# Existence of smooth vector field on Riemannian Manifold property

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)

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!