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1. • CommentRowNumber1.
   • CommentAuthorUrs
   • CommentTimeAug 21st 2021
   • PermaLink
   Author: Urs
   Format: MarkdownItexan essentially empty stub, for the moment just to satisfy a link long requested at *[[harmonic analysis]]* <a href="https://ncatlab.org/nlab/revision/Peter-Weyl+theorem/1">v1</a>, <a href="https://ncatlab.org/nlab/show/Peter-Weyl+theorem">current</a>

   an essentially empty stub, for the moment just to satisfy a link long requested at harmonic analysis

   v1, current

2. • CommentRowNumber2.
   • CommentAuthorUrs
   • CommentTimeJan 16th 2022
   • (edited Jan 16th 2022)
   • PermaLink
   Author: Urs
   Format: MarkdownItexA reference to be included for when the editing functionality is back: * Arash Ghaani Farashahi, *Peter-Weyl Theorem for Homogeneous Spaces of Compact Groups*, Int. J. Anal. Appl., 13 (1) (2017), 22-31 ([ijaa:792](http://etamaths.com/index.php/ijaa/article/view/792)) I arrived at this in search for answer to the following question: *If $K$ is a finite group with a free isometric action on some $S^n$, does $L^2(S^n)$ contain copies of all unitary irreps of $K$?* To get a handle on this, I thought I'd consider the special case where we have a coset space realization $S^n \,=\,G/H$ for some inclusion $K \subset G$. In that case, Theorem 4.6 of the above article tells me that $L^2(G/H)$ decomposes as a direct sum of $G$-representations "$\mathcal{E}_\pi(G/H)$" that are *indexed* by those irreps $\pi$ of $G$ whose restriction to $H$ is trivial. I can't quite make out yet whether the summand $\mathcal{E}_\pi(G/H)$ is meant to consist of copies of that very irrep $\pi$, though by the very last Rem. 4.2 that's the case at least when $H$ is normal, so that the statement reduces to the ordinary PW theorem. Once I knew which irreps of $G$ the reps $\mathcal{E}_\pi(G/H)$ consist of, I'd need to know whether these branch into all irreps of $K \subset G$... [edit: Hm, from standard Peter-Weyl, shouldn't the decomposition of $L^2(G/H)$ simply be given by each $G$-irrep appearing with multiplicity the dimension of its $H$-fixed space? ]

   A reference to be included for when the editing functionality is back:

   • Arash Ghaani Farashahi, Peter-Weyl Theorem for Homogeneous Spaces of Compact Groups, Int. J. Anal. Appl., 13 (1) (2017), 22-31 (ijaa:792)

   I arrived at this in search for answer to the following question:

   If $K$ is a finite group with a free isometric action on some $S^n$, does $L^2(S^n)$ contain copies of all unitary irreps of $K$?

   To get a handle on this, I thought I’d consider the special case where we have a coset space realization $S^n \,=\,G/H$ for some inclusion $K \subset G$.

   In that case, Theorem 4.6 of the above article tells me that $L^2(G/H)$ decomposes as a direct sum of $G$-representations “$\mathcal{E}_\pi(G/H)$” that are indexed by those irreps $\pi$ of $G$ whose restriction to $H$ is trivial.

   I can’t quite make out yet whether the summand $\mathcal{E}_\pi(G/H)$ is meant to consist of copies of that very irrep $\pi$, though by the very last Rem. 4.2 that’s the case at least when $H$ is normal, so that the statement reduces to the ordinary PW theorem.

   Once I knew which irreps of $G$ the reps $\mathcal{E}_\pi(G/H)$ consist of, I’d need to know whether these branch into all irreps of $K \subset G$…

   [edit: Hm, from standard Peter-Weyl, shouldn’t the decomposition of $L^2(G/H)$ simply be given by each $G$-irrep appearing with multiplicity the dimension of its $H$-fixed space? ]

