# Spinors on orbifolds

Let $R^{n}$ be a cone over sphere $S^{n-1}$ with the metric $g = dr^2 + r^{2}g[S^{n-1}]$ ($r> 0$).

Whether it is true that the cone over $S^{n-1}/Z_{2} = RP^{n-1}$ has twice less parallel spinors, than $R^{n}$: and, if $n$ is even then the parallel spinors have one chirality (left or right)?

Let us consider a pseudo-cone over pseudo-sphere $H^{n-1}$(Lobachevsky space) with the metric $g = -dt^{2} + t^{2}g[H^{n-1}]$, $t> 0$. It is the upper light cone $V_{+}^{n}$ in the Minkovsky space $R^n = R^{1, n-1}$.

Whether it is true that the pseudo-cone over $H^{n-1}/Z_2$ has twice less parallel spinors, than $V_{+}^{n}$ (or $R^{n}$): and, if $n$ is even, the parallel spinors are of one chirality?

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I'm not sure what "has twice less parallel spinors than" means. – Ryan Budney May 27 '11 at 19:03
I think it means "the dimension of the space of parallel spinors is half of". – José Figueroa-O'Farrill May 27 '11 at 22:38

I've hesitated to answer this question because it is really not very well written. The question is really about the nature of parallel spinor fields on orbifolds $\mathbb{R}^n/G$ and $\mathbb{R}^{n-1,1}/G$ where $G$ is a particular order-2 subgroup of linear transformations preserving the inner product.

The cone is an unnecessary distraction and I suspect the reason it is mentioned is that the original question had to do with the Killing spinors on the sphere and on hyperbolic space, which Bär's cone construction relates to parallel spinors on the metric cones.

Let me answer the spherical part of the question and leave the hyperbolic as an exercise to the OP.

In the case of $\mathbb{R}^n/G$, the generator of $G$ acts by $x \mapsto -x$, whence if $n$ is odd it is not in $\mathrm{SO}(n)$. Thus the quotient is not even orientable, let alone spin. So this question really only makes sense for $n$ even, say $n = 2m$.

The parallel spinors on $\mathbb{R}^{2m}/G$ are the $G$-invariant parallel spinors on $\mathbb{R}^{2m}$, whose space of parallel spinors is isomorphic (as $\mathrm{Spin}(2m)$-modules) to the direct sum $\Delta_+ \oplus \Delta_-$ of the two irreducible half-spinor representations of $\mathrm{Spin}(2m)$. This requires lifting $G < \mathrm{SO}(2m)$ to $\mathrm{Spin}(2m)$.

There are two possible lifts, giving rise to the two inequivalent spin structures of the real projective space. The generator of $G$ lifts to $\pm \omega$, where $\omega$ is the volume element of the Clifford algebra $C\ell(2m)$, which actually sits in $\mathrm{Spin}(2m)$. Now $\omega$ acts like $\pm 1$ on $\Delta_\pm$, whence for one choice of spin structure, the space of parallel spinors is $\Delta_+$ and for the other it is $\Delta_-$.

So provided that we take $n=2m$ even, the answer to the first question (properly interpreted) is "Yes".

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Many thanks for a nice and detailed answer in Euclidean case. As far as I understand in pseudo-Euclidean case there are (at least) two possibilities. – Nastya May 30 '11 at 19:02
1) We consider $R^{1, n-1}/Z_{2}$, where the generator of $Z_{2}$ acts by $x\mapsto−x$ . The quotient is orientable only for n even, say n=2m. In this case there are two inequivalent spin structures. The dimensions of spaces of chiral parallel spinors $(n_{+}, n_{-})$ are either $(2^{m-1}),0)$ or $(0,2^{m-1})$ depending of the choice of the spin structure. (The proof is analogous to that of Euclidean case.) For n=2 we get for $(n_{+}, n_{-})$ either $(1,0)$ or $(0,1)$. – Nastya May 30 '11 at 19:02
2) The pseudo-cone over $H^{n-1}/Z_{2}$ is considered. The generator of $Z_{2}$ acts by $y\mapsto −y$, where $y$ belongs to (a realization of $H^{n-1}$ by) (n-1)- dimensional unit ball $B^{n-1}$ $(|y|< 1)$. The quotient is orientable only for n odd, say $n=2m + 1$. In this case there are two inequivalent spin structures. The dimension of the space of parallel spinors is $2^{m-1}$ for any choice of the spin structure. For n =3 (m =1) this dimension is 1. Is it correct? Thank you in advance. – Nastya May 30 '11 at 19:03

In Euclidean case one should consider $n = 4\,m$ and in pseudo-Euclidean case $n = 4\,m - 2$, where $m = 1, 2, 3, \dots$ In this case the half of parallel spinors (of the same chirality) will be cut by the factorization over group $\mathbb{Z}_2$.

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