# Dual of Kalb Ramond field [closed]

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

## closed as off-topic by AccidentalFourierTransform, Kyle Kanos, Cosmas Zachos, Aaron Stevens, sammy gerbilNov 6 '18 at 22:20

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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.

• 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. – 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)