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In Kaku's QFT textbook page 54, there is a saying:

$GL(N)$ does not have any finite-dimensional spinorial representation.

This implicates that $GL(N)$ owns infinite-dimensional spinorial representation. While in my opinion, a group's spinorial representation is the representation of its universal covering group. And the connnected compenent of $GL(n,\mathbb{R})$ ($n>2$) group is not simply connected and its fundamental group is $\mathbb{Z}_2$. So what group is its covering group?

My question:

  1. Since the connnected compenent of $GL(n,\mathbb{R})$ ($n>2$) group is not simply connected and according to Lie's theorem, there exist a simple connnected Lie group whose Lie algebra is $gl(n,\mathbb{R})$ , then what's this covering group of $GL(n,\mathbb{R})$? While I cannot imagine which group can cover the $GL(n,\mathbb{R})$.

  2. Now that $GL(N)$ owns infinite-dimensional spinorial representation, can show me explicitly, or give me some reference which have solved this problem.

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    $\begingroup$ You don't need to know the covering group - every representation of the Lie algebra induces a representation of the universal cover, so it suffices to see that the algebra possess no "spinorial" representation (though I am not fully certain what "spinorial" means for groups that are not $\mathrm{SO}(1,n)$) $\endgroup$ – ACuriousMind Jan 27 '15 at 11:43
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I) Recall that since the Lie group $SO(N)\subset GL(N)$ is a proper subgroup of $GL(N)$, then functorially speaking, an irreducible representation of $GL(N)$ is also a (possible reducible) representation of $SO(N)$, but not necessarily the other way around.

When Ref. 1 states

There are no finite-dimensional spinorial representations of $GL(N)$,

it means in this context that the finite-dimensional spinor representation of $SO(N)$ does not arise from a finite-dimensional representation of $GL(N)$.

II) For$^1$ $N>2$, the (double) covering group of the general linear group $$GL(N,\mathbb{R})~\cong~ \mathbb{R}_{>0}\times SL(N,\mathbb{R})$$ is the metalinear group $ML(N,\mathbb{R})$. The metalinear group is a subgroup of the metaplectic group $Mp(2N,\mathbb{R})$ in twice the dimension. The reason that the metaplectic group $Mp(2N,\mathbb{R})$ has no non-trivial finite-dimensional representations is closely related to a similar fact for the Heisenberg Lie algebra.

References:

  1. M. Kaku, QFT, 1993; p. 54. and p. 640.

  2. M.B. Green, J.H. Schwarz and E. Witten, Superstring theory, Vol. 2, 1986; p. 272.

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$^1$ $N=2$ is a special case, since $\pi_1(SL(2,\mathbb{R}),*)=\mathbb{Z}$, cf. e.g. this Phys.SE post and this Wikipedia page.

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  • $\begingroup$ Note for later: The above generalizes to other signatures of the metric. $\endgroup$ – Qmechanic Aug 12 '16 at 18:50

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