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In the wikipedia article on spinors a number of mathematical definitions are given of spinors which I find slightly confusing.

There are essentially two frameworks for viewing the notion of a spinor. One is representation theoretic. In this point of view, one knows a priori that there are some representations of the Lie algebra of the orthogonal group that cannot be formed by the usual tensor constructions. These missing representations are then labeled the spin representations, and their constituents spinors.

According to the answer by Korman of this question, every irreducible representation is found inside a tensor product of fundamental representations. So there can't be any missing representations. So how do we make sense of the above statement?

In this view, a spinor must belong to a representation of the double cover of the rotation group SO(n, R)

Whereas the statement above refers to lie algebras, this is referring to lie groups. Since there is a 1-1 correspondance between the two I guess this is fine. (But seeing that Spinors orginated in Physics around the idea of 'spin' is it better to think of Spinors as representation of lie groups rather than lie algebras?).

Representations of the double covers of these groups yield projective representations of the groups themselves, which do not meet the full definition of a representation.

Is there a 1-1 relationship between representations of double covers & projective representations of the original group?

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    $\begingroup$ It's no surprise you find the wiki article confusing. I know there are a few enthusiastic yet confused spinor aficionados who edit there, and that might have contributed to the general feeling of the article. $\endgroup$ – rschwieb Jul 15 '13 at 20:14
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    $\begingroup$ Watch out, it's not quite exactly true that the correspondence between Lie groups and Lie algebras is one-to-one: if you restrict to simply_connected (and connected) Lie groups, yes. Thus, several (closely-related) Lie groups have the same Lie algebra. Some repns of the Lie algebra do not "descend" to non-simply-connected quotients, such as orthogonal groups (rather than the simply-connected universal covers, the Spin groups). $\endgroup$ – paul garrett Jul 15 '13 at 20:51
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    $\begingroup$ For what it's worth, I actually do not think of spinors are elements in a representation of either a Lie group or Lie algebra, but rather, as elements in a representation of a Clifford algebra $\mathcal{Cl}(p,q)$. This then yields a representation of $\mathrm{so}(p,q)$, but IMHO the Clifford algebra picture is more fundamental. I should mention that I have almost exclusively worked with spinors in the context of physics, and that most definitely has shaped my perspective---perhaps if I first learned spinors in a mathematical context, my perspective would be different. $\endgroup$ – Jonathan Gleason Jul 13 '17 at 20:44
  • $\begingroup$ @Gleason: I've come across Clifford Algebras since I asked this question, so your comment is quite apposite; there's also Atiyahs Clifford Modules, which I hasten to add I don't know much about but look intriguing. $\endgroup$ – Mozibur Ullah Jul 16 '17 at 18:49
  • $\begingroup$ @Garrett: thanks for the warning, it's taken me a while to understand the situation that you're describing, though I'm not sure I'm there yet. $\endgroup$ – Mozibur Ullah Jul 16 '17 at 18:56
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According to the answer by Korman of this question, every irreducible representation is found inside a tensor product of fundamental representations. So there can't be any missing representations.

"The usual tensor constructions" do not include taking direct summands. They include, for example, taking duals, taking tensor products, taking direct sums, and applying Schur functors. In other words, you can apply all functors that come from functors $\text{Vect} \times ... \times \text{Vect} \to \text{Vect}$ (possibly contravariant in some variables).

Is there a 1-1 relationship between representations of double covers & projective representations of the original group?

Yes, in this case, because the double cover is the universal cover. Both are in turn identified with representations of the Lie algebra.

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  • $\begingroup$ So, the 'missing representations' are direct summands of representations formed by the usual tensor constructions? $\endgroup$ – Mozibur Ullah Jul 15 '13 at 20:13
  • $\begingroup$ @Mozibur: yes, but in order to find out what these are you have to know something about the representations (e.g. if you knew their endomorphism algebras then you could pick out the idempotents in these). To apply the usual tensor constructions you don't need to know anything other than that you are dealing with representations of some group. $\endgroup$ – Qiaochu Yuan Jul 15 '13 at 20:14
  • $\begingroup$ I think the problem related in the part on "tensorial constructions" is not that one is not allowed to take direct summands. The "fundamental representations" are the representations corresponding to the fundamental weights of the Lie algebra. For the orthogonal algebra, the spinor representations are among these fundamental representations, so there is no contradiction here. (It is true that apart from the spin representations the fundamental representations of the classical groups can be constructed tensorially out of the defining representation, but that's a different story.) $\endgroup$ – Andreas Cap Jan 2 '15 at 18:38

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