11
$\begingroup$

$\DeclareMathOperator{\SO}{\operatorname{SO}}\DeclareMathOperator{\SU}{\operatorname{SU}}$Background: In the quantum field theory literature people commonly consider "expansions" of $\SO(n)$ or $\SU(n)$ field theories in $1/n$ (see this wikipedia article). While this essentially restricts to an expansion on the level of perturbative QFT, more quantitative approaches exist for Spin-$\mathrm{O}(n)$ models, by comparison with spherical models. The general idea is that in the limit of large $n$, the field theory becomes Gaussian. I was hoping to gain additional understanding into this "large spin-dimension limit" by studying the representation theory for $\SO(n)$ in the limit of large $n$. Morally, I would like to know if/how the group and representation theory of $\SO(n)$ "simplifies" for large $n$. More precisely, I am hoping for (1) bounds on Clebsch-Gordan coefficients in terms of $1/n$ and (2) some form of "approximate commutativity". In general, however, I have the following broad question:

Question: What is known on the asymptotic behavior of $\SO(n)$ as $n\to \infty$ in terms of its representation and group theory?

$\endgroup$
2
  • $\begingroup$ There was some work by Collins and coauthors about 10 years ago that sought to say things about large $n$ asymptotics for representations of classical matrix groups, from the point of view of free probability techniques. The paper arxiv.org/abs/0911.5546 is looking at U(n) rather than SO(n) and it might be somewhat tangential to the questions you have in mind, but perhaps it might offer a route into related literature that would be helpful? $\endgroup$
    – Yemon Choi
    Nov 9, 2020 at 19:41
  • 1
    $\begingroup$ You might want to clarify what you mean by Clebsch-Gordan coefficients. In math, this usually refers to the multiplicity of an irrep $V_{\rho}$ inside a tensor product of irreps $V_{\mu}\otimes V_{\nu}$. In physics this would refer the explicit matrix elements, in some basis, of an intertwiner in ${\rm Hom}(V_{\rho},V_{\mu}\otimes V_{\nu})$. And you also have Wigner symbols which are basis independent. $\endgroup$ Nov 9, 2020 at 20:12

1 Answer 1

17
$\begingroup$

In this answer I'll focus on the representation theory of $SO(n)$ as $n \to \infty$ (rather than the group theory). Strictly speaking, I want to discuss $O(n)$ rather than $SO(n)$, but I hope that it's good enough for your purposes. There are two approaches that I would like to discuss. Both of them involve a category that "interpolates" representations of $O(n)$ as $n$ varies, and in both cases, the objects that play the role of irreducible representations are indexed by "irreducible representations of $O(n)$" as $n \to \infty$ (depending on your preferred conventions, these can be thought of as partitions of any size).

Also, I don't know what is meant by "approximate commutativity", but if you could tell me, I might be able to comment.

Approach number 1: Deligne Categories

Reference: Representation theory in complex rank, II - Etingof, see also Deligne Categories in Lattice Models and Quantum Field Theory, or Making Sense of O(N) Symmetry with Non-integer N - Binder and Rychkov

In this approach, we construct a category using a symbol $V$ which is supposed to be the usual representation of $O(n)$ on $\mathbb{C}^n$, but we don't specify what $n$ is. In more detail, consider a category whose objects are indexed by natural numbers $r$, and denoted $V^{\otimes r}$. The morphisms $\hom(V^{\otimes r}, V^{\otimes s})$ are supposed to imitate representations of orthogonal groups, where these hom-spaces are spanned by Brauer diagrams with a composition rule that involves $n$ as a parameter. For example, the endomorphism algebras are Brauer algebras, whose composition depends on a parameter $n$. The true hom-spaces in $Rep(O(n))$ are quotients (of the space with basis consisting of all Brauer diagrams) by the tensor ideal of negligible morphisms. In particular if you fix $r$ and $s$, then $\hom((\mathbb{C}^n)^{\otimes r}, (\mathbb{C}^n)^{\otimes s})$ is spanned by these Brauer diagrams, but if $n$ is too small (something like $n < r+s$), there may be linear dependence relations between the maps corresponding to those diagrams. For example, the endomorphism ring of $\mathbb{C}^n \otimes \mathbb{C}^n$ has three linearly independent basis vectors, due to the decomposition $$ \mathbb{C}^n \otimes \mathbb{C}^n = S^2(\mathbb{C}^n) \oplus \bigwedge \nolimits^2(\mathbb{C}^n) = (S^2(\mathbb{C}^n)/\mathbb{C}) \oplus \mathbb{C} \oplus \bigwedge \nolimits^2(\mathbb{C}^n), $$ where in the last term $\mathbb{C}$ denotes a trivial representation (embedded in $S^2(\mathbb{C}^n)$ via the invariant bilinear form), which makes $\mathbb{C}^n \otimes \mathbb{C}^n$ the sum of three non-isomorphic irreducibles. However, if $n=1$, then the first and last factors are zero, and the true endomorphism space is only one-dimensional.

