TITLE: Symmetric subspace of linear operators
QUESTION [6 upvotes]: This is a question that stemmed from fooling around with unitary t-designs.
Let 
\begin{equation}
  \mathbb{V} = \mathrm{span} \{\; U^{\otimes t}\; |\; U \in \mathrm{U}(d)\}
\end{equation}
Where $\mathrm{U}(d)$ is the unitary group acting on $\mathbb{C}^d$.
What can we say about $\dim \mathbb{V}$? What tools can I use to figure out the dimension or good lower bounds for it?
I've read (p. 162) about the analogous question for $\mathbb{U} = \mathrm{span}\{|\psi\rangle^{\otimes t}\;|\; |\psi\rangle \in \mathbb{C}^d\}$. In particular, we can show that $\dim \;\mathbb{U} = {d + t - 1 \choose t - 1}$. To use that approach, we need to change $\mathrm{U}(d)$. 
If we change $\mathrm{U}(d)$ to $\mathbb{C}^{d\times d}$ (i.e. all the matrices on $\mathbb{C}^d$) then is it obvious that the resulting space is the same as $\mathbb{V}$? If they are the same, then the question becomes easier, since then we have the space of all linear operators that are invariant under permutations of the $t$ registers and we can just use the formulate for $\dim \mathbb{U}$ thinking of the operators as row vectors.
(I am not sure of the tags for this, so feel free to change them)

REPLY [6 votes]: Here is another proof of the same result (it is probably the same as the one given by darij grinberg but I do not understand all the words. It has perhaps more his place as a comment, but I do not know how to write comments). In the case $t=2$, it is Claim 1 in Lemma 1.3.2 of http://arxiv.org/pdf/1002.4595v2 
Denote by $P : M_d(\mathbb C)^{\otimes t} \to M_d(\mathbb C)^{\otimes t}$ the linear map such that $P(A_1 \otimes \dots \otimes A_t) = 1/t!\sum A_{\sigma(1)} \otimes \dots \otimes A_{\sigma(t)}$, where the sum os over all permutations of $\{1,\dots ,t\}$. $P$ is the natural projection on the symmetric tensors.
Let $(H_k)$ be the assertion "$\mathbb V$ contains  the space $F_k$ spanned by $P(h_1 \otimes h_2 \otimes \dots \otimes h_t)$ for any hermitian matrices $h_1,...,h_t$ with $h_{k+1} = h_{k+2} = \dots h_t=1$" ($1$ is the identity matrix).
By polarity, $(H_k)$ is equivalent to the assertion "$\mathbb V$ contains $P(h_1 \otimes h_2 \otimes \dots \otimes h_t)$ for any hermitian matrices $h_1,...,h_t$ with $h_1 = h_2 = \dots =h_k$ and $h_{k+1} = h_{k+2} = \dots h_t=1$".
Prove $(H_k)$ by induction on $k$. Take indeed a symmetric matrix $h$, and consider $f(x) = \exp(ixh)^{\otimes t}$ and develop $f$ in power series: $f(x) = \sum_n x^n a_n$. Then $a_n$ belongs to $\mathbb V$ for all $n$. For example $a_1 = i t P(h\otimes 1 \ \dots \otimes 1)$ which proves $(H_1)$.
One moment of thinking gives you that you can write $a_k=i^k C(t,k) P(h\otimes \dots h \otimes 1 \dots \otimes 1) + $something in $F_{k-1}$, where $C(t,k)$ is the binomial coefficient and there are $k$ $h$'s and $(t-k)$ $1$'s in $h\otimes \dots h \otimes 1 \dots \otimes 1$. This allows to proceed by induction.
Thus for $k=t$, and using that $M_d(\mathbb C)$ is the linear span of the hermitian matrices, you get that $\mathbb V$ is the whole space of symmetric tensors.