3. • CommentRowNumber3.
   • CommentAuthorsamwinnick
   • CommentTimeJan 17th 2022
   • (edited Jan 17th 2022)
   • PermaLink
   Author: samwinnick
   Format: MarkdownItexI just came here to ask another question, but thought I would post a response here since I worked through some of this stuff when I was working on my master's thesis. Not sure if it helps, but here are a few facts that you might find useful: Normally, when we realize $S^n=G/H$ by $G=SO(n+1)$ and $H=SO(n)$, the important property is that the $G$ action makes $S^n$ a 2-point homogeneous space. If $X=G/H$ is 2-point homogeneous, it follows that $(G,H)$ is a Gelfand pair, and this is the key property that simplifies a lot of the representation theory. The Gelfand pair property implies that $L^2(G/H)$ decomposes into Harmonic subrepresentations (generalizing spherical harmonics), but these might not be irreducible. This is where my understanding "peters out". They are not irreducible, for example, when instead of the sphere we look at Grassmannians. These are no longer 2-point homogeneous but I think they're still Gelfand pairs (when realized as homogeneous spaces in a usual way). I'm curious as to what is the criterion that makes the harmonic subreps irreducible. If I remember correctly, this example is shown in MacDonald's symmetric functions book. Another helpful resource for this is "Sphere Packings, Lattices And Groups" (SPLAG). You may find it if you look for Gegenbauer polynomials or zonal polynomials, as these arise as reproducing kernels in the harmonic subspaces. As for your question about which irreps of G appear in $L^2(G/H)$, here is what I can say, and I hope I am right on the details here. Let $G$ be a compact group, and let $\hat G$ be a complete set of pairwise non-isomorphic (finite dimensional, but this is automatic) unitary irreps of $G$. Then for each $\rho\in\hat G$ and $\phi\otimes v\in V_\rho^*\otimes V_\rho$, define $f_{\rho,\phi\otimes v}(x) = \phi(\rho_x v)$. Then the map out of $\bigoplus_{\rho\in\hat G}V_\rho^*\otimes V_\rho$ given by $\phi\otimes v\mapsto f_{\rho,\phi\otimes v}$ is an isomorphism of unitary representations onto its image. I appear to have failed to cite the book where I learned this, but in that book, they denote the image $C_{\mathrm{alg}}(G)$, though this appears to be non-standard notation. Then $C_{\mathrm{alg}}(G)$ densely embeds into $L^2(G)$. The point is it seems (to me at least) easier to understand the technical details of $C_{\mathrm{alg}}(G)$ than those of $L^2(G)$. Then I think this leads to the answer to your question: $$C_{\mathrm{alg}}(G/H) \cong C_{\mathrm{alg}}(G)^H \cong \bigoplus_{\rho\in\hat G}V_\rho^*\otimes V_\rho^H\cong\bigoplus_{\rho\in\hat G_H}V_\rho^*\cong\bigoplus_{\rho\in\hat G_H}V_\rho$$ where $$\hat G_H = \{\rho\in\hat G\,|\,dim(V_\rho^H)=1\}.$$ Here when I write the superscript $H$ I mean things invariant under translation on the right, say. So I think the decomposition into Harmonic subreps is in general a little coarser. To understand the $L^2$ version it's just a matter of applying some topological arguments.

   I just came here to ask another question, but thought I would post a response here since I worked through some of this stuff when I was working on my master’s thesis. Not sure if it helps, but here are a few facts that you might find useful: Normally, when we realize $S^n=G/H$ by $G=SO(n+1)$ and $H=SO(n)$, the important property is that the $G$ action makes $S^n$ a 2-point homogeneous space. If $X=G/H$ is 2-point homogeneous, it follows that $(G,H)$ is a Gelfand pair, and this is the key property that simplifies a lot of the representation theory. The Gelfand pair property implies that $L^2(G/H)$ decomposes into Harmonic subrepresentations (generalizing spherical harmonics), but these might not be irreducible. This is where my understanding “peters out”. They are not irreducible, for example, when instead of the sphere we look at Grassmannians. These are no longer 2-point homogeneous but I think they’re still Gelfand pairs (when realized as homogeneous spaces in a usual way). I’m curious as to what is the criterion that makes the harmonic subreps irreducible. If I remember correctly, this example is shown in MacDonald’s symmetric functions book. Another helpful resource for this is “Sphere Packings, Lattices And Groups” (SPLAG). You may find it if you look for Gegenbauer polynomials or zonal polynomials, as these arise as reproducing kernels in the harmonic subspaces.

