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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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up vote 3 down vote accepted

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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This is useful, but what is $A$ and $B$ exactly? I still feel somehow confused. – Bombyx mori Sep 11 '12 at 23:54
sorry for being dumb but I want to ask why we can get the first one by your formula. For example we have $e^{1-s},e^{s}$ in your formula, but not the same thing in his one... – Bombyx mori Sep 12 '12 at 1:57
Why is the integral expression you give equivalent to the more standard one (found here, for example: – Potato Jul 6 '15 at 2:12

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