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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.

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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 $\otimes^t \mathrm{M}_d$. Let $\left(\otimes^t \mathrm{M}_d\right)_{\mathrm{symm}}$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$U\in \mathrm{U}_d$. 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.) 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 ) 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, contradicting our assumption on$P$. Qed. 1 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 $\otimes^t \mathrm{M}_d$. Let$\left(\otimes^t \mathrm{M}_d\right)_{\mathrm{symm}}$ 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$U\in \mathrm{U}_d$. 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 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, contradicting our assumption on$P\$. Qed.