonal matrices (with phases for two of their entries) is:

I guess you can’t mean the Killing form metric, since that’s already Euclidean.

For any Riemannnian metric on a surface, near any point we can multiply it by a smooth positive function so that the metric is flat. I think that’s what Leo was trying to say. Anyway, it’s true. (It’s just a local statement, not a global one, since the Gauss-Bonnet theorem says any metric on a sphere has curvature whose integral over the whole sphere is nonzero.)

However, one of the several metrics we’re talking about is – I believe – not a Riemannian metric. Namely, the ‘straight line distance’ between two points in SU(3) defined using a straight line in the ambient Euclidean space of 3×3 matrices.