Checking Killing fields are $G$-invariant

Question

The following is taken from John Roe’s book Elliptic Operators, Topology, and Asymptotic Methods on the bottom of page 23. I quote:

Let $E$ be a principal bundle with group $G$. Differentiating the $G$-action we find that to each element $u$ of $\mathfrak{g}$, the Lie algebra of $G$, there is an a $G$-invariant vector field $X_u$ on $E$ corresponding to $u$.

My question concerns the $G$-invariance of the vector field, which I take to mean right invariance. As I understand, we define for any $e\in E$, $$X_u|_e:=\frac{d}{dt}\bigg\vert_{t=0}e\cdot \exp(tu) \,,$$ and to check right-invariance, we must show for all $g\in G$ $$(X_u)|_{e\cdot g}=d(R_g)_e(X_u|_e) \,,$$ where $R_g \colon E\times G\to E$ is the right action. Expanding the left-hand side, we see $$(X_u)|_{e\cdot g}=\frac{d}{dt}\bigg\vert_{t=0}(e\cdot g)\cdot \exp(tu)=\frac{d}{dt}\bigg\vert_{t=0}e\cdot g\cdot \exp(tu) \,.$$ Expanding the right-hand side, we see \begin{align*} d(R_g)_e(X_u|_e)&=d(R_g)_e\left(\frac{d}{dt}\bigg\vert_{t=0}e\cdot \exp(tu)\right)\\ &=\frac{d}{dt}\bigg\vert_{t=0}R_g(e\cdot \exp(tu))\\ &=\frac{d}{dt}\bigg\vert_{t=0}e\cdot \exp(tu)\cdot g \,, \end{align*} but clearly this isn’t $G$-invariant. Did I make a mistake in the computation somewhere? This question has been asked here: Killing field associate to an element in the Lie Algebra, but no satisfactory answer was provided.

Answer 1

Ok, so it looks like $X_u$ is called the fundamental vector field, and the $G$-invariance isn't exactly what I thought it was. We have two possibilities, where we follow page 7-8 of these notes: https://empg.maths.ed.ac.uk/Activities/GT/Lect1.pdf. The two possibilities correspond to considering the pushforward of either the right or the left multiplication maps on the fundamental vector field.

1. Right: We compute \begin{align*}d(R_g)_e(X_u|_e)&=\frac{d}{dt}\bigg|_{t=0}e\cdot\exp(tu)\cdot g\\ &=\frac{d}{dt}\bigg|_{t=0}(e\cdot g)\cdot (g^{-1}\cdot\exp(tu)\cdot g)\\ &=\frac{d}{dt}\bigg|_{t=0}(e\cdot g)\cdot(\text{Ad}_{g^{-1}}(\exp(tu))\\ &=\frac{d}{dt}\bigg|_{t=0}(e\cdot g)\cdot (\exp(t\text{ } \text{ad}_{g^{-1}}(u))\\ &=\left(\text{ad}_{g^{-1}}X_{u}\right)|_{e\cdot g} \end{align*}
2. Left: We want to make a similar statement as above, involving something like $d(L_g)_e(X_u|_e)$, but the glaring problem is that we don't have left G-action on $E$. Therefore, we cannot formulate the statement on $E$, but on $G$, naturally, where left multiplication is defined.

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Take an open chart $U_\alpha$ in the base, and corresponding to a local trivialization $\psi_\alpha:\pi^{-1}(U_\alpha)\to U_\alpha\times G$, consider the $G-$equivariant map $g_\alpha:\pi^{-1}(U_\alpha)\to G$ defined by $\pi(e)=((\text{proj}_{U_\alpha}(\psi_\alpha(e)),g_\alpha(e))$. We now show the following equality:

$$d(L_{g_\alpha(e)})_{g_\alpha(e)}(u)=d(g_\alpha)_e(X_u|_e).$$ As a sanity check, note that this equality is happening at $T_{g_{\alpha}(e)}G$. We compute: \begin{align*} d(L_{g_\alpha(e)})_{g_{\alpha(e)}}(u)&=\frac{d}{dt}\bigg\vert_{t=0}L_{g_\alpha(e)}(\exp(tu))\\ &=\frac{d}{dt}\bigg\vert_{t=0}g_\alpha(e) \cdot \exp(tu)\\ &=\frac{d}{dt}\bigg\vert_{t=0}{g_\alpha}(e \cdot \exp(tu))\\ &=d(g_\alpha)_e\left(\frac{d}{dt}\bigg\vert_{t=0}e\cdot \exp(tu)\right)\\ &=d(g_\alpha)_e(X_u|_e) \end{align*} where the third equality uses equivariance of $g_\alpha$.