Let (M,g) be a Riemannian manifold and $\nabla$ a connection on $TM$. Let $X \in \mathfrak{X}(M)$ be a smooth vector field on M, and denote by $\theta$ its flow. Let $Y$ be a smooth vector field on M that is smooth along the flow of $X$ at $p$. Denote by $D_t$ the induced covariant derivative along the flow of $X$ at $p$. Under what condition(s) on $\nabla$ can we have:

$$L_X Y = D_t Y$$

Recall that :
$$L_X Y = \underset{t \rightarrow 0}{lim} \frac{d(\theta_{-t})_{\theta^{p}(t)}(Y_{\theta^{p}(t)})-Y_{p}}{t} = [X,Y]$$

Thank you !


1 Answer 1


I'm a bit confused why you take $(M,g)$ to be Riemannian but then separately impose a connection $\nabla$, which is not necessarily the Levi-Civita connection (so the metric structure is not used). Also, $\nabla$ is usually used for the covariant derivative, so what you call $D_t$ seems to be the same thing as $\nabla_X$?

I think what you're asking is when does $$\nabla_XY = \mathcal{L}_XY$$ for all $X, Y\in\mathfrak{X}(M)$? If so, never: $\nabla$ needs to satisfy $$ \nabla_{fX}Y = f\nabla_XY $$ for all $f\in C^\infty(M)$. By contrast, $$ \mathcal{L}_{fX}Y = [fX,Y] = f[X,Y]+(fX)Y-(Yf)X=f\mathcal{L}_XY +(fX)Y-(Yf)X, $$ and for non-zero $Y$ and $X$ you can always choose $f$ so that $fX$ and $Yf$ doesn't vanish.


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