Ramond equations of motion in superstring field theory
HJE
Ramond equations of motion in superstring field theory
Theodore Erler 0 1
Sebastian Konopka 0 1
Ivo Sachs 0 1
0 Theresienstrasse 37 , 80333 Munich , Germany
1 Arnold Sommerfeld Center, Ludwig-Maximilians University
We extend the recently constructed NS superstring field theories in the small Hilbert space to give classical field equations for all superstring theories, including Ramond sectors. We also comment on the realization of supersymmetry in this framework.
Superstrings and Heterotic Strings; String theory and cosmic strings
1 Introduction
2
3
4
5
6
Ramond sector of open superstring with stubs
Ramond sector of heterotic string
Ramond sectors of type II closed superstring
Supersymmetry
6.1
6.2
6.3
Perturbative construction of supersymmetry transformation
Supersymmetry in Berkovits’ superstring field theory
Supersymmetry algebra
7
Conclusions
1
Introduction
The recent construction of NS actions [1, 2] has raised the prospect of obtaining a
secondquantized, field-theoretic description of all superstring theories. The next step in this
programme is to include the Ramond sectors. As is well-known, formulating kinetic terms
for the Ramond sector is complicated by the fact that the string field must carry a definite
picture [3], which for a holomorphic Ramond state is naturally chosen to be −1/2. This is
not the right picture to form the usual string field theory kinetic term,
since in the small Hilbert space the BPZ inner product must act on states whose picture
adds up to −2. While there are some proposals for circumventing this problem [4–7], at
this time it is not clear what is the most promising way forward.
Therefore it is worth considering a simpler problem first: namely, constructing classical
field equations for all superstring theories, including Ramond sectors. This is the goal of
the present paper. With the classical equations of motion, we can
• compute of tree level amplitudes including Ramond asymptotic states around the
perturbative vacuum or any classical solution;
1
2
hΨ, QΨi,
– 1 –
• investigate the broken and unbroken supersymmetries of classical solutions
representing distinct string backgrounds;
• construct classical solutions in type II closed superstring field theory representing
nontrivial Ramond-Ramond backgrounds.
The last point is interesting, since Ramond-Ramond backgrounds are quite difficult to
describe in the first quantized RNS formalism. While solving the equations of motion of
closed string field theory is a tremendously difficult task, it does not appear to be more
difficult for Ramond-Ramond backgrounds than other types of background.
tained in [1]. The main new ingredient will be incorporating additional labels associated
with multiplication of Ramond states. Like [1, 2], our approach is based on A∞ and
L∞ algebras, and makes extensive use of associated concepts and notation. We will
review the needed apparatus as we go, but for more dedicated discussion see several recent
works [1, 2, 8, 9]. A different formulation of the equations of motion using the large Hilbert
space has already been provided for the open superstring in [4] and recently the heterotic
string in [10, 11]. Our approach has the advantage of describing type II closed superstrings
as well, and, once suitable Ramond kinetic terms are formulated, might be generalized to
give a classical Batalin-Vilkovisky action.
The essential idea behind our construction of the equations of motion is already
conΨR are Grassmann odd and carry ghost number 1; the NS field ΦN carries picture −1 while
the Ramond field ΨR carries picture −1/2.
For clarity, let us explain why the Ramond string field is a Grassmann odd object. Any
state in the Ramond sector can be built by acting oscillators on the Ramond ground state
c Θ~s e−φ/2(0)|0i,
#
1
√
2
– 2 –
(2.1)
(2.2)
(2.3)
(2.4)
Θ~s(z) = exp i X saHa (z),
"
4
a=0
~s = (s0, s1, s2, s3, s4), sa = ± 2 ,
1
and Ha, a = 0, ..., 4 realize the bosonization of the worldsheet fermions through
where Θ~s denotes the spin field [3]1
1Our conventions concerning the Ramond sector, spinors and gamma matrices follows [13].
1
√
2
ψ0 + ψ
1 = eiH0 ,
ψ2a + iψ2a+1 = eiHa , a = 1, ..., 4.
m−1
X I⊗k
k=0
n−1
X I⊗k
k=0
[bm, cn] ≡ bm
⊗ cn ⊗
I⊗m−k−1
− (−1)deg(bm)deg(cn)cn
where I is the identity operator on the state space and
(2.5)
(2.6)
(2.7)
!
(2.8)
(2.9)
We take the Ramond ground state to be Grassmann even if it is a positive chirality spinor
in the GSO(+) sector. Then all GSO(+) projected states built from acting oscillators
on (2.2) will be Grassmann even as well. A Ramond string field is created by taking linear
combinations of Ramond states with coefficients which are anticommuting spacetime fields,
as is appropriate for fermions. The anticommuting fields anticommute with each other,
and in addition we will assume that they anticommute with Grassmann odd worldsheet
operators.2 Therefore, in total a Ramond string field must be Grassmann odd, since it is
built from Grassmann even states with Grassmann odd coefficients.
Let us quickly review some notation and conventions which will be essential for our
discussion. When discussing A∞ algebras in open string field theory, it is very useful to use
a shifted grading on the open string state space called degree. The degree of a string field A,
denoted deg(A), is defined to be its Grassmann parity plus one (mod Z2). The dynamical
string fields ΦN and ΨR are Grassmann odd, but degree even. Consider a product of m
string fields:
The degree of the product bm, denoted deg(bm), is defined to be the degree of its output
minus the sum of the degrees of its inputs (mod Z2). It is useful to think of the product
bm as a linear map from the m-fold tensor product of the state space into the state space:
We will write
bn(A1, ..., Am) = bn(A1 ⊗ ... ⊗ Am),
where on the right hand side bm is regarded as an linear operator acting on tensor products
of states. Given a pair of multi-string products bm and cn, we define a “commutator”3
bm (A1, ..., Am) .
bm : H
⊗m
→ H.
!
I⊗n = I
⊗ ... ⊗
| n t{izmes }
I
is the identity operator on H
⊗n. The commutator satisfies the Jacobi identity, and [bm, cn]
defines an m + n − 1 string product. Tensor products of operators act in the natural way
H
on tensor products of states. Given two linear maps bk,ℓ and cm,n from H
⊗n → H
⊗m, respectively, the tensor product map bk,ℓ ⊗ cm,n satisfies
⊗ℓ
→ H
⊗k and
bk,ℓ ⊗ cm,n(A1 ⊗ A2 ⊗ ... ⊗ Aℓ+n) = (−1)deg(cm,n)(deg(A1)+...+deg(Aℓ))×
bk,ℓ(A1 ⊗ ... ⊗ Aℓ) ⊗ cm,n(Aℓ+1 ⊗ ... ⊗ Aℓ+n). (2.10)
2Another convention would assume that anticommuting spacetime fields commute with anticommuting
worldsheet operators. This is closer in spirit to the sign rules of [14], and in this context the A∞ and L∞
algebras we will construct would be equivalently described as super A∞ or L∞ algebras. T.E. thanks U.
Schreiber for explaining this convention.
3Commutators of multi-string products are always graded with respect to degree.
– 3 –
An A∞ algebra is defined by a sequence of degree odd multi-string products dn, n =
1, 2, 3, ... which satisfy a hierarchy of identities called A∞ relations:
[d1, dn] + [d2, dn−1] + ... + [dn−1, d2] + [dn, d1] = 0,
by spreading the picture changing operators away from the midpoint [8], which following [1]
we accomplish by postulating that the equations of motion take the form
0 = QΦN + M2(ΦN, ΦN) + m2(ΨR, ΨR) + higher orders,
0 = QΨR + M2(ΨR, ΦN) + M2(ΦN, ΨR) + higher orders,
1
3
with higher order terms that we will construct in a moment. The degree odd product m2 is
Witten’s open string star product with a sign needed to shift the grading from Grassmann
parity to degree:
m2(A, B) ≡ (−1)deg(A)A ∗ B.
The degree odd product M2 must carry picture +1 and takes the form
M2(A, B) =
Xm2(A, B) + m2(XA, B) + m2(A, XB)
where X is a BPZ even charge of the picture changing operator:
X ≡ Q · ξ,
ξ ≡
I
dz
is BPZ even and anticommutes with the eta zero mode η ≡ η0 to give 1:
[η, ξ] = 1.
M2 = [Q, µ 2],
– 4 –
(2.14)
(2.15)
(2.16)
(2.17)
(2.18)
(2.19)
(2.20)
(2.21)
An important observation of [1] is that M2 is BRST exact in the large Hilbert space,
and therefore may be formally obtained by an improper field redefinition from a free
theory [9]. Here µ 2 is called the gauge 2-product 4
1
3
µ 2(A, B) =
ξm2(A, B) − m2(ξA, B) − (−1)deg(A)m2(A, ξB) ,
4In [1] μ2 was denoted M 2 and was called the “dressed 2-product.” Our current notation and terminology
follows [2].
and is degree even. It also satisfies
products.5 However, our goal is not necessarily to provide the most general possible form
of the equations of motion. We will try, as far as possible, to mimic the construction of the
NS sector, which includes some choices which in that context were motivated by cyclicity.
if N1, N2 are NS sector string fields and R1, R2 are R sector string fields, the composite
2-product M˜ 2 is defined to satisfy
5Following the recent suggestion of Sen [7], one can try to construct a tree-level action with two Ramond
fields supplemented by a constraint, in a similar spirit as [6]. In this approach the 2-product of NS states
can be chosen to be M2 while the 2-product of an NS and R state should be chosen to be Xm2. We did
not consider this approach, though it could lead to an interesting refinement of our equations of motion.
See also comments in the conclusion.
– 5 –
M˜ 2(N1, N2) ≡ M2(N1, N2),
M˜ 2(N1, R1) ≡ M2(N1, R1),
M˜ 2(R1, N1) ≡ M2(R1, N1),
M˜ 2(R1, R2) ≡ m2(R1, R2).
Φ˜ = ΦN + ΨR ∈ H˜ ≡ HN ⊕ HR,
0 = QΦ˜ + M˜ 2 Φ˜ , Φ˜
+ higher orders.
hQ, M˜ 3i +
Q
2 = 0,
hQ, M˜ 2i = 0,
2
1 hM˜ 2, M˜ 2i = 0.
Introducing a composite string field
the equations of motion up to second order can be expressed
Projecting on the NS output (or picture −1) gives the equation of motion (2.14) and
projecting on the R output (or picture −1/2) gives the equation of motion (2.15).
Up to cubic order the A∞ relations are
(2.23)
(2.24)
(2.25)
(2.26)
(2.27)
(2.28)
(2.29)
(2.30)
(2.31)
(2.32)
The first two A∞ relations are already satisfied since Q is nilpotent and a derivation of
both m2 and M2. We will use the third A∞ relation to determine the composite 3-product
M˜ 3. First, act the third A∞ relation on three NS states, or two NS states and one R state.
