The proof below is mostly by Peter Scholze, so no point in voting it up.

1° I claim that $\dim \mathbb V=\binom{d^2+t-1}{t-1}$. In order to prove this, I will show that the subset $\mathrm{U}_d$ (this is what you call $\mathrm{U}\left(d\right)$ and is defined as the set of all $d\times d$ unitary matrices) is Zariski-dense as a subset of the *$\mathbb C$-vector space* $\mathrm{M}_d$ (which is short for $\mathrm{M}_d\left(\mathbb C\right)$).

Why is this enough? Because let us consider the tensor product $\otimes^t \mathrm{M}_d$. Let $\left(\otimes^t \mathrm{M}_d\right)_{\mathrm{symm}}$ be the subspace of this tensor product containing only the *symmetric* tensors (= tensors invariant under the canonical action of $S_t$ on $\otimes^t \mathrm{M}_d$). Physicists like to call $\otimes^t \mathrm{M}_d$ the "$t$-th symmetric power" of $\mathrm{M}_d$; for us mathematicians it is just *canonically isomorphic* to the $t$-th symmetric power of $\mathrm{M}_d$. Anyway, let $D:\mathrm{M}_d\to \left(\otimes^t \mathrm{M}_d\right)_{\mathrm{symm}}$ be the map given by $D\left(A\right)=A\otimes A\otimes ...\otimes A$ for every $A\in\mathrm{M}_d$. This map is polynomial, and the images of $\mathrm{M}_d$ and $\mathrm{U}_d$ under this map are $\left(\otimes^t \mathrm{M}_d\right)_{\mathrm{symm}}$ and $\left\lbrace U^{\otimes t}\mid U\in\mathrm{U_d}\right\rbrace$, respectively (for $\mathrm{M}_d$, this is known (look up "polarization" and "full restitution"), and for $\mathrm{U}_d$, this is clear). Hence, once we can show that $\mathrm{U}_d$ is Zariski-dense in $\mathrm{M}_d$, it will follow that $\left\lbrace U^{\otimes t}\mid U\in\mathrm{U_d}\right\rbrace$ is Zariski-dense in $\left(\otimes^t \mathrm{M}_d\right)_{\mathrm{symm}}$, so that $\left(\otimes^t \mathrm{M}_d\right)_{\mathrm{symm}}$ must be the span of $\left\lbrace U^{\otimes t}\mid U\in\mathrm{U_d}\right\rbrace$. In other words, $\left(\otimes^t \mathrm{M}_d\right)_{\mathrm{symm}}=\mathbb V$, so that $\dim\mathbb V = \dim \left(\otimes^t \mathrm{M}_d\right)_{\mathrm{symm}} = \binom{d^2+t-1}{t-1}$.` This is what we want to prove.

So it just remains to show that $\mathrm{U}_d$ is Zariski-dense in $\mathrm{M}_d$.

2° Assume the contrary. Then there is a polynomial $P$ on $\mathrm{M}_d$ such that $P$ is not identically zero, but $P\left(U\right)=0$ for every $U\in \mathrm{U}_d$.

3° A lemma from complex analysis: If $V$ is an $\mathbb R$-vector space, and if some holomorphic function $f: V\otimes\mathbb C\to\mathbb C$ is identically zero on the subset $V\otimes \mathbb R$ of $V\otimes \mathbb C$, then $f$ is zero everywhere. (This is easily deduced from the one-dimensional case $V=\mathbb R$, which is a basic fact.)

4° Consider the Lie algebra $\mathfrak{u}_d\in \mathrm{M}_d$. This Lie algebra is an $\mathbb R$-subspace and not a $\mathbb C$-vector subspace of $\mathrm{M}_d$; its $\mathbb C$-span is actually the whole $\mathrm{M}_d$ (because $\mathfrak{u}_d$ consists of all anti-Hermitian matrices). We can even say a bit more: There is an isomorphism $\mathrm{M}_d\cong \mathrm{u}_d\otimes \mathbb C$ as $\mathbb C$-vector spaces, and this isomorphism maps the subset $\mathrm{u}_d$ of $\mathrm{M}_d$ to the subset $\mathrm{u}_d\otimes \mathbb R$ of $\mathrm{M}_d\cong \mathrm{u}_d\otimes \mathbb C$. (Explicitly, this isomorphism is given by $M\mapsto \frac{M-M^{\ast}}{2}\otimes 1 + \frac{M+M^{\ast}}{2}\otimes i$.)

Now we know (from 3°) that if a holomorphic function $\mathrm{u}_d\otimes \mathbb C\to\mathbb C$ is identically zero on the subset $\mathrm{u}_d\otimes \mathbb R$, then it is zero everywhere. Thus, due to the above isomorphism, if a holomorphic function $\mathrm{M}_d\to \mathbb C$ is identically zero on the subset $\mathrm{u}_d$, then it is zero everywhere.

Consider the mapping $\exp:\mathrm{M}_d\to\mathrm{M}_d$. This mapping is holomorphic on the whole $\mathrm{M}_d$. For every $u\in\mathfrak{u}_d$, we have $\exp u\in \mathrm{U}_d$ (since $\mathfrak{u}_d$ is the Lie algebra of $\mathrm{U}_d$) and thus $P\left(\exp u\right)=0$.

The function $P\circ\exp:\mathrm{M}_d\to\mathbb C$ is holomorphic (since so are $P$ and $\exp$) and is identically zero on the subset $\mathrm{u}_d$. According to our above results, it thus is zero everywhere, so that $P$ is zero on $\exp\mathrm{M}_d$. But the set $\exp\mathrm{M}_d$ is Zariski-dense in $\mathrm{M}_d$ (in fact, it is $\mathrm{GL}_d$, but I guess it is easier to prove its Zariski-density by different means), so that $P$ must be zero everywhere, contradicting our assumption on $P$. Qed.

**Addendum:** The above proof is now complete, but it has the disadvantage of not being constructive. The culprit is the $\exp$ mapping, being non-algebraic. Fortunately, we can amend this by replacing the $\exp$ mapping by a different mapping, namely the mapping $T:\mathrm{M}_d\dashrightarrow \mathrm{M}_d$ given by

$T\left(A\right)=\left(I_d+A\right)\left(I_d-A\right)^{-1}$.

Note that this mapping $T$ is not a total mapping, but it is a dominant rational partial mapping, and all that we have used about holomorphic mappings holds for rational partial mappings as well. Instead of the result "for every $u\in\mathfrak{u}_d$, we have $\exp u\in \mathrm{U}_d$" we now must use "for every $u\in\mathfrak{u}_d$ for which $T\left(u\right)$ is defined, we have $T\left( u\right)\in \mathrm{U}_d$" (which is an easy calculation); note that $T\left(u\right)$ is defined for "most" elements $u\in\mathfrak{u}_d$ (in other words, the set of all $u\in\mathfrak{u}_d$ for which $T\left(u\right)$ is defined is a Zariski-closed subset of the set $\mathfrak{u}_d$). Otherwise the proof remains pretty much the same.

where? $\endgroup$ – Mariano Suárez-Álvarez Oct 25 '10 at 5:27