Does anyone know of a simple proof that the general linear group GL(N) does not admit spinor representations?


Thank you!



Why does GL(N) have no spinor representations?
The statement in GSW that you quote has to be interpreted properly. When they write, "Spinors form a representation of $\mathrm{SO}(N)$ which does not arise from a representation of $\mathrm{GL}(N,\mathbb{R})$", they really mean either "Spinors form a representation of ${\frak{so}}(N)$ which does not arise from a representation of ${\frak{gl}}(N,\mathbb{R})$" (i.e., it's really a statement about representations of Lie algebras) or else they mean what some sources call "multi-valued representations", i.e., representations of the simply-connected covers of the groups $\mathrm{SO}(N)$ and $\mathrm{GL}(N,\mathbb{R})$. Moreover, 'arise from' means that there is no (finite dimensional) representation of ${\frak{gl}}(N,\mathbb{R})$ that, when restricted to ${\frak{so}}(N)\subset{\frak{gl}}(N,\mathbb{R})$, contains a copy of a 'spinor representation' of ${\frak{so}}(N)$.


Either way, what it comes down to is this: If $\pi:\hat{\mathrm{SL}}(n,\mathbb{R})\to \mathrm{SL}(n,\mathbb{R})$ is the nontrivial double cover of $\mathrm{SL}(n,\mathbb{R})$, then any finite-dimensional representation $\hat\rho:\hat{\mathrm{SL}}(n,\mathbb{R})\to \mathrm{GL}(k,\mathbb{R})$ must factor through $\pi$, i.e., $\hat\rho = \rho\circ\pi$ for some (unique) representation $\rho:\mathrm{SL}(n,\mathbb{R})\to\mathrm{GL}(k,\mathbb{R})$.


The reason for this is topological: Although $\pi\_1\bigl(\mathrm{SL}(n,\mathbb{R}),I\_n\bigr)$ is nontrivial for $n\ge 2$ (being $\mathbb{Z}$ when $n=2$ and $\mathbb{Z}\_2$ when $n>2$), the group $\pi\_1\bigl(\mathrm{SL}(n,\mathbb{C}),I\_n\bigr)$ is trivial for all $n\ge 2$. Using this fact, the proof goes like this:


If $\hat\rho:\hat{\mathrm{SL}}(n,\mathbb{R})\to \mathrm{GL}(k,\mathbb{R})$ is any representation, let $\hat\rho':{\frak{sl}}(n,\mathbb{R})\to {\frak{gl}}(k,\mathbb{R})$ denote the induced Lie algebra homomorphism. This complexifies to a Lie algebra homomorphism $(\hat\rho')^{\mathbb{C}}:{\frak{sl}}(n,\mathbb{C})\to {\frak{gl}}(k,\mathbb{C})$ and, since $\mathrm{SL}(n,\mathbb{C})$ is simply-connected, this is induced by a Lie group homomorphism $\rho^{\mathbb{C}}:\mathrm{SL}(n,\mathbb{C})\to \mathrm{GL}(k,\mathbb{C})$, which restricts to a Lie group homomorphism $\rho:\mathrm{SL}(n,\mathbb{R})\to \mathrm{GL}(k,\mathbb{C})$ with associated Lie algebra homomorphism $\rho':{\frak{sl}}(n,\mathbb{R})\to{\frak{gl}}(k,\mathbb{C})$. By construction, this homomorphism $\rho'$ must be the composition of $\hat\rho':{\frak{sl}}(n,\mathbb{R})\to {\frak{gl}}(k,\mathbb{R})$ with the inclusion ${\frak{gl}}(k,\mathbb{R})\hookrightarrow {\frak{gl}}(k,\mathbb{C})$. In particular, $\rho'$ maps into ${\frak{gl}}(k,\mathbb{R})$ after all. Now it's clear that $\hat\rho = \rho\circ\pi$.


In particular, if $\mathrm{Spin}(n)\subset \hat{\mathrm{SL}}(n,\mathbb{R})$ is the pre-image under $\pi$ of $\mathrm{SO}(n)\subset \mathrm{SL}(n,\mathbb{R})$, then any finite dimensional representation of $\hat{\mathrm{SL}}(n,\mathbb{R})$ restricts to a representation of $\mathrm{Spin}(n)$ that factors through $\mathrm{SO}(n)$ and hence cannot be (or even contain) a spinor representation, since those do *not* factor through $\mathrm{SO}(n)$.


