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In quest of a generalized Callias index theorem

We give a prescription for how to compute the Callias index, using as regulator an exponential function. We find agreement with old results in all odd dimensions. We show that the problem of computing the dimension of the moduli space of self-dual strings can be formulated as an index problem in even-dimensional (loop-)space. We think that the regulator used in this Letter can be applied to this index problem.


Introduction
We do not know what six-dimensional (2, 0) theory really is. It is believed that it can sustain solitonic self-dual strings [1], although no one today knows what a (non-Abelian) self-dual string really is. But if we break the gauge group maximally to U (1) r , then we should be able to define the charges of these mysterious self-dual strings by the asymptotic behaviour of the U (1) gauge fields. One should expect these asymptotic U (1) fields to be (at least isomorphic with) a copy of the familiar abelian two-form gauge potentials (with self-dual field strengths).
It now seems to make sense to ask a question like, what is the dimension of the moduli space of self-dual strings of a given charge?
If the gauge group is SU (2) and is broken to U (1) by the Higgs vacuum expectation value (that should also determine the tension of the string), then the intuitive answer to this question is 4N where N is the U (1) charge in a suitable normalization, such that N = 1 corresponds to one self-dual string. One may argue that half the supersymmetry is broken by the string. Therefore one string should sustain 4 fermionic zero modes. Since some (half) of the supersymmety is unbroken there should also be 4 corresponding bosonic zero modes. These are naturally identified with the translational zero modes associated with the four transverse directions to the string. Furthermore, the strings being BPS, should be possible to separate at no cost of energy (thus staying in the moduli space approximation). If we take them far from each other, one may suspect that we can just add 4 bosonic zero modes from each string, to get 4N bosonic zero modes in total in a configuration of N strings [2].
It would of course be nice to have a proof of this conjecture. Could it be proven if one had some index theorem? We will not provide a full solution to this problem in this Letter. But we will make it plausible that the problem can indeed be solved by computing the index of a certain Dirac operator in loop space.
To address our index problem, we think that one can lend the methods that Callias [3] used to prove his index theorem in odd-dimensional spaces. In our case we have an even number of dimensions (namely the four transverse direction) so it is apparent that we would have to construct a new type of index. This we do in section 3.
In section 2 we recall the Callias method [3] to address index problems in open spaces, though we will modify Callias' regularization, using the more convergent exponential function to obtain the index, as the limit (here D 2 > 0 and γ =diag (1, −1)) rather than which is the regularization that Callias used. We think that using the more convergent regularization of an exponential function is interesting in itself, as it could possibly extend the Callias index theorem to a wider class of index problems. Therefore we will devote the first part of this Letter on this subject.
But let us at once say that our regulator probably has no advantages when attacking these old problems. It does not provide us with a solution for how to count the number of zero modes in a multimonopole configuration with a nonmaximally broken gauge group, where the index can not be reliable computed due to a contribution from the continuum portion of the spectrum. What we hope though, is that our regulatization can be useful when attacking our new index problem associated with the moduli space of self-dual strings.
In section 2 we obtain the index in one and three dimensions. In three dimensions we apply this on the multimonopole moduli space and re-derive the result in [4]. A recent review article on monopoles and supersymmetry is [5]. The one and three-dimensional index problems have also been studied in [6]. We then indicate how our method manages to reproduce the correct results in any odd dimensions. In section 3 we show how one at least in principle should be able to compute the dimension of the moduli space of N self-dual strings by computing a certain index.

