arXiv:0704.0018v2  [hep-th]  21 Apr 2007
In quest of a generalized Callias index theorem
Andreas Gustavsson1
Förstamajgatan 24,
S-415 10 Göteborg, Sweden
Abstract
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.
1a.r.gustavsson@swipnet.se
http://arxiv.org/abs/0704.0018v2
1 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
γe−sD
, (1)
(here D2 > 0 and γ =diag (1,−1)) rather than
D2 +M2
, (2)
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 non-
maximally 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.
2 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
Dψ = 0 (3)
where the Dirac operator D is of the form
D = γiiDi + γnφ. (4)
Here i = 1, ..., n− 1 and γµ ≡ (γi, γn) denote the Dirac gamma matrices,
{γµ, γν} = 2δµν . (5)
We define the gauge covariant derivative as iDis = i∂is +Ais 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 Aµ = (Ai, φ), D = γµiDµ, 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
γ ≡ −i−
2 γ1···n (7)
which then is hermitian, and we define the projectors
(1∓ γ) . (8)
In odd dimensions n− 1, the Dirac operator splits into two Weyl operators
D ≡ P+DP−
D† ≡ P−DP+ (9)
Because P± andD are all hermitian, it follows thatD† 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
D = P+D = DP−
D† = P−D = DP+ (10)
The index can now be defined as
dimkerD − dimkerD† (11)
Since kerD = ker
and kerD† = ker
we can express this as2
dimker
− dimker
= dimker
. (12)
where we have noted that γ = P− − P+.
Callias, Weinberg and others used the regulator
I(M2) = Tr
D2 +M2
to obtain the index as the limit M2 → 0. In this Letter we will be slightly more
general. We define
Ji(x, y) ≡ tr 〈x |γγif(D)| y〉 , (14)
for any function f (and of course D is not dimensionless, so D has to be accom-
panied by M in a suitable way). Then we notice that
W (x, y) ≡ (iγi∂xi + γµAµ(x) +M) 〈x |f (D)| y〉
= 〈x |f (D)| y〉
−iγi∂yi + γµAµ(y) +M
where (manifestly)
W (x, y) = 〈x |(D +M)f (D)| y〉 . (16)
From this, we obtain the following identity
∂xi + ∂yi
Ji(x, y) = 2tr 〈x |γDf(D)| y〉
+tr (Aµ(y)−Aµ(x)) 〈x |γγµf(D)| y〉 (17)
In odd dimensions, the second term in the right hand side vanishes as x ap-
proaches 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
i∂iJi(x, x) = 2tr 〈x |γDf(D)|x〉 (18)
2To see this that kerD = kerD†D we apply the definition of hermitian conjugate
with respect to the inner product (ψ, χ) =
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
f(D) =
D2 +M2
(however there is no unique choice of Ji). We then get
Ji(x, y) = tr
D2 +M2
−D2 +D2 +M2
D2 +M2
= −tr
D2 +M2
. (20)
provided
= 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
I(M2) =
dΩn−2r
n−2x̂iJi(x, x). (22)
where r is the radius of the sphere and dΩn−2 denotes the volume element of
the unit sphere.
If instead we wish to compute the index as the limit of
I(s) = Tr
γe−sD
. (23)
as s→ ∞, then we get
Ji(x, y) = tr
. (24)
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 Ji to
compute the index in odd dimensions.
One dimension
We choose our gamma matrices as
, γ2 =
and we have
γ = iγ1γ2 =
. (26)
The Dirac operator reads
D = iγ1∂ + γ2φ (27)
We need the square of the Dirac operator,
D2 = −∂2 + φ2 + γ∂φ. (28)
We make the choice
J1(x, y) = −tr
D2 +M2
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
J1(x, x) = −tr (γγ1γ2)
k2 + φ2 +M2
φ2 +M2
The index is now given by
(J1(+∞)− J1(−∞)) = ±1 (31)
if φ flips the sign an odd number of times when going from −∞ to +∞, and 0
otherwise.
If instead we choose
J(x, y) = tr
then we get
J(x, x) = tr (γγ1γ2)
k2 + φ2
e−s(k
2+φ2) (33)
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
A(s) ≡
e−s(k
k2 + 1
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
< θ < π
where k = |k|eiθ.
