Instanton Solutions in Open Superstring Field Theory

Open superstring field theory admits a"hybrid"formulation where N = 1 D = 4 supersymmetry is manifest for Calabi-Yau compactifications to four dimensions. Using this formulation, the half-BPS instanton solution of four-dimensional super-Yang-Mills can be easily generalized to the full open superstring field theory. In this paper, we compute the first stringy correction to the super-Yang-Mills instanton solution which involves turning on certain fields at the first massive level.


Introduction
Open superstring field theory (SFT) can be a powerful method to obtain non-perturbative information about superstring theory. The best example is tachyon condensation where tachyonic solutions to the SFT equations of motion have been shown to describe the decay of non-supersymmetric D-branes [1]. In addition to the non-supersymmetric tachyonic solutions, the equations of motion of SFT also contains spacetime supersymmetric solutions which can be studied. In particular, it would be interesting to use SFT to search for stringy generalizations of instanton solutions of super-Yang-Mills which give rise to nonperturbative effects in gauge theories.
After compactifying to four dimensions, the massless states of the open superstring include D = 4 super-Yang-Mills fields and the instanton solution is obtained by requiring the four-dimensional Yang-Mills field-strength to be self-dual. However, it is difficult to generalize the concept of self-dual field strength to SFT since the only gauge-invariant quantity that can be constructed from the string field Φ is QΦ, which vanishes on-shell. Nevertheless, one can instead define the four-dimensional instanton solution as a localized half-BPS solution to the equations of motion, i.e. a solution which is annihilated by half of the N = 1 D = 4 spacetime supersymmetries. In this paper, we will show that there is a unique generalization of this half-BPS definition of the instanton solution in SFT and will find the first stringy correction.
Since D = 4 spacetime supersymmetry plays an important role in our construction, we will use the manifestly N = 1 D = 4 super-Poincaré invariant action for SFT of [2], which is based on the D = 4 hybrid formalism of [3]. The massless contribution to this SFT action reproduces the usual N = 1 D = 4 superspace action for super-Yang-Mills theory, and the massive states are also described in terms of N = 1 D = 4 superfields. Although we will be unable to find an exact solution to the SFT equations of motion which generalizes the self-dual instanton, we will find the first stringy correction to the instanton solution which corresponds to turning on certain massive spin-2 and spin-0 fields.
In section 2 of this paper, we review self-dual instanton solutions in super-Yang-Mills theory. In section 3 we review the D = 4 hybrid formalism for the superstring and introduce the central problem of this paper -solving the half-BPS condition in open superstring field theory. Our strategy for tackling this problem is to write a series expansion for the string field and solve order by order in the expansion parameter, as explained in section 4. In section 5 we review the BPST instanton in super-Yang-Mills and generalize it to SFT using the star product. In section 6 we calculate the first stringy correction to the BPST instanton and show it corresponds to turning on certain off-shell massive fields by comparing with the vertex operators for the first massive level of the string.
2 Super-Yang-Mills Instantons N = 1 D = 4 super-Yang-Mills can be described by a vector superfield V i (x, θ,θ) where i is a gauge index and (x m , θ α ,θα) are the usual N = 1 D = 4 superspace variables for m = 0 to 3 and α,α = 1 to 2. We also define V = V i T i where T i are the generators of the gauge group. Supersymmetry transformations are generated by acting on V with where σ m are the Pauli matrices andσ 0 = σ 0 ,σ i = −σ i for i = 1, 2, 3. We want to find field configurations that are preserved by half of this supersymmetry. Naively, this condition would be formulated as where ǫ α is a supersymmetry parameter and ǫq = ǫ α q α . However, that is not quite correct. We must keep in mind that the theory has a gauge symmetry, given by (2.4) where Λ andΛ are chiral and antichiral, respectively, and L is the Lie derivative: So the correct condition is not that (ǫq)V vanishes, but that it is pure gauge. In other words, V is preserved by half of the supersymmetry if there are Λ andΛ such that Of course, one could also consider states preserved by the antichiral supersymmetry, for which the analysis is similar. To see what eq. 2.6 implies, we first write V in Wess-Zumino gauge and assume that the fermions vanish (since we are concerned with classical field configurations): Since V 3 = 0 in Wess-Zumino gauge, the gauge transformation of eq. 2.4 becomes The most general Λ can be written in components as Similarly, the most generalΛ is The gauge transformation is then (2.11) On the other hand, the supersymmetry transformation is Plugging eqs. 2.11 and 2.12 into eq. 2.6 and splitting it in components, we find the conditions: where σ mn = 1 4 (σ mσn − σ nσm ). Since σ mn is traceless, we conclude that and (σ mn ) α β F mn = 0, (2.17) which tells us that the gauge field must be anti-self-dual.

