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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 $Spin(p,q)$?

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2 Answers 2

up vote 4 down vote accepted

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.

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:Except for $SO(2)$. –  user72694 Jul 25 '13 at 12:14
    
Oops, yes, indeed, with the exception of the circle $SO(2)$. Thanks, @user72694! –  paul garrett Jul 25 '13 at 12:45

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.

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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 7:40
    
Yes. Exactly. So $Spin(n)$ is the central extension in question. –  geodude Jul 29 '13 at 13:05

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