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The exponential map $e_{m}: M(n,\mathbb{R})\rightarrow M(n,\mathbb{R})$ is defined by $e_m(\alpha)=me^\alpha,\quad e^\alpha=1+\alpha+\frac{\alpha^2}{2}+\frac{\alpha^3}{3!}+\cdots$

Now fix $q\in M(n,\mathbb{R})$ and introduce the left-invariant 1-form $w=\operatorname{Tr}(qm^{-1}dm)$ on $GL(n,\mathbb{R})$. The pull-back of this form by $e_{I}$ is a one form $e^{*}_{I}=\int^1_0 ds \, \operatorname{Tr}(e^{-s\alpha}qe^{s\alpha}d\alpha)$ claimed by Clifford Taubes.

My question is how do we get this formula. Recall that $w$ can be written as $w=\sum_{1\le i\le j\le k\le n}(q_{ij}(m^{-1})_{jk} \, dm_{ki})$

In the general setting, for a map $\psi$ between $M$ and $N$ with bases $dy_{i}$ for $TN^{*}$ and $dx_i$ for $TM^{*}$, under local homoemorphism $\phi_{U}$ to $\mathbb{R}^m$ for $M$ and $\phi_v$ for $N$ to $\mathbb{R}^n$ we should have the pull back of the one form $dy_k$ by $\psi$ be: $(\psi^{*} \, dy_k) = \sum_{1\le i\le m}\left(\frac{\partial \phi_k}{\partial x_i}\right)| \, dx_i$ where $\phi=\phi_U\circ \psi\circ \psi_v^{-1}$.

I am wondering why the first formula is true and how can we derive it from the second formula. I could not simplify it into something basic (for example I do not get why we have the integral sign). So I venture to ask at here. I asked some professors but is still stuck. I think this must boils down to something very basic. Taubes claim we can simplify this by writing it as a linear functional so $e^{*}w_q(c)=\int^1_0 ds \, \operatorname{Tr}(e^{-sa}qe^{sa})c$ (viewing $c$ as a member of the tangent plane at $\alpha$ and evaluate this)

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It follows from the fact that if $A,B$ are $n\times n$-matrices then $\frac{d}{dt}\exp(A+tB)|_{t=0}=\int_0^1 \exp((1-s)A)\,B\,\exp(sA)\,ds.\qquad(*)$ That formula can be derived from $\frac{d}{dt}(1+(A+tB)/n)^n|_{t=0}=\sum_{k=1}^n (1+A/n)^{n-k}\,B/n\,(1+A/n)^{k-1}$ by taking the limit $n\to\infty$.

edit: To finish the calculation: Formula $(*)$ means that the differential of $ M(n,\mathbb{R})\rightarrow M(n,\mathbb{R})$, $A\mapsto \exp A$, is $B\mapsto\int_0^1 \exp((1-s)A)\,B\,\exp(sA)\,ds$. To pull back the 1-form $Tr(XA^{-1}\,dA)$ (for some constant matrix $X$) we need to replace $A$ by $\exp A$ and $dA$ by $\int_0^1 \exp((1-s)A)\,dA\,\exp(sA)\,ds$. That gives $\exp^*Tr(XA^{-1}\,dA)=\int_0^1 Tr\,(X\exp(-sA)\,dA\,\exp(sA))\,ds$ $=\int_0^1 Tr\,(\exp(sA)X\exp(-sA)\,dA)\,ds.$ (my $A$ is your $m$ and my $X$ is your $q$, sorry for changing notation).

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    Why is the integral expression you give equivalent to the more standard one (found here, for example: https://en.wikipedia.org/wiki/Derivative_of_the_exponential_map)?2015-07-06