Symmetric Power Irreducible

If $k=0$, the tensor unit is one-dimensional and the action is trivial, so the representation is irreducible. Suppose $k\ge 1$. Let $e_1,\dots ,e_d$ be the standard orthonormal basis of ${\mathbb{C}}^d$. For each multi-index $\alpha =({\alpha }_1,\dots ,{\alpha }_d)\in {\mathbb{N}}^d$ with ${\sum }_j{\alpha }_j=k$, let ${\mathrm{\Omega }}_{\alpha }$ be the set of words of length $k$ in which $j$ occurs ${\alpha }_j$ times, and set $v_{\alpha }=\mathrm{\mid }{\mathrm{\Omega }}_{\alpha }{\mathrm{\mid }}^{-1\mathrm{/}2}{\sum }_{(i_1,\dots ,i_k)\in {\mathrm{\Omega }}_{\alpha }}{e}_{i_1}\otimes \cdots \otimes e_{i_k}$. The sets ${\mathrm{\Omega }}_{\alpha }$ partition the elementary tensor basis, so the $v_{\alpha }$ are orthonormal. A vector fixed by all permutations has equal coefficients on each permutation orbit ${\mathrm{\Omega }}_{\alpha }$; hence these vectors span exactly the symmetric subspace in [F2].

Put $t=(2k+1{)}^{-1}$ and $D=\mathrm{diag}(e^{it},e^{i{t}^2},\dots ,e^{i{t}^d})\in U(d)$. Each $v_{\alpha }$ is an eigenvector of $D^{\otimes k}$ with eigenvalue ${\lambda }_{\alpha }=e^{i{\sum }_j{\alpha }_j{t}^j}$. These eigenvalues are distinct: for $\alpha \mathrm{\ne }\beta$, let $\mathrm{\ell }$ be the first index with $c_{\mathrm{\ell }}:={\alpha }_{\mathrm{\ell }}-{\beta }_{\mathrm{\ell }}\mathrm{\ne }0$. Then $\mathrm{\mid }{c}_{\mathrm{\ell }}\mathrm{\mid }{t}^{\mathrm{\ell }}\ge t^{\mathrm{\ell }}$, while $\mathrm{\mid }{\sum }_{j>\mathrm{\ell }}{c}_j{t}^j\mathrm{\mid }\le k{t}^{\mathrm{\ell }+1}\mathrm{/}(1-t)=t^{\mathrm{\ell }}\mathrm{/}2$; thus the phase difference is nonzero. Also $\mathrm{\mid }{\sum }_j{c}_j{t}^j\mathrm{\mid }\le kt\mathrm{/}(1-t)=1\mathrm{/}2<2\pi$, so the phase difference is not a nonzero multiple of $2\pi$.

Let $W\mathrm{\ne }\{0\}$ be a subspace invariant under every $U^{\otimes k}$, and choose $0\mathrm{\ne }w={\sum }_{\alpha }{c}_{\alpha }{v}_{\alpha }\in W$. At least one coefficient $c_{\alpha }$ is nonzero. Because the eigenvalues in step 1.2 are distinct, the interpolation polynomial $p_{\alpha }(z)={\prod }_{\beta \mathrm{\ne }\alpha }(z-{\lambda }_{\beta })\mathrm{/}({\lambda }_{\alpha }-{\lambda }_{\beta })$ satisfies $p_{\alpha }(D^{\otimes k})w=c_{\alpha }{v}_{\alpha }$. Invariance under $D^{\otimes k}$ and linearity imply $v_{\alpha }\in W$. If $d=1$, step 1.1 has only the single multi-index $(k)$, so this already gives $W={\mathrm{Sym}}^k(\mathbb{C})$. Hence assume $d\ge 2$ in the transfer argument below.

Fix indices $i\mathrm{\ne }j$ with ${\alpha }_i>0$, let $q>0$ be sufficiently small, and set $c=(1+q^2{)}^{-1\mathrm{/}2}$ and $s=q(1+q^2{)}^{-1\mathrm{/}2}$, so $c^2+s^2=1$. The unitary rotation sending $e_i\mapsto c{e}_i+s{e}_j$, $e_j\mapsto -s{e}_i+c{e}_j$ and fixing all other $e_{\mathrm{\ell }}$ sends $v_{\alpha }$ into the symmetric subspace by [L1]. The coefficient of $v_{\beta }$ for $\beta =\alpha -e_i+e_j$ is the positive factor $\sqrt{\mathrm{\mid }{\mathrm{\Omega }}_{\beta }\mathrm{\mid }\mathrm{/}\mathrm{\mid }{\mathrm{\Omega }}_{\alpha }\mathrm{\mid }}$ times $s{c}^{{\alpha }_i+{\alpha }_j-1}P(q^2),P(z)={\sum }_{r=0}^{\min ({\alpha }_i-1,{\alpha }_j)}(-1{)}^r\left(\genfrac{}{}{0px}{}{{\alpha }_i-1}{r}\right)\left(\genfrac{}{}{0px}{}{{\alpha }_j+1}{r+1}\right)z^r\mathrm{.}$ This follows by counting which $r$ output $i$-positions came from $j$ and which $r+1$ output $j$-positions came from $i$ in the tensor expansion. Since $P(0)={\alpha }_j+1>0$ and $P$ is a polynomial, choose $q>0$ small enough that $P(q^2)\mathrm{\ne }0$. The rotated vector lies in $W$ by invariance; applying the interpolation projection from step 2.1 for weight $\beta$ isolates a nonzero multiple of $v_{\beta }$, so $v_{\beta }\in W$.

Repeating step 3.1 transfers one unit from any coordinate with ${\alpha }_i>{\beta }_i$ to one with ${\alpha }_j<{\beta }_j$. Any two nonnegative multi-indices of total $k$ are connected by finitely many such transfers. Thus $W$ contains every basis vector $v_{\beta }$ from step 1.1 and equals ${\mathrm{Sym}}^k({\mathbb{C}}^d)$. Every nonzero invariant subspace is therefore the whole representation, proving irreducibility.