Uniqueness of Hilbert space given algebra of operators Let $\mathfrak{H}$ be a Hilbert space and let $X,Y,Z$ be operators on $\mathfrak{H}$ which satisfy the same commutation relations as the Pauli matrices, e.g., $X^2=I$ and $XY=iZ$ and so on. If $I,X,Y,Z$ generate the algebra of bounded operators on $\mathfrak{H}$, then does this imply that $\mathfrak{H}\cong \mathbb{C}^2$?
How about the case where we have operators $X_1,Y_1,Z_1,X_2,Y_2,Z_2$ which satisfy the same relations as Pauli matrices on 2 qubits, e.g., $X_1,Y_2$ "acts" the same as $X\otimes I,I\otimes Y$ on $\mathbb{C}^2\otimes \mathbb{C}^2$ (have the same commutation relations)? Would that imply that $\mathfrak{H} \cong \mathbb{C}^2\otimes \mathbb{C}^2$?
More generally, given a Lie algebra/group (some commutation structure) of operators on some Hilbert space $\mathfrak{H}$ which generates the algebra of bounded operators on $\mathfrak{H}$, is such a Hilbert space unique? For the time being, let us only consider finite-dim $\mathfrak{H}$.
 A: 
Given a Lie algebra/group (some commutation structure) of operators on some finite dimensional Hilbert space Hilbert space $\mathfrak{H}$ which generates the algebra of operators on $\mathfrak{H}$, is such a Hilbert space unique?

The answer is yes. One argument to this effect is as follows.
Suppose that $\mathfrak g$ is a complex Lie algebra, $\mathfrak H$ is a Hilbert space with dimension $n$, and $\rho_1:\mathfrak g \to \mathcal B(\mathfrak H)$ is a fully-faithful (bijective) representation. In other words, $\rho$ is an isomorphism of Lie Algebras between $\mathfrak g$ and $\mathfrak{gl}(\mathfrak H)$. In order to show that the dimension of $\mathfrak H$ is uniquely determined, it suffices to show that the Lie algebras $\mathfrak {gl}_m$ and $\mathfrak{gl}_n$ are not isomorphic for $m \neq n$.
We say that an operator $A$ over $\mathfrak H$ is nilpotent if $A^k = 0$ for some integer $k$, and the minimal such integer is the (nilpotency) index of $A$.
We note that for any $\mathfrak{H}$ with dimension $n$, there exists a nilpotent operator of index $k$ if and only if $1 \leq k \leq n$.
Now, suppose that $m < n$. Suppose for the purpose of contradiction that $\rho: \mathfrak{gl}_n \to \mathfrak{gl}_m$ is an isomorphism. Let $A \in \mathfrak{gl}_n$ be nilpotent of index $k$. We have
$$
0 = \rho(0) = \rho(A^{n}) = \rho(A)^n.
$$
Because $\rho(A) \in \mathfrak{gl}_m$ is nilpotent, we have $\rho(A)^m = 0$. Thus, $A^m \neq 0$, but $\rho(A)^m = 0$. So, $\rho$ is not injective and therefore not an isomorphism.

To put things a bit more concretely for the case of the Pauli matrices: we find that for all linear combinations $A = a_0 I + a_1 X + a_2 Y + a_3 Z$, we have $A^3 = 0 \implies A^2 = 0$. Thus, it is impossible for $I,X,Y,Z$ to span $B(\mathfrak{H})$ when $\mathfrak{H}$ has dimension $3$ or higher.
Conversely, $I,X,Y,Z$ cannot be given the correct relations if they are operators over a one-dimensional space. One way to see this is to note that $A = X - iY$ satisfies $A^2 = 0$, but $A \neq 0$; this cannot occur for operators over a one-dimensional space.
