# One question about "D-branes and K-Theory"

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I am trying to understand the equation

on page 4 of http://arxiv.org/pdf/hep-th/9810188v2.pdf

Making my proper computation I am obtaining

$$\int_{S^8}ch(V)= \prod _{i=1}^{16}\int_{S^8}(e^{\lambda_{i}}+e^{-\lambda_{i}})=\int_{S^8}2^{16}\frac{p_2(V)}{6} =\int_{S^8}\frac{32768}{3}p_2(V) = integer$$

and then we obtain

$$\int_{S^8}p_2(V) = 3k$$

where $k$ is an integer.

But the equation (2.2) on page 3 of  http://arxiv.org/pdf/hep-th/9810188v2.pdf   says

$$\int_{S^8}p_2(V) = 6k$$

My question is if my computation is correct and how conciliate my results with (2.2) and (2.3) of  http://arxiv.org/pdf/hep-th/9810188v2.pdf .

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It is just not true that

$\int_{S^8}ch(V)=\prod_i \int_{S^8}(e^{\lambda_i}+e^{-\lambda_i})$

The correct formula is

$\int_{S^8}ch(V)=\sum_i \int_{S^8}(e^{\lambda_i}+e^{-\lambda_i})$

(the Chern character is the SUM of the exponentials of the Chern roots).

From there it is easy to check the formula of the paper:

$\sum_i \int_{S^8}(e^{\lambda_i}+e^{-\lambda_i})=\sum_i \int_{S^8} 2 \frac{\lambda_i^4}{4!}$

and conclude using $(\sum_i x_i)^2 = \sum_i x_i^2+2\sum_{i<j}x_i x_j$ with $x_i=\lambda_i^2$ and using the fact that $p_1(V)=\sum_i \lambda_i^2=0$.

answered Sep 10, 2015 by (5,140 points)

Many thanks for your answer.  I was confused about the relevant representation of $SO(32)$ that is used in the Witten´s paper.  I was considering the spinor representation of $SO(32)$ but Witten is using the adjoint representation of $SO(32)$.  All is clear now.  Thank you.

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Other form to derive (2.3) is as follows.

$$ch(V)= Tr(e^{{\frac {iV}{2\pi}}})=Tr(1)+Tr(\frac {iV}{2\pi})+\frac{1}{2!}Tr([\frac {iV}{2\pi}]^2)+$$

$$\frac{1}{3!}Tr([\frac {iV}{2\pi}]^3)+\frac{1}{4!}Tr([\frac {iV}{2\pi}]^4)+...$$

Taking the 8-form we have

$$ch(V)_{8-form}=\frac{1}{4!}Tr([\frac {iV}{2\pi}]^4)={\frac {1}{384}}\,{\frac {{\it Tr} \left( { V}^{4} \right) }{{\pi }^{4}}}$$

Now given that

$$p_{{1}} \left( V \right) =-\frac{1}{8}\,{\frac {{\it Tr} \left( {V}^{2} \right) }{{\pi }^{2}}}$$

$$p_{{2}} \left( V \right) ={\frac {1}{128}}\,{\frac {-2\,{\it Tr} \left( {V}^{4} \right) + \left( {\it Tr} \left( {V}^{2} \right) \right) ^{2}}{{\pi }^{4}}}$$

and using that $p_1(V)=0$ we obtain $Tr(V^2) = 0$ and then

$$p_{{2}} \left( V \right) ={\frac {1}{128}}\,{\frac {-2\,{\it Tr} \left( {V}^{4} \right) }{{\pi }^{4}}}= -{\frac {1}{64}}\,{\frac {{\it Tr} \left( {V}^{4} \right) }{{\pi }^{4}}}$$

which is equivalent to

$${\it Tr} \left( {V}^{4} \right) =-64\,p_{{2}} \left( V \right) {\pi }^{4}$$

Using this last result we obtain finally

$$ch(V)_{8-form}=-\frac{1}{6}p_{{2}} \left( V \right)$$

answered Sep 10, 2015 by (1,130 points)
edited Sep 10, 2015 by juancho

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