In this case, the commutator [M˜ 2, M˜ 2] reduces to [M2, M2], and we can take M˜ 3 = M3,
where M3 is the 3-product of the NS open superstring field theory found in [1] (whose form
will be reviewed momentarily). Therefore
M˜ 3(N1, N2, N3) = M3(N1, N2, N3),
M˜ 3(N1, N2, R1) = M3(N1, N2, R1),
M˜ 3(N1, R1, N2) = M3(N1, R1, N2),
M˜ 3(R1, N1, N2) = M3(R1, N1, N2).
If there is more than one R input, M˜ 3 will take a different form. For example, let us act
the third A∞ relation on three Ramond states:
hQ, M˜ 3i+
2
1 hM˜ 2, M˜ 2i R1 ⊗R2 ⊗R3 =
hQ, M˜ 3i+ M˜ 2 M2 ⊗
˜
I + I
where in the second step we acted M˜ 2 on the R states to produce M2 and m2. Next we
use the fact that M2 is BRST exact in the large Hilbert space:
hQ, M˜ 3i+
2
1 hM˜ 2, M˜ 2i R1 ⊗ R2 ⊗ R3 = hQ, M˜ 3 + µ 2(m2 ⊗
I + I
⊗ m2) iR1 ⊗ R2 ⊗ R3.
Since this must be zero, it is natural to identify
M˜ 3(R1, R2, R3) = −µ 2 (m2 ⊗
I + I
⊗ m2) R1 ⊗ R2 ⊗ R3.
Note that this product is in the small Hilbert space,
ηM˜ 3(R1, R2, R3) = 0,
since η turns µ 2 into m2, and the result vanishes by associativity of m2. Similar
considerations determine the remaining 3-products between NS and R states:
M˜ 3(N1, R1, R2) = m2 (µ 2(N1, R1), R2) − (−1)deg(N1)µ 2(N1, m2(R1, R2)),
M˜ 3(R1, N1, R2) = m2 (µ 2(R1, N1), R2) + m2(R1, µ 2(N1, R2)),
M˜ 3(R1, R2, N1) = −µ 2 (m2(R1, R2), N1) + m2(R1, µ 2(R2, N1)),
M˜ 3(R1, R2, R3) = −µ 2 (m2(R1, R2), R3) − (−1)deg(R1)µ 2(R1, m2(R2, R3)).
In general, when multiplying n strings there will be 2n formulae representing all ways that
NS and R states can multiply. Determining all these formulae seems like a daunting task,
but there is a trick to it which we explain in the next subsection.
– 6 –
(2.33)
(2.34)
(2.35)
(2.36)
(2.37)
(2.38)
(2.39)
(2.40)
(2.41)
(2.42)
(2.43)
(2.44)
Before we get to this, however, it is interesting to consider the product of four
Ramond states:
M˜ 4(R1, R2, R3, R4).
Since this product would contribute to the NS part of the equations of motion (2.14), its
ghost number must be −2 and its picture number must be +1. In fact, this is the first
product where the ghost number is more negative than the picture number is positive. It
is easy to see that any product built from composing Q, m2 and ξ must satisfy
ghost number ≥ −picture number.
(2.46)
HJEP1(205)9
This inequality must be violated for products of four or more Ramond states. Therefore
such products potentially present an obstruction to our solution of the A∞ relations. To
see how this problem is avoided, consider the fourth A∞ relation,
hQ, M˜ 4i + hM˜3, M˜2i = 0,
acting on four Ramond states:
0 =
=
=
hQ, M˜ 4i + hM˜3, M˜2i
hQ, M˜ 4i + M˜3
+M˜ 2
˜
M3 ⊗
Keeping careful track of the NS and R inputs of M˜ 3, this can be further expanded
0 =
hQ, M˜ 4i + m2(µ 2 ⊗ I)(m2 ⊗
I
⊗ I) + µ 2(m2 ⊗ m2) + m2(µ 2 ⊗ I)(I ⊗ m2 ⊗ I)
+m2(I ⊗ µ 2)(I ⊗ m2 ⊗ I) − µ 2(m2 ⊗ m2) + m2(I ⊗ µ 2)(I ⊗
I
⊗ m2)
−m2(µ 2 ⊗ I)(m2 ⊗
−m2(I ⊗ µ 2)(m2 ⊗
I
I
⊗ I) − m2(µ 2 ⊗ I)(I ⊗ m2 ⊗ I)
⊗ I) − m2(I ⊗ µ 2)(I ⊗ m2 ⊗ I) R1 ⊗ R2 ⊗ R3 ⊗ R4,
= [Q, M˜ 4]R1 ⊗ R2 ⊗ R3 ⊗ R4.
Therefore we can simply choose
M˜ 4(R1, R2, R3, R4) = 0.
More generally, we claim that all products with four or more Ramond states can be set to
zero. Therefore the equations of motion will be cubic in the Ramond string field.
At first this seems somewhat strange. If the equations of motion have terms which are
cubic in the Ramond string field, cyclicity would naturally imply that they should have
terms that are quartic in the Ramond string field as well. This is a clear indication that
the equations of motion cannot be derived from an action. While this was expected, one
– 7 –
might still worry that quartic Ramond terms in the equations of motion are needed to get
the correct physics. For example, the quadratic Ramond term m2(ΨR, ΨR) is not implied
by A∞ relations or gauge invariance, but is required to incorporate the backreaction of the
R field on the NS field. The difference at quartic order is that there is no 4-product of
Ramond states at the relevant ghost and picture number which is nontrivial in the small
Hilbert space BRST cohomology. Therefore, any quartic term in the Ramond string field
can be removed by field redefinition. As a cross check on our equations of motion, it will
be shown in [17] that they imply the correct tree level amplitudes.
A key ingredient in constructing the equations of motion at higher order is to realize
that multi-string products can be characterized according to their Ramond number. The
Ramond number of a product is defined to be the number of Ramond inputs minus the
number of Ramond outputs required for the product to be nonzero:
Ramond number = (number of Ramond inputs)−(number of Ramond outputs). (2.51)
Generally, products do not have well-defined Ramond number. A product of Ramond
number N has the specific property that it will be nonzero only when multiplying N or
(N + 1) Ramond states (together with possibly other NS states), in which case it will
respectively produce an NS or R state. When Ramond number is defined, we will indicate
it by a vertical slash followed by an extra index attached to the product:
HJEP1(205)9
If the product is nonzero, its Ramond number must be restricted to the range
since the number of Ramond inputs cannot exceed the total number of inputs and the
number of Ramond outputs cannot exceed one. While generically multi-string products do
not possess well-defined Ramond number, they can always be decomposed into a sum of
products which do. To see this, consider the projector
Pn(N ) : H˜⊗n
→ H˜⊗n, Pn(N )2 = Pn(N ),
which selects elements of H˜⊗n which have N Ramond factors. Given an n-string product
bn, we can define the component at Ramond number N :
bn|N = P1(0) bn Pn(N ) + P1(1) bn Pn(N + 1).
Using the resolution of the identity,
bn|N .
✻✻
✡ Ramond number
✡ number of inputs
− 1 ≤ N ≤ n,
n
I⊗n = X Pn(N ),
N=0
– 8 –
(2.52)
(2.53)
(2.54)
(2.55)
(2.56)
Using the Jacobi identity, this implies that Ramond number is additive when taking
commutators of products:
bm|M , cn|N |M+N = bm|M , cn|N .
Finally, let us mention that the products in the equations of motion always carry even
Ramond number (odd Ramond number components vanish), since picture changing operators
do not mix NS and R sector states. Products of odd Ramond number will play a role once
we consider supersymmetry in section 6.
Now let us revisit the results of the previous subsection. The BRST operator has
with the indicated Ramond projection of M2 and m2. Note that M2|0 can be derived from
the Ramond number zero projection of the gauge 2-product µ 2:
(2.57)
(2.58)
(2.59)
(2.60)
(2.61)
(2.62)
(2.63)
(2.64)
it immediately follows that bn can be expressed as the sum of component products at all
Ramond numbers:
A comment about notation: generally, we use bn|N to denote an n-string product of Ramond
number N , but this does not necessarily mean that bn|N is derived from a product bn after
projection to Ramond number N . When we do mean this, it should be clear from context.
Consider a 1-string product R1 which acts as the identity on a Ramond state and as zero
on an NS state. A product has definite Ramond number N if and only if it satisfies
The composite 2-product M˜ 2 can be written as the sum of products at Ramond number
zero and two. Comparing with equations (2.24)–(2.27), we can apparently write
Also,
The composite 3-product M˜ 3 can likewise be written as the sum of products at Ramond
number zero and two:
The Ramond number zero piece corresponds to equations (2.33)–(2.36). The Ramond
number two piece m′3|2 is seemingly more complicated, as it must produce four distinct
expressions (2.41)–(2.44) depending on how it multiplies two or three Ramond states. To
derive the 3-string products, consider the third A∞ relation:
0 = hQ, M˜ 3i +
2
1 hM˜ 2, M˜ 2i ,
=
=
Q, M3|0 + Q, m′3|2 +
Q, M3|0 + Q, m′3|2 +
M2|0, M2|0 + M2|0, m2|2 +
m2|2, m2|2 ,
1
2
M2|0, M2|0 + M2|0, m2|2 .
This is equivalent to two independent equations at Ramond number 0 and 2:
The first equation can be solved following [1]. To review, the solution is
where µ 3|0 is the gauge 3-product,
1
4
⊗ ξ ⊗
and m3|0 is the bare 3-product,
With these definitions one can show that M3|0 is in the small Hilbert space:
1
2
1
2
1
2
m3|0 = m2|0, µ 2|0 .
η, M3|0 = 0.
0 =
0 =
Q, M3|0 +
M2|0, M2|0 ,
Q, m′3|2 + M2|0, m2|2 .
M3|0 =
Q, µ 3|0 + M2|0, µ 2|0 ,
(2.65)
(2.66)
(2.67)
(2.68)
(2.69)
(2.70)
(2.71)
(2.72)
(2.73)
The only difference between M3|0 and the 3-product of [1] is the explicit restriction to
Ramond number zero. To multiply more than one Ramond state we need m′3|2. By
inspection of the Ramond number two component of the A∞ relation, we can instantly
guess the solution
Happily, this simple formula reproduces all four equations (2.41)–(2.44) for 3-products of
two or more Ramond states. To check, for example, we can compute the NRR product:
M˜ 3(N1, R1, R2) =
m2|2, µ 2|0 N1 ⊗ R1 ⊗ R2,
=
m2|2(µ 2|0 ⊗
I + I ⊗ µ 2|0) − µ 2|0(m2|2 ⊗
I + I
= (m2(µ 2 ⊗ I) − µ 2(I ⊗ m2))N1 ⊗ R1 ⊗ R2,
= m2(µ 2(N1, R1), R2) − (−1)deg(N1)µ 2(N1, m2(R1, R2)),
⊗ m2|2) N1 ⊗ R1 ⊗ R2,
which reproduces (2.41).
