Random Points on a Group

In Random Points on a Sphere (Part 1), we learned an interesting fact. You can take the unit sphere in $\mathbb{R}^n$ , 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 $G,$ 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 $\rho$ of any compact Lie group $G.$ So far we’ve only studied groups that are spheres:

• The defining representation of $\mathrm{O}(1) \cong S^0$ on the real numbers $\mathbb{R}$ gives the powers of 2.

• The defining representation of $\mathrm{U}(1) \cong S^1$ on the complex numbers $\mathbb{C}$ gives the central binomial coefficients $\binom{2n}{n}.$

• The defining representation of $\mathrm{Sp}(1) \cong S^3$ on the quaternions $\mathbb{H}$ 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 $G$ be a compact Lie group with a unitary representation $\rho$ on $\mathbb{C}^n.$ This means we have a smooth map

$\rho \colon G \to \mathrm{End}(\mathbb{C}^n)$

where $\mathrm{End}(\mathbb{C}^n)$ is the algebra of $n \times n$ complex matrices, such that

$\rho(1) = 1$

$\rho(gh) = \rho(g) \rho(h)$

and

$\rho(g) \rho(g)^\dagger = 1$

where $A^\dagger$ is the conjugate transpose of the matrix $A.$

To define a distance between points on $G$ we’ll give $\mathrm{End}(\mathbb{C}^n)$ its metric

$\displaystyle{ d(A,B) = \sqrt{ \sum_{i,j} \left|A_{ij} - B_{ij}\right|^2} }$

This clearly makes $\mathrm{End}(\mathbb{C}^n)$ into a $2n^2$ -dimensional Euclidean space. But a better way to think about this metric is that it comes from the norm

$\displaystyle{ \|A\|^2 = \mathrm{tr}(AA^\dagger) = \sum_{i,j} |A_{ij}|^2 }$

where $\mathrm{tr}$ is the trace, or sum of the diagonal entries. We have

$d(A,B) = \|A - B\|$

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 $g$ in the group. So let’s work out this distance!

I really mean the distance between $\rho(g)$ and $\rho(1),$ so let’s compute that. Actually its square will be nicer, which is why we only consider even moments. We have

$\begin{array}{ccl} d(\rho(g),\rho(1))^2 &=& \|\rho(g) - \rho(1)\|^2 \\ \\ &=& \|\rho(g) - 1\|^2 \\ \\ &=& \mathrm{tr}\left((\rho(g) - 1)(\rho(g) - 1)^\dagger)\right) \\ \\ &=& \mathrm{tr}\left(\rho(g)\rho(g)^\dagger - \rho(g) - \rho(g)^\ast + 1\right) \\ \\ &=& \mathrm{tr}\left(2 - \rho(g) - \rho(g)^\dagger \right) \end{array}$

Now, any representation $\sigma$ of $G$ has a character

$\chi_\sigma \colon G \to \mathbb{C}$

defined by

$\chi_\sigma(g) = \mathrm{tr}(\sigma(g))$

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

$d(\rho(g),\rho(1))^2 = \mathrm{tr}\left(2 - \rho(g) - \rho(g)^\dagger \right)$

But there’s another representation lurking here, the dual

$\rho^\ast \colon G \to \mathrm{End}(\mathbb{C}^n)$

given by

$\rho^\ast(g)_{ij} = \overline{\rho(g)_{ij}}$

This is a fairly lowbrow way of defining the dual representation, good only for unitary representations on $\mathbb{C}^n,$ but it works well for us here, because it lets us instantly see

$\mathrm{tr}(\rho(g)^\dagger) = \mathrm{tr}(\rho^\ast(g)) = \chi_{\rho^\ast}(g)$

This is useful because it lets us write our distance squared

$d(\rho(g),\rho(1))^2 = \mathrm{tr}\left(2 - \rho(g) - \rho(g)^\dagger \right)$

in terms of characters:

$d(\rho(g),\rho(1))^2 = 2n - \chi_\rho(g) - \chi_{\rho^\ast}(g)$

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 $G,$ 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 $R(G)$ consisting of formal differences of isomorphism classes of representations of $G$ , mod the relation

$[\rho] + [\sigma] = [\rho \oplus \sigma]$

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

We can also multiply them, by saying

$[\rho] [\sigma] = [\rho \otimes \sigma]$

and decreeing that multiplication distributes over addition and subtraction. This makes $R(G)$ into a ring, called the representation ring of $G.$

There’s a map

$\chi \colon R(G) \to C(G)$

where $C(G)$ is the ring of continuous complex-valued functions on $G.$ This map sends each finite-dimensional representation $\rho$ to its character $\chi_\rho.$ 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

$\chi_{\rho \oplus \sigma} = \chi_\rho + \chi_\sigma$

and

$\chi_{\rho \otimes \sigma} = \chi_\rho \chi_\sigma$

These facts are easy to check directly.