   As for your question about which irreps of G appear in $L^2(G/H)$, here is what I can say, and I hope I am right on the details here. Let $G$ be a compact group, and let $\hat G$ be a complete set of pairwise non-isomorphic (finite dimensional, but this is automatic) unitary irreps of $G$. Then for each $\rho\in\hat G$ and $\phi\otimes v\in V_\rho^*\otimes V_\rho$, define $f_{\rho,\phi\otimes v}(x) = \phi(\rho_x v)$. Then the map out of $\bigoplus_{\rho\in\hat G}V_\rho^*\otimes V_\rho$ given by $\phi\otimes v\mapsto f_{\rho,\phi\otimes v}$ is an isomorphism of unitary representations onto its image. I appear to have failed to cite the book where I learned this, but in that book, they denote the image $C_{\mathrm{alg}}(G)$, though this appears to be non-standard notation. Then $C_{\mathrm{alg}}(G)$ densely embeds into $L^2(G)$. The point is it seems (to me at least) easier to understand the technical details of $C_{\mathrm{alg}}(G)$ than those of $L^2(G)$. Then I think this leads to the answer to your question:

   $$C_{\mathrm{alg}}(G/H) \cong C_{\mathrm{alg}}(G)^H \cong \bigoplus_{\rho\in\hat G}V_\rho^*\otimes V_\rho^H\cong\bigoplus_{\rho\in\hat G_H}V_\rho^*\cong\bigoplus_{\rho\in\hat G_H}V_\rho$$

   where

   $$\hat G_H = \{\rho\in\hat G\,|\,dim(V_\rho^H)=1\}.$$

   Here when I write the superscript $H$ I mean things invariant under translation on the right, say. So I think the decomposition into Harmonic subreps is in general a little coarser. To understand the $L^2$ version it’s just a matter of applying some topological arguments.

4. • CommentRowNumber4.
   • CommentAuthorUrs
   • CommentTimeJan 17th 2022
   • (edited Jan 17th 2022)
   • PermaLink
   Author: Urs
   Format: MarkdownItexThanks for the comments! I'll dig out some of the references. Is the master thesis you refer to available? [edit: oh, I see, your thesis must be [this here](https://uwspace.uwaterloo.ca/handle/10012/15578)] Your expression $\bigoplus_{\rho\in\hat G}V_\rho^*\otimes V_\rho^H$ agrees with what I arrived at in the addendum at the bottom of #2, after struggling with reading the article quoted there. I wonder what this article does if not arriving at this transparent form, may need to read it in more detail. Allow me to highlight my real question, in case you have a hint for me: What I really want to know is, for $K$ a finite group acting *freely* on an $n$-sphere by isometries, whether $L^2(S^n)$ with the induced action contains all the irreps of $K$. (I.e. I am after the variant of the usual PW statement for $L^2(G)$ buth with $G$ replaced by a sphere with a free $G$-action.) I was trying to use coset realizations $S^n = G/H$ only as a stepping stone towards this question, subject to a good choice of $K,H \subset G$. But maybe there are better strategies to approach this.

   Thanks for the comments! I’ll dig out some of the references. Is the master thesis you refer to available?

   [edit: oh, I see, your thesis must be this here]

   Your expression $\bigoplus_{\rho\in\hat G}V_\rho^*\otimes V_\rho^H$ agrees with what I arrived at in the addendum at the bottom of #2, after struggling with reading the article quoted there. I wonder what this article does if not arriving at this transparent form, may need to read it in more detail.

   Allow me to highlight my real question, in case you have a hint for me:

   What I really want to know is, for $K$ a finite group acting freely on an $n$-sphere by isometries, whether $L^2(S^n)$ with the induced action contains all the irreps of $K$. (I.e. I am after the variant of the usual PW statement for $L^2(G)$ buth with $G$ replaced by a sphere with a free $G$-action.)

   I was trying to use coset realizations $S^n = G/H$ only as a stepping stone towards this question, subject to a good choice of $K,H \subset G$. But maybe there are better strategies to approach this.

