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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    $\begingroup$ 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. $\endgroup$ – Venkataramana Jul 20 '14 at 17:03
  • $\begingroup$ @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. $\endgroup$ – Joonas Ilmavirta Jul 20 '14 at 17:32
  • $\begingroup$ @JoonasIlmavirta I am sorry if my question is trivial. Why the image of $\gamma$ is geodesic? $\endgroup$ – Ali Taghavi Oct 8 '14 at 19:22
  • $\begingroup$ @JoonasIlmavirta In particular, is every torus subgroup, totally geodesic? $\endgroup$ – Ali Taghavi Oct 8 '14 at 19:38
  • $\begingroup$ @AliTaghavi, if you take any bi-invariant Riemannian metric on $G$, then the periodic geodesics passing through the identity are precisely the nontrivial homomorphisms $S^1\to G$. Every Lie subgroup is totally geodesic because of this algebraic structure. $\endgroup$ – Joonas Ilmavirta Oct 9 '14 at 7:02

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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    $\begingroup$ Do you mean $V^T=V$? $\endgroup$ – Venkataramana Jul 20 '14 at 17:02
  • $\begingroup$ Oops, yes. I'll edit. $\endgroup$ – José Figueroa-O'Farrill Jul 20 '14 at 17:08

José Figueroa-O'Farrill already answered the first question. Let me answer the two other ones:

2) If $G$ is not $S^1$ or $S^3$, the answer is yes. If this did not hold for some $\rho$, there would be a nonzero vector $a\in\mathbb C^{\dim\rho}$ that is in the kernel of $I(\rho,\gamma)$ for all $\gamma$. Let $A$ be the square matrix whose every row is $a^*$. Then $AI(\rho,\gamma)=0$ for all $\gamma$.

Now let $f:G\to\mathbb C$ be the unique smooth function that satisfies $\int_Gf(x)\rho(x)dx=A$ and $\int_Gf(x)\sigma(x)dx=0$ for all irreducible representations $\sigma$ that are not equivalent to $\rho$. The existence and uniqueness of $f$ follows from the Peter-Weyl theorem.

For any $x\in G$ and any nontrivial homomorphism $\gamma:S^1\to G$ let $$ F(x,\gamma)=\int_{S^1}f(x\gamma(t))dt $$ be the integral of $f$ over the geodesic $t\mapsto x\gamma(t)$. For any (unitary, irreducible) representation $\sigma$, a calculation gives $$ \int_G F(x,\gamma)\sigma(x)dx = \int_G f(x)\sigma(x)dx I(\sigma,\gamma). $$ By the definition of $f$, this is always zero (for both $\sigma=\rho$ and $\sigma\neq\rho$). Thus $F$ is orthogonal to all matrix elements of all irreducible representations, so $F=0$.

This means that $f$ is not identically zero but its integral over every geodesic is zero. This is impossible by theorem 1.1 of this paper.

If $G=S^1$, the answer is no for all nonconstant $\rho$. If $G=S^3$, I don't know the exact answer. On this group a function is known to integrate to zero over all geodesics if and only if it is antipodally antisymmetric ($f(-x)=-f(x)$ if we embed $S^3\subset\mathbb R^4$). This with the previous argument might tell for which $\rho$s the answer is yes; it is not yes for all of them.

3) The answer to the third question is positive.

Let $D(\rho,\gamma)=\left.\frac{d}{dt}\rho(\gamma(t))\right|_{t=0}$. Then $\frac{d}{dt}\rho(\gamma(t))=\rho(\gamma(t))D(\rho,\gamma)$. Let $1$ denote the identity matrix, since $I$ is reserved for the integral. If $A$ satisfies $AI(\rho,\gamma)=0$, then \begin{eqnarray} A &=& A(1-I(\rho,\gamma)) \\&=& A\left(1-\int_{S^1}\rho(\gamma(t))dt\right) \\&=& A\int_{S^1}(\rho(\gamma(0))-\rho(\gamma(t)))dt \\&=& A\int_{S^1}\int_t^0\frac{d}{ds}\rho(\gamma(s))dsdt \\&=& A\int_{S^1}\int_t^0\rho(\gamma(s))dsdtD(\rho,\gamma). \end{eqnarray} If we let $$ B=A\int_{S^1}\int_t^0\rho(\gamma(s))dsdt, $$ then we have $A=BD(\rho,\gamma)$. Moreover, the matrix $B$ can be explicitly expressed in terms of $A$.

Since $I(\rho,\gamma)$ is hermitean and $D(\rho,\gamma)$ skew-hermitean, the result states that the image of the derivative of a representation is exactly the kernel of the integral of the representation and vice versa. (Here, of course, the derivative and the integral mean $D$ and $I$.)

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