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Let $G$ be a simple compact connected Lie group and let $K$ be a connected closed subgroup of $G$. Let $\widehat G$ and $\widehat K$ denote the corresponding unitary duals, that is, the (equivalence classes of) irreducible representations. By the Highest Weight Theorem, elements in $\widehat G$ are in correspondence with elements in the set $P^+(G)$ of $G$-integral dominant weights. For $\lambda\in P^+(G)$, we will denote by $\pi_\lambda$ the irreducible representation of $G$ with highest weight $\lambda$.

Given $\tau\in\widehat K$, we define $\widehat G_\tau=\{\pi\in\widehat G: \textrm{Hom}_K(\pi,\tau)\neq0\}$. In other words, $\pi$ is in $\widehat G_\tau$ if and only if $\tau$ appears in the decomposition in irreducible representations of the restriction of $\pi|_K$ (the restriction of $\pi$ to $K$). Elements in $\widehat G_\tau$ are called $\tau$-spherical representations.

I wish to know if there is $\omega\in P^+(G)$, with $\omega\neq0$, satisfying the following property:

if $\lambda\in P^+(G)$ satisfies $\pi_\lambda\in\widehat G_\tau$, then $\pi_{\lambda+\omega}\in\widehat G_\tau$.

Here I list some examples coming from compact symmetric spaces of real rank one. In each case I used the obvious inclusion $K\subset G$ and standard conventions on the root system of $\mathfrak g$:

  1. $G$ arbitrary and $K=\{1\}$, $\omega =$ any choice in $P^+(G)$.
  2. $G=SU(2)$ and $K=T$ (maximal torus), $\omega = 2\varepsilon_1$ (twice the first fundamental weight).
  3. $G=SO(n+1)$ and $K=SO(n)$, $\omega=\varepsilon_1$ (the first fundamental weight).
  4. $G=SU(n+1)$ and $K=S(U(n)\times U(1))$, $\omega=\varepsilon_1-\varepsilon_{n+1}$ (the first plus the last fundamental weights).
  5. $G=Sp(n+1)$ and $K=Sp(n)$, $\omega=\varepsilon_1+\varepsilon_2$ (the second fundamental weight).
  6. $G=SO(2n)$ and $K=T$ (max. torus), $\omega= 2\varepsilon_1$ (twice the first fundamental weight).

Each of the above examples follows by the explicit branching law from $G$ to $K$, which is (of course) not available for arbitrary $K$. Clearly, in each case, any positive multiple of the indicated $\omega$ also works.

In case the answer is affirmative, it follows that $\widehat G_\tau$ can always be written as a union of subsets of the form $\{\pi_{\lambda+k\omega}: k\geq0 \}$ for some $\lambda\in P^+(G)$. (The union is not assumed finite or either disjoint.) This is the result I am looking for.

Edit: Looking for more information, here is an intermediate question:

For $G=SO(2n)$. Does a positive multiple of $\varepsilon_1$ satisfy the above property for any closed subgroup $K$ of $G$?

Either a proof or a counterexample would help me to understand this problem. I picked $SO(2n)$ because is my favorite, but we may replace $(SO(2n),\varepsilon_1)$ in the question by $(SO(2n+1),\varepsilon_1)$, $(SU(n+1),\varepsilon_1-\varepsilon_{n+1})$, or $(Sp(n+1),\varepsilon_1+\varepsilon_2)$.

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    $\begingroup$ You probably meant $K=T$, the maximal torus of $G$, in examples 1 and 2. If $K=\{1\}$ then any representation of $G$ is spherical, so the property trivially holds for any $\omega$. Also, you should exclude the dominant weight $\omega=0$, again, the property holds trivially. $\endgroup$ Commented Apr 11, 2019 at 5:19
  • $\begingroup$ Although I had written what I wanted, it was confusing and your idea is better. I have just edited. Thanks! $\endgroup$
    – emiliocba
    Commented Apr 11, 2019 at 10:04

1 Answer 1

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It is true that there exists such a $\omega$. Indeed, $every$ $\omega \neq 0$ works. Fix a representation $\pi _{\omega}$ which has a nonzero $K$ fixed vector, with $\omega \neq 0$. One can replace $G,K$ by their complexifications. Fix a Borel subgroup $B$ of $G(\mathbb C)$ and view the vectors in the representations $\pi _{\lambda}$ as algebraic sections $\phi (x)$ of a homogeneous line bundle ${\mathcal L}_{\lambda}$ on $G/B$ (the Borel-Weil theorem) . Fix a section $\phi (x)$ whose $K$ translates generate a representation of $K(\mathbb C)$ isomorphic to $\tau$.

If $\psi (x)$ is the non-zero $K$ fixed vector in $\pi _{\omega}$ viewed as a section of ${\mathcal L}_{\omega}$, then the product $\phi (x)\psi (x)$ is a section of ${\mathcal L}_{\lambda +\omega}$. But the $K$ translates of the product $\phi \psi$ generate a representation also isomorphic to $\tau$ since $\psi$ is $K$ invariant. Hence the restriction of $\pi _{\lambda +\omega}$ to $K$ contains $\tau$.

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  • $\begingroup$ So, the highest weight of any non-trivial representation in $\widehat G_{1_K}$ works ($1_K=$ trivial representation of $K$). In particular, the set of highest weight of elements in $\widehat G_{1_K}$ is a monoid ($0$ is in it, and the sum of any elements in it is again in it). Is all this right? In the affirmative case, thanks! $\endgroup$
    – emiliocba
    Commented Apr 12, 2019 at 11:24
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    $\begingroup$ @emiliocba: Yes, that is correct. I do not know a reference but the proof is obvious given the Borel-Weil theorem. $\endgroup$ Commented Apr 12, 2019 at 11:45
  • $\begingroup$ In fact, the situation is controlled by an affine semigroup $A_{G,K}$ contained in $P(G)^{+}\times P(K)^{+}$ formed by the pairs $(\lambda,\tau)$ for which the restriction of the simple $G$-module with highest weight $\lambda$ to $K$ contains the simple $K$-module with highest weight $\tau$. This branching semigroup provides a multigrading on the branching algebra of $K$-highest weight vectors in $R(G/U)$. See the work on branching algebras by Roger Howe and others. $\endgroup$ Commented Apr 13, 2019 at 5:16
  • $\begingroup$ @Victor : does it mean what I have said is incorrect? I thought I gave a proof. $\endgroup$ Commented Apr 13, 2019 at 5:39
  • $\begingroup$ @Venkataramana: of course, your proof works. I was pointing out the additional structure present, related to the monoid OP asked about, and commented on references. Viewing all line bundles $\mathcal{L}_\lambda$ simultaneously by considering $R(G/U)$ with the grading coming from the right $T$-action instead of individual spaces $\Gamma(G/B,\mathcal{L}_\lambda)$ streamlines the description and the proof. $\endgroup$ Commented Apr 13, 2019 at 15:48

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