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Let $G$ be a compact, connected Lie group and $\rho$ any of its irreducible, unitary representations. If $\gamma:S^1\to G$ is an injective homomorphism (a periodic geodesic passing through the identity), what can be said about the matrix valued integral $$ I(\rho,\gamma)=\int_{S^1}\rho(\gamma(t))dt? $$ More explicitly, I am interested in the following questions:

  • Given $\rho$, can we find a $\gamma$ so that $I(\rho,\gamma)$ is invertible?
  • Given $\rho$, can we find geodesics $\gamma_i$ and complex numbers $a_i$ so that $\sum_ia_iI(\rho,\gamma_i)$ is invertible?
  • If a square matrix $A$ (of same dimension as $\rho$) satisfies $AI(\rho,\gamma)=0$, what can be said about $A$? This condition holds if $A=B\left.\frac{d}{dt}\rho(\gamma(t))\right|_{t=0}$ for some constant matrix $B$. Are all matrices $A$ satisfying this condition of this form?

The case of the torus $(\mathbb R/\mathbb Z)^n$ is quite simple. Then $\rho$ is parametrized by $\mathbb Z^n$ and $\gamma$ by $\mathbb Z^n\setminus\{0\}$, and $I(\rho_m,\gamma_k)=1$ if $m\cdot k=0$ and $I(\rho_m,\gamma_k)=0$ otherwise. (I have normalized the measure of $S^1$ to $1$.) Also, the irreducibles have dimension one, so we do not have a true matrix problem. Is there something as neat for nonabelian groups?

Although I am looking for a general answer, any examples for any groups are welcome.

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If the representation is faithful, then there is no such geodesic. This follows from Figuera-O'Farrel's answer below, since the integral is zero unless the element $\rho (\gamma) $ is identity. –  Venkataramana Jul 20 at 17:03
    
@Venkataramana True, but the second question can have an affirmative answer even for a faithful representation. If $G=SU(3)$ and $\rho$ is the tautological representation, the geodesic $\gamma(w)=\text{diag}(w,\bar w,1)$ gives $I(\rho,\gamma)=\text{diag}(0,0,1)$. Summing three matrices like this gives an invertible one. I don't know how general this phenomenon is. –  Joonas Ilmavirta Jul 20 at 17:32

1 Answer 1

I had made a silly error in the first version of this answer.

To try to answer your first question, notice that the image of $\gamma$ is a circle subgroup $T$, say, of $G$ and that the integral $I(\rho,\gamma)$ is (with your normalisation of the measure on the circle) the projector onto the subspace $V^T$ of $T$-invariants in the restriction to $T$ of the representation $\rho: G \to GL(V)$. Hence $I(\rho,\gamma)$ is invertible if and only if $V^T=V$. (Previously I had written $V^T=0$ and hence reached the wrong conclusion.) As Venkataramana points out in the comment, this cannot happen if $\rho$ is a faithful representation, hence for simple $G$.

You could take $G = G_1 \times G_2$ and consider $\rho$ such that the $G_2$ subgroup acts trivially and have $\gamma(S^1)$ be a circle subgroup of the form $(e,\gamma_2(t)) \in G_1 \times G_2$, with $e \in G_1$ the identity element.

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Do you mean $V^T=V$? –  Venkataramana Jul 20 at 17:02
    
Oops, yes. I'll edit. –  José Figueroa-O'Farrill Jul 20 at 17:08

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