Schreiber notes

$\,$

$\,$

Some notes.

$\,$

We want to check that the the $L_\infty$-isomorphism sends IIA cocycles to IIB cocycles.

Conventions

The following computation uses the spinor conventions of CDF, II.8.1.

In particular we take $\Gamma_0^\dagger = \Gamma_0$ and $(\Gamma_{spatial})^\dagger = -\Gamma_{spatial}$.

Using $\overline{\psi} = \psi^\dagger \Gamma_0$ we get

$$\begin{aligned} \left( \overline{\psi} \Gamma_{a_1 \cdots a_p} \psi \right)^\ast & = \overline{\psi} \Gamma_{a_p \cdots a_1} \psi \\ &= (-1)^{p(p-1)/2} \overline{\psi} \Gamma_{a_1 \cdots a_p} \psi \end{aligned}$$

As a consequence, the following expressions are real

$$\begin{aligned} & \overline{\psi} \Gamma_a \psi \\ i & \overline{\psi}\Gamma_{a_1 a_2} \psi \\ i & \overline{\psi} \Gamma_{a_1 a_2 a_3} \psi \\ & \overline{\psi} \Gamma_{a_1 \cdots a_4} \psi \\ & \overline{\psi} \Gamma_{a_1 \cdots a_5} \psi \\ i & \overline{\psi} \Gamma_{a_1 \cdots a_6} \psi \\ i & \overline{\psi} \Gamma_{a_1 \cdots a_7} \psi \end{aligned}$$

Given a choice of Clifford matrices in $d = 9$ we choose Clifford matrices in dimension 10 and 11 as discussed at Majorana spinors – In dimension 11, 10 and 9:

$$\Gamma_{a \leq 8} \coloneqq \left( \array{ 0 & \gamma_a \\ \gamma_a & 0 } \right) \;\,,\;\; \Gamma_{9} \coloneqq \left( \array{ 0 & id \\ -id & 0 } \right) \;\,,\;\; \Gamma_{10} \coloneqq \left( \array{ i id & 0 \\ 0 & -i id } \right) \,.$$

Notice that $\Gamma_{a b }$ for $a \lt b \leq 9$ is the same for both chiralities, except if $b = 9$. Therefore if we write

$$\Gamma_a^B = \left\{ \array{ \Gamma_a & \vert \; a \leq 8 \\ \left( \array{ 0 & id \\ id & 0 } \right) & \vert \; a = 9 } \right.$$

then the IIB spin representation is given by elements $\exp(\omega^{a b} \Gamma_a^B \Gamma_b^B)$. Hence we set

Definition

$$\Gamma_{a \leq 8}^B \coloneqq \Gamma_a$$

$$\Gamma_9^B \coloneqq i \Gamma_9 \Gamma_{10}$$

hence

$$\Gamma_{10} = - i \Gamma_9^B \Gamma_9$$

Remark

Beware that the $\{\Gamma_a^B\}$ are not a Clifford algebra, but the even part $\exp(\omega^{a b} \Gamma_a^B \Gamma_b^B)$ is a spin representation (namely the direct sum of two copies of $\mathbf{16}$), and for odd $p$ then

$$\begin{aligned} \overline{\psi} \Gamma^B_{a_1\cdots a_{p}} \psi & = \psi^\dagger \Gamma_0 \Gamma^B_{a_1 \cdots a_p} \psi \\ & = \psi^\dagger \Gamma^B_0 \Gamma^B_{a_1 \cdots a_p} \psi \\ \end{aligned}$$

is the sum of the corresponding spin pairings in two copies of $\mathbf{16}$.

Double dimensional reduction

For $\pi_{d-1} \colon \mathbb{R}^{d-1,1\vert ...} \longrightarrow \mathbb{R}^{d-2,1\vert ...}$ an extension we decompose any element in

$$\mu \in CE(\mathbb{R}^{d-1,1\vert ...})$$

into a piece of the same degree and a piece of one degree lower, by the identification

$$\mu = \mu|_{d-1} - e^9 \wedge (\pi_{d-1})_\ast \mu \,.$$

Here $\mu|_{d-1}$ is the “dimensional reduction”, while $(\pi_{d-1})_\ast \mu$ is the “double dimensional reduction”.

Remark

If

$$\mu = \sum_{a_i = 0}^{d-1} \overline{\psi} \Gamma_{a_1 \cdots a_p}\psi \wedge e^{a_1} \wedge \cdots \wedge e^{a_p}$$

is a cochain in dimension $d$, then its double dimensional reduction is

$$(\pi_{d-1})_\ast(\mu) = (-1)^{p} p \, \sum_{a_i = 0}^{d-2} \overline{\psi} \Gamma_{a_1 \cdots a_{p-1}} \Gamma_{d-1} \psi \wedge e^{a_1} \wedge \cdots \wedge e^{a_{p-1}}$$

Notice the prefactor of $(-1)^p p$ here, which is crucial in the following.

