The Identity I am trying to prove is the one in this already asked question
The author is confident they have answered the question but I am still unclear on a few things.
First and foremost, I understand that $g_{\alpha\beta}(\gamma(t))g_{\beta\alpha}(\gamma(t))=e$.
However I don't understand how to make sense of the next line
$0=(g_{\alpha\beta}\circ\gamma)^{\prime}(t)(g_{\beta\alpha}\circ\gamma)(t)+(g_{\alpha\beta}\circ\gamma)(t)(g_{\beta\alpha}\circ\gamma)^{\prime}(t)$.
While on the superficial level I just think 'that's just the product rule'... the quantity $(g_{\alpha\beta}\circ\gamma)^{\prime}(t)$ is a vector in $T_{g_{\alpha\beta}(\gamma(t))}G$ whereas $(g_{\beta\alpha}\circ\gamma)(t)$ is an element in $G$ so how does it make sense to multiply a group element and a vector?
I am also unclear in the later lines of the equations where the author gets rid of the push forwards by just doing the associated multiplication. However I feel this might be intrinsically related to my first query ^.
And finally, I do believe they have dropped off the minus sign in the last few lines?
Some help with this would be much appreciated!
EDIT: I believe I have figured it out. I think the author was being a bit sloppy with their notation. I believe this makes sense of it all. Then one can follow through the calculation with the same steps.
Define the map $\Phi:\mathbb{R}\rightarrow G$ as \begin{align*} \Phi(t)&=(g_{\alpha\beta}\circ \gamma)(t)\cdot(g_{\beta\alpha}\circ\gamma)(t) \end{align*} So $\Phi\equiv e$. We can also express the map $\Phi$ as \begin{align*} \Phi(t)&=(m\circ \psi)(t), \end{align*} Where $m$ is the multiplication map $G\times G\rightarrow G$, $(g_1,g_2)\mapsto g_1g_2$ and $\psi:\mathbb{R}\rightarrow G\times G$ is the map \begin{align*} \psi:t\mapsto ((g_{\alpha\beta}\circ\gamma)(t),(g_{\beta\alpha}\circ\gamma)(t)). \end{align*} Since $\Phi\equiv e$, it follows that $\Phi^{\prime}(t)\equiv 0$ and thus we have \begin{align*} 0&=\Phi^{\prime}(0)\\ &=(m\circ \psi)^{\prime}(0)\\ &=m_*(\psi^{\prime}(0)) \end{align*} Now $\psi^{\prime}(0)\in T_{(g_{\alpha\beta}(m),g_{\beta\alpha}(m))}(G\times G)$ which we may identify with the pair \begin{align*} (X,Y):=((g_{\alpha\beta}\circ\gamma)^{\prime}(0),(g_{\beta\alpha}\circ\gamma)^{\prime}(0))\in T_{g_{\alpha\beta}(m)}G\oplus T_{g_{\beta\alpha}(m)}G, \end{align*} (see page 11 of Kobayahsi and Nomizu's Foundations of Differential Geometry for the details). The Leibniz rule (Pg11 - 12 of the same book) then gives \begin{align*} m_*(\psi^{\prime}(0))&=(m_1)_*(X)+(m_2)_*(Y), \end{align*} Where $m_1:G\rightarrow G$ is the map $g:\mapsto m(g,g_{\beta\alpha}(m))$ and $m_2:G\rightarrow G$ is the map $g:\mapsto m(g_{\alpha\beta}(m),g)$. So, we have \begin{align*} 0=m_*(\psi^{\prime}(0))&=(m_1)_*\left(\frac{d}{dt}\bigg|_{t=0} (g_{\alpha\beta}\circ\gamma)(t)\right)+(m_2)_*\left(\frac{d}{dt}\bigg|_{t=0}(g_{\beta\alpha}\circ\gamma)(t)\right)\\ &=\frac{d}{dt}\bigg|_{t=0}(g_{\alpha\beta}\circ\gamma)(t)g_{\alpha\beta}(m)+\frac{d}{dt}\bigg|_{t=0}g_{\alpha\beta}(m)(g_{\beta\alpha}\circ\gamma)(t). \end{align*}