Computing the Callias index in odd-dimensional spaces
For Dirac operators on open n − 1-dimensional space where n − 1 is odd, there is an index theorem by Callias [3]. This applies to Dirac equations of the form where the Dirac operator D is of the form Here i = 1, ..., n − 1 and γ µ ≡ (γ i , γ n ) denote the Dirac gamma matrices, We define the gauge covariant derivative as iD is = i∂ is + A is and all our fields are hermitian. If n − 1 is odd, the gamma matrices can be represented as One may use the n-dimesional notation A µ = (A i , φ), D = γ µ iD µ , but one must then remember that space is really n − 1 dimensional. If n − 1 is even there is no Weyl representation of the gamma matrices (because of the inclusion of the 'gamma-five'), and no index theorem of this form exists.
We define the 'gamma-five' for even n as which then is hermitian, and we define the projectors In odd dimensions n − 1, the Dirac operator splits into two Weyl operators Because P ± and D are all hermitian, it follows that D † is the hermitian conjugate of D. Also, because D is already of an off-block diagonal form, it suffices to include just one of the projectors, so we can just as well write this as The index can now be defined as Since ker D = ker D † D and ker D † = ker DD † we can express this as 2 where we have noted that γ = P − − P + . Callias, Weinberg and others used the regulator to obtain the index as the limit M 2 → 0. In this Letter we will be slightly more general. We define for any function f (and of course D is not dimensionless, so D has to be accompanied by M in a suitable way). Then we notice that where (manifestly) From this, we obtain the following identity In odd dimensions, the second term in the right hand side vanishes as x approaches y. This can be seen as being equivalent to the statement that there is no chiral anomaly in odd dimensions (by using point-splitting and inserting a Wilson line). So we get 2 To see this that ker D = ker D † D we apply the definition of hermitian conjugate with respect to the inner product (ψ, χ) = R dxψ † χ and the property of the norm, to 0 = (ψ, D † Dψ) = (Dψ, Dψ).
If we wish to compute the index as in Eq (13), then we can take (however there is no unique choice of J i ). We then get provided tr x γγ i 1 D y = 0 (21) We will see in the next few paragraphs how one can achieve this by using a principal value prescription. The virtue of expressing Eq (13) as a total divergence, is that we then can compute the index as a boundary integral over an (n − 2)-sphere at infinity as where r is the radius of the sphere and dΩ n−2 denotes the volume element of the unit sphere. If instead we wish to compute the index as the limit of as s → ∞, then we get It might seem confusing that we can have a plus sign here, when we have a minus sign in Eq (20). These peculiar signs seem to be correct though. Why we can have opposite signs should be a reflection of the fact that these expressions can not be continuously connected with each other, at least not in any obvious way (like taking M to zero and s to zero. In fact s should be taken to plus infinity as M goes to zero). We will now illustrate how one can use this J i to compute the index in odd dimensions.

One dimension
We choose our gamma matrices as and we have The Dirac operator reads We need the square of the Dirac operator, We make the choice We assume that φ(x) converges towards some constant values at x = −∞ and x = +∞. That means that we may ignore ∂φ(x) for sufficiently large |x|, where we then get The index is now given by if φ flips the sign an odd number of times when going from −∞ to +∞, and 0 otherwise. If instead we choose then we get If we compute the integral over k in the most natural way, then we get a result that vanishes in the limit s → ∞. Could there be another way of defining this integral, such that we do not get zero as the result? We notice that the integral for s > 0 is convergent only if we integrate k along a line in the complex plane which is such that it asymptotically is such that − π 2 < θ < π 2 where k = |k|e iθ . Integrating along any such line in the complex plane, we get the same value of this integral. If on the other hand we integrate over a line that asymptotically lies outside this cone, then we get a divergent integral for s > 0. But we get a convergent integral for s < 0. We then define the value of the integral for s > 0 as the analytic continuation of the same integral for s < 0. It remains to compute this convergent integral. Replacing k by ik and s by −s, we get the integral We can compute its derivative The right-hand side can obviously be analytically continued to −s, and that is how we will define A(s) where the integral representation does not converge. We can then integrate up A ′ (s), and we then need to compute We define this as the principal value. This is ad hoc -we have no argument why one should define it like this. But if we accept this, then we get A(0) = 0. We conclude that we could just as well define the integral that we had, as But this requires us to perform the integration of k in the cone where it diverges for s > 0, and then define this integral by analytic continuation. This seem to be rather ad hoc. We have three rather week arguments why one should Wick rotate. First, if we keep x − y as a small number, then we get the factor e ik(x−y) and this can act as a convergence factor only if we Wick rotate. (We illustrate this in the Appendix where we compute the corresponding integral in any complex number of dimensions.) Second, it seems to be the only way that we could produce a non-trivial answer. Third, with this prescription we will manage to reproduce the right answer in any odd number of dimensions, where we can check our result against the safer regularization used by Callias.
If we compute the integral by this prescription, then we get and we see that we indeed get the right answer.