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
A(−s) = −i
e−s(k
k2 − 1 (35)
We can compute its derivative
′(−s) = −i
−s(k2−1) = −i
s (36)
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),
A(∞) = A(0)−
e−s = A(0)−
= A(0)− π (37)
and we then need to compute
A(0) = i
k2 − 1
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
e−s(k
k2 + 1
= −π. (39)
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
eik(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
J(x, x) = tr (γγ1γ2) lim
k2 + φ2
e−s(k
2+φ2) = i
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
Fij = ǫijkDkφ (41)
We choose the convention that our fields are hermitian. It is convenient to group
the fields into ‘gauge potential’
Aµ = (Ai, φ) (42)
We define Dµ = (Di, φ) such that iDµ = i∂µ + Aµ and we let Gµν = i[Dµ, Dν ]
be the associated ‘field strength’. Then the Bogomolnyi equation reads
Gµν =
ǫµνρσGρσ . (43)
Linearizing this, we get
DµδAν =
ǫµνρσDρδAσ (44)
Contracting with γµν , we get
(1 + γ)γµνDµδAν = 0 (45)
and if we impose the background gauge condition
DµδAµ = 0 (46)
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
Dψ ≡ γµDµψ = 0 (47)
where
ψ := (1 + γ)γµδAµ. (48)
We compute
D2 = −D2i + φ2 +
iγµνGµν (49)
Inserting the Bogomolnyi configuration we can write this, thus using the fact
that Gµν is selfdual,
2 = −D2i + φ2 +
(1 + γ)iγµνGµν . (50)
and get a vanishing theorem. Namely, dimkerDD† = 0 as DD† > 0 is strictly
postive. Hence we can compute the dimension of the moduli space dim kerD ≡
dimkerD†D just by computing the index of D. To compute the index, we now
wish to compute
Ji(x, x) = tr
γγiγkDk
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 Ji, for
a sufficiently large two-sphere, only from the term
Ji(x, x) = tr
γγiγ4φ
(2π)3
k2 + φ2 + 1
iγµνGµν
−s(k2+φ2+ 1
iγµνGµν)
We thus need to perform an integral of the form
A(s) =
k2 + 1
e−s(k
2+1) (53)
If we choose the same prescription as we did in one dimension, then we get the
result
A(+∞) = π. (54)
For details of such a computation we refer to appendix A.
If we apply this result to the integral that we had, we get
Ji(x, x) =
γγiγ4φ
iγµνGµν
We expand the square root,
iγµνGµν = φ+
iγµνGµν + ... (56)
In the far distance, in a charge Q monopole configuration, we find that
γµνGµν = 2γkγ4(1− γ)
Q (57)
and so when we trace over the gamma matrices, we get
Ji(x, x) =
. (58)
If we now for instance assume SU(2) gauge group, broken to U(1), then if we
integrate i
Ji 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
A(µ) ≡ lim
k2 + µ2
e−s(k
2+µ2) (59)
if we use our regulator. Here
µ2 ≡ v2 +G (60)
(and G is an abbreviation for 1
iγµνGµν .) This should be compared to the
integral
B(µ) ≡ − lim
(−1)m
(k2 + v2 +M2)
Gm (61)
that we get using the Callias regulator. 3 In order to compare these integrals,
we rewrite them as
A(µ) = µ2m−1a
B(µ) = v−1bGm (63)
where
a = lim
ξ2 + 1
−s̃(ξ2+1)
b = − lim
(−1)m
ξ2 + 1 + M̃2
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
a = −(−1)mπ
b = (−1)m 1
) (65)
We next expand
vA(µ) = v
v2 +G
)m− 1
= v2ma+ ...+
m + ...
vB(µ) = bGm (66)
and we find that the coefficient of Gm becomes equal to
−(−1)m
) π (67)
if one uses our regularization, and equal to
(−1)m 1
) π (68)
3This integral comes from expanding
k2 + v2 +G+M2
k2 + v2 +M2
+ ... (62)
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.
3 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 Bµν , five scalar fields φ
A and corresponding Weyl fermions ψ. The
field strength Hµνρ = ∂µBνρ + ∂ρBµν + ∂νBρµ is selfdual. The supersymmetry
variation of the Weyl fermions is
ΓµνρHµνρ + Γ
µΓA∂µφ
ǫ (69)
where we use eleven-dimensional gamma matrices splitted into SO(1, 5)×SO(5),
so that in particular
{Γµ,ΓA} = 0. (70)
In a static and x5 independent field configuration, in which only φ5 =: φ is
non-zero, we find the SUSY variation
Γ0i5H0i5 + Γ
iΓA=5∂iφ
ǫ (71)
If we assume that the classical bosonic field configuration is such that
∂iφ = H0i5 (72)
then the SUSY variation reduces to
δψ = ∂iφΓ
Γ05 + ΓA=5
ǫ (73)
and we find the condition for unbroken SUSY as
1 + Γ05ΓA=5
ǫ = 0 (74)
If we use the Weyl condition
Γǫ = −ǫ (75)
of the (2, 0) supersymmetry parameter ǫ, then we can also write this as
1 + Γ1234ΓA=5
ǫ = 0. (76)
We may represent the gamma matrices as
Γµ = (Γ0,Γi,Γ5) =
1⊗ iσ2 ⊗ 1, γi ⊗ σ1 ⊗ 1, γ ⊗ σ1 ⊗ 1
ΓA = 1⊗ iσ2 ⊗ σA (77)
where σ1,2,3 are the Pauli sigma matrices, γ = γ1234. Then the condition for
unbroken SUSY is
(1 + γ ⊗ σ) ǫ = 0 (78)
where σ = σ1234 = σA=5.
