Why is $df|_p(X)$ called the “gradient of $f$” in differential geometry?

In undergraduate applied vector calculus, the gradient is simply the one-dimensional array of partial derivatives of a function $f$: $(\delta_1f,...,\delta _df)$

In differential geometry (at least in the general relativity course I'm taking), the gradient of $f$ (a function on a manifold $M$) is defined as a map $$(df)_p: T_pM \to \mathbb R,$$ $$(df)_p(X) := X(f),$$ where $T_pM$ is the tangent space on point $p \in M$.

Why do these definitions intuitively capture the same concept of "gradient"?

• Gradient means "A rate of inclination or declination of a slope" (from Wiktionary), which is very well captured by $\nabla$. – Henricus V. Feb 18 '17 at 6:24
• If $f$ defines a surface, then the "gradient" vector points "uphill", along the "gradient of height". – Omnomnomnom Feb 18 '17 at 6:30

If you fix a Riemannian metric $g$ on $M$, then the vector field $\nabla f$ determined by $$g_p(\nabla f, X) = X_p(f) = (df)_p(X)$$ is a direct generalization of the gradient in calculus. Specifically, if $M = \mathbb{R}^n$ and $g$ is the usual scalar product then $\nabla f$ is just the vector of partial derivatives.
Referring to $(df)_p$ itself as the gradient of $f$ is a slight abuse of notation.
Edit Here is what happens when $M = \mathbb{R}^n$. If $X = \partial_i$, then $$g(\nabla f, \partial_i) = \partial_i (f);$$ since these are an orthonormal basis of $T_p M$ at every $p$, it follows that $$\nabla f = \sum_{i=1}^n g(\nabla f, \partial_i) \partial_i = \sum_{i=1}^n (\partial_i f) \partial_i,$$ so the components of $\nabla f$ are indeed just the $\partial_i f$.
As Travis points out, $\nabla f$ depends on the choice of $g$, while $df$ does not. This makes $df$ a more natural object.
• Do you mean dot product? Also, could you explain a little bit more in detail why $\nabla f$ would then be the vector of partial derivatives? – user56834 Feb 18 '17 at 6:58
• It's perhaps worth mentioning explicitly that the gradient $\nabla f$ depends on the choice of $g$, whereas the differential $df$ itself does not. – Travis Feb 18 '17 at 11:45
• @Programmer2134 OK. It comes down to the fact that the partial derivatives $\partial_i$ are orthonormal with respect to the standard Riemannian metric (yes, it's essentially the dot product) – user399601 Feb 18 '17 at 16:29