Integrals of representations over geodesics 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.
 A: 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.
A: 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 $I$.)
