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?

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    $\begingroup$ 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). $\endgroup$ – peter a g Mar 3 at 3:08
  • $\begingroup$ "Your object belongs to the dual of the notion of a vector field" might have been more idiomatic of me... $\endgroup$ – peter a g Mar 3 at 3:17
  • $\begingroup$ 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. $\endgroup$ – 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.)

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