33-652: String Theory Assignment 7

Peter Battaglino 11. April, 2005

Problem 1: Differential forms

A p-form is a totally antisymmetric tensor, so it has \binom{D}{p} independent components, or

\frac{D!}{p!(D − p)!}  (1)

independent components. Since any p-form has the representation

ω = \frac{1}{p!}A_{a_{1}a_{2}...a_{p}}dx^{a_{1}} ∧ dx^{a_{2}} ∧ ⋯ ∧ dx^{a_{p}},  (2)

it can be represented as a linear combination of wedge products of p basis one-forms. Examine the form

dx^{a_{1}} ∧ dx^{a_{2}} ∧ ⋯ ∧ dx^{a_{p}}.

In order for this form to be different from zero, each of the {a_{i}} must take on a different value, since

dx^{a_{i}} ∧ dx^{a_{i}} = 0  (3)

for all a_{i}. Since each of the a_{i}’s can run over the index set {1, …, D}, we see that if we had more than D forms wedging to create our basis form, then at least two of them would have to have the same index. We can then push one of them through to the other, and by (3) we see that the form we have constructed vanishes. Hence the maximum possible value for p is D. Denoting the space of p-forms on a manifold M by Ω_{p}(M), we have the exterior derivative d as a mapping d : Ω_{p}(M) → Ω_{p+1}(M), whose action on a p-form ω_{p} is

dω_{p} = \frac{1}{p!}∂_{a_{p+1}}A_{a_{1}a_{2}...a_{p}}dx^{a_{p+1}} ∧ dx^{a_{2}} ∧ ⋯ ∧ dx^{a_{p}}.  (4)

If we act on this new p + 1-form with d a second time, we obtain

d^{2}ω_{p} = \frac{1}{p!}∂_{a_{p+2}}∂_{a_{p+1}}A_{a_{1}a_{2}...a_{p}}dx^{a_{p+2}} ∧ dx^{a_{p+1}} ∧ dx^{a_{1}} ∧ dx^{a_{2}} ∧ ⋯ ∧ dx^{a_{p}}.  (5)

Since the partial derivatives commute, we can commute them and relabel indices to arrive at

d^{2}ω = −d^{2}ω,  (6)

which proves that d^{2} = 0 when acting on any p-form. Considering Maxwell’s equations, we can write the gauge field A as

A = A_{μ}dx^{μ}.  (7)

We then have

dA = ∂_{ν}A_{μ}dx^{ν} ∧ dx^{μ}
= ∂_{ν}A_{μ}(dx^{ν} ⊗ dx^{μ} − dx^{μ} ⊗ dx^{ν})
= (∂_{μ}A_{ν} − ∂_{ν}A_{μ})dx^{μ} ⊗ dx^{ν}
= F_{μν}dx^{μ} ⊗ dx^{ν}
= F,  (8)

where F is the field strength two-form. Now let’s examine the components of dA:

dA = ∂_{μ}A_{ν}dx^{μ} ∧ dx^{ν}
= (∂_{t}A_{x} − ∂_{x}Φ)dt ∧ dx
+ (∂_{t}A_{y} − ∂_{y}Φ)dt ∧ dy
+ (∂_{t}A_{z} − ∂_{z}Φ)dt ∧ dz
+ (∂_{x}A_{y} − ∂_{y}A_{x})dx ∧ dy
+ (∂_{z}A_{x} − ∂_{x}A_{z})dz ∧ dx
+ (∂_{y}A_{z} − ∂_{z}A_{y})dy ∧ dz
= E_{x}dt ∧ dx + E_{y}dt ∧ dy + E_{z}dt ∧ dz
+ B_{z}dx ∧ dy + B_{y}dz ∧ dx + B_{x}dy ∧ dz.  (9)

Taking the exterior derivative of the above expression, we have

d^{2}A = ∂_{y}E_{x}dy ∧ dt ∧ dx + ∂_{z}E_{x}dz ∧ dt ∧ dx
+ ∂_{x}E_{y}dx ∧ dt ∧ dy + ∂_{z}E_{y}dz ∧ dt ∧ dy
+ ∂_{x}E_{z}dx ∧ dt ∧ dz + ∂_{y}E_{z}dy ∧ dt ∧ dz
+ ∂_{t}B_{x}dt ∧ dy ∧ dz + ∂_{x}B_{x}dx ∧ dy ∧ dz
+ ∂_{t}B_{y}dt ∧ dz ∧ dx + ∂_{y}B_{y}dy ∧ dz ∧ dx
+ ∂_{t}B_{z}dt ∧ dx ∧ dy + ∂_{z}B_{z}dz ∧ dx ∧ dy
= (∂_{z}E_{y} − ∂_{y}E_{z} + ∂_{t}B_{x})dt ∧ dy ∧ dz
+ (∂_{x}E_{z} − ∂_{z}E_{x} + ∂_{t}B_{y})dt ∧ dz ∧ dx
+ (∂_{y}E_{x} − ∂_{x}E_{y} + ∂_{t}B_{z})dt ∧ dx ∧ dy
+ (∂_{x}B_{x} + ∂_{y}B_{y} + ∂_{z}B_{z})dx ∧ dy ∧ dz
= 0.  (10)