With some experience from [1], it is not difficult to guess the general form of the
products to all orders. Let us give the answer first, and then we can prove it. The
composite (n + 2)-string product M˜ n+2 can be decomposed
M˜ n+2 = Mn+2|0 + mn+2|2.
′
(2.74)
As anticipated before, products with four or more Ramond states can be set to zero. In
addition we will need to introduce supplemental bare products and gauge products. In total
we have four kinds of product:
gauge products µ n+2|0:
degree even, picture# = n + 1,
Ramond#= 0,
HJEP1(205)9
products
bare products
Mn+1|0: degree odd,
picture# = n,
m′n+2|2:
mn+2|0:
which are determined recursively by the equations:
The recursive procedure for constructing the products, gauge products, and bare
products is illustrated in figure 1. Note that these equations are nearly the same as those
from [1] determining the NS open superstring field theory. The only major difference is
the appearance of a new set of products m′n+2|2 for multiplying 2 or 3 Ramond states.
To prove these formulas it is helpful to work with the coalgebra representation of A∞
algebras. See also [1, 2, 8, 9]. Consider the tensor algebra generated by taking formal sums
of tensor products of string fields:
T H = H
⊗0
⊕ H ⊕ H
⊕ H
⊗2
⊗3
We promote an m-string product into a linear operator on the tensor algebra called a
coderivation, which we denote with the same symbol in boldface:
ξmn+2|0 −
X mn+2|0 I⊗k
⊗ ξ ⊗
bm → bm.
Ramond#= 0
Ramond#= 0,
!
,
(2.75)
(2.76)
(2.77)
(2.78)
(2.79)
(2.80)
(2.81)
bm =
n−m
X I⊗k
k=0
⊗n<m it gives zero. A commutator of products [bm, cn] can also be promoted
to a coderivation.
This coderivation turns out to be equal to the commutator of the
coderivations bm and cn (graded with respect to degree):
[bm, cn] → [bm, cn].
If a sequence of degree odd multi-string products dn, n = 1, 2, 3, .. define an A∞ algebra,
the A∞ relations can be expressed
[d1, dn] + [d2, dn−1] + ... + [dn−1, d2] + [dn, d1] = 0,
The main utility of introducing coderivations is that they can be added to each other. For
example, given the products of an A∞ algebra we can define a coderivation
. . .
m6|0
constructing all products which appear the NS+R equations of motion using intermediate bare
products and gauge products.
By definition, the coderivation bm acts on the n-string component of the tensor algebra as
(2.82)
(2.83)
(2.85)
(2.86)
(2.87)
This single object encapsulates all products of the A∞ algebra. To get the product dn, we
act d on the n-string component of the tensor algebra and look at the result in the 1-string
component of the tensor algebra. Moreover, the A∞ relations (2.11) are equivalent to the
statement that d is nilpotent:
To recover the A∞ relations (2.11), we act this equation on n-string component of the
tensor algebra and look at the result in the 1-string component of the tensor algebra.
Let us promote the products, gauge products and bare products to coderivations:
d = X dn.
∞
n=0
[d, d] = 0.
Mn+1|0 → Mn+1|0,
mn+2|2 → m′n+2|2,
′
µ n+2|0 → μn+2|0,
mn+2|0 → mn+2|0,
and define generating functions:
Note that
M(t) =
m′(t) =
X tnMn+1|0,
X tnm′n+2|2,
m(t) =
X tnmn+2|0,
∞
∞
∞
∞
μ(t) =
X tnμn+2|0.
n=0
M(0) = Q,
m′(0) = m2|2,
Substituting the generating functions and expanding in powers of t, it is straightforward
to show that equations (2.75)–(2.78) are equivalent to:
Here the operation ξ◦ is defined by its action on an n-string product
ξ ◦ bn =
1
n + 1
n−1
k=0
ξbn + (−1)deg(bn)bn X I⊗k
⊗ ξ ⊗
I⊗n−1−k .
!
This operation defines a homotopy operator for the eta zero mode, in the sense that
(2.88)
(2.89)
(2.90)
(2.91)
(2.92)
(2.93)
(2.94)
(2.95)
(2.96)
(2.97)
(2.98)
(2.99)
(2.100)
(2.101)
(2.102)
where b is an arbitrary coderivation, and η is the coderivation representing the η zero mode.
Let A(t) or B(t) stand for M(t), m′(t) or m(t). We have
since Q, m2|0, m2|2 mutually anticommute. Now note that the differential equation (2.95)–
(2.97) imply
[η, ξ ◦ b] + ξ ◦ [η, b] = b,
[A(0), B(0)] = 0,
d
dt
[A(t), B(t)] = [[A(t), B(t)], μ(t)].
Since this equation is homogeneous in [A(t), B(t)], which vanishes at t = 0, we conclude
In other words, M(t), m′(t) and m(t) are nilpotent and mutually anticommute. Next
note that
since Q, m2|0 and m2|2 are in the small Hilbert space. Equations (2.95)–(2.98) together
with (2.103) imply
d
dt
= [[η, A(t)], μ(t)] − [A(t), m(t)] + [A(t), ξ ◦ [η, m(t)]],
= [[η, A(t)], μ(t)] + [A(t), ξ ◦ [η, m(t)]].
(2.105)
(2.106)
Suppose A(t) = m(t). Then this equation is homogeneous in [η, m(t)], and since this
vanishes at t = 0 we conclude [η, m(t)] = 0. Therefore
Since this equation is homogeneous in [η, A(t)], which vanishes at t = 0, we conclude
In other words, all products and bare products are in the small Hilbert space. Finally,
consider the coderivation representing the composite products in the equations of motion
d
dt
∞
n=0
[η, A(t)] = [[η, A(t)], μ(t)].
[η, A(t)] = 0.
˜
M ≡
X M˜n+1 = M(1) + m′(1).
hη, M˜i = 0,
h M˜, M˜i = 0.
The above results immediately imply that
The first equation says that the composite products are in the small Hilbert space, and
the second equation says that they satisfy A∞ relations. This completes the construction
of the Neveu-Schwarz and Ramond equations of motion for the open superstring based on
Witten’s open string star product.
3
Ramond sector of open superstring with stubs
In preparation for studying the closed superstring, in this section we provide a more
general construction of the open superstring equations of motion which does not require the
associativity of Witten’s open string star product. Specifically, we build the equations of
motion by inserting picture changing operators a set of elementary products at picture zero:
M1(0)
≡ Q,
M2(0),
M3(0),
M4(0), ... .
(3.1)
(2.103)
(2.104)
(2.107)
(2.108)
(2.109)
(2.110)
We assume that these products have odd degree, live in the small Hilbert space, and
satisfy A∞ relations. For example, we could define the elementary 2-string product M2(0)
by attaching “stubs” to Witten’s open string star product [2, 18] (hence the title of this
section). In the following we will need to introduce a multitude of products with different
picture and Ramond numbers. We denote the number of inputs, the picture number and
the Ramond number of a product as follows:
❄
Mn(p+)1|2r,
✻ ✻
☛ picture number
✡ Ramond number
✡ number of inputs
product Mn(0+)1.
The product Mn(0+)1|2r is defined to be the Ramond number 2r projection of elementary
The goal is to construct the NS+R equations of motion,
0 = QΦ˜ + M˜ 2 Φ˜, Φ˜
+ M˜ 3 Φ˜ , Φ˜, Φ˜
+ higher orders,
(3.2)
(3.3)
(3.4)
(3.5)
(3.6)
(3.7)
(3.8)
(3.9)
with the sum terminating when the Ramond number exceeds the number of inputs. The
picture number is correlated with the Ramond number so that the NS component of the
equations of motion will have picture −1 and the R component will have picture −1/2.
The composite 2-product is a sum of two terms:
M˜ 2 = M2(1)|0 + M2(0)|2.
The term with Ramond number 2 is the elementary 2-product M2(0) acting on two Ramond
states. The term with Ramond number zero will be defined analogously to (2.17):
where Φ˜ = ΦN + ΨR and M˜ n+1 are degree odd composite products which appropriately
multiply NS and R states. We require that the composite products live in the small Hilbert
space and satisfy A∞ relations. The composite products can be decomposed into a sum of
products of definite Ramond and picture number,
M˜ n+1 = Mn(n+)1|0 + Mn(n+−11)|2 + Mn(n+−12)|4 + ...,
We also have
where the gauge product µ (21)|0 is defined
M2(1)|0 =
XM2(0)|0 + M2(0)|0(X ⊗
So far, the equations of motion are precisely the same as in the previous section with the
replacement of Witten’s associative star product with M2(0).
At higher orders the non-associativity of M2(0) begins to play a significant role. The
higher order products at zero Ramond number were already described in [2], so let us
focus on products with nonzero Ramond number. The composite 3-product is the sum of
two terms:
The A∞ relations imply that M3(1)|2 must satisfy
Pulling a Q out of this equation we have
0 = hQ, M3(1)|2i + hM2(1)|0, M2(0)|2i .
0 = hQ, M3(1)|2 −
hM2(0)|2, µ (21)|0ii .
We conclude that M3(1)|2 satisfies
M3(1)|2 = hQ, µ (31)|2i + hM2(0)|2, µ (21)|0i .
Here we introduce a new gauge 3-product µ (31)|2 which is to be defined so that M3(1)
|2 is in
the small Hilbert space. In the previous section we could consistently set this product to
zero — indeed, in the previous section all gauge products had vanishing Ramond number.
If we postulate that
hη, µ (31)|2i = M3(0)|2,
product M3(1)|2 will be in the small Hilbert space:
where M3(0)
|2 is the Ramond number 2 projection of the elementary 3-product M3(0), the
η, M3(1)|2] = −
hQ, M3(0)|2i + hM2(0)|2, M2(0)|0i ,
= −
= 0,
hQ, M3(0)i +
2
1 hM2(0), M2(0)i
2
,
(3.10)
(3.11)
(3.12)
(3.13)
(3.14)
(3.15)
(3.16)
(3.17)
(3.18)
from M3(0)|2:
as follows from the A∞ relations for the elementary products. In summary, the product
M3(1)|2 can be constructed by climbing a “ladder” of products and gauge products, starting
M3(0)|2 = given,
1
4
µ (31)|2 =
ξM3(0)|2 − M3(0)|2(ξ ⊗
I
⊗
⊗ ξ ⊗
⊗
I
M3(1)|2 = hQ, µ (31)|2i + hM2(0)|2, µ (21)|0i ,
where the second step inverts (3.14). The Ramond number zero piece of the composite
3-product can be found by climbing a similar ladder, as described in [2].