Three dimensions and magnetic monopoles
The physics problem that we will consider in three dimensions, is to compute number of zero modes of the Bogomolnyi equation We choose the convention that our fields are hermitian. It is convenient to group the fields into 'gauge potential' be the associated 'field strength'. Then the Bogomolnyi equation reads Linearizing this, we get Contracting with γ µν , we get and if we impose the background gauge condition which is to say that zero modes are orthogonal to gauge variations with respect to the moduli space metric, then we can write this linearized equation as a Dirac equation We compute Inserting the Bogomolnyi configuration we can write this, thus using the fact that G µν is selfdual, and get a vanishing theorem. Namely, dim ker DD † = 0 as DD † > 0 is strictly postive. Hence we can compute the dimension of the moduli space dim ker D ≡ dim ker D † D just by computing the index of D. To compute the index, we now wish to compute We assume that asymptotically φ approaches a constant value at infinity. This corresponds to a gauge choice where we have a Dirac string singularity. Some further examination reveals that we get a non-negligible contribution to J i , for a sufficiently large two-sphere, only from the term We thus need to perform an integral of the form If we choose the same prescription as we did in one dimension, then we get the result For details of such a computation we refer to appendix A.
If we apply this result to the integral that we had, we get We expand the square root, In the far distance, in a charge Q monopole configuration, we find that and so when we trace over the gamma matrices, we get If we now for instance assume SU (2) gauge group, broken to U (1), then if we integrate i 2 J i over S 2 , we get the index 2Q. The number of bosonic zero modes is twice the index, i.e. −4Q in our conventions [4,5].

(2m + 1) dimensions
In 2m + 1 dimensions we get the integral if we use our regulator. Here (and G is an abbreviation for 1 2 iγ µν G µν .) This should be compared to the integral that we get using the Callias regulator. 3 In order to compare these integrals, we rewrite them as We compute a according the prescription introduced above in one and three dimensions, that is by Wick rotating ξ and continue analytically in s. (Details are in appendix A.) We can compute b using residue calculus (introducing a regulator so that we can close the contour on a semi-circle at infinity). The result is We next expand and we find that the coefficient of G m becomes equal to if one uses our regularization, and equal to 3 This integral comes from expanding in powers of G as a geometric series [4].
if one uses the Callias regularization. We see that the two expressions coincide for all m.
We have now showed that if we use our prescription of Wick rotating k to compute the integrals over the exponential, then we get the right answer for all cases that can be safely computed using a regulator that is less convergent. We are inclined to think that our prescription for how to compute the integral, will also work for index problems where the Callias regulator diverges. But we have no proof. It is perhaps not so obvious that more general index problems can be formulated. In the next section we will give one example of a more general type of index problem.

