$\DeclareMathOperator{\SO}{\operatorname{SO}}\DeclareMathOperator{\SU}{\operatorname{SU}}$**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](https://en.wikipedia.org/wiki/1/N_expansion)). 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](https://en.wikipedia.org/wiki/Spherical_model#cite_note-1). 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?



Representation theory of $\operatorname{SO}(n)$ for large $n$
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](https://arxiv.org/abs/1407.0373), 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](https://arxiv.org/abs/1911.07895)


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](https://arxiv.org/abs/1302.5859)


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 $n \to \infty$.


