Consider the group $\text{Spin}(n)$ with $n$ even, i.e. the double cover of $SO(n)$, with Lie group homomorphism $\lambda:\text{Spin}(n)\rightarrow SO(n)$. We know that the spin group admits a faithful representation on $\mathbb{C}^N$ for some $N$ induced by the isomorphism:
$$\mathbb{C}\text{l}(n)\cong \text{End}(\mathbb{C}^N)$$
where $\mathbb{C}\text{l}(n)$ is the standard complex Clifford algebra of $\mathbb{C}^n$. Denote this representation by $\kappa$, and further recall that a vector $x\in \mathbb{R}^n$ acts on $\psi\in\mathbb{C}^n$ via the composition of the canonical injection $\gamma:\mathbb{R}^4\rightarrow \mathbb{C}\text{l}(n)\cong\text{Cl}(n)\otimes_\mathbb{R}\mathbb{C}$ with the above isomorphism. Then $\kappa$ satisfies the property:
$$
\kappa(g)\cdot(x\cdot \psi)=(\lambda(g)\cdot x)\cdot (\kappa(g)\cdot \psi)
$$
My question is then this, does the above property uniquely determine spin representation up to isomorphism? More concretely, if $\rho:\text{Spin}(n)\rightarrow \text{Aut}(\mathbb{C}^N)$ is a faithful representation which satisfies the above property, do we have that $\rho$ and $\kappa$ are related by a conjugation map, i.e. does there exist an isomorphism $f:\mathbb{C}^N\rightarrow \mathbb{C}^N$ such that:
$$f(\kappa(g)\cdot \psi)=\rho(g)\cdot f(\psi)$$
I am asking because in the case of $n=4$, we have that $N=4$, and there are two ways I know of to obtain a faithful representation of $\text{Spin}(n)$ on $\mathbb{C}^4$. The first is the way outlined above, i.e. by making use of the isomorphism between the Clifford algebra and the endomorphism algebra. The other is to identify $\text{Spin}(n)$ with $SU(2)\times SU(2)$, which then acts naturally on $\mathbb{C}^4$ by multiplying with block diagonal matrices. both representations satisfy the above property, and I believe that they should be the same representation since many authors use the latter representation when discussing the spin representation in the case of $n=4$.
Any help would be greatly appreciated.
Best Answer
To clarify, the question is: for fixed $\kappa : \mathbb C\mathrm l(n) \cong \mathrm{End}(\mathbb C^N)$ does the equation $$ f(g)\cdot(\kappa(x)\cdot\psi) = \kappa(\lambda(g)\cdot x)\cdot(f(g)\cdot\psi) $$ for all $g \in \mathrm{Spin(n)}$, $x \in \mathbb R^n$, $\psi \in \mathbb C^N$, and $\lambda(g)\cdot x = gxg^{-1}$ imply that $f = \kappa|_{\mathrm{Spin(n)}}$?
The answer is yes. First, rewrite as $$ f(g)\kappa(x) = \kappa(gxg^{-1})f(g) $$ and apply $\kappa^{-1}$ to get $$ g'x = gxg^{-1}g',\quad g' = \kappa^{-1}(f(g)). $$ Then we can write $$ Gx = xG,\quad G = g^{-1}g'. $$ Because $\mathrm{Cl}(n)$ is generated by $\mathbb R^n$ this means that $G$ is in the center of $\mathrm{Cl}(n)$; but because $n$ is even that means that $G$ is a scalar. So there is some $a \in \mathbb R$ such that $g' = ag$. But we have to have $(1)' = 1$, so $a = 1$ and $g' = g$. In other words, $f = \kappa|_{\mathrm{Spin}(n)}$.