The M-brane cocycles

Let’s be careful and go back to the source: from DAuria-Fre 82, equation (3.27) we have (with our convention that $d e^{10} = \overline{\psi}\Gamma^{10}\psi$) that

$$d \left(\overline{\psi}\Gamma_{a_1 \cdots a_5} \wedge e^{a_1} \wedge \cdots \wedge e^{a_5}\psi\right) = 15 \left(\overline{\psi}\Gamma_{a_1 a_2}\psi \wedge e^{a_1} \wedge e^{a_2} \right) \wedge \left(\overline{\psi}\Gamma_{a_1 a_2}\psi \wedge e^{a_1} \wedge e^{a_2} \right) \,.$$

Hence if we set

$$\mu_{M2} \coloneqq i c \, \overline{\psi}\Gamma_{a_1 a_2}\psi \wedge e^{a_1} \wedge e^{a_2}$$

for any $c \in \mathbb{R}-\{0\}$ then

$$\mu_{M5} = -\frac{c^2}{15} \left(\overline{\psi}\Gamma_{a_1 \cdots a_5}\psi \wedge e^{a_1} \wedge \cdots \wedge e^{a_5}\right) \,.$$

Now for T-duality to come out right, below it turns out that we need

$$\begin{aligned} \mu_{M2} &= i \tfrac{c'}{2} \sum_{a_i = 0}^{10} \overline{\psi} \Gamma_{a_1} \Gamma_{a_2} \psi \wedge e^{a_1} \wedge e^{a_2} \\ \mu_{M5} &= - \tfrac{c'}{5 \cdot 4 \cdot 3 \cdot 2} \sum_{a_i = 0}^{10} \overline{\psi} \Gamma_{a_1 \cdots a_5} \psi \wedge e^{a_1} \wedge \cdots \wedge e^{a_5} \end{aligned}$$

for some global prefactor $c'$

So we should solve for $c$ and $c'$.

Matching them on $\mu_{M2}$ gives that

$$c ' = 2 c \,.$$

Then matching that on $\mu_{M5}$ gives

$$-\frac{2 c}{5 \cdot 4 \cdot 3 \cdot 2} = -\frac{c^2}{5 \cdot 3}$$

hence

$$\frac{c}{4 } = c^2$$

hence there is a unique solution:

$$c = \tfrac{1}{4} \;\;\;\; \Leftrightarrow \;\;\; c' = \tfrac{1}{2}$$

The IIA D-brane cocycles

We get the IIA cocycle for F1, D0, D2 and D4 from double dimensional reduction of the above M-brane cocycles.

$$\begin{aligned} \mu_{D0} &= c'\overline{\psi}\Gamma_{10}\psi \\ \mu_{F1}^{IIA} & = (\pi_{10})_\ast(\mu_{M2}) \\ & = i c' \sum_{a = 0}^{9} \overline{\psi} \Gamma_a \Gamma_{10} \wedge e^a \\ \mu_{D2} & = \mu_{M2}|_{8+1} \\ & = \tfrac{i c'}{2} \sum_{a_i = 0}^{9} \overline{\psi} \Gamma_{a_1} \Gamma_{a_2} \psi \wedge e^{a_1} \wedge e^{a_2} \\ \mu_{D4} & = (\pi_{10})_\ast(\mu_{M5}) \\ & = + \tfrac{c'}{4 \cdot 3 \cdot 2} \sum_{a_i = 0}^9 \overline{\psi} \Gamma_{a_1 \cdots a_4} \Gamma_{10}\psi \wedge e^{a_1} \wedge \cdots \wedge e^{a_4} \end{aligned} \,.$$

In all of the following the global prefactor $c'$ is being suppressed.

Problem: For the (unique!) solution $c' = 1/2$ from above, then $\mu_{D0}$ is $(1/2) \overline{\psi}\Gamma_{10}\psi$, hence half of what it needs to be to give the IIA-extension. This breaks the dd-reduction. Needs a fix. But what?

$$\begin{aligned} d \mu_{D4} & = h_3 \wedge \mu_{D2} \\ \tfrac{c'}{3 \cdot 2} \left(\overline{\psi} \Gamma_{a_1 a_2 a_3 a \Gamma_{10}} \psi \right) \wedge \left( \overline{\psi} \Gamma^a \psi \right) & = - \tfrac{(c')^2}{2} \left( \overline{\psi} \Gamma_{a_1 a_2} \psi \right) \wedge \left( \overline{\psi} \Gamma_a \Gamma_{10} \right) \end{aligned}$$

Deriving the IIB D-brane cocycles

Now let’s check that a IIB cocycle like $\mu_{D1}$ will come out as it should under sending the above data through the $L_\infty$-T-duality construction.