Let us proceed to quartic order. The composite 4-product can be written as the sum
of three terms:
(3)
(2)
˜
In the previous section the Ramond number 4 contribution could be set to zero. Now it
will not vanish. Projecting the fourth A∞ relation,
onto Ramond number 4 implies
Plugging in the expression for M3 |2 gives
(1)
Q, M4(1)|4i + hh
⊗ ξ ⊗
I
⊗
⊗
I
⊗ ξ ⊗
⊗
I
⊗
I
Q, µ (41)|4i + h (0)
The last term is zero since [M2 |2, M2
|2 would be a 3-product at Ramond number four,
which must vanish identically. We therefore conclude that
(1)
M4 |4 =
h
Q, µ (41)|4i + h (0)
M2 |2, µ 3 |2 ,
(1) i
(1)
(1)
where the gauge 4-product µ 4 |4 is defined so that M4 |4 is in the small Hilbert space. If
h
(1) i
= −
= −
Therefore, the product M4 |4 does not vanish and can be constructed by climbing a
“ladder” of products and gauge products, starting from M4 |4:
(0)
(3.19)
(3.20)
(3.21)
(3.22)
(3.23)
(3.24)
(3.25)
(3.26)
With sightly more sophisticated use of the Jacobi identity, we can follow this procedure
again to construct M4(2)|2 by climbing the “ladder:”
M4(0)|2 = given,
µ (41)|2 =
ξM4(0)|2 − M4(0)|2(ξ ⊗
I
⊗
⊗
I + I
⊗ ξ ⊗
⊗
I + I
⊗ ξ ⊗
I + I
⊗
⊗
M4(1)|2 = hQ, µ (41)|2i + hM2(0)|2, µ (31)|0i + hM2(0)|0, µ (31)|2i + hM3(0)|2, µ (21)|0i ,
µ (42)|2 =
ξM4(1)|2 − M4(1)|2(ξ ⊗
I
⊗
⊗
I + I
⊗ ξ ⊗
⊗
I + I
⊗ ξ ⊗
I + I
⊗
⊗
1 hQ, µ (42)|2i + hM2(1)|0, µ (31)|2i + hM2(0)|2, µ (32)|0i + hM3(1)|2, µ (21)|0i .
The number of steps in the ladder increases with the number of picture changing operators
we need to insert in the product. The Ramond number zero piece of the composite
4product was already constructed in [2].
Having provided a few examples, let us describe the general construction. We introduce
a list of products and gauge products as follows:
2
5
1
5
2
where the integers N, p, r take the ranges
products :
gauge products :
M N(p+)1|2r,
µ (Np++12)|2r,
The restriction on the range of r is a little too generous — many products and gauge
products on this list vanish because the Ramond number exceeds the number of inputs.
(The first example is M3(0)|4 = 0). For convenience we include them on the list anyway.
The restriction on the picture number p comes from the fact that the equations of motion
require that products have picture number one less than the number of inputs minus half
the Ramond number. Since every step on the ladder to construct a product increases the
picture, we do not require products whose picture number exceeds this bound. We can
alternatively parameterize the list of products and gauge products by three integers d, p, r:
products :
gauge products :
Md(+p)p+r+1|2r,
µ (dp++p1+)r+2|2r,
d, p, r ≥ 0.
The integer d can be interpreted as the picture number deficit —specifically, the amount
of picture that is missing from a product (with a given Ramond number and number of
I
I
I
I
(3.29)
M5(3)|2 M5(3)|0
M3(2)|0
M4(2)|2 M4(2)|0
M5(2)|4
M5(2)|2 M5(2)|0
M2(1)|0
M3(1)|0 M3(1)|0
M4(1)|4
M4(1)|2 M4(1)|0
M5(1)|4
M5(1)|2 M5(1)|0
Q
M2(0)|2
M2(0)|0
M3(0)|2 M3(0)|0
M4(0)|4
M4(0)|2 M4(0)|0
M5(0)|4
M5(0)|2 M5(0)|0
HJEP1(205)9
in the equations of motion, and the elementary products at picture zero are boxed. At each order
in the string field, the products in the equations of motion are distinguished according to their
Ramond number, and at each Ramond number the product is derived by climbing a “ladder” of
products and gauge products starting from an elementary product at picture zero. The “ladders”
are shown above as a stack of products connected by vertical arrows. Each vertical arrow indicates
deriving a product with one higher unit of picture from a product of one lower unit of picture by
substituting (3.39) followed by (3.40). Climbing each ladder only requires the elementary products
and knowledge of products already constructed at lower orders in the string field.
inputs) before it can appear in the equations of motion. The products, gauge products,
and bare products introduced in the previous section correspond to a subset of this list:
Mn+1|0 = Mn(n+)1|0,
mn+2|2 = Mn(n+)2|2,
′
If we assume the associativity of Witten’s star product, the remaining products and gauge
products can be set to zero. Extrapolating from the first few orders, one can guess that
the products and gauge products are determined recursively by a pair of equations,
µ (dp++p1+)r+2|2r =
d + 1
d + p + r + 3
ξMd(+p)p+r+2|2r − Md(+p)p+r+2|2r
I⊗k
⊗ ξ ⊗
Id+p+r+1−k ,
d+p+r+1
X
k=0
Md(+p+p+1)r+2|2r =
1
p
X
d
X
r
p + 1 p′=0 d′=0 r′=0
X hMd(′p+′)p′+r′+1|2r′ , µ (dp+−pp+′ +r−1)d′−p′−r′+2|2(r−r′)i .
The procedure for constructing the products and gauge products from these equations is
illustrated in figure 2.
Now we need to show that that the composite products M˜ n+1 which appear in the
equations of motion are in the small Hilbert space and satisfy A∞ relations. We do this
(3.38)
!
(3.39)
(3.40)
√ I
2
dz
|z|=1 2πi Θ~s e−φ/2(z)ǫ~s,
where the index on s1 indicates that this operator is a 1-string product. The object ǫ~s is
the supersymmetry parameter — a constant Grassmann odd spinor. To keep the notation
simple, we leave the dependence of s1 on the supersymmetry parameter implicit.
We
make a GSO(+) projection in both NS and R sectors, so the supersymmetry parameter
must have positive chirality. Then, following the Grassmann assignments in section 2, the
supersymmetry operator s1 will be Grassmann even (and degree even).7 It is natural to
identify the linearized supersymmetry transformation of the NS field as
Now that we have equations of motion including all NS and R sectors, it is interesting
to understand in what sense they are supersymmetric. Here we will limit our discussion
to the open superstring using Witten’s associative star product, leaving generalizations to
future work. For heterotic and type II closed superstrings, spacetime supersymmetry can
be described in a different way as part of the gauge symmetry at the interacting level [7].
For other discussions of supersymmetry in open string field theory, see [
4, 15, 20
].
We will work with open string field theory on a BPS D-brane with sixteen
supersymmetries. As described in [3], the simplest way to realize a supersymmetry transformation
is by integrating the zero-momentum fermion vertex in the −1/2 picture:
(6.1)
(6.2)
(6.3)
(6.4)
(6.5)
(6.6)
δΦN = s1ΨR.
Likewise, the linearized supersymmetry transformation of the R field should be proportional
to the NS field. However, to get the pictures to line up we need a supersymmetry operator
of picture +1/2. This can be defined by integrating the zero momentum fermion vertex at
picture +1/2:
S1 ≡
√ I
2
dz
i∂Xµ Θ~s′ eφ/2(z)Γ~µs′,~s + bηΘ~s e3φ/2(z) ǫ~s,
where Γ~µs,~s′ are 10 dimensional gamma matrices. This operator is also Grassmann even.
The linearized supersymmetry transformation of the Ramond field will therefore be
We can write both supersymmetry transformations in a single equation using the composite
string field Φ˜ = ΦN + ΨR
where
δΨR = S1ΦN.
δΦ˜ = S˜1Φ˜ ,
S˜1 = S1|−1 + s1|1,
with the indicated Ramond projections of S1 and s1. Now the question is how this
supersymmetry generalizes to the nonlinear theory.
7The factor √2 is inserted to obtain the canonical normalization of the supersymmetry algebra.
Before we get to this, however, let us mention a few essential properties of s1 and S1.
Both operators are in the small Hilbert space and BRST invariant
This is in fact necessary for the supersymmetry transformation to be a symmetry of the
linearized equations of motion. Moreover, since both operators are zero modes of weight
one primary fields, they are derivations of Witten’s open string star product:
It will also be useful to introduce a Grassmann odd supersymmetry operator in the large
and is also a derivation of the star product. Note the relationship between the
supersymmetry operators S1, σ1 and s1 is somewhat analogous to the relation between the products
M2, µ 2, and m2.
6.1
Perturbative construction of supersymmetry transformation
In this subsection we give a perturbative construction of the supersymmetry
transformation. It turns out that the final form of the supersymmetry transformation is easiest
to understand in a different set of field variables, described in the next subsection, and
some readers may wish to skip ahead. (See, in particular, equation (6.90)) The
following derivation, however, has the advantage that it likely generalizes to supersymmetry
transformations in other forms of superstring field theory.
Our task is to find a sequence of degree even products in the small Hilbert space,
S˜1,
S˜2,
S˜3, ... ,
so that the supersymmetry transformation can be written
δΦ˜ = S˜1Φ˜ + S˜2 Φ˜ , Φ˜
+ S˜3 Φ˜ , Φ˜, Φ˜
Let us explain what it means for this transformation to be a symmetry. Consider a pair
of coderivations M˜ and S˜ representing the products in the equations of motion and
supersymmetry transformation:
˜
M ≡
S˜ ≡
∞
n=0
∞
X M˜ n+1,
X S˜n+1,
n=0
(6.7)
(6.8)
(6.10)
(6.11)
(6.12)
(6.13)
(6.14)
(6.15)
(6.16)
where M˜ n+1 and related products in the equations of motion are taken from section 2. We
also introduce a group-like element8
1
1 − Φ
˜ = 1T H˜ + Φ˜ + Φ˜ ⊗ Φ˜ + Φ˜ ⊗ Φ˜ ⊗ Φ˜ + ... .
