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  Do Killing spinors know global information?

+ 7 like - 0 dislike

The conformal Killing spinor equations on $R\times S^3$ in Minkowski signature are \begin{equation} \nabla_\mu \epsilon=\pm \frac{i}{2}\gamma_\mu\gamma^0\gamma^5\epsilon \end{equation} whose solution is \begin{equation} \epsilon=e^{ix^0\gamma^5/2}\epsilon_0 \end{equation} where $\epsilon_0$ is a constant spinor. Note that we identify $S^3$ with $SU(2)$ and use the left-invariant vector fields of $SU(2)$ as an orthonormal frame. For more detail, please see around Eq.2.20 in the following paper.


When the superconformal index is interpreted as a partition function on $S^1\times S^3$ in the Euclidean signature, the conformal Killing spinors are modified by the Wick rotation $\epsilon =e^{-x^0\gamma^5/2}\epsilon_0$ as in the following papers.

http://arxiv.org/pdf/1104.4482v3 http://arxiv.org/abs/1104.4470

However, $\epsilon =e^{-x^0\gamma^5/2}\epsilon_0$ is not well-defined on the temporal circle $S^1$.

This issue also happens on the Killing spinor on $AdS_3$. The metric of $AdS_3$ in the Minkowski signature can be written as \begin{equation} ds^2=-\cosh^2 \rho dt^2 + \sinh^2\rho d\phi^2+d\rho^2~. \end{equation} The Killing spinors on this coordinate takes the form \begin{equation} e^{\frac 12\rho\gamma_3}e^{-\frac {i }2 (\phi+ t)\gamma_1}\epsilon_0 \end{equation} as in http://arxiv.org/abs/hep-th/9310194 . However, when you consider the elliptic genus $Tr(-1)^Fq^{L_{0}-c/24} \overline{q}^{\overline L_{0} -\overline c/24} y^{J}$, the bulk duals which satisfy the E.O.M of the supergravity theory are the Euclidean thermal $AdS_3$ and the family of its $SL(2,Z)$ transformations such as the BTZ BH. Then, again the Killing spinors on the Euclidean thermal $AdS_3$ take the form \begin{equation} e^{\frac 12\rho\gamma_3}e^{(-\frac {i\phi }2 + \frac{t}{2})\gamma_1}\epsilon_0 \end{equation} which are also not well-defnined on the temporal circle.

The Killing spinor equations themselves are local, so does the Killing spinor know only local information? If so, how you distinguish the Killing spinors in the NS from in the R boundary condition on 2-torus?

This post has been migrated from (A51.SE)
asked Oct 20, 2011 in Theoretical Physics by Satoshi Nawata (345 points) [ no revision ]
retagged Apr 19, 2014 by dimension10

1 Answer

+ 15 like - 0 dislike

Spinor fields are sections of a spinor bundle, so you have to be careful when you work with them as if they were functions. For a spinor field, the notions of being parallel, Killing, conformal Killing,... make perfect sense globally as equations on sections of the spinor bundle, but you have to specify which bundle.

Spinor bundles are associated vector bundles to a spin bundle, which is a lift of the oriented orthonormal frame bundle. Lifts need neither exist or, if they do exist, be unique, hence there are manifolds on which you cannot define a spin bundle and manifolds on which you have more than one such bundle. The obstruction for the existence of a spin structure is orientability and the vanishing of the second Stiefel-Whitney class of the tangent bundle. If a manifold M admits a spin structure, then it may admit more than one: they are classified by $H^1(M,\mathbb{Z}_2)$ which is isomorphic to the set of group homomorphisms from the fundamental group to $\mathbb{Z}_2$. Roughly speaking this measures how you can consistently assign signs to noncontractible loops.

The circle has fundamental group $\mathbb{Z}$ and since there are two homomorphisms $\mathbb{Z} \to \mathbb{Z}_2$, there are two different spin structures, which in string theory are usually called NS and R, much to the amusement of spin geometers everywhere.

Hence the lesson is that before you can even talk about <insert your favourite spinor equation> you need to say what your spinors are; that is, which spinor bundle they are sections of.

A rough analogy (which can be made precise in this case) is that you have equations and then boundary conditions and both are necessary in order to define the problem. The analogue of the boundary conditions is specifying the spinor bundle. This is indeed the case for the circle: where the spinor field will either change by a sign or not as you move along the circle.

It is not uncommon for manifolds admitting inequivalent spin structures, that there should be parallel, Killing,... spinor fields relative to one of the spin structures, but not relative to others. In fact, this is the generic situation.

In summary, the answer to the question in the title is emphatically Yes.

Further remarks

This may answer the OP's question in the comment to an earlier version of this answer.

One has to be careful to conclude that a spinor field does not obey the right periodicity conditions. Indeed, one must remember that there is a "gauge" symmetry whenever one deals with sections of associated vector bundles to principal bundles, and that is the freedom to perform a local $G$-transformation, where $G$ is structure group of the bundle. In the case of the spinor bundles, the structure group is the relevant Spin group. Hence it could be that the discrepancy in sign is simply an artefact of the choice of frame and can be cured by a local Spin transformation. With apologies for referring to my own work, an illustrative example of this occurs at the end of §3.2.2, particularly around equation (32), in my paper with Gutowski and Sabra on 4- and 5-dimensional preons: arXiv:0705.2778 [hep-th].

I hope that this helps.

This post has been migrated from (A51.SE)
answered Oct 20, 2011 by José Figueroa-O'Farrill (2,315 points) [ no revision ]
Thank you very much for your comment. What if a solution of the Killing spinor equation does not satisfy the boundary condition you want to impose? Does it mean that the theory cannot be supersymmetric? But I think one can put the ${\cal N}=4$ SCFT on $S^1\times S^3$ in the Euclidean signature. In addition, does the existence of a solution depend on the signature of a metric?

This post has been migrated from (A51.SE)
Thank you very much for your further remarks. However, local G-transformations are allowed only in a local supersymmetric theory, namely supergravity. In the case of $S^1\times S^3$, the theory has rigid supersymmetry. So one cannot gauge away the factor in front of a constant spinor.

This post has been migrated from (A51.SE)
Satoshi-san, my statement has nothing to do with supergravity or gauge theories, it's a statement about spin geometry. When you write the (conformal) Killing spinor equation as you have done, you are trivialising the spinor bundle and thinking of spinor fields are functions with values in a spinor representation. You could choose a different trivialisation of the spinor bundle: the spinor field, as a section of the spinor bundle, does not change, but its description as a function with values in the spinor representation does. This clearly makes no reference to any underlying theory.

This post has been migrated from (A51.SE)
Perhaps the following simple example illustrates what I mean. Consider parallel spinors on flat euclidean space, say in two dimensions. If you choose to write them relative to the frame associated to the standard flat coordinates (write $ds^2 = dx^2 + dy^2$ and then the frame is $(\partial_x,\partial_y)$) parallel spinors are constant. However if you change coordinate to polar coordinates $ds^2 = dr^2 + r^2 d\theta^2$ and you choose the corresponding coordinate frame, then the parallel spinor is no longer constant, but is related to one by a local spin transformation.

This post has been migrated from (A51.SE)
(continued) The geometrical object (the parallel spinor) is the same, but its expression as a function with values in the spinor representation is different. Basically it comes down to the fact that sections are not really functions.

This post has been migrated from (A51.SE)

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