Four dimensions and self-dual strings
To introduce the notation, we first consider the free Abelian tensor multiplet theory in 1 + 5 dimensions. The on-shell field content is a two-form gauge potential B µν , five scalar fields φ A and corresponding Weyl fermions ψ. The field strength H µνρ = ∂ µ B νρ + ∂ ρ B µν + ∂ ν B ρµ is selfdual. The supersymmetry variation of the Weyl fermions is where we use eleven-dimensional gamma matrices splitted into SO(1, 5)×SO(5), so that in particular In a static and x 5 independent field configuration, in which only φ 5 =: φ is non-zero, we find the SUSY variation If we assume that the classical bosonic field configuration is such that then the SUSY variation reduces to and we find the condition for unbroken SUSY as If we use the Weyl condition of the (2, 0) supersymmetry parameter ǫ, then we can also write this as We may represent the gamma matrices as where σ 1,2,3 are the Pauli sigma matrices, γ = γ 1234 . Then the condition for unbroken SUSY is where σ = σ 1234 = σ A=5 . We have found that if then half SUSY is unbroken. This equation is the Bogomolnyi equation for selfdual strings [1]. We are interested in finding the number of parameters needed to describe solutions of this equation. We can linearize it and get the equation for the bosonic zero modes, that we have gathered into a matrix For this to work we must also assume the background gauge condition Now this linearized equation Eq (80) does not make any reference to the gauge field. So there is no way that we could count the number of parameters of a multi-string configuration just using this equation. This should of course not be a surprise. The strings that we have in the Abelian theory are not solutions of the field equations. They have to be inserted by hand, that is we need to insert delta function sources by hand, in the same spirit as for Dirac monopoles.
To be able to count the number of zero modes, we must consider some interacting theory which (at the classical level) has solitonic string solutions.
To pass to non-Abelian theory we begin by rewriting the Abelian theory in loop space. Loop space consists of parametrized loops C: s → C µ (s). We introduce the Abelian 'loop fields' [7] A µs = B µν (C(s))Ċ ν (s) With these definitions, a short computation reveals that A µs transforms as a vector and φ µs a contra-variant vector under diffeomorphisms in loop space induced by diffeomorphisms in space-time. One may then extend these transformation properties to any diffeomorphism in loop space. Space-time diffeomorphism and reparametrizations of the loops then get unified and are both diffemorphisms in loop space. The only thing to remember is what is kept fixed under the variation. If it is the parameter of the loop, or the loop itself.
The field strength becomes In terms of these fields, the Bogomolnyi equation will read 4 We pass to the non-Abelian theory by letting these loop fields become non-Abelian, in the sense that A µs = A a µs λ a (s) where λ a (s) are generators of a loop algebra associated to the gauge group [7]. We introduce a covariant derivative Local gauge transformations act as Given a loop C, we automatically get a tangent vectorĊ µ (s) that makes no reference to space-time. We can therefore impose the loop space constraintṡ for each s, and also for some subtle field φ(s; C) on loop space. As a consequence, we find that which is the condition of reparametrization invariance. With the assumption made that λ a (s) are generators of a loop algebra, we find that the constraint can also be written as A local gauge variation of this constraint is The last term vanishes by the constraint. The first term gives us the constraint Eq (92) that we must impose on the gauge parameter Λ = dsΛ a (s, C)λ a (s).
We have now introduced non-local non-Abelian fields with infinitely many components. It is also likely that consisteny of the theory requires an infinite set of constraints on these fields. Maybe then, it could be that we may in the end descend to a finite degrees of freedom. But this is just a speculation. The problem appears to be difficult and ill-defined -How should one define a degree of freedom in a strongly coupled non-local theory?
The non-Abelian generalization of the Bogomolnyi equation should be given by [7] F is,jt = ±ǫ ijkl D k(s φ lt) .
This equation is gauge invariant and invariant under the residual SO(4) Lorentz group that is preserved by the strings. We can not think of any reasonable modification of this equation that would preserve these symmetries, so on this grounds alone one could suspect this equation to be correct. Of course this is not the only requirement that the BPS condition imposes. We also get conditions on the 0s and the 5s components. But these BPS equations will be of no interest to us right now. We will show below that the linearized Bogomolnyi equation can be written as We will also see below that we (presumably) can actually drop the symmetrization in s and t in this equation. The fields transform in the adjoint representation of the loop algebra, by which we mean that φ is χ t = [φ is , χ t ]. We define the Dirac operator and the projectors We can now formulate an index problem, in an even-dimensional (loop-)space. The even-dimensional space in this case is given by the 4-dimensional transverse space to the strings, and the index is given by where Since D s and P ± are hermitian, it is manifest that D † s defined this way will be the hermitian conjugate of D s , thus justifying the notation.
Computing the index alone is not sufficient in order to obtain the dimension of the moduli space of self-dual strings. We also need a vanishing theorem that says that dim ker D † s = 0. Linearizing the Bogomolnyi equation, we get Contracting by γ ij , we get To see that the linearized BPS equation can be written like this, one must use the constraint We can avoid having explicit ± signs by introducing the other chiraly matrix at our disposal, namely σ that lives in a different vector space than γ. We can then hide the ± signs in the tensor product then we can write the zero mode equation as Let us analyze the second term in this equation. It is given by We should not count variations that are gauge variations as bosonic zero modes. We can insure this by demanding the zero modes to be orthogonal to gauge variations, with respect to the metric on the moduli space, This leads to the background gauge condition 5 To really understand what is going on, one should apply (1 ± γσ) on everything, on ψs and on Ds. Then one notices that That is, we can trade ∓γ for σ, once we apply (1 ± γσ) on everything. This is what we really should do, but to keep the notation simple, we do not spell this out.
This condition implies that the gauge variation of the zero modes vanishes, To see this, we make a gauge variation δ Λ δA is = D is Λ, δ Λ φ is = φ is Λ, and ask which gauge parameters Λ will respect the background gauge condition. Inserting this gauge variation into the background gauge condition, we get For this to work nicely, it seems that we must constrain the non-locality of our loop field such that ∂ i (s ∂ it) < 0. Then the only solution to this equation is Λ = 0. In other words all gauge variations of the zero modes have to vanish.
Furthermore we want the variation to preserve the orthogonality between A is and φ is , If we make a gauge variation of this, then we get the condition We conclude that the zero mode equation can be written as We are interested in counting the number of such modes in a background of k BPS strings. We compute (Here D 2 ≡ D s D s ≡ ds 2π D s D s , and analogously for the other fields or operators.) In a BPS configuration, we get is Furthermore, in the subspace where 1 + γσ = 0, we find that is a strictly negative operator, hence has no zero modes. This means that we have a vanishing theorem, dim ker D † = 0.