5. • CommentRowNumber5.
   • CommentAuthorsamwinnick
   • CommentTimeJan 17th 2022
   • PermaLink
   Author: samwinnick
   Format: MarkdownItexI don't know the answer to your question, but here are a few thoughts. I'm guessing you want $K$ to be small, perhaps a dense free subgroup of $SO(n+1)$, like a group used in the $n=2$ case pertinent to the Banach-Tarski paradox or for approximating a qubit unitary. Is this the setting? I'm guessing $K$ is not so large to act transitively, because intuitively that probably would break the free condition. So you can't realize $S^n$ as $K/H$ for some $H\leq K$. So my only thought would be to try to use Frobenius reciprocity. Assuming you can realize $S^n$ as $G/H$ with $(G,H)$ a Gelfand pair with $K\leq G$, and you know how the harmonic subspaces break down under the action of $G$, depending on the specific relationship of $G$ and $K$, I would think Frobenius reciprocity might at least be able to tell you something how the irreducible components of $L^2(S^n)$ under the action of $G$ decompose further under the action of $K$. It's been a while since I thought about this stuff though. I'm kind of curious what kinds of groups $K$ you are considering, and the relevance of the sphere. You truly want to consider functions on the entire sphere right? Yeah that's my thesis. I hope it helps, but just to let you know, you won't find the answer to your specific question about free actions there. The stuff I mentioned here is outlined in 4.4 and 4.5 of my thesis. Also 4.1,2,3 is about a lesser known combinatorial generalization that does away entirely with groups, where you nevertheless get something like $L^2(X)$ that breaks down into harmonic subspaces.

   I don’t know the answer to your question, but here are a few thoughts.

   I’m guessing you want $K$ to be small, perhaps a dense free subgroup of $SO(n+1)$, like a group used in the $n=2$ case pertinent to the Banach-Tarski paradox or for approximating a qubit unitary. Is this the setting? I’m guessing $K$ is not so large to act transitively, because intuitively that probably would break the free condition. So you can’t realize $S^n$ as $K/H$ for some $H\leq K$. So my only thought would be to try to use Frobenius reciprocity. Assuming you can realize $S^n$ as $G/H$ with $(G,H)$ a Gelfand pair with $K\leq G$, and you know how the harmonic subspaces break down under the action of $G$, depending on the specific relationship of $G$ and $K$, I would think Frobenius reciprocity might at least be able to tell you something how the irreducible components of $L^2(S^n)$ under the action of $G$ decompose further under the action of $K$. It’s been a while since I thought about this stuff though. I’m kind of curious what kinds of groups $K$ you are considering, and the relevance of the sphere. You truly want to consider functions on the entire sphere right?

   Yeah that’s my thesis. I hope it helps, but just to let you know, you won’t find the answer to your specific question about free actions there. The stuff I mentioned here is outlined in 4.4 and 4.5 of my thesis. Also 4.1,2,3 is about a lesser known combinatorial generalization that does away entirely with groups, where you nevertheless get something like $L^2(X)$ that breaks down into harmonic subspaces.

6. • CommentRowNumber6.
   • CommentAuthorUrs
   • CommentTimeJan 17th 2022
   • (edited Jan 17th 2022)
   • PermaLink
   Author: Urs
   Format: MarkdownItexThanks. Right, I should have been more specific: I am thinking of "small" $K$, such as finite $K$. In fact, to start with I'd be happy to know the answer just for finite cyclic groups. I am looking at twisted equivariant complex K-theory (let me say "KU-theory" not to clash with the finite group $K$ we are talking about). Here the twists are required to be $PU(\mathcal{H})$-bundles which are "stable" in that over each $K$-fixed point $x$ of their base space the induced projective representation $K_x \to PU(\mathcal{H})$ is the projectivization of a unitary rep $\widehat K_x \to \mathrm{U}(\mathcal{H})$ with the property that it contains all irreps of $\widetilde K$ (on which $S^1 \to \widehat K$ acts canonically) in infinite multiplicity. By invoking the general abstract classification result for principal bundles, it is fairly straightforward to see that -- as soon as $K$ acts freely isometrically on any sphere $S^n$ of dimension $n \gt 3$ -- these "stable" equivariant $PU(\mathcal{H})$-bundles over a single $K$-fixed point have the same classification as plain $PU(\mathcal{H})$-bundles over the "blowup" of the fixed point singularity to the "spherical space form" $S^n/K$. While straightforward to see on abstract grounds, this identification used to be (for me, at least) a little mysterious on the level of cocycles. But if it is true that $L^2( S^n )$ contains all $K$-irreps with infinite multiplicity, then I have, I think, a nicely transparent argument for this.