This is a fancy way of writing

∇ × E + \frac{∂B}{∂t} = 0,  (11)
∇ · B = 0.  (12)

Now consider the Hodge dual of a form, defined as

⋆dx^{a_{1}} ∧ ⋯ ∧ dx^{a_{p}} ≡ \frac{1}{(D − p)!}ϵ_{a_{p+1}···a_{D}}^{a_{1}···a_{p}}dx^{a_{p+1}} ∧ ⋯ ∧ dx^{a_{D}}.  (13)

To construct the vector curl in three dimensional Euclidean space, we examine the Hodge dual of the exterior derivative of a one-form. Let a one-form ω defined as

ω = ω_{a}dx^{a}  (14)

be given. Its exterior derivative is given by

dω = (\frac{∂ω_{z}}{∂y} − \frac{∂ω_{y}}{∂z} ) dy ∧ dz + (\frac{∂ω_{x}}{∂z} − \frac{∂ω_{z}}{∂x} ) dz ∧ dx + (\frac{∂ω_{y}}{∂x} − \frac{∂ω_{x}}{∂y} ) dx ∧ dy.  (15)

Now consider the Hodge dual of the form dx^{a} ∧ dx^{b}, given by

⋆dx^{a} ∧ dx^{b} = ϵ^{abc}dx^{c}.  (16)

Applying this to equation (15), we see that

⋆dω = (\frac{∂ω_{z}}{∂y} − \frac{∂ω_{y}}{∂z} ) dx + (\frac{∂ω_{x}}{∂z} − \frac{∂ω_{z}}{∂x} ) dy + (\frac{∂ω_{x}}{∂x} − \frac{∂ω_{x}}{∂y} ) dz.  (17)

Now, for each point p on a manifold M, we can consider the metric to be a map between the tangent space T_{p}M and the dual space T_{p}^{∗}M of that manifold. Denote this operation by ♯, so that ♯ dx^{α} = ∂/∂x^{α} and ♯ ∂/∂x^{α} = dx^{α}. From the above equations we see that the curl of a vector field V is given (in a really retarded way) by

∇ × V = ♯ ⋆ d ♯ V.  (18)

Problem 14.4: Separated Dp-branes and an Op-plane

a. Show the two strings obtained by the orientifold symmetry. Since the arguments p^{+}, p⃗ of the ground states are always present, let us omit them for brevity. The ground states are of four types:

|[ij]⟩, |[ij̄]⟩, |[īj]⟩, |[īj̄]⟩.  (19)

Each class contains N^{2} ground states since i and j run from 1 to N and ī and j̄ run from ¯1 to N̄. Define an expected action of Ω_{p} on the ground states in (19). Show that your choice satisfies Ω_{p}^{2} = 1 acting on the ground states.

The two strings and their orientifolding partners are depicted as follows:

We expect the following action of Ω_{p} on the ground states in (19):

Ω_{p}|[ij]⟩ = |[j̄ī]⟩, Ω_{p}|[ij̄]⟩ = |[j̄i]⟩, Ω_{p}|[īj]⟩ = |[jī]⟩, Ω_{p}|[īj̄]⟩ = |[ji]⟩.  (20)

It is obvious that Ω_{p}^{2} = 1 when acting on any of the ground states.

b. What are the possible interactions between strings in the four types of sectors built on the states (19)?

The following interactions between strings can take place:

\begin{matrix}[ij] ∗ [jk] = [ik] & [ij] ∗ [jk̄] = [ik̄] & [ij̄] ∗ [j̄k] = [ik] & [ij̄] ∗ [j̄k̄] = [ik̄] \\ [īj] ∗ [jk] = [īk] & [īj] ∗ [jk̄] = [īk̄] & [īj̄] ∗ [j̄k] = [īk] & [īj̄] ∗ [j̄k̄] = [īk̄]. \end{matrix}  (21)

c. String products have a component along the ground state, as in

|[ij̄]⟩ ∗ |[j̄k]⟩ = |[ik]⟩ + · · · .  (22)

Write the other possible ground state products. Test the consistency of your definition of Ω_{p} by acting with Ω_{p} on both sides of the equations giving ground state products. Ω_{p} acts on a product via

Ω_{p}(|A⟩ ∗ |B⟩) = (Ω_{P} |B⟩) ∗ (Ω_{p}|A⟩).  (23)

