How do I prove the following:

Let $M$ be a Riemannian manifold. Let $\omega: M\rightarrow T^{*}M$ be a smooth 1 form on $M$. Then, there exists a unique smooth vector field $Y$ on $M$, such that

$\omega(X)=\langle X,Y\rangle$ for every smooth vector field $X$ on $M$.

I'm not sure how to construct such a smooth vector field $Y$. I initially thought of using Riesz representation theorem, but I wasn't sure how to proceed.

May I have hints? (without using Christoffel symbols)


1 Answer 1


On every tangent space you can indeed use the Riesz representation theorem to construct a one-form fulfilling the condition at that particular point. Using this, you can define a section of the cotangent bundle in a pointwise manner. What remains to be shown is that this section is actually smooth. For this you only have to understand the linear algebraic situation: the explicit formula for the value at a given point $p \in M$ only contains the values of the Riemannian metric and the given vector field $Y$ at $p$. Convince yourself of this!


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