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

$\displaystyle{\int} \colon C(G) \to \mathbb{C}$

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

$\displaystyle{\int} \overline{\chi_\sigma} \chi_\rho = \dim(\mathrm{hom}(\sigma,\rho))$

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

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

$\displaystyle{\int} \chi_\rho$

is an integer because we can take $\sigma$ to be the trivial representation in the previous formula, giving $\chi_\sigma = 1.$

Thus, the map

$R(G) \stackrel{\chi}{\longrightarrow} C(G) \stackrel{\int}{\longrightarrow} \mathbb{C}$

actually takes values in $\mathbb{Z}.$

Now, our distance squared function

$2n - \chi_\rho - \chi_{\rho^\ast} \in C(G)$

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

$2n - [\rho] - [\rho^\ast]$

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 $\mathrm{SU}(3),$ 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 $\mathrm{SU}(3).$ A much easier, less thrilling but still interesting example is $\mathrm{SO}(3).$ This is the 3-dimensional real projective space $\mathbb{R}\mathrm{P}^3,$ which we can think of as embedded in the 9-dimensional space of $3\times 3$ 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.

This entry was posted on Friday, July 13th, 2018 at 7:43 pm and is filed under mathematics. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

41 Responses to Random Points on a Group

duetosymmetry says:

13 July, 2018 at 10:45 pm

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 $S^n$ to consider distances in the $L^2$ 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 $S^n$ . But, every $S^n$ is a homogeneous space that comes from the quotient of groups, namely $S^n \sim O(n+1)/O(n)$ .

Is there a representation-theoretic approach to computing moments of distances in $S^n$ based on being a group quotient? Is this something that might come from induced or restricted representations?

Reply
- John Baez says:
  
  14 July, 2018 at 2:06 am
  
  Yes, I suspect one could compute the even moments of the distances on other spheres $S^n$ using the representation theory of $\mathrm{O}(n+1).$ 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.
  
  Reply
  - nad says:
    
    14 July, 2018 at 7:28 am
    
    Yes, I suspect one could compute the even moments of the distances on other spheres S^n using the representation theory of $\mathrm{O}(n+1)$ . 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?
    
    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.
    
    Reply
BARBARESCO Frederic says:

14 July, 2018 at 6:12 am

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

Reply
- John Baez says:
  
  14 July, 2018 at 7:14 am
  
  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.
  
  Reply
BARBARESCO Frederic says:

14 July, 2018 at 7:23 am

See in paper of Marle:
7.3. Examples of Generalized Gibbs States
7.3.1. Action of the Group of Rotations on a Sphere

Reply
- John Baez says:
  
  14 July, 2018 at 3:38 pm
  
  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.
  
  Reply
Allen Knutson says:

14 July, 2018 at 4:06 pm

If G is compact connected, we can use character theory as follows to compute the dimension of the invariant space in $V^{\otimes n}$ ; it’s the constant term in the Laurent polynomial

$\prod_{\beta \in \Delta_+} (1-t^\beta) \text{ch}(V)^n,$

where that product is the Weyl denominator.

For the SU(2) case, we’re looking at the 0th minus 2nd coefficient of $(-x^{-1}+2-x)^n = (-x^{-1})^n (1-x)^{2n}$ , so ${2n\choose n} - {2n\choose n+2}$ , 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.)

Reply
- John Baez says:
  
  15 July, 2018 at 1:24 am
  
  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.
  
  Reply
  - Allen Knutson says:
    
    16 July, 2018 at 4:35 pm
    
    One way to understand WCF is as the Euler characteristic of the BGG resolution. The $w$ th term in the BGG resolution has, as character, $t^{w\cdot \lambda} / \text{Den}$ where $w\cdot \lambda = w(\lambda+\rho)-\rho$ (so, only dominant if $w=1$ ) and $\text{Den}$ is the Weyl denominator from above. Hence, if you multiply by $\text{Den}$ and look at multiplicities of dominant weights only, you see the $G$ -multiplicities.
    
    Reply
    - Allen Knutson says:
      
      24 July, 2018 at 1:31 pm
      
      In the SO(3) case, do you get ${2n+1 \choose n} =$ 1, 3, 10, 35, 126, 462, 1716, 6435, 24310, 92378, … ?
  - Allen Knutson says:
    
    24 July, 2018 at 7:36 pm
    
    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.
    