The equations of motion and the supersymmetry transformation can be expressed in
the form:
˜
M
1
1 − Φ
˜ = 0,
δ
1
1 − Φ
˜ = S˜
1
1 − Φ
˜
.
We claim that the products S˜n+1 generate a symmetry if
hM˜ , S˜i = 0.
(6.17)
(6.18)
(6.19)
(6.20)
(6.22)
(6.23)
(6.24)
(6.25)
(6.26)
To see this, suppose Φ˜ is a solution to the equations of motion. Then calculate the equations
of motion on the transformed string field Φ˜ + δΦ˜ :
˜
M
1
1 − Φ
˜ + δ
1
1 − Φ
˜
= M˜S˜
1
1 − Φ
˜ = S˜ M˜
1
1 − Φ
Therefore the transformation is a symmetry. Note that the condition [M˜ , S˜] = 0 is
somewhat stronger than the statement that the transformation maps solutions into solutions.
It places a nontrivial condition on the off-shell form of the supersymmetry transformation.
In fact, (6.19) is the nearest we can come to the statement that the transformation is a
symmetry of the action — all that is missing is a symplectic structure which would allow
us to define an action and impose cyclicity.
Equation (6.19) implies that the products S˜n+1 satisfy a hierarchy of identities:
hM˜ 1, S˜n+1i + hM˜ 2, S˜ni + ... + hM˜ n+1, S˜1i = 0,
n = 0, 1, 2, ... ,
(6.21)
where M˜ 1 = Q. We have already discussed S˜1 which appears in (6.6), and it provides a
solution to
The next step is to derive the 2-product S˜2 from the identity
hQ, S˜1i = 0.
hQ, S˜2i + hM˜ 2, S˜1i = 0.
S˜2 = S2|−1 + s2|1.
Since S˜2 is part of a supersymmetry transformation, it can be split into a sum of products
with odd Ramond number:
Equation (6.23) breaks up into two independent equations at Ramond number −1 and 1:
0 =
0 =
Q, S2|−1] + M2|0, S1|−1 ,
Q, s2|1 + M2|0, s1|1 + m2|2, S1|−1 .
8See [9] for more detailed discussion of coderivations, cohomomorphisms, and group-like elements.
It turns out to be convenient to introduce separate symbols for these two terms:
Check that this is in the small Hilbert space:
s2|1 = s1|1, µ 2|0 + m2|2, σ1|−1 .
s(I)
2 |1 ≡ s1|1, µ 2|0 ,
s(II)
2 |1 ≡
m2|2, σ1|−1 .
where we used conservation of Ramond number and the fact that s1 is a derivation of m2.
This completes the definition of the 2-product S˜2 in the supersymmetry transformation.
To establish a pattern it is helpful to continue on to the 3-product S˜3, which must satisfy
To solve the first equation, we can pull a Q out of either M2 or S1. The solution we prefer
is to pull a Q out of M2, obtaining
We can check that this is in the small Hilbert space:
S2|−1 = S1|−1, µ 2|0 .
η, S2|−1 = S1|−1, m2|0 = [S1, m2]|−1 = 0,
where we used conservation of Ramond number and the fact that S1 is a derivation of m2.
To solve for the Ramond number 1 component of S˜2, we pull a Q in the natural way out
of the two terms in (6.26)
(6.27)
(6.28)
(6.29)
(6.30)
(6.31)
(6.32)
(6.33)
(6.34)
(6.35)
(6.36)
+ m2|2, m2|2, σ1|−1
+
m2|2, µ 2|0 , s1|1 ,
Let us look at the component of this equation at Ramond number 3:
hQ, S˜3i + hM˜ 2, S˜2i + hM˜ 3, S˜1i = 0.
hQ, S˜3|3i + [m2|2, s2|1] + m′3|2, s1|1 = 0.
Substituting (6.29) for s2|1 and (2.72) for m′3|2,
0 = hQ, S˜3|3i + m2|2, s1|1, µ 2|0
= hQ, S˜3|3i ,
where the additional terms either cancel or vanish identically because the Ramond number
exceeds the number of inputs. From this we conclude that the Ramond number 3
component of S˜3 can be set to zero. Therefore S˜3 must be a sum of products at Ramond number
−1 and 1
S˜3 = S3|−1 + s3|1,
M2|0, S2|−1
M2|0, S2|−1
+
+
1
2
M3|0, S1|−1 ,
Q, µ 3|0, S1|−1
M2|0, S2|−1
Q, µ 3|0
+
M2|0, µ 2|0 , S1|−1 ,
M2|0, S2|−1
+
M2|0, S1|−1 , µ 2|0 ,
just like S˜1 and S˜2. Let’s look at the Ramond number −1 component of the identity (6.33):
0 =
=
=
=
=
Q, S3|−1
Q, S3|−1
Q, S3|−1
1
− 2
Q, S3|−1
Q, S3|−1 − 2
+
+
+
+
1
2
1
2
1
This suggests we identify
Check that this is in the small Hilbert space:
S3|−1 =
S1|−1, µ 3|0
+ S2|−1, µ 2|0
.
η, S3|−1
S1|−1, m3|0
+ S2|−1, m2|0
Q, µ 3|0, S1|−1
M2|0, S2|−1
+
Q, S2|−1], µ 2|0 ,
S1|−1, µ 3|0
+ S2|−1, µ 2|0
.
1
2
,
,
=
=
=
= 0.
1
2
1
2
1
2
1
2
1
2
1
2
1
2
+
+
1
2
+
h
+
m2|2, S1|−1, µ 2|0
+
m2|2, µ 2|0 , S1|−1 ,
Q, s3|1
h
M2|0, s2 |1
(I) i
+
Q, µ 3|0, s1|1
1 h
− 2
M2|0, s2 |1
(I) i
+
1
2
M2|0, s1|1 , µ 2|0 ,
m2|2, S1|−1 , µ 2|0 ,
1 h
2
(I) i
M2|0, s2 |1 − 2
1 hh
(I) i
Q, s2 |1 , µ 2|0 ,
i
Finally, let’s look at the Ramond number 1 component of (6.33):
M2|0, s2|1
m2|2, S2|−1
M3|0, s1|1
+
′
m3|2, S1|−1 ,
(6.37)
(6.38)
(6.39)
(6.40)
(6.41)
S1|−1, m2|0, µ 2|0
+ S2|−1, m2|0
,
m2|0, S2|−1
+ S2|−1, m2|0
1
2
(II) i
M3|0, s1|1
M2|0, s2 |1 + m2|2, S2|−1 + m3|2, S1|−1 ,
′
Q, µ 3|0 , s1|1
+
M2|0, µ 2|0 , s1|1 ,
0 =
Q, s3|1
Q, s3|1
Q, s3|1
h
h
M2|0, s2 |1 +
(I) i
M2|0, s2 |1 +
(I) i
+
+
h
h
M2|0, s2 |1
(II) i
M2|0, s2 |1
(II) i
1
2
(II)
1
Q, s3|1 − 2
Q, s3|1
Q, µ 3|0, s1|1
+
+ M2|0, s2 |1 −
(II)
Q, s2 |1 , µ 2|0 ,
s1|1, µ 3|0
+ s
h (I)
2 |1, µ 2|0
i
h (II)
s
2
|1, µ 2|0
i
.
Therefore we can identify
(I)
with
s(3I)|1 =
1
2
s(II)
Thus we see a pattern where the supersymmetry product at Ramond number 1 breaks
up into a product denoted with (I) and a product denoted with (II), each determined by
independent recursions. Let us check that s3|1 is in the small Hilbert space:
=
s1|1, m2|0 , µ 2|0 +
m2|2, σ1|−1 , m2|0 −
m2|2, s1|−1 , µ 2|0 ,
= [s1, m2]|1, µ 2|0 − 2
= 0.
1
[m2, m2]|2, σ1|−1 + m2|2, [σ1, m2]|−1 ,
This completes the definition of the 3-product S˜3 in the supersymmetry transformation.
Now we can guess the form of the supersymmetry transformation at higher orders.
The (n + 1)st product can be written
Components with higher Ramond number can be set to zero. In addition, sn+2|1 can be
written as a sum
The products are determined recursively by the equations,
S˜n+1 = Sn+1|−1 + sn+1|1.
sn+2|1 = s(nI+)2|1 + s(nI+I)2|1.
Sn+2|−1 =
s(nI+)2|1 =
s(nI+I)3|1 =
1
n + 1
n + 1
1
1
n + 1
n
X
k=0
n
k=0
n
k=0
X h (I)
X h (II)
Sk+1|−1, µ n−k+2|0 ,
sk+1|1, µ n−k+2|0 ,
sk+2|1, µ n−k+2|0 ,
i
i
starting from S1|−1 in (6.3), s(I)
1 |1 = s1|1 in (6.1), and s(II)
2 |1 =
m2|2, σ1|−1 . Now let’s
prove that these products have the required properties.
We promote the products to coderivations and define generating functions
S(t) =
X tnSn+1|−1,
s(I)(t) =
X tns(I)
n+1|1,
s(II)(t) =
X tns(II)
n+2|1.
∞
n=0
∞
n=0
∞
n=0
(6.42)
(6.43)
(6.44)
(6.45)
(6.46)
(6.47)
(6.48)
(6.49)
(6.50)
Using the generating function of the gauge products (2.91), the recursive equations (6.47)–
(6.49) can be reexpressed
d
dt
dt
S(t) = [S(t), μ(t)],
d s(I)(t) = hs(I)(t), μ(t)i ,
d s(II)(t) = hs(II)(t), μ(t)i ,
The generating functions for the products in the equations of motion and supersymmetry
transformation take the form
HJEP1(205)9
M˜ (t) = M(t) + tm′(t),
S˜(t) = S(t) + s(I)(t) + ts(II)(t).
The differential equations for the generating functions imply a set of equations:
(6.51)
(6.52)
(6.53)
(6.54)
(6.55)
(6.56)
(6.57)
(6.58)
(6.59)
(6.60)
(6.61)
(6.62)
(6.63)
dt
d hM˜ (t), S˜(t)i = hhM˜ (t), S˜(t)i, μ(t)i+hm′(t), S˜(t)i+h M˜(t), s(II)(t)i ,
d hm′(t), S˜(t)i+h M˜(t), s(II)(t)i =hhm′(t), S˜(t)i+hM˜ (t), s(II)(t)i, μ(t)i+2hm′(t), s(II)(t)i,
dt
d hm′(t), s(II)(t)i = hhm′(t), s(II)(t)i, μ(t)i .
Start with the last equation. Note that [m′(0), s(II)(0)] vanishes because
m2|2, s(2II)|1i = [m2|2, [m2|2, σ1|−1]] = 0.
hm′(t), s(II)(t)i = 0.