The D1-brane

So we write

$$\begin{aligned} \mu_{D1} & = \mu_{D1}|_{8+1} - e^9 (\pi_9)_\ast \mu_{D1} \end{aligned}$$

and need to match this to the corresponding 1-forms that we get from the IIA data above. This gives

$$\begin{aligned} \mu_{D1} & = (\pi_9)_\ast \mu_{D2} - e^9 \wedge (\mu_{D0}|_{8+1}) \\ & = i \sum_{a = 0}^8 \overline{\psi}\Gamma_a \Gamma_9 \psi e^{a} - e^9 \wedge \overline{\psi} \Gamma_{10} \psi \\ & = i \sum_{a = 0}^8 \overline{\psi}\Gamma^B_a \Gamma_9 \psi e^{a} + i e^9 \wedge \overline{\psi} \Gamma_9^B \Gamma_9 \psi \\ & = i \sum_{a = 0}^9 \overline{\psi} \Gamma_a^B \Gamma_9 \psi \wedge e^a \end{aligned}$$

where in the third line we inserted $\Gamma_{10} = -i \Gamma_9^B \Gamma_9$ from def. .

This matches the expression in (Sakaguchi 00, equation (2.10)) if we identify our $\Gamma_9$ with his $\sigma_1$ (which makes sense).

The D3-brane

Similarly:

$$\begin{aligned} \mu_{D3} &= -e^9 \wedge (\mu_{D2}|_{8+1}) + (\pi_9)_\ast (\mu_{D4}) \\ & = -i \tfrac{1}{2} \sum_{a_i = 0}^8 \overline{\psi}\Gamma_{a_1 a_2} \psi \wedge e^{a_1}\wedge e^{a_2} \wedge e^9 - \tfrac{1}{2\cdot 3} \sum_{a_i = 0}^8 \overline{\psi} \Gamma_{a_1\cdots a_3} \Gamma_{9}\Gamma_{10}\psi \wedge e^{a_1}\wedge \cdots \wedge e^{a_3} \\ & = -\tfrac{1}{2} \sum_{a_i = 0}^8 \overline{\psi}\Gamma_{a_1 a_2} (\Gamma_9^B \Gamma_9 \Gamma_{10}) \psi \wedge e^{a_1}\wedge e^{a_2} \wedge e^9 + \tfrac{1}{2\cdot 3} \sum_{a_i = 0}^8 \overline{\psi} \Gamma_{a_1\cdots a_3} \Gamma_{9}\Gamma_{10}\psi \wedge e^{a_1}\wedge \cdots \wedge e^{a_3} \\ & = -\tfrac{1}{2 \cdot 3} \left( 3 \sum_{a_i = 0}^8 \overline{\psi}\Gamma^B_{a_1 a_2} \Gamma_9^B (\Gamma_9 \Gamma_{10})\psi \wedge e^{a_1}\wedge e^{a_2} \wedge e^9 + \sum_{a_i = 0}^8 \overline{\psi} \Gamma^B_{a_1\cdots a_3} \Gamma_{9}\Gamma_{10}\psi \wedge e^{a_1}\wedge \cdots \wedge e^{a_3} \right) \\ & = -\tfrac{1}{2 \cdot 3} \sum_{a_i = 0}^9 \overline{\psi} \Gamma^B_{a_1 a_2 a_3} (\Gamma_9 \Gamma_{10}) \psi \end{aligned}$$

where in the third line we inserted $-i = \Gamma_9^B \Gamma_9 \Gamma_{10}$ from def. .

This matches with Sakaguchi 00, equation (2.11).

References

• Riccardo D'Auria, Pietro Fré, Geometric Supergravity in $D=11$ and its hidden supergroup, Nuclear Physics B201 (1982) 101-140 (pdf)

• Leonardo Castellani, Riccardo D'Auria, Pietro Fré, Supergravity and Superstrings - A Geometric Perspective, World Scientific (1991)

The IIA $D$-brane cocycles are in section 6.1 of

• C. Chrysso‌malakos, José de Azcárraga, J. M. Izquierdo and C. Pérez Bueno, The geometry of branes and extended superspaces, Nuclear Physics B Volume 567, Issues 1–2, 14 February 2000, Pages 293–330 (arXiv:hep-th/9904137)

The IIB $D$-brane cocycles are in section 2 of

• Makoto Sakaguchi, section 2 of IIB-Branes and New Spacetime Superalgebras, JHEP 0004 (2000) 019 (arXiv:hep-th/9909143)