   Thanks. Right, I should have been more specific: I am thinking of “small” $K$, such as finite $K$. In fact, to start with I’d be happy to know the answer just for finite cyclic groups.

   I am looking at twisted equivariant complex K-theory (let me say “KU-theory” not to clash with the finite group $K$ we are talking about). Here the twists are required to be $PU(\mathcal{H})$-bundles which are “stable” in that over each $K$-fixed point $x$ of their base space the induced projective representation $K_x \to PU(\mathcal{H})$ is the projectivization of a unitary rep $\widehat K_x \to \mathrm{U}(\mathcal{H})$ with the property that it contains all irreps of $\widetilde K$ (on which $S^1 \to \widehat K$ acts canonically) in infinite multiplicity.

   By invoking the general abstract classification result for principal bundles, it is fairly straightforward to see that – as soon as $K$ acts freely isometrically on any sphere $S^n$ of dimension $n \gt 3$ – these “stable” equivariant $PU(\mathcal{H})$-bundles over a single $K$-fixed point have the same classification as plain $PU(\mathcal{H})$-bundles over the “blowup” of the fixed point singularity to the “spherical space form” $S^n/K$.

   While straightforward to see on abstract grounds, this identification used to be (for me, at least) a little mysterious on the level of cocycles. But if it is true that $L^2( S^n )$ contains all $K$-irreps with infinite multiplicity, then I have, I think, a nicely transparent argument for this.

7. • CommentRowNumber7.
   • CommentAuthorsamwinnick
   • CommentTimeJan 17th 2022
   • (edited Jan 17th 2022)
   • PermaLink
   Author: samwinnick
   Format: MarkdownItexMaybe you can use the inclusion $$\bigoplus_{\rho\in\hat G_H}\mathbb{C}z_{\rho,x_0} = L^2(H\backslash G/H) = \,^H L^2(S^n)\subseteq L^2(S^n)$$ where $z_{\rho,x_0}$ are related to the Gegenbauer polynomials $z_\rho$ via $z_{\rho,x_0}(x)=z_\rho(x_0\cdot x)$. The reason for using this basis is that each summand on the left-hand-side is the left-$H$-invariant subspace of a (harmonic) subrepresentation of $L^2(S^n)$. By the inclusion, it would suffice to show that every irreducible character $\chi$ of $K$ is of the form $\chi=z_{\rho,x_0}$ for some irrep $\rho$ of $G$. The Gegenbauer polynomials are given by an explicit recurrence relation so this is something that could be practically checked for a given $K$. I couldn't figure out how to use the freeness of the $K$ action. It implies that $\,^K L^2(S^n)$ is one dimensional and $V_\rho^K$ is 0 dimensional for all nontrivial $\rho\in\hat G_H$, but I'm not sure what to do with that. --- I admit I don't understand the part of your last response where you told me why you were wondering this, but that is not your fault because I don't at the moment know K-theory. A couple questions though, because now I am wondering where you are coming from with this.. When you write "$PU(\mathcal{H})$-bundle'', is this in the sense of "$G$-bundles''? Are any of the other objects involved are the "$B$'' or "$E$'' of this bundle $E\to B$? What does $K$ act on in this picture? You write "$K_x$''. This makes me think maybe $K$ is the bundle (the "$E$'') and $K_x$ is the fiber over $x$. But then again this sort of notation is used for stabilizers so I am a bit confused by that. Also, out of curiosity, given that you are interested in projective representations of finite groups and you denote your vector space by $\mathcal{H}$, I'm wondering if this is by any chance coming from thinking about projective reps of $\mathbb{Z}/d\times\mathbb{Z}/d$ or of $\mathrm{Sp}(\mathbb{Z}/d\times\mathbb{Z}/d)$? Like, is this stemming from something to do with Heisenberg groups or something?

   Maybe you can use the inclusion

   $$\bigoplus_{\rho\in\hat G_H}\mathbb{C}z_{\rho,x_0} = L^2(H\backslash G/H) = \,^H L^2(S^n)\subseteq L^2(S^n)$$

   where $z_{\rho,x_0}$ are related to the Gegenbauer polynomials $z_\rho$ via $z_{\rho,x_0}(x)=z_\rho(x_0\cdot x)$. The reason for using this basis is that each summand on the left-hand-side is the left-$H$-invariant subspace of a (harmonic) subrepresentation of $L^2(S^n)$. By the inclusion, it would suffice to show that every irreducible character $\chi$ of $K$ is of the form $\chi=z_{\rho,x_0}$ for some irrep $\rho$ of $G$. The Gegenbauer polynomials are given by an explicit recurrence relation so this is something that could be practically checked for a given $K$.