We have found that if
Hijk = ǫijkl∂lφ (79)
then half SUSY is unbroken. This equation is the Bogomolnyi equation for self-
dual 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
γi∂iχ = 0 (80)
for the bosonic zero modes, that we have gathered into a matrix
χ ≡ γijδBij + γδφ. (81)
For this to work we must also assume the background gauge condition
∂iBij = 0. (82)
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 7→ Cµ(s). We
introduce the Abelian ‘loop fields’ [7]
Aµs = Bµν(C(s))Ċ
ν (s)
φµs = φ(C(s))Ċµ(s)
µs = ψ(C(s))Ċµ(s) (83)
With these definitions, a short computation reveals that Aµs transforms as a vec-
tor and φµs 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
Fµs,νt = Hµνρ(C(s))Ċ
ρ(s)δ(s− t) (84)
In terms of these fields, the Bogomolnyi equation will read4
Fis,jt = ǫijkl∂k(sφlt). (85)
We pass to the non-Abelian theory by letting these loop fields become non-
Abelian, in the sense that Aµs = A
a(s) where λa(s) are generators of a loop
algebra associated to the gauge group [7]. We introduce a covariant derivative
Dµs = ∂µs +Aµs. (86)
Local gauge transformations act as
δΛAµs = DµsΛ
µs = [φµs,Λ]. (87)
Given a loop C, we automatically get a tangent vector Ċµ(s) that makes no
reference to space-time. We can therefore impose the loop space constraints
Ċµ(s)Aµs = 0 (88)
for each s, and also
φµs = Ċµ(s)φ(s;C) (89)
for some subtle field φ(s;C) on loop space. As a consequence, we find that
µs = 0. (90)
These constraints are covariant under diffeomorphisms of space-time and reparametriza-
tions of loops. They are invariant also under local gauge transformations, pro-
vided that the gauge parameter is subject to the condition
µ(s)∂µsΛ = 0 (91)
which is the condition of reparametrization invariance. With the assumption
made that λa(s) are generators of a loop algebra, we find that the constraint
can also be written as
[Aµs, φ
µt] = 0 (92)
A local gauge variation of this constraint is
[DµsΛ, φ
µt] + [Aµs, [φ
µt,Λ]] = [∂µsΛ, φ
µt] + [[Aµs,Λ], φ
µt] + [Aµs, [φ
µt,Λ]]
= [∂µsΛ, φ
µt] + [Λ, [φµt, Aµs]] (93)
The last term vanishes by the constraint. The first term gives us the constraint
Eq (92) that we must impose on the gauge parameter
dsΛa(s, C)λa(s). (94)
4We denote by ∂is the usual functional derivative with respect to C
µ(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]
Fis,jt = ±ǫijklDk(sφlt). (95)
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
Di(s + σφi(s
χt) = 0 (96)
We will also see below that we (presumably) can actually drop the symmetriza-
tion in s and t in this equation. The fields transform in the adjoint represen-
tation of the loop algebra, by which we mean that φisχt = [φis, χt]. We define
the Dirac operator
Ds = γi (Dis + σφis) (97)
and the projectors
(1∓ γσ) , (98)
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
dimkerDs − dimkerD†s (99)
where
Ds = P+Ds = DsP−
D†s = P−Ds = DsP+. (100)
Since Ds and P± are hermitian, it is manifest that D†s defined this way will be
the hermitian conjugate of Ds, 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 dimkerD†s = 0.
Linearizing the Bogomolnyi equation, we get
2D[isδAjt] = ±ǫijkl (Dksδφlt + φksδAlt) (101)
Contracting by γij , we get
γijD̃isχjt = 0 (102)
where we have defined
D̃is ≡ Dis ∓ γφis
χis ≡ δAis ∓ γδφis (103)
To see that the linearized BPS equation can be written like this, one must use
the constraint
γijφisδφjt = 0. (104)
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
γ ⊗ σ = ±1 (105)
which amounts to
D̃is ≡ Dis + σφis
χis ≡ δAis + σδφis (106)
without any ±.5 If we define
χs ≡ γiχis (108)
then we can write the zero mode equation as
γiD̃isχt + D̃
sχit = 0. (109)
Let us analyze the second term in this equation. It is given by
DisδAit + φ
sδφit
φisδAit +D
sδφit
(110)
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,
(δΛAis, δAit) + (δΛφis, δφjt) = 0 (111)
This leads to the background gauge condition
sδAit + φ
sδφit = 0. (112)
5To really understand what is going on, one should apply (1± γσ) on everything, on ψs
and on Ds. Then one notices that
∓γ (1∓ γσ) = σ (1∓ γσ) . (107)
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,
δΛδAis = 0 = δΛδφis (113)
To see this, we make a gauge variation δΛδAis = DisΛ, δΛφis = φisΛ, and
ask which gauge parameters Λ will respect the background gauge condition.
