show/hide this revision's text 3 edited body

"Then the claim would be immediate by semisimplicity if one can show that every irreducible representation of $G(\mathbb{C})$ (or perhaps of Lie groups in some more general class than these) occurs as a subquotient of a tensor power of a faithful one. How might one prove the latter? (Can one prove the latter for compact real groups in a manner similar to the proof for finite groups, and then pass to semisimple complex groups by the unitary trick?)"

Yes. The analysis is not so pretty, but it is elementary. Let $K$ be a compact Lie group, $V$ a faithful representation, and $W$ any other representation. Just as in the finite group case, $\mathrm{Hom}_G(W,V^{\otimes \mathrm{Hom}_K(W,V^{\otimes N}) \cong (W^* \otimes V^{\otimes N})^K$, and the dimension of the latter is $\int_K \overline{\chi_W} \cdot \chi_V^{N}$, where $\chi_V$ and $\chi_W$ are the characters of $V$ and $W$, and the integral is with respect to Haar measure. Let $d_V$ and $d_W$ be the dimensions of $V$ and $W$.

We now come to a technical nuisance. Let $Z$ be those elements of $K$ which are diagonal scalars in their action on $V$; this is a closed subgroup of $S^1$. For $g$ not in $Z$, we have $|\chi_V(g)| < d_V$. We first present the proof in the setting that $Z = { e }$.

Choose a neighborhood $U$ of ${ e }$ small enough to be identified with an open disc in $\mathbb{R}^{\dim K}$. On $U$, we have the Taylor expansion $\chi_V(g) = d_V \exp(- Q(g-e) + O(g-e)^3)$, where $Q$ is a positive definite quadratic form; we also have $\chi_W(g) = d_W + O(g-e)$. Manipulating $\int_U \overline{\chi_W} \chi_V^N$ should give you $$\frac{d_W \pi^{\dim K/2}}{\det Q} \cdot d_V^N \cdot N^{-\dim K/2}(1+O(N^{-1/2}))$$ Meanwhile, there is some $D<d_V$ such that $|\chi_V(g)| < D$ for $g \in K \setminus U$. So the integral of $\overline{\chi_W} \chi_V^N$ over $K \setminus U$ is $O(D^N)$, which is dominated by the $d_V^N$ term in the $U$ integral.

We deduce that, unless $d_W=0$, we have $\mathrm{Hom}_K(W, V^{\otimes N})$ nonzero for $N$ sufficiently large.

If $Z$ is greater than ${e }$, then we can decompose $W$ into $Z$-isotypic pieces. Let $\tau$ be the identity character of the scalar diagonal matrices, and let the action of $Z$ on $W$ be by $\tau^k$. (If $\tau$ is finite, then $k$ is only defined modulo $|Z|$; just fix some choice of $k$). Then we want to consider maps from $W$ to $V_N:=V^{k+Nd_V} (\det \ )^{-N}$. $V_N$ is constructed so that $\overline{\chi_W} \chi_{V_N}$ is identically $d_W$ on $Z$; one then uses the above argument with a neighborhood of $Z$ replacing a neighborhood of the origin.
show/hide this revision's text 2 added 5 characters in body

"Then the claim would be immediate by semisimplicity if one can show that every irreducible representation of $G(\mathbb{C})$ (or perhaps of Lie groups in some more general class than these) occurs as a subquotient of a tensor power of a faithful one. How might one prove the latter? (Can one prove the latter for compact real groups in a manner similar to the proof for finite groups, and then pass to semisimple complex groups by the unitary trick?)"

Yes. The analysis is not so pretty, but it is elementary. Let $K$ be a compact Lie group, $V$ a faithful representation, and $W$ any other representation. Just as in the finite group case, $\mathrm{Hom}_G(W,V^{\otimes N}) \cong (W^* \otimes V^{\otimes N})^K$, and the dimension of the latter is $\int_K \overline{\chi_W} \cdot \chi_V^{N}$, where $\chi_V$ and $\chi_W$ are the characters of $V$ and $W$, and the integral is with respect to Haar measure. Let $d_V$ and $d_W$ be the dimensions of $V$ and $W$.

We now come to a technical nuisance. Let $Z$ be those elements of $K$ which are diagonal scalars in their action on $V$; this is a closed subgroup of $S^1$. For $g$ not in $Z$, we have $|\chi_V(g)| < d_V$. We first present the proof in the setting that $Z = { e }$.