   I couldn’t figure out how to use the freeness of the $K$ action. It implies that $\,^K L^2(S^n)$ is one dimensional and $V_\rho^K$ is 0 dimensional for all nontrivial $\rho\in\hat G_H$, but I’m not sure what to do with that.

   ---

   I admit I don’t understand the part of your last response where you told me why you were wondering this, but that is not your fault because I don’t at the moment know K-theory. A couple questions though, because now I am wondering where you are coming from with this.. When you write “$PU(\mathcal{H})$-bundle”, is this in the sense of “$G$-bundles”? Are any of the other objects involved are the “$B$” or “$E$” of this bundle $E\to B$? What does $K$ act on in this picture? You write “$K_x$”. This makes me think maybe $K$ is the bundle (the “$E$”) and $K_x$ is the fiber over $x$. But then again this sort of notation is used for stabilizers so I am a bit confused by that. Also, out of curiosity, given that you are interested in projective representations of finite groups and you denote your vector space by $\mathcal{H}$, I’m wondering if this is by any chance coming from thinking about projective reps of $\mathbb{Z}/d\times\mathbb{Z}/d$ or of $\mathrm{Sp}(\mathbb{Z}/d\times\mathbb{Z}/d)$? Like, is this stemming from something to do with Heisenberg groups or something?

8. • CommentRowNumber8.
   • CommentAuthorUrs
   • CommentTimeJan 17th 2022
   • PermaLink
   Author: Urs
   Format: MarkdownItexThanks for the Gegenbauer suggestion. I'll think about it. Regarding the equivariant bundles: Yes, I am referring to equivariant principal bundles with structure group $PU(\mathcal{H})$ and, yes, $K_x \subset K$ was meant to denote the stabilizer group of a point in the base space. Generally, for structure group $\Gamma$, this is about $\Gamma$-principal bundles $P \xrightarrow{\;p\;} X$ such that both the total space $P$ and the base space $X$ are equipped with $K$-actions, and the bundle projection is $K$-equivariant and, finally, the $\Gamma$-action on $P$ commutes with the $K$-action (in the special case of interest here). Given any such $K$-equivariant $\Gamma$-principal bundle, its structure over a point $x \in X$ is encoded by a group homomorphism $K_x \to \Gamma$: In terms of the above total space this homomorphism assigns to $k \in K_x$ the unique $\gamma(k) \in \Gamma$ whose commuting action on the fiber $P_x$ over $x$ cancels that of $k$. Regarding your last question: This touches on a point that is quite interesting and maybe underappreciated: From the point of abstract twisted equivariant K-theory, projective representations appear because $PU(\mathcal{H})$ happens to act well on spaces of Fredholm operators which happen to be classifying spaces for abstractly defined K-theory, and that's it. But it is indeed true that the moment one starts playing with actual examples and applications, then notions and constructions familiar from quantum physics show up all over the place. So quantum structures are not (for the most part) the motivation for twisted equivariant K-theory. But in the other direction, a fair bit of quantum structures appear -- or find a natural place -- in discussion of twisted equivariant K-theory. In fact, that's where I am headed here: we think there is a big story to be revealed concerning the twisted equivariant K-theory of configuration spaces of points in relation to monodromy braid representations and thus, if you wish, topological quantum computation. I am eager to start taling about that, but first I need to get some nitty-gritty technical details sorted out... :-)

   Thanks for the Gegenbauer suggestion. I’ll think about it.

   Regarding the equivariant bundles:

   Yes, I am referring to equivariant principal bundles with structure group $PU(\mathcal{H})$ and, yes, $K_x \subset K$ was meant to denote the stabilizer group of a point in the base space.

   Generally, for structure group $\Gamma$, this is about $\Gamma$-principal bundles $P \xrightarrow{\;p\;} X$ such that both the total space $P$ and the base space $X$ are equipped with $K$-actions, and the bundle projection is $K$-equivariant and, finally, the $\Gamma$-action on $P$ commutes with the $K$-action (in the special case of interest here).