    Reply
    - John Baez says:
      
      26 July, 2018 at 5:58 am
      
      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 $k a^n$ asymptotically. I wonder what $a$ is!
    - allenknutson says:
      
      6 August, 2018 at 3:19 am
      
      The ratio starts at 6 and seems to keep increasing, at least out to n=40, where it’s gotten to about 8.1.
    - John Baez says:
      
      6 August, 2018 at 4:48 am
      
      Hmm, maybe a logarithmic correction?
    - Greg Egan says:
      
      6 August, 2018 at 4:48 am
      
      You might already know this, but this sequence satisfies the recurrence relation:
      
      $y_1 =1$
      $y_2 = 6$
      $\displaystyle{y_n=\frac{72 (2-n) (n+1) y_{n-2}+\left(17 n^2+33 n-14\right) y_{n-1}}{(n+2) (n+3)}}$
      
      In the large-n limit, this is satisfied by:
      
      $\frac{y_n}{y_{n-1}} \to 9$
      
      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.
    - John Baez says:
      
      6 August, 2018 at 5:24 am
      
      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
      
      $\displaystyle{ \lim_{n \to \infty} \frac{y_{n+1}}{y_n} = 9 }$
      
      is coming from, but I feel it must ultimately be something simple.
    - Greg Egan says:
      
      6 August, 2018 at 5:50 am
      
      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:
      
      $y_n=\alpha^2 y_{n-2}$
      $y_{n-1}=\alpha y_{n-2}$
      
      in the recurrence formula, and then taking the limit as $n\to \infty$ to get a quadratic in $\alpha$ :
      
      $\alpha^2 = -72 + 17\alpha$
      
      which has roots $\alpha=8,9$ .
    - John Baez says:
      
      6 August, 2018 at 5:54 am
      
      8 is the dimension of SU(3), and 9 is the complex dimension 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.
    - John Baez says:
      
      6 August, 2018 at 6:16 am
      
      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 $p$ on [0,a] that vanishes only at the endpoints, a very high moment
      
      $\displaystyle{ \int_0^a x^n p(x) \, dx }$
      
      should grow sort of like $O(a^n),$ though not quite that fast. I think it should at least grow faster than $O(b^n)$ for any $b < a.$
      
      In our problem, the maximum distance between points of SU(3) should play the role of the number $a.$
    - Greg Egan says:
      
      6 August, 2018 at 7:04 am
      
      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.
    - John Baez says:
      
      6 August, 2018 at 3:21 pm
      
      Excellent!
    - Travis Willse says:
      
      6 August, 2018 at 10:25 pm
      
      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
      
      $n \mapsto \mathcal{I}^6[(-1)^n \cdot A151366])[n],$
      
      where $\mathcal{I}$ is the inverse binomial transform. A151366 is “Number of walks within $\mathbb{N}^2$ (the first quadrant of $\mathbb{Z}^2$ ) 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 $\rho$ gives that this behavior holds for those groups, too: The sequence is something like $\mathcal{I}^{\dim \rho}[b_n]$ , with negatives appearing in some places, where $\{b_n\}$ are the dimensions of the invariant subspaces of the nth tensor powers of $\rho$ (surely this is already known in some guise to people who know more representation theory than I do). The sequences $\{b_n\}$ 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 $C_n$ and $D_n$ agree to high order, at least for small $n$ . For $n = 2,$ both sequences start 1, 4, 17, 76, but the next terms for $C_2$ and $D_2$ are, respectively, 355 and 356. I don’t immediately see a good explanation for this phenomenon.
    - John Baez says:
      
      6 August, 2018 at 10:59 pm
      
      That’s incredibly cool—thanks, Travis! More mysteries.
      
      What’s $D_2?$ Do you mean $\mathrm{SO}(4),$ or maybe
      
      $\mathrm{Spin}(4) \cong \mathrm{SU}(2) \times \mathrm{SU(2)}?$
      
      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.
    - Travis Willse says:
      
      8 August, 2018 at 10:27 am
      
      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
      
      $\displaystyle{ \int_{G \times H} (\rho \otimes \sigma)^n = \sum_{k = 0}^n {n \choose k} \int_G \rho^{\otimes k} \int_H \sigma^{\otimes (n - k)} } .$
      
      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 $(-1)^n \mathcal{I}^{2n}[1, 0, 1, 0, ...]$ .
      
      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 $\otimes^{2n} V$ 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 $G_2 \hookrightarrow B_3$ . G2 preserves (in fact is characterized by stabilizing) a single certain 3-form up to scale, but B3 preserves no 3-form, so applying $\mathcal{I}^7$ 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.).
Greg Egan says:

6 August, 2018 at 9:33 am

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 $\mathrm{SU}(n)$ 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:

$6-2\mathrm{Re}(\text{Tr}(g))$

If g has 3 identical eigenvalues of $\exp(\pm 2\pi i/3)$ , we have $\mathrm{Re}(\mathrm{Tr}(g)) = -\frac{3}{2}$ , 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.