Inserting this gauge variation into the background gauge condition, we get
sDit + φ
Λ = 0. (114)
For this to work nicely, it seems that we must constrain the non-locality of our
loop field such that ∂i
∂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
Ais and φis,
(Ais, δφit) + (δAis, φit) = 0 (115)
If we make a gauge variation of this, then we get the condition
(δΛAis, δφit) + (δAis, δΛφit) = 0 (116)
which amounts to
φisδAit +D
sδφit = 0. (117)
We conclude that the zero mode equation can be written as
Dsχt = 0 (118)
where
Ds = γi (Dis + σφis) (119)
We are interested in counting the number of such modes in a background of k
BPS strings. We compute
D2 = (Dis)
2 + (φis)
γij (Fis,js + γσǫijklDksφls) (120)
(Here D2 ≡ DsDs ≡
DsDs, and analogously for the other fields or opera-
tors.) In a BPS configuration, we get is
2 = (Dis)
2 + (φis)
ij (1 + γσ)Fis,js (121)
Furthermore, in the subspace where 1 + γσ = 0, we find that
D2 = (Dis)
2 + (φis)
2 (122)
is a strictly negative operator, hence has no zero modes. This means that we
have a vanishing theorem, dimkerD† = 0.
A small comment
The zero mode equation was really
D(sχt) = 0 (123)
where we should symmetrize in s and t. That means that we should rather
consider
DsD(sχt) =
(DsDsχt +DsDtχs)
(DsDsχt +DtDsχs + [Ds, Dt]χs) . (124)
If now D[sDt] = 0 and Dsχs = 0, then we get
DsDsχt = 0 (125)
The latter condition, Dsχs = 0 is of course a consequence of D(sχt) = 0 with
s = t. The former condition reads
0 = D[sDt]
= Di[sDit] + φi[sφit] + σDi[sφit] (126)
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
dsD†sDsχt = 0 (127)
because Ds is anti-self-adjoint with respect to the inner product
(ψs, χt) =
ψ†s(C)χt(C)
(128)
on loop space. We can also go in the opposite direction. Assuming that Eq
(127) holds, we get
χt, D
sDsχt
= (Dsχt, Dsχt) (129)
and we conclude that (123) implies
Dsχt = 0 (130)
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
I(s) = Tr
(131)
when s→ ∞. We define the quantity
Jis(C,C
′) = tr
γσγiγk (Dks + σφks)
(132)
(it should be clear that the two s’s involved in this formula are totally unrelated)
and find that
I(s) =
DC∂isJ is(C,C) (133)
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
DxC (134)
Then we can write I(s) as an integral over a large three-sphere at spatial infinity,
∂Jis(C)
∂Ci(s)
dΩ3x̂
DxCJis(C,C) (135)
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,
Fis,jt = Hijl(x)Ċ
l(s)δ(s− t)
φks = vĊk(s) (136)
Without doing any computations, we can guess what the outcome of the
index calculation should be. A term like
ǫijkl
DxCtr (Fis,jt(C)Fks,jt(C)) (137)
could certainly arise somewhere (in odd dimensions a corresponding term van-
ished since there is no chiral anomaly in odd dimensions). In our case this term
vanishes identically by the Bogomolnyi equation and the constraint6
Fis,jtDisφjt = 0. (139)
Then there can be a term
ǫijkl
DxCtr
Fis,jtφks
(140)
6For U(1) fields this would read
Fis,jt∂isφjt ∼ Hijk(C(s))∂iφ(C(s))Ċ
k(s)Ċj(s)δ(s − t)2 ≡ 0. (138)
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 propor-
tional to
ǫijklHijk(x) (141)
That means that the index should be given by some numerical constant, times
the magnetic charge
H. (142)
A Integrals over the exponential
The integral we will analyze here is
a(s) =
k2 + 1
−s(k2+1)eiǫk (143)
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
a(0) =
k2 + 1
eiǫk (144)
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 eiπ/2, and by −1 we
mean eiπ . Then we get
a(0) = −i2ζ+1
k2 − 1
e−ǫk (145)
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
a(0) = (−1)ζπ 1− (−1)
. (146)
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 = k2 we can put it on the form of two gamma functions. The result is that
′(−s) = −eiπ(ζ+
1 + (−1)2ζ
−ζ− 1
s (147)
which we can trivially continue analytically to +s, and then integrate up. The
result is
a(+∞) = −π 1 + (−1)
cos(πζ)
+ π(−1)ζ 1− (−1)
. (148)
References
[1] P. S. Howe, N. D. Lambert and P. C. West, “The self-dual string soliton,”
Nucl. Phys. B 515, 203 (1998) [arXiv:hep-th/9709014].
[2] D. S. Berman and J. A. Harvey, “The self-dual string and anomalies in
the M5-brane,” JHEP 0411, 015 (2004) [arXiv:hep-th/0408198].
[3] C. Callias, “Index Theorems On Open Spaces,” Commun. Math. Phys.
62, 213 (1978).
