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We can define a representation of a Lie group and get the induced representation of the Lie algebra. Let $G$ act on $V$ and $W$, $\mathfrak{g}$ be the Lie algebra associated to $G$ and $X \in \mathfrak{g}$. Then the action on $V$ is defined as $\displaystyle X(v)=\left.\frac{d}{dt}\right|_{t=0}\gamma_t(v)$ where $\gamma_t$ is a path in $G$ with $\gamma'_0 = X$. Then: $$ \begin{align*} X(v \otimes w) & = \left.\frac{d}{dt}\right|_{t=0}\gamma_t(v)\otimes \gamma_t(w) \\ & \stackrel{?}{=}\left(\left.\frac{d}{dt}\right|_{t=0}\gamma_t(v)\right)\otimes w + v\otimes \left(\left.\frac{d}{dt}\right|_{t=0}\gamma_t(v)\right) \\ & = X(v)\otimes w + v \otimes x(w)\end{align*}$$ My question is: why does it split?

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    $\begingroup$ You are effectively taking the derivative of a product: $D(fg)=(Df)g+f(Dg)$ $\endgroup$ – Jyrki Lahtonen Aug 25 '11 at 20:40
  • $\begingroup$ I figured it out but it wont let me answer. I'll post it in 7 hours. Thanks for the edit Theo. $\endgroup$ – Sven Aug 25 '11 at 20:52
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$$ \begin{align*} X(v \otimes w) & = \left.\frac{d}{dt}\right|_{t=0}\gamma_t(v)\otimes \gamma_t(w) \\ & = \lim_{t \to 0} \,\,\, \frac{\gamma_t(v)\otimes \gamma_t(w)-v \otimes w}{t} \\ & = \lim_{t \to 0} \,\,\, \frac{\gamma_t(v)\otimes ( \gamma_t(w)-w+w)-v \otimes w}{t} \\ & = \lim_{t \to 0} \,\,\, \frac{\gamma_t(v)\otimes (\gamma_t(w)-w)+ \gamma_t(v)\otimes w-v \otimes w}{t} \\ & = \lim_{t \to 0} \,\,\, \frac{\gamma_t(v)\otimes (\gamma_t(w)-w)+ (\gamma_t(v)-v)\otimes w}{t} \\ & = \lim_{t \to 0} \,\,\, \frac{v \otimes (\gamma_t(w)-w)}{t}+ \lim_{t \to 0} \,\,\, \frac{(\gamma_t(v)-v)\otimes w}{t} \\ & =\left(\left.\frac{d}{dt}\right|_{t=0}\gamma_t(v)\right)\otimes w + v\otimes \left(\left.\frac{d}{dt}\right|_{t=0}\gamma_t(w)\right) \\ & = X(v)\otimes w + v \otimes X(w)\end{align*}$$

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