1
$\begingroup$

Suppose $\theta\colon G\times M\to M$ is a transitive smooth left action of a compact Lie group $G$ on a manifold $M$ and $\pi\colon G\to M\cong G/K$ the corresponding smooth submersion for some closed subgroup $K$. Then $\pi$ is equivariant with respect to $\theta$ and left multiplication on $G$, i.e. for all $g,h\in G$

$$\theta(h,\pi(g)) = \pi \circ L_h(g)\qquad \Big(=\pi \circ R_g(h)\Big).$$

If $\langle\cdot\ , \cdot\rangle$ is a bi-invariant Riemannian metric on $G$ then $\pi$ induces a $G$-invariant metric $\gamma$ on $M$, turning $\pi$ into a Riemannian submersion. The vertical bundle $V$ (where each fibre is the nullspace of $d\pi_g$) and the metric $\langle\cdot\ , \cdot\rangle$ define a horizontal bundle $H$ by letting $H_g = V_g^\perp$ with respect to $\langle\cdot\ , \cdot\rangle_g$.

Given $X\in \mathrm{Lie}(G)$ (the set of left-invariant vector fields), consider the induced fundamental vector field $\mathcal{K}(X)\in\mathfrak{X}(M)$ defined for $p=\pi(g)\in M$ by

$$\mathcal{K}(X)|_p= d_t|_{t=0} \theta(\exp^G(tX),p)=d\pi_g\big(d(R_g)_e(X_e)\big).$$

Question: Is it correct that the horizontal lift of $\mathcal{K}(X)$ is the vector field $\tilde{Z}\in\mathfrak{X}(G)$ given by $$\tilde{Z}_g = \big(d(R_g)_e(X_e)\big)^H$$ i.e., the horizontal part of the right (!) translation of the tangent vector $X_e\in T_e G$? Initially, I thought it should be the horizontal part of $X$ itself which is only true at $g=e$ it seems.

Even though taking the horizontal part commutes with left translation (by construction of the induced metric on $M$), the vector field $\tilde{Z}$ is not left-invariant (i.e. not in $\mathrm{Lie}(G)$) because in general $L_h\circ R_g\neq R_{hg}$. Neither is $\tilde{Z}$ right-invariant because taking the horizontal part does not commute with right translations. This somehow feels strange to me.

Improve this question
$\endgroup$
4
  • $\begingroup$ The main point (I think) is that the left action of $\mathfrak g=\operatorname {Lie}(G)$ on $G$ itself induces right invariant vector fields. $\endgroup$
    – Sebastian Goette
    Apr 30, 2020 at 16:45
  • $\begingroup$ The left action "left multiplication on G", i.e. $L\colon G\times G\to G, (g,h)\mapsto L_g(h)$ induces the right-invariant fundamental vector fields $\mathcal{K}^L(X)|_g= \frac{d}{dt} L(\exp^G(tX),g) = d(R_g)_e(X_e)$. Is this what you mean? I do see how that is related to my question. However, what I find so strange is that in my case above I start with a left-invariant vector field on $G$, calculate the fundamental vector field downstairs on $M$, lift it back upstairs to $G$ and get something that is neither left- nor right-invariant. Is my calculation for $\tilde{Z}$ correct at least? $\endgroup$
    – Sven Pistre
    Apr 30, 2020 at 17:12
  • $\begingroup$ The action of $K$ is from the right, so the subbundle of $TG$ generated by this action is left invariant, same for its complement. Projecting something right-invariant onto a left invariant subbundle can give something with no invariance at all. Or try to look at it from a geometric point of view: the horizontal lift of the Killing field has the same norm as the Killing field itself. If the horizontal lift was somehow invariant, it would have constant norm. But there are examples $G/K$ with nonzero Euler characteristic, so each vector field must vanish somewhere. $\endgroup$
    – Sebastian Goette
    May 1, 2020 at 13:49
  • $\begingroup$ The first part of your comment is what confused me. Thanks for clearing this up for me. I like your geometric argument! Do you want to turn this comment into an answer so I can accept it? $\endgroup$
    – Sven Pistre
    Jun 12, 2020 at 12:21

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.