[m2|2, S1|−1] + [Q, [m2|2, σ1|−1]] = 0.
hm′(t), S˜(t)i + h M˜(t), s(II)(t)i = 0.
h M˜(t), S˜(t)i = 0.
hM˜ , S˜i = 0,
The last equation then implies
Therefore (6.56) implies
Therefore Setting t = 1, we have in particular
The next to last equation (6.56) is now homogeneous in [m′(t), S˜(t)] + [ M˜(t), s(II)(t)]. We
know that [m′(0), S˜(0)] + [ M˜(0), s(II)(0)] = 0 because
Finally, consider the first equation (6.55), which is now homogeneous in [M˜ (t), S˜(t)]. The
commutator [ M˜(0), S˜(0)] vanishes since the supersymmetry operators are BRST invariant.
Start with the last equation. Note that [m(0), s(II)(0)] vanishes because
m2|0, s(2II)|1i = [m2|0, [m2|2, σ1|−1]] =
1
2
The last equation then implies
which proves that the products S˜n+1 generate a symmetry. To prove that the
transformation acts in the small Hilbert space, consider the following set of equations:
d hS˜(t), m(t)i + hη, s(II)(t)i
dt
d hη, S˜(t)i = hhη, S˜(t)i , μ(t)i + hS˜(t), m(t)i + hη, s(II)(t)i , (6.64)
= hhS˜(t), m(t)i + hη, s(II)(t)i , μ(t)i + 2 hs(II)(t), m(t)i ,
dt
d hs(II)(t), m(t)i = hhs(II)(t), m(t)i , μ(t)i .
The next to last equation (6.65) is now homogeneous in [S˜(t), m(t)] + [η, s(II)(t)]. We know
that [S˜(0), m(0)] + [η, s(II)(0)] = 0 because
[S1|−1 + s1|1, m2|0] + [η, [m2|2, σ1|−1] = [S1, m2]|0 + [s1, m2]|1 = 0.
Therefore (6.65) implies
hS˜(t), m(t)i + hη, s(II)(t)i = 0.
Finally, consider the first equation (6.64), which is now homogeneous in [η, S˜(t)]. Since
[η, S˜(0)] = 0 we conclude
hη, S˜(t)i = 0,
so the products S˜n+1 are in the small Hilbert space. This completes the construction of
the supersymmetry transformation.
6.2
Supersymmetry in Berkovits’ superstring field theory
Following [8, 9], our analysis should imply a natural form for the supersymmetry
transformation of the NS+R equations of motion in Berkovits’ open superstring field
theory [4, 21, 22]. To derive this supersymmetry transformation, as an intermediate step
we must perform a field redefinition from our string field Φ˜ to a new string field ϕ˜ whose
equations of motion and constraint are only polynomial [9]. The field redefinition is defined
by the cohomomorphism
Gˆ (t) = P exp
dt′ μ(t′)
,
Z t
0
where the path ordered exponential is defined in sequence of increasing t′. In particular,
we can express the relation between Φ˜ and ϕ˜ as
1
˜
(6.65)
(6.66)
(6.67)
(6.68)
(6.69)
(6.70)
(6.71)
(6.72)
(6.73)
where Gˆ ≡ Gˆ(1). Equivalently, we can write
ϕ˜ = G hΦ˜ i ,
(6.74)
where G[Φ˜ ] is given by
G hΦ˜ i
≡ π1Gˆ
1
1 − Φ
,
+higher orders,
1
2
= Φ˜ + µ 2|0 Φ˜ , Φ˜ +
µ 3|0 Φ˜ , Φ˜, Φ˜ +µ 2|0 µ 2|0 Φ˜ , Φ˜ , Φ˜ +µ 2|0 Φ˜ , µ 2|0 Φ˜, Φ˜
and π1 denotes the projection onto the 1-string component of the tensor algebra. Note
that this field redefinition is only defined in the large Hilbert space — it is not compatible
with the small Hilbert space constraint on the string field. In [9], such a field redefinition
was called an improper field redefinition. Once we have transformed from Φ˜ to ϕ˜, we must
perform a second transformation to map from ϕ˜ into Berkovits’ superstring field theory.
However, it will be useful to work out the supersymmetry transformation for ϕ˜ first.
Let us find the equations of motion and constraint for ϕ˜. The generating functions
given in (2.88)–(2.90) can be written
(6.75)
(6.76)
(6.77)
(6.78)
(6.79)
(6.80)
(6.81)
(6.82)
(6.83)
since these expressions give a solution to the differential equations (2.95)–(2.97) with the
correct initial conditions. This implies that the composite products of the equations of
motion can be expressed:
M˜ = M(1) + m′(1) = Gˆ−1 Q + m2|2 Gˆ.
Moreover, it follows from the computation in [8] that
The equations of motion and constraint for Φ˜ can be expressed together in the form
M(t) = Gˆ(t)−1Q Gˆ(t),
m′(t) = Gˆ(t)−1m2|2 Gˆ(t),
m(t) = Gˆ(t)−1m2|0 Gˆ(t),
η = Gˆ −1 η − m2|0 Gˆ.
M˜ − η
1
1 − Φ
Q − η + m2 1 − ϕ˜
= 0.
1
(Q − η)ϕ˜ + ϕ˜ ∗ ϕ˜ = 0.
Multiplying by Gˆ , using (6.79) and (6.80), and replacing Φ˜ with the new string field ϕ˜,
we find:
Simons-like equations
After projecting onto the 1-string component of the tensor algebra, we find familiar
ChernThe field redefinition implies that we can break ϕ˜ into NS and Ramond components
where the NS field ϕN is Grassmann odd and has picture −1 and the R field ψR is
Grassmann odd and has picture −1/2. Then (6.83) is equivalent to four equations:
ϕ˜ = ϕN + ψR,
0 = QψR,
0 = QϕN + ψR ∗ ψR,
where the commutator of string fields is computed with the star product and is graded with
respect to Grassmann parity. The first two equations should be interpreted as equations of
motion, since they are the image of the equations of motion for Φ˜ after the field redefinition.
The second two equations should be interpreted as equations of constraint, since they are
the image of the small Hilbert space constraint on Φ˜ after the field redefinition. Note
that the equations of motion for ϕ˜ are quadratic the the Ramond string field, while the
equations of motion for Φ˜ are cubic in the Ramond field. This happens because the field
redefinition (6.75) is at most linear in the Ramond field, and, in conjunction with the
quadratic term in the Ramond field in the equations of motion for ϕ˜, this produces cubic
terms in the Ramond field in the equations of motion for Φ˜ .
Now let’s derive the new form of the supersymmetry transformation. The differential
equations (6.51)–(6.53) imply that the supersymmetry transformation can be written in
the form
where
S˜ = Gˆ −1˜s Gˆ.
˜s ≡ S1|−1 + s1|1 + [σ1|−1, m2|2].
δϕN = s1ψR + [ψR, σ1φN],
δψR = S1ϕN + σ1(ψR ∗ ψR).
(6.84)
(6.85)
(6.86)
(6.87)
(6.88)
(6.89)
(6.90)
(6.91)
(6.92)
(6.93)
(6.94)
This implies that the supersymmetry transformation of ϕ˜ can be expressed
δ
1
= ˜s
1
.
Explicitly in terms of the NS and R components,
We can check that ˜s is a symmetry of the equations of motion:
˜s, Q + m2|2 = S1|−1 + s1|1 + [σ1|−1, m2|2], Q + m2|2 ,
= − S1|−1, m2|2 + S1|−1, m2|2 + s1|1, m2|2 +
σ
1|−1, m2|2 , m2|2 ,
= 0.
The terms either cancel or vanish because the Ramond number exceeds the number of
inputs. We can also check that ˜s preserves the constraints:
˜s, η − m2|0 = S1|−1 + s1|1 + σ
1|−1, m2|2 , η − m2|0 ,
= − s1|−1, m2|2 − S1|−1, m2|0 − s1|1, m2|0 −
1|−1, m2|2 , m2|0 ,
= − s1, m2 |1 − S1, m2 |−1 − 2
= 0.
1 σ
σ1, m2 |−1, m2|2 ,
This vanishes since s1, S1 and σ1 are derivations of the star product and because the star
product is associative. Conjugating by Gˆ, this provides an alternative proof that the
supersymmetry transformation S˜ preserves the equations of motion for Φ˜ and is consistent
with the small Hilbert space constraint.
Now that we have the supersymmetry transformation for ϕ˜, it is straightforward to
translate into Berkovits’ superstring field theory. Berkovits’ superstring field theory uses a
Grassmann even NS field Φ in the large Hilbert space and at picture and ghost number 0,
ghost number 1 [4]. These are related to ϕN and ψR through9
and a Grassmann odd R string field Ω in the small Hilbert space at picture −1/2 and at
ϕN = (ηeΦ)e−Φ,
ψR = eΦΩe−Φ.
With this identification, the constraints (6.87) and (6.88) reduce to identities, and the
equations of motion (6.85) and (6.86) translate to
η e−ΦQeΦ
= Ω2,
QΩ + e−ΦQeΦ, Ω = 0.
The Berkovits equations of motion are invariant under infinitesimal gauge invariances
δeΦ = eΦ(v + [Ω, Λ]) + ueΦ,
δΩ = [Ω, v] + η QΛ + e−ΦQeΦ, Λ .
The NS gauge parameters u, v are Grassmann even and carry ghost and picture number
zero; u is BRST closed while v is in the small Hilbert space. The Ramond gauge parameter
Λ is Grassmann odd and in the large Hilbert space, and carries ghost number −1 and picture
number +1/2. The supersymmetry transformation is
δΩ = η e−ΦS1eΦ ,
(δeΦ)e−Φ = σ1 eΦΩe−Φ .
(6.100)
Plugging into (6.96) gives the supersymmetry transformation ˜s for the string field ϕ˜, and
further mapping gives the original supersymmetry transformation S˜ for the string field Φ˜.
One interesting potential application of these results is to check whether or not classical
solutions in open superstring field theory are supersymmetric. However, the
supersymmetry transformation we have derived is not necessarily the most convenient for this purpose.
Specifically, our supersymmetry transformation is natural for an NS string field at picture
9The string field we call Ω here is called iΩ in [4].