The Deligne category $\underline{Rep}(O(n))$ is the Karoubian envelope of the category we've described (whose objects are $V^{\otimes n}$, morphisms are linear combinations of Brauer diagrams, and whose composition depends polynomially on the parameter $n$). I want to emphasise that $n$ can be taken to be any element of the ground ring/field, and does not have to be an integer (in contrast to the definition of orthogonal groups). This Karoubian envelope is a formal completion that allows us to take direct sums and direct summands.

This should be thought of by analogy to compact Lie groups where every irreducible can be found as a summand of $V^{\otimes n} \otimes (V^*)^{\otimes m}$ for some $m,n \in \mathbb{N}$, where $V$ is a faithful representation of the Lie group in question. For us, $V$ (mimicking $\mathbb{C}^n$) is self-dual because we're working with orthogonal groups, so including tensor powers of $V^*$ is redundant.

The indecomposable objects of the Deligne category are indexed by primitive idempotents in the Brauer algebras, which means they can be thought of as being indexed by "irreducible representations of $O(n)$ as $n \to \infty$". These objects can have a dimension associated to them using the formalism of tensor categories, which turns out to be a polynomial interpolating the usual dimensions of irreducible objects. For example, $\bigwedge \nolimits^2 (V)$ is a summand of $V \otimes V$ whose dimension is ${n \choose 2}$.

When $n$ is a positive integer, there's a functor from $\underline{Rep}(O_n)$ to $Rep(O(n))$ which sends $V^{\otimes r}$ to $(\mathbb{C}^n)^{\otimes r}$ and each Brauer diagram to the map of $O(n)$ representations that it ordinarily defines. The main way that $n$ enters the story is via composition of morphisms, and idempotents in the Brauer algebras (which have some kind of normalising factor that is a rational function of $n$).

Since you've specifically asked about asymptotics as $n \to \infty$, I think this might be relevant to you. Calculation of particular Clebsch-Gordan coefficients might be difficult because they ultimately depend on a choice of basis of the representations in question, while we are working with hom-spaces that are insensitive to choices of bases in representations. However, if you are able to phrase your exact question in terms of $\underline{Rep}(O(n))$, it seems likely that you could easily show that the coefficients that come our are rational functions of $n$.

Approach number 2: Stable Representation Categories.

Reference: Stability patterns in representation theory - Sam and Snowden

In this approach, one works explicitly with the group $O(\infty)$. If $GL(\infty)$ is the group of $\infty \times \infty$ matrices differing from the identity matrix in finitely many entries, then $GL(\infty)$ acts on $\mathbb{C}^\infty$ in the usual way. Then $O(\infty)$ is the subgroup of $GL(\infty)$ preserving a non-degenerate symmetric bilinear form on $\mathbb{C}^\infty$. This group is "too big" to understand all its representations, so we restrict to the category $Rep(O(\infty))$ of subquotients of direct sums of tensor powers of $\mathbb{C}^\infty$.

This is analogous to the statement about all irreducibles of a compact Lie group appearing in $V^{\otimes n} \otimes (V^*)^{\otimes m}$ for some $m,n \in \mathbb{N}$. In the case of $O(n)$, $\mathbb{C}^n$ is self-dual so there's no need to consider $(\mathbb{C}^n)^*$. While this reasoning doesn't apply to $O(\infty)$ (e.g. it's infinite dimensional so it's not a Lie group, $\mathbb{C}^\infty$ isn't self-dual because it's infinite dimensional), hopefully it's sufficient to convince you that $Rep(O(\infty))$ is a sensible category to consider.

