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

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