(6.95)
(6.96)
(6.97)
(6.98)
(6.99)
−1, but usually analytic solutions in Berkovits superstring field theory are closely related
to an NS string field at picture 0.10 For this reason, we will find it convenient to consider
a “dual” formulation of open superstring field theory using the string field
where ϕ∗N is a Grassmann odd NS field at picture 0 and ghost number 1 and ψR∗ is a
Grassmann odd Ramond field at picture −1/2 and ghost number 1. The “dual” fields
satisfy the equations
ϕ˜∗ = ϕ∗N + ψR,
∗
0 = Qϕ∗N + ϕN ∗ ϕN,
∗ ∗
When ψR∗ = 0, these are the equations of motion of the NS sector of the modified cubic
superstring field theory [24, 25]. Note that the new field equations for ϕ˜∗ are identical to
the previous field equations for ϕ˜ after the “duality” map:
Q ↔ −η,
picture ↔ −(1 + picture).
We indicate the “dual” fields with a star ∗. We have
0 =
Q − η + m2 1 − ϕ˜∗
.
1
Taking the “dual” of the supersymmetry transformation for ϕ˜ we arrive at a new
supersymmetry transformation for ϕ˜∗
where
1
δ
1 − ϕ˜∗ = ˜s∗
1
1 − ϕ˜∗
,
˜s∗ ≡ s1|−1 + S1|1 −
Relating this “dual” formulation to Berkovits’ superstring field theory requires a slight
change of perspective on the Ramond sector. In the previous paragraph we described the
Ramond sector of the Berkovits theory using a string field Ω in the small Hilbert space.
Now it will be more natural to use a string field Ω∗ which is not in the small Hilbert space
but is BRST closed. The Ramond field Ω∗ is Grassmann odd and carries ghost number 1
and picture −1/2. On shell, the Ramond fields Ω and Ω∗ are related by
Ω = e−ΦΩ∗eΦ, (on shell).
This relation is not meaningful off-shell since it is not consistent with the assumption that
Ω∗ is BRST closed while Ω is in the small Hilbert space. The “dual” fields ϕ∗N and ψR∗ can
be then be related to Berkovits’ superstring field theory through
ϕ∗N = e−ΦQeΦ,
ψR∗ = e−ΦΩ∗eΦ
10Often it is possible to “dualize” analytic solutions into a form which is natural at picture −1 [23].
(6.101)
(6.102)
(6.103)
(6.104)
(6.105)
(6.106)
(6.107)
(6.108)
(6.109)
(6.110)
(6.111)
In this way, (6.102) and (6.103) reduce to identities while (6.104) and (6.105) are equivalent
to the equations of motion of Berkovits’ superstring field theory expressed in the form
Q (ηeΦ)e−Φ
+ (Ω∗)2 = 0,
ηΩ∗ − (ηeΦ)e−Φ, Ω∗ = 0.
In these variables, the “dual” supersymmetry transformation of the Berkovits theory takes
the form
in (6.109).
HJEP1(205)9
δ∗Ω∗ = Q (s1eΦ)e−Φ ,
e−Φδ∗eΦ = σ1 e−ΦΩ∗eΦ .
Plugging into (6.111) produces to the supersymmetry transformation of ϕ∗N and ψR∗ given
Note that the supersymmetry transformation δ and the “dual” supersymmetry
transformation δ∗ assume different off-shell degrees of freedom, and therefore really apply to
different string field theories. On shell, however, they should be equivalent. In fact, one
can show that they are equal up to an infinitesimal gauge transformation:
δ e
∗ Φ = δeΦ + eΦ[Ω, Λ],
δ∗Ω = δΩ + η QΛ + e−ΦQeΦ, Λ ,
(on shell),
(on shell),
where the Ramond gauge parameter is given by
Λ = −e−Φσ1eΦ.
This shows that the two supersymmetry transformations are physically equivalent.
We are now ready to discuss supersymmetry of classical solutions. On a BPS D-brane
there are not many classical solutions whose existence is well-established. The only known
solutions represent marginal deformations of the reference boundary superconformal field
theory. We will consider specifically a transverse displacement of the reference D-brane,
for which it is sufficient to consider the solution [26, 27]
eΦ = 1 + F X
1
1 + B 1−F 2 J F,
in the context of Berkovits’ superstring field theory. As is appropriate for a classical
background, the Ramond string field is zero. We use the algebraic notation for analytic
solutions introduced in [28], following the conventions of [29].
The string field F is a
function of K which we can take to be the square root of the SL(2, R) vacuum. The fields
X and J are defined
X ≡ λ ξe−φcψ+,
J ≡ QX = λ i√2c∂X+ + γψ+ ,
where λ is the marginal parameter describing the displacement of the D-brane, and the
index + indicates a lightcone direction whose spatial component is transverse to the
Dbrane.11 The Berkovits solution implies an expression for the NS field ϕN at picture −1
11Technically, this solution translates the D-brane and switches on a timelike Wilson line of corresponding
magnitude. However, as described in [30], the timelike Wilson line is physically trivial. An alternative
approach would use the solution of [31], which does not require excitation of spurious primaries in the
timelike factor of the X
µ and ψµ BCFTs.
(6.112)
(6.113)
(6.114)
(6.115)
(6.116)
(6.117)
(6.118)
and the “dual” NS field ϕ∗N at picture zero:
ϕN = (ηeΦ)e−Φ = F (ηX)
ϕ∗N = e−ΦQeΦ = F J
1
1 + B 1−F 2 J F.
1
1 + B 1−F 2 J + F 2X
δ∗Ω∗ = Q
s1eΦ e−Φ ,
= eΦs1 e−ΦQeΦ e−Φ,
= eΦ (s1ϕ∗N) e−Φ.
Note that the picture zero field is simpler, which is why the “dual” supersymmetry
transformation δ∗ is more convenient. The Berkovits solution also implies a solution Φ˜ of the
original equations of motion in the small Hilbert space, which up to second order in the
marginal operator takes the form
HJEP1(205)9
ΦN = F (ηX)F − F (ηX)B
1 − F 2
K
J F − F (ηX)F 2XF − µ 2 F (ηX)F, F (ηX)F
Translation of a D-brane does not break any supersymmetries. Therefore we should be able
to show that this solution is supersymmetric. The only nontrivial supersymmetry
transformation to compute is for the fermion. Using the “dual” supersymmetry transformation
we find
(6.119)
(6.120)
(6.121)
(6.122)
(6.123)
(6.124)
Thus we need to compute s1ϕ˜∗N. This is zero because s1 is a derivation of the star product
and because
s1K = 0,
s1B = 0,
s1J = 0.
The first equation follows because s1 is the zero mode of a weight 1 primary, and the last
two equations follow because there are no poles in the OPE between the −1/2 picture
fermion vertex and either b or J . Therefore the solution is supersymmetric. However, the
solution is not identically invariant under the supersymmetry transformation δ which is
natural at picture −1. From (6.115) one finds that the supersymmetry transformation of
the fermion is
δΩ = η QΛ + e−ΦQeΦ, Λ .
with Λ = −e−Φσ1eΦ. While σ1 annihilates K and B, it does not annihilate J , and the
Ramond gauge parameter Λ is not zero. Translating back into our original equations of
motion, this means that the supersymmetry transformation S˜ constructed in the previous
section leaves the solution Φ˜ in (6.121) invariant up to a gauge transformation.
While it is interesting to know the supersymmetry transformation on a BPS D-brane, it
is not really enough to give a full account of the role of supersymmetry in open superstring
field theory. This is because at least half of the supersymmetries are spontaneously broken
by the reference boundary superconformal field theory, and broken supersymmetries can
in principle be restored upon expanding around a nontrivial classical solution. The most
dramatic example of this is tachyon condensation on a non-BPS D-brane [32], where all 32
supersymmetries are restored at the tachyon vacuum. This can be seen as follows. While
we do not presently know the explicit form of broken supersymmetry transformations in
superstring field theory, in any case they will generate at most an infinitesimal deformation
of the tachyon vacuum. But since the kinetic operator around the tachyon vacuum has
no cohomology, any infinitesimal deformation can be removed by gauge transformation,
which shows that the tachyon vacuum is invariant under all supersymmetries. (In fact,
this argument shows that the tachyon vacuum is invariant under all symmetries of the
closed string background). For more general BPS solutions the explicit realization of this
story will be more nontrivial, but progress along these lines may be possible following [30].
Supersymmetry algebra
Now that we have a supersymmetry transformation, it is interesting to compute the
supersymmetry algebra. For this purpose it is easier to work with the polynomial string field
ϕ˜ in (6.74) rather than the original string field Φ˜ in the small Hilbert space. Since the
supersymmetry transformation of ϕ˜ is given by the coderivation ˜s, we compute
(6.125)
(6.126)
(6.127)
(6.128)
(6.129)
(6.130)
(6.131)
(6.132)
The operators Pµ , πµ and pµ are all derivations of the star product. Moreover, they satisfy
analogous to the relation between M2, µ 2 and m2. To compute the supersymmetry algebra
we need the following commutators between supersymmetry operators:
string field Φ˜ by making a similarity transformation with Gˆ.
where the prime indicates that the operator is defined with a second independent
supersymmetry parameter ǫ~′s. Once we find ˜s, ˜s′], we can easily recover [S˜, S˜′] for the original
The commutator of supersymmetry transformations should produce the momentum
operator:
and
It will also be useful to consider a “momentum operator” at picture −1
P
µ ≡
I
dz
pµ ≡ √
1 I
2 |z|=1 2πi ψµ e−φ(z),
π
µ ≡ √
ξψµ e−φ(z).
dz
dz
P
µ = [Q, πµ ],
s1, s′1 = −2 p ǫ, ǫ′ ,
S1, s′1 = s1, S1′ = −2 P ǫ, ǫ′ ,
where for short we denote
P ǫ, ǫ′
ǫ~s C~s,~s′ Γ~µs′,~s′′ ǫ′ ′′ Pµ ,
~s
(6.133)
and similarly for p(ǫ, ǫ′) and π(ǫ, ǫ′), where C is the charge conjugation matrix.
Now we are ready to compute the supersymmetry algebra. Plugging in (6.90) and
expanding cross-terms gives
[˜s, ˜s′] = S1|−1, s′1|1 + s1|1, S′1|−1 + S1|−1, σ′
1|−1, m2|2
+
1|−1, m2|2 , S′1|−1
+ s1|1, σ′
1|−1, m2|2 , s′1|1 +
To extract the momentum operator, rewrite the first two terms
S1|−1, s′1|1 + s1|1, S′1|−1 = [S1, s′1] + [s1, S′1] − S1|1, s′1|−1 − s1|−1, S′1|1 ,
= −4 P(ǫ, ǫ′) − S1|1, s′1|−1 − s1|−1, S′1|1 ,
= −2 P(ǫ, ǫ′) − 2[Q, π(ǫ, ǫ′)] − S1|1, s′1|−1 − s1|−1, S′1|1 ,
where P(ǫ, ǫ′) is the coderivation corresponding to P (ǫ, ǫ′) and π(ǫ, ǫ′) is the coderivation
corresponding to π(ǫ, ǫ′). In the third step we chose to express part of the translation
operator in the form [Q, πµ ], for reasons that will be clear shortly. Substituting we find
[˜s, ˜s′] = −2 P(ǫ, ǫ′) − 2[Q, π(ǫ, ǫ′)] − S1|1, s′1|−1 − s1|−1, S′1|1 + S1|−1, σ′
1|−1, m2|2
σ
1|−1, m2|2 , S′1|−1 + s1|1, σ′
1|−1, m2|2
σ
1|−1, m2|2 , s′1|1
1|−1, m2|2 , σ′
1|−1, m2|2 .
