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In Clifford Taubes' differential geometry(if you doubt I might be wrong in my transcription, please check section 5.6) , he seems to be claiming that $$i^{*}w_{q}=e_{I}^{*}w_{q}=\int^{1}_{0}Tr(e^{-as}qe^{as}da)ds$$ is 0 when we are working on $SU(n)$ if we assume $q$ is Hermitian, of if $q=irI$ with $r\in \mathbb{R}$. And it is nowhere 0 if $\overline{q^{T}}=iq$ and $Tr(q)=0$. This seems to be clear when $q=irI$, but I do not know how to prove the other cases. My guess is we can show $$Tr(e^{-as}qe^{as}c)=0$$ directly for suitable $c\in TSU(n)$ and choice of basepoint $\alpha$ (say the origin or $I$).

However I do not know how to work with $Tr(B)$ when $\overline{B^{T}}=iB$, $Tr(B)=0$, since in this case using $a=I$ we immediately immediately have $Tr(e^{-s}qe^{s})=Tr(q)=0$ since $s\in \mathbb{R}$. This seems contradicting the author's result.

To provide the general context, we consider $i:SU(n)\rightarrow GL(n,\mathbb{C})$ and pull-back the differential form $w_{q}|m=Tr(qm^{-1}dm)$ from $GL(n,\mathbb{C})$ to $SU(n)$. Since the exponential map $e_{m}(a)=me^{a}$ maps the ball around of any element $m$ differeomorphically to $m$'s open neighborhood. Taubes seems assuming $i^{*}w_{q}=e_{I}^{*}w_{q}$ without giving a proof.

To see how we get this formula,my previous question provides some context.

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Thank you. This helps. –  Bombyx mori Sep 12 '12 at 4:00

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