The key difference between representations of $O(n)$ and representations of $O(\infty)$ is that there are fewer maps of representations than you might expect. For example, for $O(n)$, you have an embedding of the trivial representation in $\mathbb{C}^n \otimes \mathbb{C}^n$ given essentially by scalar multiples of the bilinear form preserved by $O(n)$. So for the usual bilinear form $\langle (x_i), (y_i) \rangle = \sum_i x_i y_i$, the invariant vector would be $\sum_i e_i \otimes e_i$ (where $e_i \in \mathbb{C}^n$ is the $i$-th basis vector). This fails for $O(\infty)$ because $\sum_i e_i \otimes e_i$ is now an infinite sum and so doesn't define an element of $\mathbb{C}^\infty \otimes \mathbb{C}^\infty$. Nevertheless there is a quotient map from $\mathbb{C}^\infty \otimes \mathbb{C}^\infty$ to the trivial representation given by applying the invariant bilinear form. So the trivial representation is a quotient, but not a submodule of $\mathbb{C}^\infty \otimes \mathbb{C}^\infty$, which illustrates that $Rep(S(\infty))$ is not semisimple (in constrast to the representation categories of orthogonal groups which it is imitating).

What I suspect is of particular interest to you is the existence of "specialisation functors" $\Gamma_n: Rep(O(\infty)) \to Rep(O(n))$. This functor admits a fairly concrete realisation given roughly as follows. Fix an $n$-dimensional subspace $\mathbb{C}^n$ of $\mathbb{C}^\infty$ on which the bilinear form restricts to another nondegenerate symmetric bilinear form. Take the orthogonal complement, which is abstractly $\mathbb{C}^\infty$, but retains an action of $H_n = O(\infty)$ (which you might write as $O(\infty-n)$ to emphasise that it acts on coordinates whose indices are shifted by $n$). Then $O(n)$ and $H_n$ may be viewed as commuting subgroups of $O(\infty)$, and so the subspace of $H_n$-invariants retains an action of $O(n)$. In fact this is precisely the functor in question; $\Gamma_n(V) = V^{H_n}$. Note in particular that $\Gamma_n(\mathbb{C}^\infty) = \mathbb{C}^n$.

So this could be relevant to your situation by considering a construction in $Rep(O(\infty))$, and then applying a specialisation functor $\Gamma_n$ to land in $Rep(O(n))$. This, in a certain sense "becomes faithful as $n \to \infty$". So the construction on the level of $Rep(O(\infty))$ would be the "$n \to \infty$ limit". One reason why this might not be what you are looking for is that $n$ is not a parameter in $Rep(O(\infty))$, and maps in $Rep(O(n))$ that depend on $n$ in a nontrivial way typically don't interpolate to $Rep(O(\infty))$, so I'm not sure how you would obtain any asymptotics as $n \to \infty$.

$\endgroup$
7
  • $\begingroup$ Thank you Christopher for your elaborate answer! Many of these categorical constructions are far beyond my comfort zone, but I am quite interested. But let me ask you one simple question: While I see that you can use above considerations to construct representation categories that nicely (?) interpolate between representations of O(n) for different n, what is this all good for? What kind of properties of the representations can you extract by these methods? $\endgroup$ Nov 9, 2020 at 20:15
  • 3
    $\begingroup$ Probably the most important consequence, in my opinion, is that it relates the tensor product multiplicities of irreducible representations for different rank groups (the functors I mentioned are all monoidal). What happens is that the tensor multiplicities in the "universal category" truncate to the multiplicities in the finite categories, and for any particular multiplicity, the truncation operation does nothing if $n$ is large enough. So for example, the multiplicity of the exterior square of $\mathbb{C}^n$ inside the tensor square is 1, provided that $n > 1$. $\endgroup$ Nov 9, 2020 at 21:19
  • 1
    $\begingroup$ Although there's not enough room to go into detail in this comment, here is a short list of things one can learn: how representations restrict from $GL(n)$ to $O(n)$, associated character formulae, calculating certain special cases of tensor product multiplicities, and certain homological algebra constructions. There are also analogous versions of both types of categories for classical groups and the symmetric group for which similar deductions can be made. $\endgroup$ Nov 9, 2020 at 21:51
  • $\begingroup$ Thanks a lot. Could you provide me with a reference regarding the relation to tensor product multiplicities? I don't see how that follows from what you wrote above (as I said, I am far from fluent in categorical methods). $\endgroup$ Nov 10, 2020 at 9:40
  • $\begingroup$ In subsection 1.4 of arxiv.org/abs/1209.3509 it is stated that the Grothendieck ring of the stable representation category is the ring of symmetric functions, $\Lambda$ (the introduction of the paper works with symplectic groups, but the same holds for orthogonal groups). Note that the Schur function $s_\lambda$ corresponds to the Schur functor associated to $\lambda$ applied to $V$, which is not the same as the irreducible representation of $O(\infty)$ indexed by $\lambda$ which corresponds to $s_{[\lambda]}$ (although they are related by an upper-triangular matrix). $\endgroup$ Nov 10, 2020 at 19:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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