A small comment
The zero mode equation was really where we should symmetrize in s and t. That means that we should rather consider If now D [s D t] = 0 and D s χ s = 0, then we get The latter condition, D s χ s = 0 is of course a consequence of D (s χ t) = 0 with s = t. The former condition reads which we would like to impose as a constraint. Restricting to the abelian case this is condition is of course true as 0 ≡ ∂ i[s ∂ |i|t] . If we can impose this as a constraint on the non-abelian fields, then we have now seen that the zero mode equation Eq (123) implies that because D s is anti-self-adjoint with respect to the inner product on loop space. We can also go in the opposite direction. Assuming that Eq (127) holds, we get and we conclude that (123) implies with no symmetrization in s, t.

How to compute the index
We should now be able to compute an index associated to self-dual strings, as the limit when s → ∞. We define the quantity (it should be clear that the two s's involved in this formula are totally unrelated) and find that We can separate the functional integral over parametrized loops C into several pieces. We can keep a point on the loops C(s) = x fixed, and separate it as Then we can write I(s) as an integral over a large three-sphere at spatial infinity, where thus x = C(s).
If we assume that the gauge group is maximally broken to a product of U (1)'s by the Higgs vacuum expectation values, then we should have U (1) loop fields at spatial infinity.
If we assume that the gauge group is SU (2) and that it is broken to U (1), then we need only the asymptotic form of the U (1) fields at spatial infinity, F is,jt = H ijl (x)Ċ l (s)δ(s − t) φ ks = vĊ k (s) Without doing any computations, we can guess what the outcome of the index calculation should be. A term like ǫ ijkl D x Ctr (F is,jt (C)F ks,jt (C)) (137) could certainly arise somewhere (in odd dimensions a corresponding term vanished since there is no chiral anomaly in odd dimensions). In our case this term vanishes identically by the Bogomolnyi equation and the constraint 6 F is,jt D is φ jt = 0.
that should arise in a very similar way as the corresponding term arose for monopoles. If we insert the asymptotic U (1) fields, this term becomes proportional to That means that the index should be given by some numerical constant, times the magnetic charge A Integrals over the exponential The integral we will analyze here is for any complex number ζ. (The ǫ > 0, say, will be taken towards zero. It arose from ǫ = x − y and we keep it here just as a convergence factor.) We first compute In order to make this integral converge for any ζ, we should Wick rotate k to ik, and henceforth we will always mean by i the branch e iπ/2 , and by −1 we mean e iπ . Then we get and this integral we evaluate as a principal value. That means to evaluate the residues along the real axis and multiply them not by 2πi, but by half of it, that is, by πi. We get Next we turn to our integral a(s). It is easier to first compute the derivative. We should still work with the Wick rotated integral. Making the substitution ξ = k 2 we can put it on the form of two gamma functions. The result is that a ′ (−s) = −e iπ(ζ+ 1 2 ) 1 2 1 + (−1) 2ζ Γ ζ + 1 2 which we can trivially continue analytically to +s, and then integrate up. The result is a(+∞) = −π 1 + (−1) 2ζ 2 1 cos(πζ) + π(−1) ζ 1 − (−1) 2ζ 2 . (148)