i've been studying string theory for 4 days, i have a Kalb Ramond $B_{(2)}$ of this kind (from a $5^2 _2$ solution [1]) and i want evaluate its dual but i don't obtain the right result:

The variables are $(x^0,r,\theta,x^3,x^4,x^5,x^6,x^7,x^8,x^9)$.

Defining the function $H=H(r)=h_0 +\sigma \log \frac{\mu}{r} $

$$B_{(2)} =-\frac {\theta \sigma} {(H^2+\theta^2 \sigma^2)} dx^8 \wedge dx^9 .$$

The metrics is $$ds^2_{string}=H(dr^2+r^2d\theta^2)+\frac{H}{(H^2+\theta^2 \sigma^2)}dx^2_{89}+dx^2_{034567}. $$

The dilaton is useless but i report it for completeness:

$$e^{2\phi} = \frac{H}{(H^2+\theta^2 \sigma^2)}.$$

I also have the expression of the evaluation i want to do : $B_{(6)}$ field which is the dual of $B_{(2)}$, it is:

$$B_{(6)}=-\frac{(H^2+\theta^2 \sigma^2)}{H} dx^0 \wedge dx^3 \wedge dx^4 \wedge dx^5 \wedge dx^6 \wedge dx^7. $$

I also know that $*_{10}dB_{(2)} =dB_{(6)}$

From all of this how can i evaluate $B_{(6)}$? $ $ It is very important because in S duality $B_{(6)}$ is mapped into $C_{(6)}$.

[1] Exotic Branes in string theory, J.de Boer and M. Shigemori, https://arxiv.org/abs/1209.6056


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This is a standard operation in exterior calculus. Given a $p$-form $\phi$, written as,

$$\phi = \frac{1}{p!}\phi_{i_1\dotsi_p} \beta^{i_1} \wedge \dots \wedge \beta^{i_p}$$

where $\beta^i$ are a basis of one-forms, we have that the Hodge dual of the $p$-form is an $(n-p)$-form in an $n$-dimensional space:

$$\star \phi = \frac{\sqrt{|\det g|}}{p!(n-p)!}\varepsilon_{i_1 \dots i_n} g^{i_1 j_1} \dots g^{i_pj_p} \phi_{j_1 \dots j_p} \beta^{i_{p+1}} \wedge \dots \wedge \beta^{i_n}.$$

In other words, the dual amounts to raising all indices of the components of the form, and contracting with the Levi-Civita symbol. In your case, $g$ is the metric of $10$-dimensional spacetime. Evaluating the dual of a two form will give you an eight form.

  • $\begingroup$ Yes i know very well this but, in this case, what i want to evaluate is a six form, as you read from above. It is called the dual but it is not properly the hodge star dual. $\endgroup$ – siderius Oct 29 '18 at 20:46

The complete answer is the following:

  1. You evaluate the exterior derivative of $B_{(2)}$
  2. The Hodge star contributes with a determinant as written in the previous answer. It will be: $ \star \phi_{i_1 i_2}=\sqrt{-detg}\ \epsilon_{i_0 i_3 i_4 i_5 i_6 i_7 i_8 i_9 i_1 i_2} H^{i_0 i_3 i_4 i_5 i_6 i_7 i_8 i_9} $

where $i_1 i_2$ are fixed indices and $\epsilon_{i_0 i_3 i_4 i_5 i_6 i_7 i_8 i_9 i_1 i_2}$ is the Levi Civita symbol.

  1. Now lower the indices to $H^{i_0 i_3 i_4 i_5 i_6 i_7 i_8 i_9}$
  2. We repeat this for each component of $\phi$
  3. We know that $H$ is the exterior derivative of something so we have to integrate and verify that the integral found behaves correctly under exterior derivative (this is similar to the calculation of a primitive)

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