[4] E. J. Weinberg, “Parameter Counting For Multi - Monopole Solutions,”
Phys. Rev. D 20, 936 (1979).
[5] E. J. Weinberg and P. Yi, “Magnetic monopole dynamics, supersymmetry,
and duality,” Phys. Rept. 43, 65 (2007) [arXiv:hep-th/0609055].
[6] M. Hirayama, “Supersymmetric Quantum Mechanics And Index Theo-
rem,” Prog. Theor. Phys. 70, 1444 (1983).
[7] A. Gustavsson, “A reparametrization invariant surface ordering,” JHEP
0511, 035 (2005) [arXiv:hep-th/0508243].
A. Gustavsson, “The non-Abelian tensor multiplet in loop space,” JHEP
0601, 165 (2006) [arXiv:hep-th/0512341].

  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
γe−sD
, (1)
(here D2 > 0 and γ =diag (1,−1)) rather than
D2 +M2
, (2)
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 non-
maximally 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.
2 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
Dψ = 0 (3)
where the Dirac operator D is of the form
D = γiiDi + γnφ. (4)
Here i = 1, ..., n− 1 and γµ ≡ (γi, γn) denote the Dirac gamma matrices,
{γµ, γν} = 2δµν . (5)
We define the gauge covariant derivative as iDis = i∂is +Ais 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 Aµ = (Ai, φ), D = γµiDµ, 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
γ ≡ −i−
2 γ1···n (7)
which then is hermitian, and we define the projectors
(1∓ γ) . (8)
In odd dimensions n− 1, the Dirac operator splits into two Weyl operators
D ≡ P+DP−
D† ≡ P−DP+ (9)
Because P± andD are all hermitian, it follows thatD† 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
D = P+D = DP−
D† = P−D = DP+ (10)
The index can now be defined as
dimkerD − dimkerD† (11)
Since kerD = ker
and kerD† = ker
we can express this as2
dimker
− dimker
= dimker
. (12)
where we have noted that γ = P− − P+.
Callias, Weinberg and others used the regulator
I(M2) = Tr
D2 +M2
to obtain the index as the limit M2 → 0. In this Letter we will be slightly more
general. We define
Ji(x, y) ≡ tr 〈x |γγif(D)| y〉 , (14)
for any function f (and of course D is not dimensionless, so D has to be accom-
panied by M in a suitable way). Then we notice that
W (x, y) ≡ (iγi∂xi + γµAµ(x) +M) 〈x |f (D)| y〉
= 〈x |f (D)| y〉
−iγi∂yi + γµAµ(y) +M
where (manifestly)
W (x, y) = 〈x |(D +M)f (D)| y〉 . (16)
From this, we obtain the following identity
∂xi + ∂yi
Ji(x, y) = 2tr 〈x |γDf(D)| y〉
+tr (Aµ(y)−Aµ(x)) 〈x |γγµf(D)| y〉 (17)
In odd dimensions, the second term in the right hand side vanishes as x ap-
proaches 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
i∂iJi(x, x) = 2tr 〈x |γDf(D)|x〉 (18)
2To see this that kerD = kerD†D we apply the definition of hermitian conjugate
with respect to the inner product (ψ, χ) =
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
f(D) =
D2 +M2
(however there is no unique choice of Ji). We then get
Ji(x, y) = tr
D2 +M2
−D2 +D2 +M2
D2 +M2
= −tr
D2 +M2
. (20)
provided
= 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
I(M2) =
dΩn−2r
n−2x̂iJi(x, x). (22)
where r is the radius of the sphere and dΩn−2 denotes the volume element of
the unit sphere.
If instead we wish to compute the index as the limit of
I(s) = Tr
γe−sD
. (23)
as s→ ∞, then we get
Ji(x, y) = tr
. (24)
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 Ji to
compute the index in odd dimensions.
One dimension
We choose our gamma matrices as
, γ2 =
and we have
γ = iγ1γ2 =
. (26)
The Dirac operator reads
D = iγ1∂ + γ2φ (27)
We need the square of the Dirac operator,
D2 = −∂2 + φ2 + γ∂φ. (28)
We make the choice
J1(x, y) = −tr
D2 +M2
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
J1(x, x) = −tr (γγ1γ2)
k2 + φ2 +M2
φ2 +M2
The index is now given by
(J1(+∞)− J1(−∞)) = ±1 (31)
if φ flips the sign an odd number of times when going from −∞ to +∞, and 0
otherwise.
If instead we choose
J(x, y) = tr
then we get
J(x, x) = tr (γγ1γ2)
k2 + φ2
e−s(k
2+φ2) (33)
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
A(s) ≡
e−s(k
k2 + 1
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
< θ < π
where k = |k|eiθ.