The other possible ground state products are

|[ij]⟩ ∗ |[jk]⟩ = |[ik]⟩ + ⋯
|[ij]⟩ ∗ |[jk̄]⟩ = |[ik̄]⟩ + ⋯
|[ij̄]⟩ ∗ |[j̄k̄]⟩ = |[ik̄]⟩ + ⋯
|[īj]⟩ ∗ |[jk]⟩ = |[īk]⟩ + ⋯
|[īj]⟩ ∗ |[jk̄]⟩ = |[īk̄]⟩ + ⋯
|[īj̄]⟩ ∗ |[j̄k]⟩ = |[īk]⟩ + ⋯
|[īj̄]⟩ ∗ |[j̄k̄]⟩ = |[īk̄]⟩ + · · · .  (24)

We act on the above equations using (23) to get

|[k̄j̄]⟩ ∗ |[j̄ī]⟩ = |[k̄ī]⟩ + ⋯
|[kj̄]⟩ ∗ |[j̄ī]⟩ = |[kī]⟩ + ⋯
|[kj]⟩ ∗ |[jī]⟩ = |[kī]⟩ + ⋯
|[k̄j̄]⟩ ∗ |[j̄i]⟩ = |[k̄i]⟩ + ⋯
|[kj̄]⟩ ∗ |[j̄i]⟩ = |[ki]⟩ + ⋯
|[k̄j]⟩ ∗ |[ji]⟩ = |[k̄i]⟩ + ⋯
|[kj]⟩ ∗ |[ji]⟩ = |[ki]⟩ + ⋯,  (25)

which are consistent.

d. Find Ω_{p} action on α_{n}^{i} by using Ω_{p} X^{i}(τ, σ) Ω_{p}^{−1} = X^{i}(τ, π − σ). Find Ω_{p} action on α_{n}^{a} using

Ω_{p} Ẋ^{a}(τ, σ) Ω_{p}^{−1} = −Ẋ^{a}(τ, π − σ).  (26)

Verify that for any arbitrary product R of oscillators of both types, Ω_{p} R Ω_{p}^{−1} = (−1)^{N_{⊥}}R, where N_{⊥} is the total number of R.

We use the method of the previous homework problem on orientifolds and D-branes to get that

Ω_{p} α_{n}^{i} Ω_{p}^{−1} = (−1)^{n}α_{n}^{i},  (27)

where the (−1)^{n} comes from a factor of cos(nπ). To obtain the action of Ω_{p} on the transverse oscillators, we use the expansion for Ẋ^{a}

Ẋ^{a} = −i\sqrt{2α′} ∑_{n∈ℤ} α_{n}^{a}e^{−inτ} sin nσ,  (28)

which turns out to be the same for a string stretched between two parallel D-branes as for one D-brane. We now apply relation (26) to get

Ω_{p} α_{n}^{a} Ω_{p}^{−1} = (−1)^{n}α_{n}^{a}.  (29)

Thus each α_{n}^{(i,a)} contributes a (−1)^{n}, so we see that for a product of oscillators of both kinds R with number N^{⊥}, the action of Ω_{p} is given by (−1)^{N_{⊥}}. e. Describe the orientifold spectrum in terms of the spectrum of a single Dp-brane.

We can build a general state from any product R of oscillators in the following manner:

|Ψ_{R}⟩ = ∑_{ij} (r_{ij}R|[ij]⟩ + r_{ij̄}R|[ij̄]⟩ + r_{īj}R|[īj]⟩ + rī ; j̄R|[īj̄]⟩) .  (30)

Acting on both sides with Ω_{p}, we obtain

Ω_{p}|Ψ_{R}⟩ = ∑_{ij}(−1)^{N_{⊥}} (r_{ij}R|[j̄ī]⟩ + r_{ij̄}R|[jī]⟩ + r_{īj}R|[j̄i]⟩ + rī ; j̄R|[ji]⟩) .  (31)

If we require that our theory be Ω_{p} invariant, then we obtain the conditions

r_{ij} = (−1)^{N_{⊥}}rj̄ ; ī
r_{ij̄} = (−1)^{N_{⊥}}r_{jī}
r_{īj} = (−1)^{N_{⊥}}r_{j̄i}.  (32)

The first of these relations states that the matrix rī ; j̄ is completely determined once we fix the N^{2} parameters in r_{ij} (one is the ± transpose of the other). The other two conditions state that the r_{ij̄} and r_{īj} matrices are either both symmetric or both antisymmetric. For the case of N_{⊥} odd, we see that the antisymmetric condition holds. This tells us that there are in total N^{2} + 2 × (N/2)(N − 1) = 2N(N − 1) independent parameters that can be used to build states. When N_{⊥} is even, the symmetric condition holds and we then have N^{2} + 2 × (N/2)(N + 1) = 2N(N + 1) independent parameters. This was what we wanted to show.