String Field Theory Instantons
To describe the SFT instantons, we'll use the hybrid formalism developed in [2] which describes a compactified superstring with manifest four dimensional super-Poicaré symmetry. We take the four-dimensional space to be Euclidean. Also, we will only consider states that are independent of the structure of the compactified manifold. The hybrid formalism contains five free bosons (x m , ρ) and eight free fermions (θ α ,θα, p α ,pα) with OPE's and a field theory for the six-dimensional compactification manifold which is described by theĉ = 3 N = 2 superconformal generators T C , G + C , G − C , J C . The N = 4 superconformal generators are defined in terms of these free fields by x m , and the notation e −iρ+iHc G − C means the contour integral of e −iρ+iHc around G − C . In principle, instantons in SFT can be found using the methods of the previous section and are defined to be field configurations whose superymmetry variation is pure gauge: (3.9) However, unlike the super-Yang-Mills case, there is no gauge choice where V 3 = 0. So the gauge transformation is now an infinite series in V: where the products are to be interpreted as star ( * ) products, as defined in [4]. The gauge parameters Λ andΛ in eq. 3.10 have to satisfỹ G + (Λ) = 0 (3.11) and G + (Λ) = 0, (3.12) where the notation G + (Λ) denotes the contour integral of G + aroundΛ. Although we will be unable to analytically solve eq. 3.9 because of the infinite number of terms, one can solve eq. 3.9 perturbatively by expanding around the linearized massless solutions to V , Λ andΛ. In the rest of this paper, we will use this perturbative method to find the first stringy corrections to the super-Yang-Mills instanton solution.

Series Expansion of the Half-BPS Condition
To perform a perturbative expansion, first write the string field V as a power series in some parameter λ: Similarly, write Λ = λΛ 0 + λ 2 Λ 1 + λ 3 Λ 2 + . . .
We can then break up eq. 3.9 in powers of λ: and so on. The goal is to first determine V 0 and then recursively determine V i+1 from V i . Finding V 0 is relatively easy because eq. 4.4 is linear and thus the massless and massive states decouple. Actually, since we are working with off-shell string theory, the expressions "massless" and "massive" need some clarification. By "massless" we mean any state that depends only on the worldsheet zero modes of x m , θ α andθα and not on their derivatives nor on p α . That is, they are the states that would be massless if they were on-shell. All remaining states will be called "massive".
Let us choose a gauge such that where V 0,massive contains the massive components. It is trivial to adapt the computations of section 2 to show that and Later we will describe an explicit solution of this condition for the case of a U (2) gauge group.
What about the massive part? We will argue here that massive on-shell states cannot contribute to the instanton solution since they are not localized in spacetime. Note that in Euclidean space, the solution to the linearized equation of motion So the linearized condition of eq. 4.4 together with the requirement that the solution is localized in spacetime implies that A specific choice for the compensating gauge transformation is Equipped with this, we can now try to tackle eq. 4.5. We will find that due to the complexity of the star product, V 0 * Λ contains massive components even though V 0 andΛ contain only massless components. So unlike V 0 , V 1 will contain massive components.