   Given any such $K$-equivariant $\Gamma$-principal bundle, its structure over a point $x \in X$ is encoded by a group homomorphism $K_x \to \Gamma$: In terms of the above total space this homomorphism assigns to $k \in K_x$ the unique $\gamma(k) \in \Gamma$ whose commuting action on the fiber $P_x$ over $x$ cancels that of $k$.

   Regarding your last question:

   This touches on a point that is quite interesting and maybe underappreciated: From the point of abstract twisted equivariant K-theory, projective representations appear because $PU(\mathcal{H})$ happens to act well on spaces of Fredholm operators which happen to be classifying spaces for abstractly defined K-theory, and that’s it. But it is indeed true that the moment one starts playing with actual examples and applications, then notions and constructions familiar from quantum physics show up all over the place.

   So quantum structures are not (for the most part) the motivation for twisted equivariant K-theory. But in the other direction, a fair bit of quantum structures appear – or find a natural place – in discussion of twisted equivariant K-theory.

   In fact, that’s where I am headed here: we think there is a big story to be revealed concerning the twisted equivariant K-theory of configuration spaces of points in relation to monodromy braid representations and thus, if you wish, topological quantum computation. I am eager to start taling about that, but first I need to get some nitty-gritty technical details sorted out… :-)

9. • CommentRowNumber9.
   • CommentAuthorUrs
   • CommentTimeJan 18th 2022
   • (edited Jan 18th 2022)
   • PermaLink
   Author: Urs
   Format: MarkdownItexComing back to the Gegenbauer suggestion in #7: I am afraid that this would not give a representation of any non-trivial subgroup of $G$, but let me check that I am reading you correctly. I am thinking: At most the normalizer $N_G(H)$ of $H$ in $G$ still acts from the left on $L^2(H \backslash G / H) \,=\, L^2(G/H)^H$, of which at most $W_G(H) \coloneqq N_G(H)/H$ acts non-trivially. For the case at hand where $H \subset G$ is $SO(n) \subset SO(n+1)$ this would mean that only the trivial group still acts. Indeed, in our case, $H\backslash G / H \,\simeq\, [-\pi/2, + \pi/2]$ is the interval of distances between $x_0$ and its antipode. This is the right domain for Gegenbauer polynomials, but it's also manifestly not acted on non-trivially by any subgroup of $SO(n+1)$.

   Coming back to the Gegenbauer suggestion in #7: I am afraid that this would not give a representation of any non-trivial subgroup of $G$, but let me check that I am reading you correctly. I am thinking:

   At most the normalizer $N_G(H)$ of $H$ in $G$ still acts from the left on $L^2(H \backslash G / H) \,=\, L^2(G/H)^H$, of which at most $W_G(H) \coloneqq N_G(H)/H$ acts non-trivially. For the case at hand where $H \subset G$ is $SO(n) \subset SO(n+1)$ this would mean that only the trivial group still acts.

   Indeed, in our case, $H\backslash G / H \,\simeq\, [-\pi/2, + \pi/2]$ is the interval of distances between $x_0$ and its antipode. This is the right domain for Gegenbauer polynomials, but it’s also manifestly not acted on non-trivially by any subgroup of $SO(n+1)$.

10. • CommentRowNumber10.
    • CommentAuthorUrs
    • CommentTimeJan 18th 2022
    • (edited Jan 18th 2022)
    • PermaLink
    Author: Urs
    Format: MarkdownItex[ comment removed ]

    [ comment removed ]

11. • CommentRowNumber11.
    • CommentAuthorDavidRoberts
    • CommentTimeJan 18th 2022
    • PermaLink
    Author: DavidRoberts
    Format: MarkdownItexIs that action of G via the inclusion of S(H) in S(O)?

    Is that action of G via the inclusion of S(H) in S(O)?

12. • CommentRowNumber12.
    • CommentAuthorUrs
    • CommentTimeJan 18th 2022
    • PermaLink
    Author: Urs
    Format: MarkdownItexSorry, that comment #10 didn't quite type-check.

    Sorry, that comment #10 didn’t quite type-check.