Problem 15.3: Properties of the string charge Q⃗

a. Consider a string at some fixed time t_{0} and a region R of space that contains a portion of this string: the string enters the region R at a point x⃗_{i} and leaves R at a point x⃗_{f} (assume there is no compactification of space). Calculate the string charge Q⃗ = ∫_{R} d^{d}xj⃗^{0} contained in R at time t_{0}. Use your result to show that the total string charge associated with a closed string is zero.

We know that j⃗^{0} is given by

j⃗^{0} = \frac{1}{2} ∫ dσ δ (x⃗ − X⃗(t, σ)) \frac{∂X⃗(t, σ)}{∂σ}.  (33)

Thus, Q⃗ is given by

Q⃗ = ∫_{R} d^{d}x \frac{1}{2} ∫ dσ δ (x⃗ − X⃗(t_{0}, σ)) \frac{∂X⃗(t_{0}, σ)}{∂σ}
= \frac{1}{2} ∫ dσ ∫_{R} d^{d}x δ (x⃗ − X⃗(t_{0}, σ)) \frac{∂X⃗(t_{0}, σ)}{∂σ}
= \frac{1}{2} ∫ dσ \frac{∂X⃗(t_{0}, σ)}{∂σ} \{ \begin{matrix}1, & X⃗(t_{0}, σ) ∈ R \\ 0, & X⃗(t_{0}, σ) ∉ R \end{matrix} \}
= \frac{1}{2} (X⃗(t_{0}, x⃗_{f}) − X⃗(t_{0}, x⃗_{i})) .  (34)

Thus we see that the charge associated with a closed string is 0 because x⃗_{i} and x⃗_{f} are identical when the region encloses the entire string.

b. A more abstract proof that Q⃗ is zero for any localized configuration of closed strings requires showing that ∇ · j⃗^{0} = 0 implies ∫ d^{d}x j⃗^{0} = 0.

First we note that the following identity holds for any two functions ϕ and ψ:

∇ · (ϕψj⃗^{0}) = ϕ∇ψ · j⃗^{0} + ψ∇ϕ · j⃗^{0} + ϕψ∇ · j⃗^{0}.  (35)

Integrating ∇ · (ϕψj⃗^{0}) over all space and using the divergence theorem, we have

∫_{space} d^{d}x ∇ · (ϕψj⃗^{0}) = ∮_{∂(space)} ϕψj⃗^{0} · d⃗σ = 0,  (36)

where we used the fact that j⃗^{0} is localized in space. We can then use our previously-derived identity, dropping off the term containing ∇ · j⃗^{0}, which vanishes, and using ϕ = 1 and ψ = x^{i}:

∫ d^{d}x \frac{∂x^{i}}{∂x^{j}} ê^{j} · j⃗^{0} = ∫ d^{d}x δ_{j}^{i} j^{0,j} = ∫ d^{d}x j^{0,i} = 0,  (37)

which holds for all i, so we’ve proven what we wanted to prove.

c. Assume now that one space coordinate x is curled up into a circle of radius R, and consider a closed string wrapped around this circle. Calculate the string charge Q⃗. Explain why the answer is not zero, and compare with the result obtained in (15.72).

The string charge is given by

\frac{1}{2} (X⃗(t_{0}, x⃗_{f}) − X⃗(t_{0}, x⃗_{i})) .  (38)

Now this evaluates to \frac{1}{2}2πRm = mπR, where m is the number of times the string is wrapped around the compactified dimension. This is nonzero because there can be two different string coordinates associated with identified points in space if some of the dimensions are compactified.

Problem 17.3: Charge carried by winding strings

The zero mode expansion of a string with winding number ℓ is given by

X^{m}(τ, σ) = x^{m}(τ),  X(τ, σ) = ℓRσ,  (39)

where the index m is used to denote all string coordinates except for X^{25} ≡ X. Consider the coupling term of the Kalb-Ramond field to the string:

S = − ∫ dτdσ \frac{∂X^{μ}}{∂τ} \frac{∂X^{ν}}{∂σ} B_{μν}(X).  (40)

Calculate the terms in S that couple B_{m,25} = −B_{25,m} to the string trajectory x^{a}(τ).

The term which couples B_{m,25} to the string trajectory is given by

(∂_{τ} X^{m}∂_{σ}X − ∂_{τ} X∂_{σ}X^{m}) B_{m,25},  (41)

which simplifies to

ℓR\frac{dx^{m}}{dτ} B_{m,25}.  (42)

Note that this is strikingly similar to the term which couples a Maxwell field to the trajectory of a particle,

S_{EM} = \frac{e}{c} ∫_{P} dτ A_{μ} (x(τ)) \frac{dx^{μ}}{dτ} (τ).  (43)

Making the identification (e/c)A_{μ} ↔ (ℓR)B_{m,25} (up to a sign), we see that the electric charge carried by the string is proportional to its winding number ℓ.