# Differential as a Vector Field

Let $$M$$ be a smooth manifold, and let $$f :M \to \mathbb{R}$$ be a $$C^\infty$$ function. Is there a reason that the map: $$p \mapsto (p,df_p)$$ is not a vector field? It seems to be a smooth assignment of a $$p$$-tangent vector at every point. What is it I am missing, and why is a metric/inner product structure needed to define the gradient vector field?

• Tangent vectors arise by taking the derivative of maps $\mathbb R \to M$. So, canonically, your object is dual notion to that of a vector field (which is why you would need e.g., an inner product to get a vector field out of it). – peter a g Mar 3 at 3:08
• "Your object belongs to the dual of the notion of a vector field" might have been more idiomatic of me... – peter a g Mar 3 at 3:17
• Ah yes, that's a gaffe on my part! Totally didn't see that. The tangent vector thus comes when you can identify the dual with the space itself. – rubikscube09 Mar 3 at 3:19

$$df_p$$ is not a tangent vector. It is a linear map $$T_pM\to\mathbb{R}$$: feed it a tangent vector at $$p$$, and it gives you a number (the directional derivative of $$f$$ with respect to that vector). An inner product then lets you turn it into a tangent vector since there will be a unique vector that this linear map is given by taking the inner product with.
(What $$p\mapsto (p,df_p)$$ is is a section of the dual vector bundle to the tangent bundle, also known as a $$1$$-form.)