Trouble understanding the derivative of a Lie Group representation along a curve.

Question

I am studying covariant derivatives on associated vector bundles via Nakahara's Geometry, Topology and Physics. In doing so, I have come across a issue, specifically in going between the last two lines of expression $(10.54)$. The context is a bit hard to explain in a single question, so I will try to formulate a sort of minimal working example for those who might not have access to the book.

Let $G$ be a Lie group, $M$ a manifold, $\gamma:I\to M$ a path and $g:M\to G$ a smooth function. I have shown that:

$$ \frac{d g(\gamma(t))}{dt}=-\mathcal{A}(X)g(\gamma(t)) $$ Where $\mathcal{A}$ is a certain $\mathfrak{g}$-valued one-form and $X$ is a tangent vector to $\gamma$ at $\gamma(0)$. Now let $\rho:G\to GL(V)$ be a representation of $G$ on a vector space $G$ and take $v\in V$. I am trying to show that:

$$ \left(\frac{d}{dt}(\rho\circ g\circ\gamma)(t)\right)(\rho\circ g\circ\gamma)(t)v\biggr|_{t=0} = -\rho(\mathcal{A}(X))v $$ But I am quite unsure of how to proceed. The obvious solution would be to show that $\rho$ "commutes with derivatives", in some sense, but I don't really know how to formalise that idea. Any help would be much appreciated.

Answer 1

I need to assume that $g(\gamma(0)) = e$. Compute $$\frac{{\rm d}}{{\rm d}t}\bigg|_{t=0} (\rho\circ g\circ \gamma)(t)v = {\rm d}\rho_{g(\gamma(0))} ((g\circ \gamma)'(0))v = \rho_\ast((g\circ \gamma)'(0))v = \rho_\ast(-\mathcal{A}(X))v = -\rho_\ast(\mathcal{A}(X))v.$$

It's very simple if you understand all the abuses of notation happening here.

• On the first equal sign, I use the chain rule.
• On the second equal sign, I use the assumption $g(\gamma(0)) = e$, noting that the author denotes both the Lie group representation $\rho\colon G \to {\rm GL}(V)$ and the induced representation $\rho_\ast\colon \mathfrak{g} \to \mathfrak{gl}(V)$ on the level of Lie algebras by the same letter. The relation is $\rho_\ast = {\rm d}\rho_e$, but he writes $\rho$ for $\rho_\ast$ as well.
• On the third equal sign, I use the given ODE that $g\circ \gamma$ satisfies together with $g(\gamma(0)) = e$ again, so $(g\circ \gamma)'(0) = -\mathcal{A}(X)e$, and since $\mathcal{A}(X) \in \mathfrak{g}$, writing $\mathcal{A}(X)h$ with $h \in G$ can only formally mean ${\rm d}(R_h)_e(\mathcal{A}(X))$ --- so $\mathcal{A}(X)e = \mathcal{A}(X)$.
• On the last equal sign, I use that $\rho_\ast$ is linear, so the negative sign comes out.

That's all there is to it.