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$\DeclareMathOperator\End{End}$Over the past few months, I have taught myself the classification of reductive groups, and continued to non-abelian (as well as a small venture to non-compact) Harmonic Analysis.

I am now trying to put everything that I learn together into something coherent. Let's, therefore, take the case of compact real Lie groups. By Chevalley the category of compact real Lie groups is equivalent to the category of $\mathbb{R}$-anisotropic reductive linear algebraic groups whose connected components have $\mathbb{R}$-points. In particular, all of the representations of a compact real Lie group are algebraic.

Therefore, by the Theorem of the Highest Weight, we have a classification of the unitary dual (the set of irreducible unitary representations) of a real compact Lie group $G$.

From the Harmonic Analysis perspective: $$L^2(G)\cong\bigoplus_{\pi\in\hat{G}} \End(\pi)(\cong \bigoplus_{\pi\in\hat{G}} \pi\otimes\pi^*).$$ It remains, therefore, to choose a basis of each $\End(\pi)$, for each $\pi$ guaranteed by the Theorem of the Highest Weight.

I looked at the examples in Folland's book on Harmonic Analysis, and did not see any mention of the Theorem of the Highest Weight. This seemed to have been done completely ad hoc.

I was also told in an old question of mine to take a look at Zonal Spherical Functions, in the particular case that $K=1$. I must confess, I find Zonal Spherical Functions to be quite confusing. I suspect that the point is that these are the method by which Harish Chandra proved the Plancherel Theorem (the non-compact variant of Peter-Weyl), but it's not clear to me how to use this in practice.

But either way, I end up wondering: what is the point of the Theorem of the Highest Weight? Does it provide a basis for each $\End(\pi)$? If it does, then what is it? If it doesn't, and we end up using some other method for finding the basis for each $\End(\pi)$ (a method that appears to ad hoc show that we have exhausted all of the irreducible unitary representations), then what good is the Theorem of the Highest Weight?

To put it succinctly: what utility does the Theorem of the Highest Weight provide, and how, if at all, does it fit into the picture of finding a basis for each $\End(\pi)$?

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    $\begingroup$ It parametrizes the irreps of $G$, yes. Does it tell me how find a nice basis of each irrep? It may very well do that, but I don't understand how. Can you expand on that? $\endgroup$
    – Andrew NC
    Dec 22, 2020 at 18:54
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    $\begingroup$ Your questions seem very often to focus on explicit bases for representation spaces. Why do you think that this is desireable? I think at least one powerful approach to harmonic analysis is to think of it in as basis-free a way as possible, so that being able to describe representations without pinning down bases is an asset, not a liability. $\endgroup$
    – LSpice
    Dec 22, 2020 at 18:57
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    $\begingroup$ In large part, you—or at least I—would invoke the theorem of the highest weight as an answer to the question "what do the irreducible representations of a compact Lie group look like?" Now the representation theory of such groups is, from my point of view, completely understood (especially since we can explicitly construct the representation with a given highest weight), so it can be used as an ingredient in less well understood theories. $\endgroup$
    – LSpice
    Dec 22, 2020 at 19:40
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    $\begingroup$ @AndrewNC There are 3 different approaches to constructing representations from their highest weight that are used in introductory textbooks. LSpice mentioned the Borel-Weil theorem which is useful for geometrical purposes. Then there is a construction via Verma modules, which is more suited for algebraists as the finite-dimensional representations arise as quotients of infinite-dimensional representations. And then finally there is case-by-case approach through fundamental representations and Cartan products. ... $\endgroup$
    – Vít Tuček
    Dec 22, 2020 at 20:57
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    $\begingroup$ ... For each type of group you can start with one (or two) basic representations and then you prove that any highest weight representation appears in a tensor power of the basic representation. This is the so called Cartan component and unfortunately, the projection to it is not usually described. (And frankly, I am not sure it is possible to do it in general.) People usually don't go into discussing bases for the representation spaces. There is a theory of crystal bases and Gelfand-Tsetlin theory, which is probably more suitable for the purpose of writing down the matrix coefficients. $\endgroup$
    – Vít Tuček
    Dec 22, 2020 at 21:01

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$\DeclareMathOperator\SL{SL}$One direct response to the question of "what does the theorem of the highest weight give us?" is that the highest weight completely determines the eigenvalue by which Casimir acts on that irreducible. (And all of the center of the universal enveloping algebra, as well ….) (This illustrates a semi-tangible form of Harish-Chandra's isomorphism describing the structure of that center.)

The highest weight does also approximately (to my mind) tell how to find a spanning set for the irreds (in general). In small cases it can give a basis, but I think in general the specification of a basis is significantly subtler, … keywords "crystal basis" …. Names include Gelfand, Kashiwara, Lusztig, et al., and relatively recent results (perhaps showing my limited awareness here) from Brubaker, Bump, Friedberg … Chinta, Gunnells, … et al.

One "disappointing" result I do remember is that the ideal structure in Verma modules turns out to be more complicated than the most optimistic conjecture. So it is non-trivial to describe the linear dependence of images of the highest weight vector under lowering operators … unfortunately. For $\SL_2$ it certainly turns out well, and I think for $\SL_3$, but I believe already for $\SL_4$ there is a (complicated) counter-example ….

