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I have two descriptions of the lie algebra of a lie group. Both come with their version of the lie bracket

  1. $\mathfrak{g}=T_eG$ as the tangent space at the identity comes with the lie bracket $ad(X_e)(Y_e)$.
  2. $\mathfrak{g}=\mathfrak{X}^{inv}(G)$ as the left invariant vector fields comes with the lie bracket $[X,Y]$.

After identifying both descriptions of the lie algebras, I wish to show their brackets coincide: $ad(X_e)(Y_e)=[X,Y]_e$. These notes explain this but I do not follow his second to last step: enter image description here How do I arrive at the sum

$\frac{\partial^2}{\partial s\partial t}|_{0}f(expt(tX)exp(sY))+\frac{\partial^2}{\partial s\partial t}|_{0}f(expt(sY)exp(-tX))$?

I tried to see if it follows from some Leibniz rule or multivariable calculus, but I fail to reproduce his result. I have seen a different proof of this result in Lee where it uses the pullback of the flow, but I fail to see how this applies here.

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Morally, it is just an instance of the Leibniz rule where, computing $\frac{\partial}{\partial t}|_{t=0}(e^{tX}e^{sY}e^{-tX})$, you apply $\frac{\partial}{\partial t}|_{t=0}$ first to the factor $\exp(tX)$ and then to the factor $\exp(-tX)$.

Adding more detail, let us recall the canonical isomorphism of vector bundles, covering the identity map $\operatorname{id}_{G\times G}$, $$T(G\times G)\overset{\sim}{\longrightarrow}TG\times TG$$ which maps $\frac{\partial}{\partial t}|_{t=0}(\gamma_1(t),\gamma_2(t))$ to $(\frac{\partial}{\partial t}|_{t=0}\gamma_1(t),\frac{\partial}{\partial t}|_{t=0}\gamma_2(t))$, for any pair of smooth curves $\gamma_1,\gamma_2\colon I\to G$.

Understanding the previous isomorphism,one gets, in particular $$\tfrac{\partial}{\partial t}|_{t=0}(e^{tX}e^{sY},e^{-tX})=(\tfrac{\partial}{\partial t}|_{t=0}e^{tX}e^{sY},\tfrac{\partial}{\partial t}|_{t=0}e^{-tX})=(\tfrac{\partial}{\partial t}|_{t=0}e^{tX}e^{sY},0_e)+(0_{e^{sY}},\tfrac{\partial}{\partial t}|_{t=0}e^{-tX})=\tfrac{\partial}{\partial t}|_{t=0}(e^{tX}e^{sY},e)+\tfrac{\partial}{\partial t}|_{t=0}(e^{sY},e^{-tX}).$$

Finally, the latter allows to compute $$\tfrac{\partial}{\partial t}|_{t=0}f(e^{tX}e^{sY}e^{-tX})=\tfrac{\partial}{\partial t}|_{t=0}f(m(e^{tX}e^{sY},e^{-tX}))=(f\circ m)_*\tfrac{\partial}{\partial t}|_{t=0}(e^{tX}e^{sY},e^{-tX})=(f\circ m)_*\tfrac{\partial}{\partial t}|_{t=0}(e^{tX}e^{sY},e)+(f\circ m)_*\tfrac{\partial}{\partial t}|_{t=0}(e^{sY},e^{-tX})=\tfrac{\partial}{\partial t}|_{t=0}f(m(e^{tX}e^{sY},e))+\tfrac{\partial}{\partial t}|_{t=0}f(m(e^{sY},e^{-tX}))=\tfrac{\partial}{\partial t}|_{t=0}f(e^{tX}e^{sY})+\tfrac{\partial}{\partial t}|_{t=0}f(e^{sY}e^{-tX}),$$ where $m\colon G\times G\to G$ denotes the group multiplication.

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