Why a tensor product of $2\times 2$ unitaries cannot implement a $3\times 3$ unitary? 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 but did not receive any answer.
 A: 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.
