This is from P53 of Differential Geometry by Taubes. Fix some $q\in \textrm M (n;\mathbb R)$. Set the 1-form $\omega_q$ on $\textrm{Gl}(n;\mathbb R)$ $$\omega_q|_m=\textrm{tr}(qm^{-1}\textrm d m),$$ where $\textrm d m$ denotes the matrix whose $(i,j)$-th entry is $\textrm dm_{i,j}$. The exponential map $e_\iota$ is $$e_\iota: \textrm M (n;\mathbb R) \to \textrm{Gl}(n;\mathbb R),m\mapsto e^m.$$ Then Taubes asserts that the pullback of $\omega_q$ by $e_\iota$ is $$e_\iota^*\omega_q=\int^1_0 \mathrm{d}s~ \text{tr}(e^{-sa}qe^{sa}\textrm da).$$ That's where I stuck. Why the pullback is like that?
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I don't understand what the $1$-form $\omega_q$ is. Please be more clear about this. – Amitai Yuval Jul 02 '17 at 23:43
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@AmitaiYuval $\textrm d m$ denotes the matrix whose $(i,j)$-th entry is $\textrm dm_{i,j}$. – user150248 Jul 03 '17 at 02:13
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did you find the answer? – wilkersmon Feb 06 '18 at 16:38
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@wilkersmon No, I did not find the answer. – user150248 Feb 07 '18 at 02:25
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@user150248 take a look at this: https://math.stackexchange.com/questions/194344/how-to-write-down-the-pull-back-of-a-differential-form-by-exponential-map – wilkersmon Feb 14 '18 at 00:20