$\DeclareMathOperator\SU{SU}\DeclareMathOperator\SO{SO}\DeclareMathOperator\GL{GL}$I asked a similar question on math stack exchange here, but I wonder if it may be better received here.

Let $n$ be even, then the standard complex Clifford algebra admits an isomorphism: $$ \mathbb{C}\text{l}(n)\cong \text{End}(\mathbb{C}^N) $$ where $N=2^{n/2}$. We call $\mathbb{C}^N$ the space of Dirac spinors. Let $\gamma:\mathbb{R}^n\rightarrow\text{Cl}(n)$ be the canonical injection, and $f:\text{Cl}(n)\rightarrow \mathbb{C}\text{l}(n)\cong \text{Cl}(n)\otimes_\mathbb{R}\mathbb{C}$ be the inclusion map. The above isomorphism then induces a blinear map called spinor multiplication with a vector: $$ \begin{align*} \mathbb{R}^n\times \mathbb{C}^N&\longrightarrow \mathbb{C}^N\\ (x,\psi)&\longmapsto x\cdot \psi =(f\circ \gamma(x))\cdot \psi \end{align*} $$ where we identify $f\circ \gamma(x)$ with it's unique endomorphism under the aforementioned isomorphism. Identifying $\text{Spin}(n)$ as a subset of the real Clifford algebra then yields a faithful representation of $\text{Spin}(n)$ on $\mathbb{C}^N$, called the spinor representation. We denote this representation by $\kappa$, and one can easily show that: $$ \kappa(g)\cdot(x\cdot \psi)=(\lambda(g)\cdot x)\cdot(\kappa(g)\cdot \psi) $$ where $\lambda:\text{Spin}(n)\rightarrow \SO(n)$ is the double covering homomorphism.

My question is then this, does the above property uniquely determine the spinor representation up to isomorphism? I.e. if we have two faithful representations of $\text{Spin}(n)$ on $\mathbb{C}^N$ satisfying the property above, denoted by $\kappa$ and $\rho$, does there exist an isomorphism $F:\mathbb{C}^N\rightarrow \mathbb{C}^N$ satisfying: $$ \begin{align*} F(\kappa(g)\cdot \psi)=\rho(g)\cdot F(\psi) \end{align*} $$ for all $\psi\in \mathbb{C}^N$.

My motivation for this question is mostly due to the case where $n=4$. When $n=4$, we have that the space of Dirac spinors is given by $\mathbb{C}^4$, then we can obtain a representation of $\text{Spin}(4)$ on $\mathbb{C}^4$ as above. Indeed, if we choose an orthonormal basis for $\mathbb{R}^4$, it is not difficult to construct a faithful representation of $\text{Cl}(n)$ on $\mathbb{C}^4$, which then induces the isomorphism between the complex Clifford algebra. In particular, such a representation, when restricted to $\text{Spin}(n)$ has image in the subgroup of block diagonal matrices who's entries lie in $\SU(2)$. By construction, this representation satisfies the property above

However, we can also construct a representation of $\text{Spin}(4)$ in a different way, due to the exceptional isomorphism: $$ \begin{align*} \text{Spin}(4)\cong \SU(2)\times \SU(2) \end{align*} $$ We can then embed $\SU(2)\times \SU(2)$ in $\GL_4(\mathbb{C})$ by mapping $\SU(2)\times \SU(2)$ to block diagonal matrices, which naturally acts on $\mathbb{C}^4$. By identifying $\mathbb{R}^4$ with the real subspace of $2\times 2$ complex matrices spanned by the Pauli spin matrices, we can specify how $\mathbb{R}^4$ acts on $\mathbb{C}^4$. This action can be made to be the same as the one obtained by construction a faithful representation of $\text{Cl}(n)$ on $\mathbb{C}^4$. This representation of $\text{Spin}(4)$ on $\mathbb{C}^4$ then also respects Clifford multiplication with a vector.

Both of these representations are seemingly used interchangeably, though the latter is used more frequently when one wishes to discuss the specific case of $n=4$. Due to this, I suspect that the representations should be isomorphic, but I am unsure of how to show it.

If they are not isomorphic, then how do we justify using the latter representation when the former is derived in such generality?

  • $\begingroup$ This follows from the classification of irreducible representations of SU(2) for n=4. $\endgroup$ Jul 29, 2023 at 0:07
  • $\begingroup$ @KentaSuzuki sorry my representation isn’t the best, could you elaborate? $\endgroup$
    – Chris
    Jul 29, 2023 at 1:57

1 Answer 1


For each integer $n\ge1$ the Lie group $SU(2)$ has a unique irreducible representation $V_n$ of dimension $n$, namely the symmetric power $\mathrm{Sym}^{n-1}(\mathbb C^2)$. Thus $Spin(4)\cong SU(2)\times SU(2)$ has irreducible representations of the form $V_{(m,n)}:=V_m\boxtimes V_n$ of dimension $mn$, where $m,n\ge1$. A four-dimensional representation of $Spin(4)$ is one of: $V_{(4,1)}$, $V_{(2,2)}$, $V_{(1,4)}$, $V_{(3,1)}+V_{(1,1)}$, $V_{(1,3)}+V_{(1,1)}$, $V_{(1,2)}+V_{(1,2)}$, $V_{(1,2)}+V_{(2,1)}$, $V_{(2,1)}+V_{(2,1)}$, $V_{(2,1)}+2V_{(1,1)}$, $V_{(1,2)}+2V_{(1,1)}$, or $4V_{(1,1)}$. Only $V_{(1,2)}+V_{(2,1)}$ is faithful, which must be the spinor representation.

  • $\begingroup$ What does $\boxtimes$ denote here? Do you know of a source which goes into this? Does this imply that any faithful representations of $SU(2)\times SU(2)$ on $\mathbb{C}^4$ must be isomorphic? Also, I'm assuming this doesn't generalize as I thought ? $\endgroup$
    – Chris
    Jul 29, 2023 at 4:30
  • $\begingroup$ @Chris FYI, regarding the general use of $\boxtimes$, please see the Math SE post What is the operation $\boxtimes$?. $\endgroup$ Jul 29, 2023 at 4:46
  • $\begingroup$ Also, how is this representation of $\text{Spin}(4)$ irreducible? Don't the projections onto the upper and lower block diagonal matrices determine sub representations of on $\mathbb{C}^2$? $\endgroup$
    – Chris
    Jul 29, 2023 at 4:59
  • $\begingroup$ The spinor representation is reducible, but I write it as a sum of irreducible representations. $\endgroup$ Jul 29, 2023 at 11:19
  • $\begingroup$ @KentaSuzuki ok thanks that makes sense. Do you know where I can find a source on this result? $\endgroup$
    – Chris
    Jul 29, 2023 at 17:40

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