BPST Instanton
To gain intuition, it will be useful to consider an explicit super-Yang-Mills instanton solution and see how it can be extended to string field theory. For an SU (2) gauge group, the k = 1 instanton can be written in a simple form which is the BPST instanton [5]: x m 0 and ρ are two fixed parameters representing the center and the size of the instanton, andη i mn are the 't Hooft symbols [6]: where i = 1, 2, 3 is an SU (2) index and m, n = 1, 2, 3, 4 are Euclidean spacetime indices. Note that the t'Hooft symbols are anti-self-dual in their Lorentz indices: To show that eq. 5.1 does indeed represent an instanton, we have to show that the corresponding field strength, whereas the quadratic part of the field strength is Using the identity we obtain Then the field strength is and the anti-self-duality of the field strength follows from the anti-self-duality ofη imn .
To connect this with what we were doing in the previous section, let us expand v i m in powers of 1/ρ 2 : That is, we are working with a large instanton expansion which has the form of 4.1 where the expansion parameter is λ = 1 ρ 2 . Now we would like to find the corrections to this instanton solution in superstring field theory -that is, a solution to eq. 2.6 where we take the massless part of the string field to have the form of the BPST instanton. Note that the gauge group should now be U (2) rather than SU (2). Since the star product in non-commutative, the massive levels of the string field will have singlet contributions, even if the massless level doesn't. We formally where the products are to be interpreted as ⊗ products, and ⊗ denotes the truncation of the star product to the massless level (see appendix A). To give meaning to division by x m , we invoke the expansion in 1/ρ 2 of eq. 5.10 where and r ⊗ in the exponent means we multiply r times with the ⊗ product. Also, x 2 means the normal ordered product -that is, no contractions between the x's. Note that V i massless = −θσ mθ v i m is a solution to the massless part of eq. 2.6. Since, as shown in the last section, the massive part of V 0 is trivial, where V 0 0 denotes the U (1) component of the U (2) gauge group. In the next section we'll compute the first massive level of V 1 .

The First Stringy Correction
We now want to compute the first stringy correction to the BPST instanton, which is the first massive level of V 1 . To that end, we must solve the half-BPS equation at the quadratic level (i.e. O(1/ρ 4 ) terms) which corresponds to eq. 4.5. Although V 0 , Λ 0 andΛ 0 are massless, the star products in the commutator will contain massive terms which will imply that V 1 has a nontrivial massive part.
The half-BPS solution is not unique -if V 1 is a solution to eq. 4.5, then V ′ 1 = V 1 + q 2 Ω is also a solution for any field Ω since (ǫq)q 2 Ω = 0. We can ask if there is a half-BPS solution which also satisfies the SFT equations of motion. The answer is yes, and we will show that this solution is unique if we also require that it vanishes at infinity, i.e. it is localized in spacetime like the Yang-Mills instanton. In other words, q 2 Ω will be uniquely determined if we require that V ′ 1 satisfies the SFT equations of motion and is localized in spacetime. This first stringy correction to the BPST instanton will be interpreted as turning on some off-shell massive fields. To see what kinds of fields these are, we will compare our result with the first massive states of the superstring which on-shell consists of a massive spin-two multiplet and two massive scalar multiplets [7,8]. We will find that our solution corresponds to turning on a spin-two and a scalar field.
In subsection 6.1 we find the general solution to eq. 4.5 when V 0 is given by the linearized BPST solution. In subsection 6.2 we find the unique half-BPS solution that also satisfies the equations of motion and vanishes at infinity. And in subsection 6.3 we compare our solution with the vertex operator for the first massive level.

