Let $\{v\_1, \dotsc, v\_m\} \in \mathbb{C}^{2^n}$ be a set of orthonormal vectors. Define a map $R\_m$ from $2^n \times 2^n$ to $m \times m$ matrices as follows:
$$R\_m(M) := \sum\_{i,j=1}^m (v\_i^\*M v\_j) E\_{ij}$$
where $E\_{ij} := e\_i e\_j^\*$ is the all-zeroes matrix with entry 1 at position $(i,j)$. In other words, $R\_m$ performs a unitary change of basis and then takes the top-left $3 \times 3$ block of the resulting matrix.


For any $n \geq 2$, show that no matter how the vectors $\{v\_i\}$ are chosen,
$$\mathrm{U}(3) \not\subseteq \bigl\{ R\_3(U\_1 \otimes \dotsb \otimes U\_n) : U\_i \in \mathrm{U}(2) \bigr\}.$$
That is, as matrices $U\_i$ range over all possible $2 \times 2$ unitaries, their tensor product will never contain the set of all $3 \times 3$ unitaries in some basis.


More generally, this should hold even if $2 \times 2$ unitaries are replaced by $d \times d$ unitaries and $m = 3$ is replaced by any $m > d$. The motivation for this question comes from "encoded universality" in quantum computing.


**Note:** I originally asked this question on [math.stackexchange](https://math.stackexchange.com/questions/1112208/why-a-tensor-product-of-2-times-2-unitaries-cannot-implement-a-3-times-3-uni) but did not receive any answer.



Why a tensor product of $2\times 2$ unitaries cannot implement a $3\times 3$ unitary?
When the (general) question is rephrased in less basis-dependent language, I believe that it translates to this: Let $\mathrm{U}(d)$ act on $V = \mathbb{C}^d$ in the usual way, and consider the $n$-fold product $G = \mathrm{U}(d)\times \mathrm{U}(d)\times\cdots\times \mathrm{U}(d)$ acting by the usual tensor product on 
$$
V^{\otimes n} = \mathbb{C}^d\otimes \mathbb{C}^d\otimes\cdots\otimes \mathbb{C}^d \simeq \mathbb{C}^{d^n}.
$$
This is an irreducible $G$-representation, and hence there is, up to a constant multiple, a unique $G$-invariant Hermitian inner product on $V^{\otimes n}$. 


Given any subspace $W\subset V^{\otimes n}$, let $\pi\_W:V^{\otimes n}\to W$ be the projection onto $W$ that is orthogonal with respect to this $G$-invariant inner product, and define a (smooth) mapping $R\_W:G\to \mathrm{End}(W)$ by
$$
R\_W(g\_1,\ldots,g\_n)(w) = \pi\_W\bigl((g\_1\otimes g\_2\otimes\cdots\otimes g\_n\bigr)(w)\bigr)
$$ 
for $(g\_1,\ldots,g\_n)\in G$.


The OP asks whether there exists a subspace $W\subset V^{\otimes n}$ of dimension $m>d$ such that $R\_W(G)$ contains the group $\mathrm{U}(W)\subset\mathrm{End}(W)$ of unitary transformations of $W$ (when it is endowed with the inner product that it inherits as a subspace of $V^{\otimes n}$). The answer is *no*, and the following argument provides a proof. 


Suppose that such a $W$ existed, and let $H\subset G$ be the (closed) subset 
$$
H = \left(R\_W\right)^{-1}\bigl(\mathrm{U}(W)\bigr).
$$
Then I claim that $H$ is a subgroup of $G$, in fact, $H$ is the subgroup of $G$ that preserves the subspace $W\subset V^{\otimes n}$. The reason is that, if $R\_W(g\_1,\ldots,g\_n):W\to W$ is unitary (i.e., preserves lengths), then $\bigl(g\_1\otimes g\_2\otimes\cdots\otimes g\_n\bigr)(w)$ must lie in $W$ for all $w\in W$. 


Since $H$ is the compact subgroup of $G$ that preserves the subspace $W$, it follows immediately that 
$$
R\_W:H\to \mathrm{U}(W)
$$
is, in fact, a group homomorphism, and a surjective one, at that. Thus, the induced Lie algebra homomorphism
$$
r\_W:{\frak{h}}\to {\frak{u}}(W)
$$
is also surjective. In particular, since ${\frak{u}}(W) = {\frak{z}}\oplus {\frak{su}}(W) \simeq \mathbb{R}\oplus {\frak{su}}(m)$ and ${\frak{su}}(m)$ is simple (since $m>d\ge 2$), there exists an injective Lie algebra homomorphism $\phi: {\frak{su}}(m)\to {\frak h}$ so that the composition $r\_W\circ\phi: {\frak{su}}(m) \to {\frak{su}}(W)\subset {\frak{u}}(W)$ is a Lie algebra isomorphism with ${\frak{su}}(W)$.


However, this is not possible, because ${\frak{h}}$ is embedded as a subalgebra 
$$
{\frak{h}}\subset {\frak{u}}(d)\oplus {\frak{u}}(d)\oplus \cdots \oplus {\frak{u}}(d)
= {\frak{g}},
$$
but there can be no nontrivial Lie algebra homomorphism from ${\frak{su}}(m)$ to ${\frak{u}}(d)$ when $m>d$, since ${\frak{su}}(W)$ is simple, and thus, such a homomorphisms would have to be injective, which is impossible for dimension reasons.