Reply
- John Baez says:
  
  6 August, 2018 at 3:34 pm
  
  Great! For $\mathrm{SU}(n)$ with $n$ even, it’s somehow obvious that the point farthest from the identity is the matrix $-1.$ Since 3 is odd, $-1 \notin \mathrm{SU}(3).$ 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 two points in $\mathrm{SU}(3)$ 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 $\mathbb{Z}_3$ is ‘bumpy’ in exactly the same way: it’s triangular. And it’s a subgroup of $\mathrm{SU}(3)$ , namely the center. But I never thought of $\mathrm{SU}(3)$ as looking sort of ‘triangular’.
  
  Maybe I should have, given our work on hypocycloids.
  
  I guess the next really fun case is $\mathrm{SU}(5).$ There are 5 cube roots of unity but I guess the farthest points from the identity in $\mathrm{SU}(5)$ are the diagonal matrix with all diagonal entries $\exp(4\pi i/5)$ and the diagonal matrix with all diagonal entries $\exp(-4\pi i/5).$
  
  Reply
  - allenknutson says:
    
    8 August, 2018 at 1:34 am
    
    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.)
    
    Reply
    - John Baez says:
      
      8 August, 2018 at 6:24 pm
      
      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?
    - Greg Egan says:
      
      9 August, 2018 at 1:34 am
      
      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
    - Greg Egan says:
      
      9 August, 2018 at 4:42 am
      
      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, $\alpha, \beta$ , then the squared distance between two such matrices is:
      
      $\displaystyle{6-2 \cos \left(\alpha _1-\alpha _2+\beta _1-\beta _2\right)-2 \cos \left(\alpha _1-\alpha _2\right)-2 \cos \left(\beta _1-\beta _2\right)}$
    - John Baez says:
      
      9 August, 2018 at 6:30 am
      
      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 that give 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.
    - Greg Egan says:
      
      9 August, 2018 at 6:51 am
      
      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 think they’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.)
    - John Baez says:
      
      9 August, 2018 at 7:36 am
      
      Greg wrote:
      
      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?
      
      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 any distance-preserving bijection between them (allowing rescaling to increase one’s chance of success). The bijection you’re describing is the ‘reasonable’ one.
      
      I don’t think they’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?
      
      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
      
      $g(v,w) = -\textrm{tr}( \rho(v) \rho(w) )$
      
      where $\rho$ 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.
Greg Egan says:

6 August, 2018 at 12:07 pm

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:

$\mathbb{E}(d_{\mathrm{SU}(2)}^{2n}) \to \frac{4}{\sqrt{\pi }} \frac{4^n}{n^{3/2}}$

With a bit of numerical tinkering, I conjecture that:

$\mathbb{E}(d_{\mathrm{SU}(3)}^{2n}) \to 27\pi^2 \frac{9^n}{n^4}$

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.

Reply
- Greg Egan says:
  
  7 August, 2018 at 12:24 am
  
  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:
  
  $\mathbb{E}(d_{\mathrm{SU}(3)}^{2n}) \to 243\pi^2 \frac{9^n}{n^4}$
  
  Reply
- Greg Egan says:
  
  7 August, 2018 at 12:46 am
  
  Also, we’re measuring norms here a bit differently on $\mathrm{SU}(2)$ than when we were treating it as a unit sphere, because:
  
  $\displaystyle{ \|I_2\|^2 = \mathrm{tr}(I_2 I_2^\dagger) = 2}$
  
  With that norm, the even moments have extra factors of $2^n$ times the Catalan numbers, and the maximum squared distance between elements of the group is 8. So the asymptotic form is:
  
  $\displaystyle{\mathbb{E}(d_{\mathrm{SU}(2)}^{2n}) \to \frac{4}{\sqrt{\pi }} \frac{8^n}{n^{3/2}}}$
  
  Reply
Greg Egan says:

9 August, 2018 at 12:18 pm

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 $S_3$ where you permute the diagonal entries, giving the equilateral triangles here that are SU(3)/conjugation.

The colours here encode the trace.

Reply
- duetosymmetry says:
  
  9 August, 2018 at 1:34 pm
  
  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.
  
  Reply
Greg Egan says:

9 August, 2018 at 11:04 pm

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 $\alpha, \beta$ for two of their entries) is:

$\displaystyle{d^2 = 6-2 \cos \left(\alpha _1-\alpha _2+\beta _1-\beta _2\right)-2 \cos \left(\alpha _1-\alpha _2\right)-2 \cos \left(\beta _1-\beta _2\right)}$

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

Reply
- John Baez says:
  
  10 August, 2018 at 2:04 pm
  
  For any Riemannnian metric $g$ on a surface, near any point we can multiply it by a smooth positive function $f$ so that the metric $f g$ 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.
  
  Reply

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