Half-BPS solutions
In this section we will solve the equation given the linearized BPST solution for V 0 The massless part of V 1 is simply As noted before, we can add any q 2 -exact term to a solution V 1 and still have a solution.
For the massive part V 1,massive , we'll show that the solution for the first massive level is unique up to a q 2 -exact term, and up to gauge transformations. If, in addition to eq. 6.1, we require that the solution satisfies the equations of motion and vanishes at infinity, we will find that the q 2 -exact term is fixed uniquely. From now on, we'll use V 1 instead of V 1,massive to denote the first massive level since we'll only deal with this level in this section. Similarly, Λ 1 andΛ 1 will denote the first massive level of the gauge parameters.
First we have to calculate the commutator on the right-hand side of eq. 6.1, which involves evaluating a star product. Then we substitute the general form of the string field and gauge parameters at the first massive level. This is given by the fields that have conformal weight 1 at zero momentum, and which have mass squared M 2 = 2 when on-shell.
The commutator will contain terms that are not ǫq-exact, which must therefore be cancelled by the gauge terms. This will fix Λ 1 andΛ 1 , up to ǫq-exact gauge parameters (that is, up to ǫq(Λ ′ 1 +Λ ′ 1 ) where Λ ′ 1 andΛ ′ 1 are gauge parameters). Once the gauge parameters are fixed, we have determined ǫqV 1 , which means we have determined V 1 up to a q 2 -exact term.
At the first massive level, the commutator [V 0 ,Λ 0 ] satisfies where σ 0 is i times the 2 × 2 identity. This is because (V i 0 * Λ j 0 ) = −(Λ j 0 * V i 0 ) at this level. Details on the calculation of the star product can be found in appendix A, and one finds at the first massive level that where y m = x m + i 2 θσ mθ . The general form of the string field at the first massive level is [7] and the gauge transformations are We can use the gauge freedom to fix V 1 in the form Note that there are no terms proportional to d ord in either V 1 or in the commutator. Fixing the d andd terms to zero in 6.13, the gauge parameters take the form where we have rescaled E α andĒα by a factor of four for later convenience. Equation 6.1 can be decomposed into five equations by separating the terms proportional to ∂θ α , ∂θα, Π m , ∂ρ and ∂H C . The first two will fix E α andĒα, up to ǫq-exact terms. The other equations, in turn, will determine V 1 .
The terms proportional to ∂θ α and ∂θα give which are solved by where E ′ α andĒ ′α are solutions to the homogeneous equations, i.e. they satisfy As we'll see shortly, E ′ α andĒ ′α are also ǫq-exact. The terms proportional to Π m give the equation In the second equality we used the fact that σ mnηi mn = 0 and σ (m σ n) α β = −2δ mn δβ α . So eq. 6.23 becomes Finally, the equations for C and B are In order for equations 6.25-6.27 to be consistent, we must verify that the righthand sides are ǫq-exact. Note that any field of the form ǫψ(y,θ) is ǫq-exact: ǫψ(y,θ) = ǫq(θψ(y,θ)). It's then easy to verify that V α and D 2Vα are ǫq-exact, as are any derivatives of these quantities, since q anti-commutes with D andD. Thus, the equations are consistent as long as we require that D αĒ ′α +DαE ′ α , D αD2 E ′ α andDαD 2Ē′α are ǫq-exact. One might suspect that the E ′ α andĒ ′α terms simply give a gauge transformation of V 1 . This is indeed the case, as we'll now show. First, note that requiring D αD2 E ′ α = ǫq(F ) implies, by the first eq. in 6.21, thatĒ ′α = − 1 2 ǫq(DαF ) = ǫq(Ē ′′ α ). Similarly,DαD 2Ē′α = ǫq(F ) implies E ′ α = ǫq(E ′′ α ). Note that E ′′ α andĒ ′′ α are only defined up to q 2 (something). Also, they satisfy eqs. 6.21 and 6.22 up to q 2 (something) -in other words, we can choose E ′′ α andĒ ′′ α such that they satisfy those conditions. That means they only change Λ 1 and Λ 1 by ǫq-exact gauge parameters, which in turn means that they only contribute pure gauge terms to V 1 .
Thus, from eqs. 6.25-6.27, we have uniquely determined ǫq(V 1 ) up to a gauge transformation. This means we also have determined V 1 , up to a gauge transformation and up to q 2 (something). The particular solution found by simply eliminating ǫq in eqs. 6.25-6.27 (after writing the right-hand sides as ǫq(something) in the way described in the previous paragraphs) isṼ where the x products, again, are normal ordered, i.e. no contractions.