Choose a neighborhood $U$ of ${ e }$ small enough to be identified with an open disc in $\mathbb{R}^{\dim K}$. On $U$, we have the Taylor expansion $\chi_V(g) = d_V \exp(- Q(g-e) + O(g-e)^3)$, where $Q$ is a positive definite quadratic form; we also have $\chi_W(g) = d_W + O(g-e)$. Manipulating $\int_U \overline{\chi_W} \chi_V^N$ should give you $$\frac{d_W \pi^{\dim K/2}}{\det Q} \cdot d_V^N \cdot N^{-\dim K/2}(1+O(N^{-1/2}))$$ Meanwhile, there is some $D<d_V$ such that $|\chi_V(g)| < D$ for $g \in K \setminus U$. So the integral of $\overline{\chi_W} \chi_V^N$ over $K \setminus U$ is $O(D^N)$, which is dominated by the $d_V^N$ term in the $U$ integral.

We deduce that, unless $d_W=0$, we have $\mathrm{Hom}_K(W, V^{\otimes N})$ nonzero for $N$ sufficiently large.

If $Z$ is greater than $0$, {e }$, then we can decompose $W$ into $Z$-isotypic pieces. Let $\tau$ be the identity character of the scalar diagonal matrices, and let the action of $Z$ on $W$ be by $\tau^k$. (If $\tau$ is finite, then $k$ is only defined modulo $|Z|$; just fix some choice of $k$). Then we want to consider maps from $W$ to $V_N:=V^{k+Nd_V} (\det \ )^{-N}$. $V_N$ is constructed so that $\overline{\chi_W} \chi_{V_N}$ is identically $d_W$ on $Z$; one then uses the above argument with a neighborhood of $Z$ replacing a neighborhood of the origin.
show/hide this revision's text 1

"Then the claim would be immediate by semisimplicity if one can show that every irreducible representation of $G(\mathbb{C})$ (or perhaps of Lie groups in some more general class than these) occurs as a subquotient of a tensor power of a faithful one. How might one prove the latter? (Can one prove the latter for compact real groups in a manner similar to the proof for finite groups, and then pass to semisimple complex groups by the unitary trick?)"

Yes. The analysis is not so pretty, but it is elementary. Let $K$ be a compact Lie group, $V$ a faithful representation, and $W$ any other representation. Just as in the finite group case, $\mathrm{Hom}_G(W,V^{\otimes N}) \cong (W^* \otimes V^{\otimes N})^K$, and the dimension of the latter is $\int_K \overline{\chi_W} \cdot \chi_V^{N}$, where $\chi_V$ and $\chi_W$ are the characters of $V$ and $W$, and the integral is with respect to Haar measure. Let $d_V$ and $d_W$ be the dimensions of $V$ and $W$.

We now come to a technical nuisance. Let $Z$ be those elements of $K$ which are diagonal scalars in their action on $V$; this is a closed subgroup of $S^1$. For $g$ not in $Z$, we have $|\chi_V(g)| < d_V$. We first present the proof in the setting that $Z = { e }$.

Choose a neighborhood $U$ of ${ e }$ small enough to be identified with an open disc in $\mathbb{R}^{\dim K}$. On $U$, we have the Taylor expansion $\chi_V(g) = d_V \exp(- Q(g-e) + O(g-e)^3)$, where $Q$ is a positive definite quadratic form; we also have $\chi_W(g) = d_W + O(g-e)$. Manipulating $\int_U \overline{\chi_W} \chi_V^N$ should give you $$\frac{d_W \pi^{\dim K/2}}{\det Q} \cdot d_V^N \cdot N^{-\dim K/2}(1+O(N^{-1/2}))$$ Meanwhile, there is some $D<d_V$ such that $|\chi_V(g)| < D$ for $g \in K \setminus U$. So the integral of $\overline{\chi_W} \chi_V^N$ over $K \setminus U$ is $O(D^N)$, which is dominated by the $d_V^N$ term in the $U$ integral.

We deduce that, unless $d_W=0$, we have $\mathrm{Hom}_K(W, V^{\otimes N})$ nonzero for $N$ sufficiently large.

If $Z$ is greater than $0$, then we can decompose $W$ into $Z$-isotypic pieces. Let $\tau$ be the identity character of the scalar diagonal matrices, and let the action of $Z$ on $W$ be by $\tau^k$. (If $\tau$ is finite, then $k$ is only defined modulo $|Z|$; just fix some choice of $k$). Then we want to consider maps from $W$ to $V_N:=V^{k+Nd_V} (\det \ )^{-N}$. $V_N$ is constructed so that $\overline{\chi_W} \chi_{V_N}$ is identically $d_W$ on $Z$; one then uses the above argument with a neighborhood of $Z$ replacing a neighborhood of the origin.