EDIT: to clarify, e.g., as prompted by @LSpice's remark, … given a choice of positive and negative roots, any repn (finite dimensional or not) with a (unique) highest weight vector is spanned by all the images of that vector under the "lowering operators", that is, the operators in the universal enveloping algebra coming from the negative root-spaces in the Lie algebra.

For $\SL_2$, it is easy and standard to see that the images of the highest weight vector under the (essentially unique) lowering operator are a basis … with the eventually-too-lower image being $0$. (The lowest weight vector ….) There are no non-obvious relations.

Even with $\SL_3$, it is non-trivial to see that the "reasonable" relations among images under negative root-space (lowering) operators are all there are. This amounts to something like showing that the only submodules of a Verma modular with given highest weight (Verma module being the universal module with given highest weight) are again Verma modules. This optimistic idea proves tooooo optimistic, with entailed complications.

EDIT_2: Jacques Dixmier's "Enveloping Algebras" (original in French, too) I believe gives a citation for the failure of submodules of Verma modules to be isomorphic to Verma modules, for $SL_4$.

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    $\begingroup$ How does one use the highest weight to determine a spanning set? Are you thinking of moving to different weights via the negative root spaces? $\endgroup$
    – LSpice
    Dec 23, 2020 at 2:47
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    $\begingroup$ Just to thank Paul for these interesting details ... in a project which I might eventually get round to (once Godot shows up), I need to do some calculations of norms of elements of Fourier algebras and for SU(2) I more or less managed to do this using an explicit formula for the natural o.n. basis of Sym(C^2) -- one needs an o.n. basis on which the Casimir acts by scalars. So it sounds like I might be able to push this for SU(3) but should not expect it to suffice for SU(4)... $\endgroup$
    – Yemon Choi
    Dec 23, 2020 at 15:04
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The theorem of the highest weight tells you the structure of all finite-dimensional representations of simple (complex) Lie groups. In particular, it tells you what is the set $\widehat{G}$ that you are summing over. For some purposes this itself might be enough, i.e. you get that $L^2(G)$ is a completion of direct sum and so each $L^2$ function can be expressed as a series of some basis functions. Compare with the prototype case of $L^2(S^1)$ with Fourier series vs $L^2(\mathbb{R})$ with Fourier transform.

If you ask what are actually these basis functions, then the highest weight theorem will help you as long as it comes with a reasonable construction of irreducible representations, since the matrix coefficients $$ \{m_{u,v}: g \mapsto \langle g u | v \rangle \,\big|\, u, v\in \mathbb{V}, \mathbb{V}\in\widehat{G}\} $$ form a basis of $L^2(G).$

Sure, you can do some ad hoc construction and never really talk about the group structure. In the same way you can talk about Fourier series and never mention that those exponentials there are actually group characters of $U(1).$ But this perspective allows you to subsume Fourier series, Fourier transform, Discrete Fourier transform and much more in one theory. Similarly, for compact simple Lie groups, the theorem of the highest weight and the Peter-Weyl theorem unlock a sort of recipe how to go about "Fourier transform" for $L^2(G).$

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    $\begingroup$ I understand the Peter-Weyl theorem and its relationship to Fourier Analysis, and I agree with the set up that you lay out. In the case that $G$ is abelian, $End(\pi)$ is one dimensional, and finding a basis is easy. What I am asking is: how, if at all, does the theorem of the highest weight help you in finding a basis of matrix coefficients? Does it help at all? In all of the examples that I have seen where matrix coefficients were computed for compact groups, I did not see a mention of the Theorem of the Highest Weight... $\endgroup$
    – Andrew NC
    Dec 22, 2020 at 18:53
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    $\begingroup$ @AndrewNC has previously seemed dissatisfied with this description of a basis for the matrix-coefficient space, so I think that, despite the repeated mentions of a basis for $\operatorname{End}_{\mathbb C}(\pi)$, the question is really about how to find a basis for $\pi$ (although it is not clear to me that that's desireable). $\endgroup$
    – LSpice
    Dec 22, 2020 at 19:00
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    $\begingroup$ @AndrewNC The theorem doesn't help you. Good constructive proof of one of the implications of the theorem might help you. $\endgroup$
    – Vít Tuček
    Dec 22, 2020 at 19:22
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    $\begingroup$ @AndrewNC I am finding your series of questions increasingly hard to relate to or understand. Please consider that the tone of comments such as "does it serve any purpose whatsoever?" is not helpful, since I (and perhaps others) do not understand what you think of as a worthwhile purpose. For instance, you never told us on an earlier question if e.g. the character theory of finite groups counts as a meaningful application of NC harmonic analysis in your view $\endgroup$
    – Yemon Choi
    Dec 22, 2020 at 20:20
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    $\begingroup$ @AndrewNC It does. It classifies irreducible representations. Moreover, it is possible to prove it in the way that makes the representations quite explicit. Combined with effective form of Peter-Weyl theorem this allows one to follow a general recipe when constructing Fourier analysis on any particular compact Lie group. $\endgroup$
    – Vít Tuček
    Dec 22, 2020 at 20:53

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