Equations of motion
We now want to verify that we can find a half-BPS solution which satisfies the equation of motion, given by [2]G The solutionṼ 1 written above does not satisfy this equation, but we can find a field V 1 that does, and which differs fromṼ 1 by a q 2 -exact term. This solution V 1 is unique if we also require that it vanishes at infinity. Calculating the anticommutator between G + (V 0 ) andG + (V 0 ) again involves evaluating a star product: and we have used that ∂ ααDα V 0 = 0.
As before, let's separate the terms proportional to Π m , d α ,dα, ∂ρ and ∂H C . Note that acting with G + orG + on the equation gives zero, so not all terms are independent. For example, we don't need to check the terms proportional to ∂θ α and ∂θα. Details on the star product can again be found in appendix A, and on the right-hand side of 6.29 we find: Terms proportional to Π m : Terms proportional to d α : Terms proportional todα: Terms proportional to ∂(H C − 3ρ): And on the left-hand side of 6.29 we find: Terms proportional to Π m : Terms proportional to d α : Terms proportional todα: Terms proportional to i∂(ρ − H C ): We conclude thatṼ 1 does not satisfy the equations of motion. However, the following V 1 can be defined which differs fromṼ 1 by a q 2 -exact term and which satisfies the equations of motion: We can still add to V 1 a term q 2 Φ where Φ satisfiesG + (G + (Φ)) = 0, i.e. Φ describes an on-shell massive field. But as explained below eqn. 4.9, any non-zero on-shell massive field will diverge exponentially when x → ∞. So in order to have an instanton solution that vanishes at infinity, we should set Φ to zero. Note that the stringy contribution V 1 of eqn. 6.41 naively diverges quadratically in x m when x → ∞, but this is expected since it was found using an expansion in 1 ρ 2 . After summing over all orders in 1 ρ 2 , one expects that the stringy contribution will vanish at infinity like the super-Yang-Mills instanton. However, adding an on-shell massive contribution to V 1 would diverge exponentially when x → ∞ and could never cancel after summing over all orders in 1 ρ 2 . We therefore have a unique solution.

Meaning of the massive fields
The first massive level of the on-shell string field describes two massive scalar multiplets and a massive spin-two multiplet. The on-shell field can be gauge-fixed to the form where F andF are the chiral and antichiral superfields for the scalar multiplets, andV m is the superfield for the spin-two multiplet. In components, where X,X, Y andȲ are massive real bosons, ξ α andξα are massive spinors, v mn is a massive symmetric traceless tensor, C m is a massive vector, and χ and λ are massive spin-3/2 fermions [7,8]. This is to be compared to our result of eq. 6.41. Since V 1 is not on-shell, it doesn't take the form 6.42. Nevertheless, we can still identify the components of V 1 with certain components of F ,F andV m . Comparing eqn. 6.43 and 6.44 with eqn. 6.41, one finds that the non-zero component fields are We can therefore interpret the correction to the BPST instanton as turning on the U (1) component of a massive scalar field and a massive spin-two field.
The coefficient is given by the correlator with the conformal transformations The correlators are given by θ 2θ2 e −iρ+iH C = 1 (A.8)

A.1 Massless level
We will denote the massless level of the star product by ⊗. In the pθ sector, there are no non-trivial massless contributions. For example, it is immediate that For the x m , however, there are non-trivial (higher derivative/higher order in α ′ ) terms, even at the massless level. Computing the product of two functions of x m is slightly tricky because x m itself is not a well-behaved operator in the worldsheet theory. Rather, the actual operators are worldsheet derivatives of x m and plane waves e ik·x . The solution is to assume that the functions of x m we are interested in admit Fourier transform representations and to compute the star product of functions x m in terms of the products of exponentials, that can be obtained by the usual methods. That is, We need then to calculate the product e ik 1 x * e ik 2 x , which amounts to calculating the coefficient e −ikx , e ik 1 x , e ik 2 x : So the ⊗ product is We conclude that