Schreiber notes-

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Some notes.

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

Contents

The M-brane cocycles

We set

μ M2icψ¯Γ a 1a 2ψe a 1e a 2 \mu_{M2} \coloneqq i c \, \overline{\psi}\Gamma_{a_1 a_2}\psi \wedge e^{a_1} \wedge e^{a_2}

for any c{0}c \in \mathbb{R}-\{0\} and

μ M5=c 215(ψ¯Γ a 1a 5ψe a 1e a 5). \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.

μ D0 =ψ¯Γ 10ψ μ F1 IIA =(π 10) *(μ M2) =2ic a=0 9ψ¯Γ aΓ 10e a μ D2 =μ M2| 8+1 =ic a i=0 9ψ¯Γ a 1Γ a 2ψe a 1e a 2 μ D4 =(π 10) *(μ M5) =+c 23 a i=0 9ψ¯Γ a 1a 4Γ 10ψe a 1e a 4. \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

Γ˜ a B{Γ a B=Γ a |a8 12cΓ 9 B |a=9 \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

Γ˜ B a{Γ B a=Γ a |a8 2cΓ 9 B |a=9 \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

c 2 B =±(π 9) *μ F1 A =2icψ¯Γ 9Γ 10ψ =±ψ¯(2cΓ 9 B)ψ =ψ¯Γ B 9ψ \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

μ D1 =(π 9) *μ D2e 9(μ D0| 8+1) =2ic a=0 8ψ¯Γ aΓ 9ψe ae 9ψ¯Γ 10ψ =2ic a=0 8ψ¯Γ a BΓ 9ψe a+ie 9ψ¯Γ 9 BΓ 9ψ =2ic a=0 8ψ¯Γ a BΓ 9ψe a+2ice 9ψ¯(12cΓ 9 B)Γ 9ψ =i2c a=0 9ψ¯Γ˜ a BΓ 9ψe a \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 Γ 10=iΓ 9 BΓ 9\Gamma_{10} = -i \Gamma_9^B \Gamma_9 from def. \ref{9thBClifford},

The D3-brane

Similarly:

μ D3 =e 9(μ D2| 8+1)+(π 9) *(μ D4) =ic a i=0 8ψ¯Γ a 1a 2ψe a 1e a 2e 9+4c 23 a i=0 8ψ¯Γ a 1a 3Γ 9Γ 10ψe a 1e a 3 =c a i=0 8ψ¯Γ a 1a 2(Γ 9 BΓ 9Γ 10)ψe a 1e a 2e 9+4c 23 a i=0 8ψ¯Γ a 1a 3Γ 9Γ 10ψe a 1e a 3 =2c 233 a i=0 8ψ¯Γ a 1a 2 B(12cΓ 9 B)(Γ 9Γ 10)ψe a 1e a 2e 9+4c 23 a i=0 8ψ¯Γ a 1a 3 BΓ 9Γ 10ψe a 1e a 3 =??? \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=Γ 9 BΓ 9Γ 10-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.