# Lie derivative characterisation of the Levi-Civita connection

In this note, equation (6) the author claimed that

$(M,g)$ is a Riemannian manifold. Let $X,Y,K$ be vector fields defined in the vicinity of a point $p\in M$. Then the condition that $\nabla$ is Levi-Civita connection means that $$L_K|_p(M\ni x\mapsto g_x(X_x,Y_x)\in\Bbb R)=g_p((\nabla_KX)_p,Y_p)+g_p(X_p,(\nabla_KY)_p)$$

I have rearranged the notations to make them look rigorous. In the original notations the equation looks like

$$L_K\langle X,Y\rangle=\partial_K\langle X,Y\rangle=\langle \nabla_KX,Y\rangle+\langle X,\nabla_KY\rangle$$

Could anybody provide any references for a proof? The reference the author gave seems inaccessible (I couldn't find it online). Many thanks!

EDIT In effect I'm looking for a proof for the characterisation of Killing fields found on Wikipedia, i.e., $X$ preserves metric (which I think is equivalent to $L_K|_p(M\ni x\mapsto g_x(X_x,Y_x)\in\Bbb R)=0$) means

$$g(\nabla _{{Y}}X,Z)+g(Y,\nabla _{{Z}}X)=0\,$$

EDIT AGAIN Sorry I was so dumb, the equation I asked about in the main text is just the compatibility of the Levi-Civita connection. And in fact, this expression doesn't quite match the one characterising the Killing form (in the previous EDIT): for the one in the main text it's $\nabla_KX$ and for the one in the EDIT it's $\nabla_XK$. I have to give it quite a lot of thoughts how to make use of the relation $\nabla_KX-\nabla_XK=[K,X]$ etc.

• Do you mean $\partial_K \langle X, Y \rangle$? – Robert Lewis May 25 '17 at 8:57
• @RobertLewis yeah of course, the directional derivative wrt $K$. Sorry for the oversight. – Vim May 25 '17 at 8:58

For completeness, here is a proof; I haven;t read the note you linked to in much detail but it appears to prove it along the same lines.

Consider the operator $A_{X}\equiv \mathcal{L}_{X}-\nabla_{X}$ in terms of a vector field $X$ on some manifold $\mathcal{M}$ where $\mathcal{L}_{X},\nabla_{X}$ denote Lie and covariant differentiation respectively. This operator inherits properties from $\mathcal{L}_{X}$ and $\nabla_{X}$, namely it is a derivation on tensor fields that commutes with contractions. It also satisfies $A_{X}f=0$, for any function $f \in \mathcal{F}(M)$. In particular, for any vector fields $X,Y,Z$ on $\mathcal{M}$:

$$A_{X}[g(Y,Z)] = (A_{X}g)(Y,Z) + g(A_{X}Y,Z) + g(Y,A_{X}Z) \;=\; 0.$$

For a metric compatible connection ($\nabla_{X} g = 0$), we have $A_{X}g = \mathcal{L}_{X}g$, so the above becomes:

$$-(\mathcal{L}_{X}g)(Y,Z) = g(A_{X}Y,Z) + g(Y,A_{X}Z)\qquad\qquad(*)$$

But also, for any vector field $Y$, we have

$$A_{X}Y \;=\; \mathcal{L}_{X}Y - \nabla_{X}Y \;=\; [X,Y] - \nabla_{X}Y$$

which, if the connection is torsion free, reduces to

$$A_{X}Y = - \nabla_{Y}X$$

since the torsion tensor is given by $T(X,Y) = \nabla_{X}Y - \nabla_{Y}X - [X,Y]=0$. Then, we may rewrite $(*)$ as

$$(\mathcal{L}_{X}g)(Y,Z) = g(\nabla_{Y}X,Z) + g(\nabla_{Z}X,Y)$$

This is true for any vector field $X$, but if $K$ is Killing, $\mathcal{L}_{K}g=0$ and this becomes:

$$g(\nabla_{Y}K,Z) + g(\nabla_{Z}K,Y) = 0$$

which is Killing's equation, as you have written above.

• Thanks. Could you please elaborate a bit on why $A_Xf=0$ for functions? Actually I can't quite make sense of $\nabla_Xf$. – Vim May 25 '17 at 12:13
• Yes - so $A_{X}f=\nabla_{X}f - \mathcal{L}_{X}f$. But the covariant derivative of a function $f$ with respect to a vector field $X$ is just $X$ acting on $f$: $\nabla_{X}f=Xf$. The same is true for Lie derivative: $\mathcal{L}_{X}f=Xf$. Hence $A_{X}f=Xf-Xf=0$. – AloneAndConfused May 25 '17 at 12:17
• Sorry for my ignorance, but would you also explain a bit why $$A_{X}[g(Y,Z)] = (A_{X}g)(Y,Z) + g(A_{X}Y,Z) + g(Y,A_{X}Z).$$ I know $$(\nabla_Xg)(Y,Z)=X(g(Y,Z))-g(\nabla_XY,Z)-g(\nabla_XZ,Y)$$ but does $\mathcal L$ share this property too? (Is it what you call "a derivation on tensor fields that commutes with contractions"? Thought I don't quite understand what "tensor fields that commute with contractions" mean) – Vim May 25 '17 at 12:25
• Never mind, I found a proof on Lee's Intro to Smooth Manifolds. Thanks again! – Vim May 25 '17 at 12:34
• That's okay! And out of interest that is exactly what I mean by "commute with contractions" and the Lie derivative has that property also. More here in fact: math.stackexchange.com/questions/1304244/… – AloneAndConfused May 25 '17 at 12:36