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
A(−s) = −i
e−s(k
k2 − 1 (35)
We can compute its derivative
′(−s) = −i
−s(k2−1) = −i
s (36)
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),
A(∞) = A(0)−
e−s = A(0)−
= A(0)− π (37)
and we then need to compute
A(0) = i
k2 − 1
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
e−s(k
k2 + 1
= −π. (39)
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
eik(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
J(x, x) = tr (γγ1γ2) lim
k2 + φ2
e−s(k
2+φ2) = i
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
Fij = ǫijkDkφ (41)
We choose the convention that our fields are hermitian. It is convenient to group
the fields into ‘gauge potential’
Aµ = (Ai, φ) (42)
We define Dµ = (Di, φ) such that iDµ = i∂µ + Aµ and we let Gµν = i[Dµ, Dν ]
be the associated ‘field strength’. Then the Bogomolnyi equation reads
Gµν =
ǫµνρσGρσ . (43)
Linearizing this, we get
DµδAν =
ǫµνρσDρδAσ (44)
Contracting with γµν , we get
(1 + γ)γµνDµδAν = 0 (45)
and if we impose the background gauge condition
DµδAµ = 0 (46)
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
Dψ ≡ γµDµψ = 0 (47)
where
ψ := (1 + γ)γµδAµ. (48)
We compute
D2 = −D2i + φ2 +
iγµνGµν (49)
Inserting the Bogomolnyi configuration we can write this, thus using the fact
that Gµν is selfdual,
2 = −D2i + φ2 +
(1 + γ)iγµνGµν . (50)
and get a vanishing theorem. Namely, dimkerDD† = 0 as DD† > 0 is strictly
postive. Hence we can compute the dimension of the moduli space dim kerD ≡
dimkerD†D just by computing the index of D. To compute the index, we now
wish to compute
Ji(x, x) = tr
γγiγkDk
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 Ji, for
a sufficiently large two-sphere, only from the term
Ji(x, x) = tr
γγiγ4φ
(2π)3
k2 + φ2 + 1
iγµνGµν
−s(k2+φ2+ 1
iγµνGµν)
We thus need to perform an integral of the form
A(s) =
k2 + 1
e−s(k
2+1) (53)
If we choose the same prescription as we did in one dimension, then we get the
result
A(+∞) = π. (54)
For details of such a computation we refer to appendix A.
If we apply this result to the integral that we had, we get
Ji(x, x) =
γγiγ4φ
iγµνGµν
We expand the square root,
iγµνGµν = φ+
iγµνGµν + ... (56)
In the far distance, in a charge Q monopole configuration, we find that
γµνGµν = 2γkγ4(1− γ)
Q (57)
and so when we trace over the gamma matrices, we get
Ji(x, x) =
. (58)
If we now for instance assume SU(2) gauge group, broken to U(1), then if we
integrate i
Ji 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
A(µ) ≡ lim
k2 + µ2
e−s(k
2+µ2) (59)
if we use our regulator. Here
µ2 ≡ v2 +G (60)
(and G is an abbreviation for 1
iγµνGµν .) This should be compared to the
integral
B(µ) ≡ − lim
(−1)m
(k2 + v2 +M2)
Gm (61)
that we get using the Callias regulator. 3 In order to compare these integrals,
we rewrite them as
A(µ) = µ2m−1a
B(µ) = v−1bGm (63)
where
a = lim
ξ2 + 1
−s̃(ξ2+1)
b = − lim
(−1)m
ξ2 + 1 + M̃2
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
a = −(−1)mπ
b = (−1)m 1
) (65)
We next expand
vA(µ) = v
v2 +G
)m− 1
= v2ma+ ...+
m + ...
vB(µ) = bGm (66)
and we find that the coefficient of Gm becomes equal to
−(−1)m
) π (67)
if one uses our regularization, and equal to
(−1)m 1
) π (68)
3This integral comes from expanding
k2 + v2 +G+M2
k2 + v2 +M2
+ ... (62)
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.
3 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 Bµν , five scalar fields φ
A and corresponding Weyl fermions ψ. The
field strength Hµνρ = ∂µBνρ + ∂ρBµν + ∂νBρµ is selfdual. The supersymmetry
variation of the Weyl fermions is
ΓµνρHµνρ + Γ
µΓA∂µφ
ǫ (69)
where we use eleven-dimensional gamma matrices splitted into SO(1, 5)×SO(5),
so that in particular
{Γµ,ΓA} = 0. (70)
In a static and x5 independent field configuration, in which only φ5 =: φ is
non-zero, we find the SUSY variation
Γ0i5H0i5 + Γ
iΓA=5∂iφ
ǫ (71)
If we assume that the classical bosonic field configuration is such that
∂iφ = H0i5 (72)
then the SUSY variation reduces to
δψ = ∂iφΓ
Γ05 + ΓA=5
ǫ (73)
and we find the condition for unbroken SUSY as
1 + Γ05ΓA=5
ǫ = 0 (74)
If we use the Weyl condition
Γǫ = −ǫ (75)
of the (2, 0) supersymmetry parameter ǫ, then we can also write this as
1 + Γ1234ΓA=5
ǫ = 0. (76)
We may represent the gamma matrices as
Γµ = (Γ0,Γi,Γ5) =
1⊗ iσ2 ⊗ 1, γi ⊗ σ1 ⊗ 1, γ ⊗ σ1 ⊗ 1
ΓA = 1⊗ iσ2 ⊗ σA (77)
where σ1,2,3 are the Pauli sigma matrices, γ = γ1234. Then the condition for
unbroken SUSY is
(1 + γ ⊗ σ) ǫ = 0 (78)
where σ = σ1234 = σA=5.