In the simplest supersymmetry algebra, the terms after −2 P(ǫ, ǫ′) would cancel.
Unfortunately they do not cancel, and we have to make sense of them. The reason why the
extra terms are present is that we are dealing with an on-shell supersymmetry algebra.
This is not surprising, since the off-shell fermionic and bosonic degrees of freedom in the
string field do not match. For example, at mass level 0 we have 16 fermion fields but only
11 boson fields, including the gauge field, transverse scalars, and an auxiliary field. In
the current context, on-shell supersymmetry implies that that the supersymmetry algebra
should be expressible as
˜s, ˜s′ = −2 P(ǫ, ǫ′) + hQ + m2|2, λ˜i ,
h
η − m2|0, λ˜i = 0.
where λ˜ is a coderivation which is consistent with the constraint on the field ϕ˜:
Now we show that the supersymmetry algebra can be expressed in this form. Continuing
from (6.136), we write
[˜s, ˜s′] = −2 P(ǫ, ǫ′) + Q, −2 π(ǫ, ǫ′) −
σ1|1, s′1|−1 − s1|−1, σ1′|1
1|−1, m2|2
σ
1|−1, m2|2 , σ′
1|−1, m2|2 .
(6.134)
HJEP1(205)9
(6.135)
(6.136)
(6.137)
(6.138)
(6.139)
Here we used the fact that s1, σ1 and π(ǫ, ǫ′) are derivations of m2. Therefore (6.139)
[˜s, ˜s′] = −2 P(ǫ, ǫ′) + Q + m2|2, −2 π(ǫ, ǫ′) − s1|−1, σ1′|1 −
σ1|1, s′1|−1
1|−1, m2|2 , σ′
1|−1, m2|2 .
Now we have to see what to do with the terms on the second two lines. For the second line
note that we can rewrite
S1|−1, σ′
1|−1, m2|2
(6.142)
and for the third line in (6.141)
1|−1, m2|2 , σ′
1|−1, m2|2
m2|2, σ
1|−1, σ′
using the fact that m2|2 is nilpotent. Taken all together, this implies that the
supersymmetry algebra takes the form
[˜s, ˜s′] = −2 P(ǫ, ǫ′) + Q + m2|2, −2 π(ǫ, ǫ′) − s1|−1, σ1′|1
− σ1|1, s′1|−1 + σ
and therefore the coderivation λ˜ in (6.137) is
λ˜ ≡ −2 π(ǫ, ǫ′) −
σ1|1, s′1|−1 − s1|−1, σ1′|1 + σ
Conjugating by Gˆ , this implies that the supersymmetry algebra for the original field Φ˜ in
the small Hilbert space takes the form:
Here we pulled a Q out of the second and third terms in (6.136) and an m2|2 out of the
sixth and seventh terms in (6.136) using the fact that s1|1 annihilates m2|2. Now we can
rewrite the term on the third line of (6.139):
m2|2, s1|1, σ1′|−1 + σ
= m2|2, s1, σ1′ + σ1, s′1 − s1|−1, σ1′|1 − σ1|1, s′1|−1 ,
m2, s1, σ1′ + σ1, s′1 |2
−[m2|2, s1|−1, σ1′|1 + σ1|1, s′1|−1 ,
σ1|1, s′1|−1 .
hS˜, S˜′i = −2 P(ǫ, ǫ′) +hM˜ , Λ˜ i ,
Λ˜ = Gˆ−1λ˜ Gˆ
(6.141)
(6.143)
(6.144)
(6.145)
(6.146)
(6.147)
= h
m2|0, λ˜i .
and we use the fact that the gauge products carry zero momentum and therefore commute
with Pµ . Now we have to show that λ˜ is consistent with the constraint on the string field.
Compute:
[η, λ˜] = −2 p(ǫ, ǫ′) − s1|1, s′1|−1 − s1|−1, s′1|1
+ s1|−1, σ′
1|−1, s′1|−1, m2|2 ,
= −2 p(ǫ, ǫ′) − [s1, s′1]|0 − s1|−1, σ1′|1, m2|0
1|−1, s′1|1, m2|0 ,
s1|−1, σ1′|1 , m2|0 −
σ1′|1, s1|−1, m2|0
1|−1, s′1|1 , m2|0 − s′1|1, σ
1|−1, m2|0 ,
m2|0, s′1|1, σ1|−1 + σ1′|1, s1|−1
σ1′|1, s1, m2 |−1 − s′1|1, σ1, m2 |−1 ,
m2|0, −2 π(ǫ, ǫ′) + s′1|−1, σ1|1 + σ1′|1s1|−1 + σ
In the first step we computed the action of η on λ˜; in the second step we used conservation
of Ramond number in addition to the derivation property of σ1 and s1; in the third step we
canceled the first two terms using (6.131) and used the Jacobi identity; in the fourth step
we collected terms and used Ramond number conservation; in the fifth step we used the
derivation property of s1 and σ1; in the sixth step we added terms which vanish because of
associativity of m2 and the derivation property of s1 and σ1; in the sixth step we substituted
the definition of λ˜. Note that we needed to express part of the translation operator in the
form [Q, π(ǫ, ǫ′)] so that λ˜ would be consistent with this constraint.
So far we have been working at the level of coderivations on the tensor algebra, but it
is helpful to bring things down to earth and express the supersymmetry algebra in terms of
the NS and R string fields ϕN and ψR. Acting equation (6.137) on the group-like element
of ϕ˜, projecting onto the 1-string component of the tensor algebra, and separating NS and
R components produces an expression of the form
[δ, δ′]ϕN = −2 P (ǫ, ǫ′)ϕN +
[δ, δ′]ψR = −2 P (ǫ, ǫ′)ψR +
gauge
gauge
transformation N
transformation R
on shell
trivial
on shell
trivial
N
R
.
The extra terms represent an infinitesimal gauge transformation and a piece which vanishes
assuming the equations of motion. An infinitesimal gauge transformation of the polynomial
equations of motion (6.83) takes the general form
gauge
gauge
transformation N
transformation R
= QλN + [ψR, λR],
= QλR,
where the NS and R gauge parameters λN and λR are subject to the constraints
ηλN = [ϕN, λN],
ηλR = [ϕN, λR] + [ψR, λN].
,
(6.148)
(6.149)
(6.150)
(6.151)
(6.152)
(6.153)
(6.154)
The particular gauge parameters which appear the supersymmetry algebra are
λN ≡ −2 π(ǫ, ǫ′)ϕN − (σ1s′1 − σ1′s1)ϕN − [σ1ϕN, σ1′ϕN],
λR ≡ −2 π(ǫ, ǫ′)ψR − (s1σ1′ − s′1σ1)ψR + σ1[ψR, σ1′ϕN] − σ1′[ψR, σ1ϕN].
(6.155)
One can check that the gauge parameters satisfy the constraints, which is basically a
consequence of the fact that λ˜ satisfies (6.138). Now let’s write down the on-shell trivial
terms. For short, let us write the NS and Ramond Euler-Lagrange functions
EN ≡ QϕN + ψR ∗ ψR,
ER ≡ QψR.
Then
R
= −2 π(ǫ, ǫ′)EN − (σ1s′1 − σ1′s1)EN + [σ1′ϕN, σ1EN] − [σ1ϕN, σ1′EN],
= −2 π(ǫ, ǫ′)ER − (s1σ1′ − s′1σ1)ER + σ1[σ1′EN, ψR] + σ1[ER, σ1′ϕN]
− σ1′[σ1EN, ψR] − σ1′[ER, σ1ϕN].
Once we impose the equations of motion, we obtain a supersymmetry algebra of the
expected form modulo gauge transformations.
7
In this paper we have constructed consistent classical field equations for all superstring
theories, and for the open superstring given an explicit analysis of supersymmetry. A
proof that our field equations imply the correct tree-level amplitudes will be provided in
upcoming work [17]. Let us conclude by discussing future directions.
Though we don’t know how to write a fully satisfactory action for the Ramond sector, it
should be possible to formulate a tree-level action which includes two Ramond string fields
(typically, at picture −1/2 and −3/2), which are afterwards related by imposing a
“selfdual” constraint on classical solution space [6]. See [33] for recent discussion. One version of
this idea was recently suggested in [7], and would be particularly natural to implement using
the methods of this paper. However, the required products in the equations of motion will
be different from those introduced here; in a sense they will be more complicated, since even
at a given Ramond number the products will differ depending on the number of Ramond
states being multiplied. However, this is probably not an insurmountable complication. It
would be particularly nice if an action with constraint could be realized for type II closed
superstring field theory, as it would give a potentially interesting gauge invariant observable
for Ramond-Ramond backgrounds. However, it remains to be seen whether an action with
constraint helps in defining the quantum theory.
One important question is whether recent developments in superstring field theory can
help in understanding higher genus amplitudes in superstring perturbation theory. The
(6.157)
(6.158)
(6.159)
(6.160)
conservative approach to this problem requires first constructing a satisfactory classical
action, and then quantizing following the methodology of the Batalin-Vilkovisky
formalism [34]. However, given present limitations in the Ramond sector, a more pragmatic
approach may be to construct a 1PI effective superstring field theory, as suggested by
Sen [7, 35]. The main question in this respect is whether the methods developed here
and in previous works can be adapted to handle spurious singularities which appear in
superconformal ghost correlators at higher genus [36]. It may also be helpful to clarify
the relation between our construction of vertices and the method of “vertical integration”
introduced in [37] and further developed in [38]. We hope to return to these questions soon.
Acknowledgments
T.E. would like to thank S. Hellerman and A. Sen for conversations. This work was
supported in parts by the DFG Transregional Collaborative Research Centre TRR 33 and
the DFG cluster of excellence Origin and Structure of the Universe.
Open Access.
Attribution License (CC-BY 4.0), which permits any use, distribution and reproduction in
any medium, provided the original author(s) and source are credited.
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