13. • CommentRowNumber13.
    • CommentAuthorsamwinnick
    • CommentTimeJan 18th 2022
    • (edited Jan 18th 2022)
    • PermaLink
    Author: samwinnick
    Format: MarkdownItexHmm. Yeah, I guess it would be pretty strange if there were a group whose characters are Gegenbauer polynomials anyway. Looks like the approach I suggested will not work. Let $X=G/H$ and suppose for each irrep $\rho$ of $K$ we can find a subgroup $J\leq G$ such that $\rho$ appears in $L^2(J\backslash X)$. Then since each $L^2(J\backslash X)\subseteq L^2(X)$ we would have shown that each $\rho$ appears in $L^2(X)$. But in the case of $G=SO(n+1)$ and $H=SO(n)$, even allowing for $J\lneq G$ other than $J=H$, this will not work by a similar argument to #9 by the subgroup structure of $G$. (Also note that choosing smaller subgroups $J'\leq J$ is counterproductive since $L^2(J'\backslash X)\subseteq L^2(J\backslash X)$.) I guess it still leaves the option of writing $X=G'/H'$ for different $G'$ and $H'$ but I kind of doubt this is the right approach, especially since the freeness of the $K$-action is not coming into play anywhere.

    Hmm. Yeah, I guess it would be pretty strange if there were a group whose characters are Gegenbauer polynomials anyway. Looks like the approach I suggested will not work. Let $X=G/H$ and suppose for each irrep $\rho$ of $K$ we can find a subgroup $J\leq G$ such that $\rho$ appears in $L^2(J\backslash X)$. Then since each $L^2(J\backslash X)\subseteq L^2(X)$ we would have shown that each $\rho$ appears in $L^2(X)$. But in the case of $G=SO(n+1)$ and $H=SO(n)$, even allowing for $J\lneq G$ other than $J=H$, this will not work by a similar argument to #9 by the subgroup structure of $G$. (Also note that choosing smaller subgroups $J'\leq J$ is counterproductive since $L^2(J'\backslash X)\subseteq L^2(J\backslash X)$.) I guess it still leaves the option of writing $X=G'/H'$ for different $G'$ and $H'$ but I kind of doubt this is the right approach, especially since the freeness of the $K$-action is not coming into play anywhere.

14. • CommentRowNumber14.
    • CommentAuthorUrs
    • CommentTimeJan 19th 2022
    • PermaLink
    Author: Urs
    Format: MarkdownItexNo problem, thanks for chatting about it. As it goes, making some noise here served to export enough entropy out of my system that I think I have now solved the problem that made me write #2 :-)

    No problem, thanks for chatting about it. As it goes, making some noise here served to export enough entropy out of my system that I think I have now solved the problem that made me write #2 :-)

15. • CommentRowNumber15.
    • CommentAuthorUrs
    • CommentTimeJan 22nd 2022
    • PermaLink
    Author: Urs
    Format: MarkdownItexJust to record another reference for when the editing functionality is back: * Karl H. Hofmann and Sidney A. Morris, *The Structure of Compact Groups*, De Gruyter Studies in Mathematics **25** (2020) ([doi:10.1515/9783110695991](https://doi.org/10.1515/9783110695991)) This textbook is one of the few (?) which makes explicit the decomposition of $L^2(G)$ into irreps -- they call it the "Fine Structure Theorem" (3.28).

    Just to record another reference for when the editing functionality is back:

    • Karl H. Hofmann and Sidney A. Morris, The Structure of Compact Groups, De Gruyter Studies in Mathematics 25 (2020) (doi:10.1515/9783110695991)

    This textbook is one of the few (?) which makes explicit the decomposition of $L^2(G)$ into irreps – they call it the “Fine Structure Theorem” (3.28).

16. • CommentRowNumber16.
    • CommentAuthorzskoda
    • CommentTimeAug 22nd 2023
    • PermaLink
    Author: zskoda
    Format: MarkdownItex[[representative function|Representative function]]s on a compact topological group form a dense subspace of the space of all continuous functions. <a href="https://ncatlab.org/nlab/revision/diff/Peter-Weyl+theorem/3">diff</a>, <a href="https://ncatlab.org/nlab/revision/Peter-Weyl+theorem/3">v3</a>, <a href="https://ncatlab.org/nlab/show/Peter-Weyl+theorem">current</a>

    Representative functions on a compact topological group form a dense subspace of the space of all continuous functions.

    diff, v3, current