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I am tripping over something elementary (I think).

Given a smooth map $f\colon M\to N$ between smooth manifolds, the differential of $f$ at $p\in M$ is defined as \begin{align}\mathrm{d}_pf \colon T_pM &\to T_{f(p)}N\\ X&\mapsto X(-\circ f)\end{align} The gradient of $f\colon M\to \mathbb{R}$ at $p$ is, if I understand correctly, just the previous definition with $T_{f(p)}\mathbb{R}\cong\mathbb{R}$. But this is were I get confused, because immediately after the gradient of $f$ is defined by $$\mathrm{d}_pf(X):=X(f), \qquad \text{ with }X\in T_pM.$$ I understand this, since $\mathrm{d}_pf\colon T_pM\to\mathbb{R}$ (linearly) it is a covector and the right hand side works, but how can I see this as a special case of the above? (i.e. why does the $``\, -\ \circ\,''$ get dropped?)

And how do the elements of $T_{f(p)}\mathbb{R}$ act on real functions? (perhaps this is more like what I am looking for).

prt13463
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    Remember that $X(f)$ is a real number. If you want to think of it this way, it's the tangent "vector" at $f(p)\in\Bbb R$ that assigns to a function $\phi\colon\Bbb R\to\Bbb R$ the number $X(f)\phi'(f(p))$. Note that this looks a lot like the chain rule for differentiating $\phi\circ f$ at $p$. – Ted Shifrin Jul 30 '17 at 20:06

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The isomorphism $$\iota_n:T_p\mathbb{R}^n \to \mathbb{R}^n$$ is given by $X \mapsto \left(X(\pi_1),\cdots,X(\pi_n)\right)$, where $\pi_i$ are the projections on the $i$-th coordinate.

In the particular case of $\mathbb{R}$, we then have that the isomorphism $$\iota_1:T_p\mathbb{R} \to \mathbb{R}$$ is given by $X \mapsto X(\mathrm{Id})$, where $\mathrm{Id}$ is the identity. Therefore,

$(\iota_1\circ d_pf)(X)=\iota_1((d_pf)(X))=\iota_1(X (-\circ f))=X(\mathrm{Id} \circ f)=X(f).$

Aloizio Macedo
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  • Thank you. I think I understand now. Writing $\mathrm{d}pf(X):=X(f)$ is a (slight, but very much real) abuse of notation. $X(f)$ is not an element of $T{f(p)}\mathbb{R}$, but rather the image of $\mathrm{d}pf(X):=X(-\circ f)$ under the isomorphism $T{f(p)}\mathbb{R}\to\mathbb{R}$. – prt13463 Jul 30 '17 at 20:44
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    @prt13463 Yes, you are correct. Differential topology is a very big abuse of notation in many places (specially things concerning tangent space). But this is for good reason: if notation were not abused, it would be overencumbered. That being said, I don't think that it is very common to visualize the gradient as $d_pf$ per se. The definition is usually directly by $(df)(X)=Xf$ (i.e., as a $1$-form) because it is the way we usually think about it. What I did above is just expose that "under the correct identifications", this coincides with $d_pf$. – Aloizio Macedo Jul 30 '17 at 20:54
  • I completely agree, but every once in a while I look at the equations and like to put every object where it belongs, to actually become ``aware'' of the abuses of notation. For example, after a bit, I understood the abuse $f\equiv f\circ x^{-1}$ ($x$ a chart map), but this one really threw me off for some reason. Thanks again :) – prt13463 Jul 30 '17 at 21:01