I'm currently self-studying the basic concepts of Riemannian Manifolds, and I'm stuck on the proof of the existence of the Levi-Civita connection that is presented in the book "Semi-Riemannian Geometry" by O'Neill, which is a coordinate-free approach.

Suppose $(M,g)$ is a Riemannian manifold, and let $\nabla:\mathfrak{X}(M)\times\mathfrak{X}(M)\to\mathfrak{X}(M)$ be a connection. So far, I've been able to prove that, if $\nabla$ verifies the conditions of the Levi-Civita connection, then it must satisfy the Koszul Formula:

$ \langle\nabla_{X}Y,Z\rangle=\dfrac{1}{2}(X\langle Y, Z\rangle+Y\langle Z, X\rangle-Z\langle X, Y\rangle -\langle Y,[X, Z]\rangle-\langle Z,[Y, X]\rangle+\langle X,[Z, Y]\rangle). $

So now, I'm trying to prove existence by using the formula: more precisely, with this result that I proved previously:

If $\omega:\mathfrak{X}(M)\to \mathcal{F}(M)$ is a differential $1$-form, then there exists a unique vector field $V$ such that for any other vector field $X$, we have $\omega(X)=\langle V,X\rangle$.

My reasoning is the following: fix two vector fields $X,Y$, and let $\omega_{X,Y}:\mathfrak{X}(M)\to \mathcal{F}(M)$ be the map

$\omega_{X,Y}(Z)=\dfrac{1}{2}(X\langle Y, Z\rangle+Y\langle Z, X\rangle-Z\langle X, Y\rangle -\langle Y,[X, Z]\rangle-\langle Z,[Y, X]\rangle+\langle X,[Z, Y]\rangle).$

If I were able to prove that $\omega_{X,Y}$ is $\mathcal{F}(M)$-linear (which, knowing the Theorem to be true, it must be), then by using the preceding result I can define a unique vector field $\nabla_{X}Y$ such that $\omega_{X,Y}(Z)=\langle \nabla_{X}Y,Z \rangle$. Thus, I would have a well defined map $\nabla$ satisfying the Koszul Formula, and it would be (after some property checking) the Levi-Civita connection.

The problem is that, while it's easy to see that $\omega_{X,Y}(Z_{1}+Z_{2})=\omega_{X,Y}(Z_{1})+\omega_{X,Y}(Z_{2})$, I haven't been able to prove that $\omega_{X,Y}(fZ)=f\omega_{X,Y}(Z)$. In reality, what I've gotten is that

$\omega_{X,Y}(fZ)=f\omega_{X,Y}(Z)-\dfrac{1}{2}\langle (Xf)Y+(Yf)X,Z \rangle$,

but I'm not sure that the second summand is $0$. Is my attempt correct so far, or am I missing something?

Thank you in advance!

  • 1
    $\begingroup$ Lie groups/algebra? $\endgroup$
    – Narasimham
    Apr 13 '20 at 19:07
  • $\begingroup$ This appeared to me as part of something unrelated to Lie groups (even though my plan is to study isometric actions of a Lie group over a Riemannian manifold in some time). $\endgroup$ Apr 13 '20 at 19:11

Just checked and $\omega_{X,Y}$ is in fact $C^\infty$ linear. It seems like the terms you picked up were from the 4th and 6th term, the first and second terms also give extra terms:

$$X(\langle Y,fZ\rangle ) = X(f\langle Y,Z\rangle) =X(f)\langle Y,Z \rangle + fX(\langle Y,Z\rangle)$$

Similarly for the second term, these will exactly cancel the other two terms that you found.

Also this is cool because thats the first time I've seen this being proven coordinate free!

  • 2
    $\begingroup$ I see! That was the mistake, i forgot that $X$ is a derivation, so I didn't get those terms. Thank you!! $\endgroup$ Apr 13 '20 at 19:36
  • $\begingroup$ Of course no problem! $\endgroup$
    – Basel J.
    Apr 13 '20 at 19:41

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