# Schreiber notes-

$\,$

$\,$

Some notes.

$\,$

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

# Contents

## The M-brane cocycles

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 $c \in \mathbb{R}-\{0\}$ and

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

## 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} &= \overline{\psi}\Gamma_{10}\psi \\ \mu_{F1}^{IIA} & = (\pi_{10})_\ast(\mu_{M2}) \\ & = 2 i c \sum_{a = 0}^{9} \overline{\psi} \Gamma_a \Gamma_{10} \wedge e^a \\ \mu_{D2} & = \mu_{M2}|_{8+1} \\ & = i c \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^2}{3} \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} \,.

## Deriving the IIB D-brane cocycles

We compute the IIB D-brane cocycles form the above cocycle by appying the T-duality isomorphism.

### Rescaled metric

Define a rescaled metric by

$\tilde \Gamma_a^B \coloneqq \left\{ \array{ \Gamma_a^B = \Gamma_a & \vert \; a \leq 8 \\ \tfrac{1}{2 c} \Gamma_9^B & \vert \; a = 9 } \right.$

and hence

$\tilde \Gamma^a_B \coloneqq \left\{ \array{ \Gamma^a_B = \Gamma^a & \vert \; a \leq 8 \\ - 2 c \Gamma_9^B & \vert \; a = 9 } \right.$

### The IIB circle

\begin{aligned} c_2^{B} & = \pm (\pi_9)_\ast \mu_{F1}^A \\ & = \mp 2 i c \overline{\psi} \Gamma_9 \Gamma_{10} \psi \\ & = \pm \overline{\psi} (2 c \Gamma_9^B) \psi \\ & = \overline{\psi} \Gamma_B^9 \psi \end{aligned}

### The D1-brane

\begin{aligned} \mu_{D1} & = (\pi_9)_\ast \mu_{D2} - e^9 \wedge (\mu_{D0}|_{8+1}) \\ & = 2 i c \sum_{a = 0}^8 \overline{\psi}\Gamma_a \Gamma_9 \psi e^{a} - e^9 \wedge \overline{\psi} \Gamma_{10} \psi \\ & = 2 i c \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 \\ & = 2 i c \sum_{a = 0}^8 \overline{\psi}\Gamma^B_a \Gamma_9 \psi e^{a} + 2 i c e^9 \wedge \overline{\psi} (\tfrac{1}{2c}\Gamma_9^B) \Gamma_9 \psi \\ & = i 2c \sum_{a = 0}^9 \overline{\psi} \tilde \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. \ref{9thBClifford},

### The D3-brane

Similarly:

\begin{aligned} \mu_{D3} &= -e^9 \wedge (\mu_{D2}|_{8+1}) + (\pi_9)_\ast (\mu_{D4}) \\ & = -i c \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{4 c^2}{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} \\ & = c \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{4 c^2}{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{2 c^2}{3} 3 \sum_{a_i = 0}^8 \overline{\psi}\Gamma^B_{a_1 a_2} (\tfrac{1}{2c} \Gamma_9^B) (\Gamma_9 \Gamma_{10})\psi \wedge e^{a_1}\wedge e^{a_2} \wedge e^9 + \tfrac{4 c^2}{3} \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} \\ & = ??? \end{aligned}

where in the third line we inserted $-i = \Gamma_9^B \Gamma_9 \Gamma_{10}$ from def. \ref{9thBClifford}.

Last revised on November 1, 2016 at 05:48:58. See the history of this page for a list of all contributions to it.