We have found that if
Hijk = ǫijkl∂lφ (79)
then half SUSY is unbroken. This equation is the Bogomolnyi equation for self-
dual 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
γi∂iχ = 0 (80)
for the bosonic zero modes, that we have gathered into a matrix
χ ≡ γijδBij + γδφ. (81)
For this to work we must also assume the background gauge condition
∂iBij = 0. (82)
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 7→ Cµ(s). We
introduce the Abelian ‘loop fields’ [7]
Aµs = Bµν(C(s))Ċ
ν (s)
φµs = φ(C(s))Ċµ(s)
µs = ψ(C(s))Ċµ(s) (83)
With these definitions, a short computation reveals that Aµs transforms as a vec-
tor and φµs 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
Fµs,νt = Hµνρ(C(s))Ċ
ρ(s)δ(s− t) (84)
In terms of these fields, the Bogomolnyi equation will read4
Fis,jt = ǫijkl∂k(sφlt). (85)
We pass to the non-Abelian theory by letting these loop fields become non-
Abelian, in the sense that Aµs = A
a(s) where λa(s) are generators of a loop
algebra associated to the gauge group [7]. We introduce a covariant derivative
Dµs = ∂µs +Aµs. (86)
Local gauge transformations act as
δΛAµs = DµsΛ
µs = [φµs,Λ]. (87)
Given a loop C, we automatically get a tangent vector Ċµ(s) that makes no
reference to space-time. We can therefore impose the loop space constraints
Ċµ(s)Aµs = 0 (88)
for each s, and also
φµs = Ċµ(s)φ(s;C) (89)
for some subtle field φ(s;C) on loop space. As a consequence, we find that
µs = 0. (90)
These constraints are covariant under diffeomorphisms of space-time and reparametriza-
tions of loops. They are invariant also under local gauge transformations, pro-
vided that the gauge parameter is subject to the condition
µ(s)∂µsΛ = 0 (91)
which is the condition of reparametrization invariance. With the assumption
made that λa(s) are generators of a loop algebra, we find that the constraint
can also be written as
[Aµs, φ
µt] = 0 (92)
A local gauge variation of this constraint is
[DµsΛ, φ
µt] + [Aµs, [φ
µt,Λ]] = [∂µsΛ, φ
µt] + [[Aµs,Λ], φ
µt] + [Aµs, [φ
µt,Λ]]
= [∂µsΛ, φ
µt] + [Λ, [φµt, Aµs]] (93)
The last term vanishes by the constraint. The first term gives us the constraint
Eq (92) that we must impose on the gauge parameter
dsΛa(s, C)λa(s). (94)
4We denote by ∂is the usual functional derivative with respect to C
µ(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]
Fis,jt = ±ǫijklDk(sφlt). (95)
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
Di(s + σφi(s
χt) = 0 (96)
We will also see below that we (presumably) can actually drop the symmetriza-
tion in s and t in this equation. The fields transform in the adjoint represen-
tation of the loop algebra, by which we mean that φisχt = [φis, χt]. We define
the Dirac operator
Ds = γi (Dis + σφis) (97)
and the projectors
(1∓ γσ) , (98)
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
dimkerDs − dimkerD†s (99)
where
Ds = P+Ds = DsP−
D†s = P−Ds = DsP+. (100)
Since Ds and P± are hermitian, it is manifest that D†s defined this way will be
the hermitian conjugate of Ds, 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 dimkerD†s = 0.
Linearizing the Bogomolnyi equation, we get
2D[isδAjt] = ±ǫijkl (Dksδφlt + φksδAlt) (101)
Contracting by γij , we get
γijD̃isχjt = 0 (102)
where we have defined
D̃is ≡ Dis ∓ γφis
χis ≡ δAis ∓ γδφis (103)
To see that the linearized BPS equation can be written like this, one must use
the constraint
γijφisδφjt = 0. (104)
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
γ ⊗ σ = ±1 (105)
which amounts to
D̃is ≡ Dis + σφis
χis ≡ δAis + σδφis (106)
without any ±.5 If we define
χs ≡ γiχis (108)
then we can write the zero mode equation as
γiD̃isχt + D̃
sχit = 0. (109)
Let us analyze the second term in this equation. It is given by
DisδAit + φ
sδφit
φisδAit +D
sδφit
(110)
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,
(δΛAis, δAit) + (δΛφis, δφjt) = 0 (111)
This leads to the background gauge condition
sδAit + φ
sδφit = 0. (112)
5To really understand what is going on, one should apply (1± γσ) on everything, on ψs
and on Ds. Then one notices that
∓γ (1∓ γσ) = σ (1∓ γσ) . (107)
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,
δΛδAis = 0 = δΛδφis (113)
To see this, we make a gauge variation δΛδAis = DisΛ, δΛφis = φisΛ, and
ask which gauge parameters Λ will respect the background gauge condition.
