Firstly, just to clarify my notation:

Let $Cl(V,q)$ denote the Clifford Algebra of a quadratic vector space $(V,q)$ and denote by $Cl(V,q)_{0\vert 1}$ the even/odd part in the $\mathbb{Z}_2$-grading of $Cl(V,q) = Cl(V,q)_0 \oplus Cl(V,q)_1$ of the Clifford-algebra.

Now for the subgroups $Pin(V,q) \subset Cl(V,q)$ and $Spin(V,q)\subset Cl(V,q)_0$ is is defined:

A pinor representation is the restriction of an irreducible representation of $Cl(V,q)$ onto $Pin(V,q)$. Similary a spinor representation is the restriction of an irreducible representation of $Cl(V,q)_0$ onto $Spin(V,q)$.

My question is: What is the reason in defining pinor/spinor-representations as the restrictions of Clifford algebra representations, rather then just as usual group-representations of the groups themselves?

Remark: "Physical" explanations (as: 'The so defined spinor fields wouldn't behave like spinors, since...') are also very welcome.

  • $\begingroup$ Where is your definition taken from? It would be helpful to add some reference. $\endgroup$ – KonKan Oct 1 '16 at 20:34
  • $\begingroup$ It is taken from the lecture notes to this course on spin geometry. $\endgroup$ – Peter Wildemann Oct 3 '16 at 11:42
  • $\begingroup$ Relevant link. $\endgroup$ – user301988 Oct 3 '16 at 22:02

I suspect that there is a matter of terminology involved here.

When speaking about "spinors" we do not exactly mean the elements of the group $Spin(V,q)$ but rather the elements of special vector spaces upon which Clifford algebras act. In other words, the terminology spinors imply the elements of some specific (Clifford algebra, Pauli matrices etc )-module.

On the other hand, you are right in the fact that these spinor representations are not the only representations of the $Spin(V,q)$ group. In fact these ones are faithful. They are the ones obtained by restriction from the action of the Clifford algebra on the spinor space (spinor module). You can have a more detailed look here for the definitions and the elementary properties implied (look mainly at the definitions and the proposition of the first page).

  • $\begingroup$ The pointer to the faithfulness actually answers my question pretty well, thanks! $\endgroup$ – Peter Wildemann Oct 3 '16 at 11:41

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