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Following on this question about how to characterise Spinors mathematically:

First, given a universal cover $\pi:G' \rightarrow G$ of a lie group $G$, is it correct to say we can always lift representations of $G$ to those of $G'$ essentially by pre-composition by $\pi$? (presumably, modulo questions about the exact smooth structure of the space $End (V)$ for a representation $V$ of $G$).

Secondly, there are representations of $G'$ that do not descend to $G$.

Is it fair to call these Spinor representations, since when $G$ is the connected component of the identity of $SO(p,q)$, its universal cover is the double cover $Spin(p,q)$?

Community
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Mozibur Ullah
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3 Answers3

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Yes, yes, and yes.

In more detail: issues about smooth vectors are much subordinate in any case, and prove trivial, so (smooth-vector?) repns of the "lower" group "lift" to the covering group. Not at all a problem.

Yes, at least as evidenced by many examples, about half the finite-dimensional repns of spin groups do not descend to the corresponding orthogonal groups, for example. Some other classical groups happen to be simply-connected, in contrast.

And, yes, to the last question, as a matter of usage or convention or tradition.

The underlying point-of-interest is that it is not obvious that the universal covering of all these not-simply-connected Lie groups is just a two-fold-cover. Weyl found this.

paul garrett
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In quantum mechanics, complex state vectors are always defined up to a "phase". What matters is the direction of a vector, any given $\psi$ is physically equivalent to $z\psi$, where $z$ is a complex number (of norm $1$ for probability normalization). That, is, physics lies in a complex projective space.

A linear representation induces always a projective representation, of course. The converse is not always true. But it is true that a projective representation induces a linear representation, in general, of a larger group, called "central extension". (If you know what a central charge is, it is exactly that, in mathematical language.) For Lie groups, extensions by a discrete group are covering groups. If the group in question is simply connected, then, this problem is trivial.

$SO(n)$ is not simply connected. So there are projective (or physical) representation of $SO(n)$ that cannot be written as linear representations, but still have physical significance.

In a way, spinors are representations of the orthogonal group. Only, not linear.

geodude
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  • Should that be 'representations of the special orthogonal group? $SO(n)$ is not simply-connected, so it has a universal cover $Spin(n)$, so we have the sequence $Z/2 \rightarrow Spin(n) \rightarrow SO(n)$. Then a projective representation of $SO(n)$ lifts up to a linear representation of $Spin(n)$? – Mozibur Ullah Jul 27 '13 at 07:40
  • Yes. Exactly. So $Spin(n)$ is the central extension in question. – geodude Jul 29 '13 at 13:05
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Note: I only intend this as a comment on the answer by @user12291 (paul garrett), but I don't have enough reputation to do so.

If I'm not mistaken, the answer to the third question is only "yes" in some cases (though arguably it is true in the cases of interest in physics). More explicitly, if $p,q\geq 3$, then $Spin(p,q)$ is indeed a double cover of (the identity component of) $SO(p,q)$, but the latter has maximal compact subgroup $SO(p)\times SO(q)$, which means that

$$ \pi_1(SO(p,q))\cong \pi_1(SO(p))\times\pi_1(SO(q))\cong \mathbb{Z}_2\times\mathbb{Z}_2 $$

(with the latter iso. coming from the $p,q\geq 3$ condition); thus the fact that $Spin(p,q)$ is a double cover implies that

$$ \pi_1(Spin(p,q))\cong\mathbb{Z}_2. $$

The point being: $Spin(p,q)$ is, as said above, a double cover of $SO(p,q)$ but it is not the universal cover in such cases.

(I'm fairly confident in this, though not entirely sure, so feel free to correct me if I'm wrong.)

OperaticDreamland
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