Inserting this gauge variation into the background gauge condition, we get
sDit + φ
Λ = 0. (114)
For this to work nicely, it seems that we must constrain the non-locality of our
loop field such that ∂i
∂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
Ais and φis,
(Ais, δφit) + (δAis, φit) = 0 (115)
If we make a gauge variation of this, then we get the condition
(δΛAis, δφit) + (δAis, δΛφit) = 0 (116)
which amounts to
φisδAit +D
sδφit = 0. (117)
We conclude that the zero mode equation can be written as
Dsχt = 0 (118)
where
Ds = γi (Dis + σφis) (119)
We are interested in counting the number of such modes in a background of k
BPS strings. We compute
D2 = (Dis)
2 + (φis)
γij (Fis,js + γσǫijklDksφls) (120)
(Here D2 ≡ DsDs ≡
DsDs, and analogously for the other fields or opera-
tors.) In a BPS configuration, we get is
2 = (Dis)
2 + (φis)
ij (1 + γσ)Fis,js (121)
Furthermore, in the subspace where 1 + γσ = 0, we find that
D2 = (Dis)
2 + (φis)
2 (122)
is a strictly negative operator, hence has no zero modes. This means that we
have a vanishing theorem, dimkerD† = 0.
A small comment
The zero mode equation was really
D(sχt) = 0 (123)
where we should symmetrize in s and t. That means that we should rather
consider
DsD(sχt) =
(DsDsχt +DsDtχs)
(DsDsχt +DtDsχs + [Ds, Dt]χs) . (124)
If now D[sDt] = 0 and Dsχs = 0, then we get
DsDsχt = 0 (125)
The latter condition, Dsχs = 0 is of course a consequence of D(sχt) = 0 with
s = t. The former condition reads
0 = D[sDt]
= Di[sDit] + φi[sφit] + σDi[sφit] (126)
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
dsD†sDsχt = 0 (127)
because Ds is anti-self-adjoint with respect to the inner product
(ψs, χt) =
ψ†s(C)χt(C)
(128)
on loop space. We can also go in the opposite direction. Assuming that Eq
(127) holds, we get
χt, D
sDsχt
= (Dsχt, Dsχt) (129)
and we conclude that (123) implies
Dsχt = 0 (130)
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
I(s) = Tr
(131)
when s→ ∞. We define the quantity
Jis(C,C
′) = tr
γσγiγk (Dks + σφks)
(132)
(it should be clear that the two s’s involved in this formula are totally unrelated)
and find that
I(s) =
DC∂isJ is(C,C) (133)
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
DxC (134)
Then we can write I(s) as an integral over a large three-sphere at spatial infinity,
∂Jis(C)
∂Ci(s)
dΩ3x̂
DxCJis(C,C) (135)
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,
Fis,jt = Hijl(x)Ċ
l(s)δ(s− t)
φks = vĊk(s) (136)
Without doing any computations, we can guess what the outcome of the
index calculation should be. A term like
ǫijkl
DxCtr (Fis,jt(C)Fks,jt(C)) (137)
could certainly arise somewhere (in odd dimensions a corresponding term van-
ished since there is no chiral anomaly in odd dimensions). In our case this term
vanishes identically by the Bogomolnyi equation and the constraint6
Fis,jtDisφjt = 0. (139)
Then there can be a term
ǫijkl
DxCtr
Fis,jtφks
(140)
6For U(1) fields this would read
Fis,jt∂isφjt ∼ Hijk(C(s))∂iφ(C(s))Ċ
k(s)Ċj(s)δ(s − t)2 ≡ 0. (138)
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 propor-
tional to
ǫijklHijk(x) (141)
That means that the index should be given by some numerical constant, times
the magnetic charge
H. (142)
A Integrals over the exponential
The integral we will analyze here is
a(s) =
k2 + 1
−s(k2+1)eiǫk (143)
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
a(0) =
k2 + 1
eiǫk (144)
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 eiπ/2, and by −1 we
mean eiπ . Then we get
a(0) = −i2ζ+1
k2 − 1
e−ǫk (145)
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
a(0) = (−1)ζπ 1− (−1)
. (146)
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 = k2 we can put it on the form of two gamma functions. The result is that
′(−s) = −eiπ(ζ+
1 + (−1)2ζ
−ζ− 1
s (147)
which we can trivially continue analytically to +s, and then integrate up. The
result is
a(+∞) = −π 1 + (−1)
cos(πζ)
+ π(−1)ζ 1− (−1)
. (148)
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