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I don't understand the derivation of Equation 2.14$$\mathrm{d}f\left(\frac{d}{d\lambda}\right)=\frac{df}{d\lambda} \tag{2.14}$$ in Carroll's Lecture Notes on General Relativity (http://ned.ipac.caltech.edu/level5/March01/Carroll3/Carroll2.html). He says the one-form $\mathrm{d}f$ is the gradient of a function $f$ and that the action of the one-form $\mathrm{d}f$ “on a vector $\frac{d}{d\lambda}$ is exactly the directional derivative of the function.” Schutz (Geometrical Methods of Mathematical Physics, p53) gives the same equation, but says it defines the gradient one-form $\mathrm{d}f$. Is the equation a definition or can it be derived in some way? I'm pretty much a beginner trying to learn this stuff and probably feel more at ease with component notation, such as $\frac{d}{d\lambda}=\frac{dx^{\mu}}{d\lambda}\frac{\partial}{\partial x^{\mu}}$ for a tangent vector and $\mathrm{d}f=\frac{\partial f}{\partial x^{\nu}}dx^{\nu}$ for a one-form (hope I've got those right).

Can I just add that if I assume $\mathrm{\mathrm{d}x^{i}}\left(\frac{\partial}{\partial x^{j}}\right)=\delta_{j}^{i}$ I can indeed derive the equation. However, a few pages further on (p56) Schutz appears to derive $\mathrm{\mathrm{d}x^{i}}\left(\frac{\partial}{\partial x^{j}}\right)=\delta_{j}^{i}$ from $\mathrm{d}f\left(\frac{d}{d\lambda}\right)=\frac{df}{d\lambda}$. That's what I find puzzling. Is there an explanation/derivation of $\mathrm{d}f\left(\frac{d}{d\lambda}\right)=\frac{df}{d\lambda}$ without the assumption of $\mathrm{\mathrm{d}x^{i}}\left(\frac{\partial}{\partial x^{j}}\right)=\delta_{j}^{i}$?

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    $\begingroup$ If you plug your component notation (as definitions) in there, the equation is also true, so it is true by definition. I'm not sure what exactly your question is. $\endgroup$
    – ACuriousMind
    Oct 25 '15 at 18:00
  • $\begingroup$ I've edited my question to try to make my confusion clearer. $\endgroup$
    – Peter4075
    Oct 25 '15 at 18:18
  • $\begingroup$ So to speak.... $\endgroup$
    – Peter4075
    Oct 25 '15 at 18:30
  • $\begingroup$ If you can see it, here's the link to p56 of the Schutz book: books.google.co.uk/… $\endgroup$
    – Peter4075
    Oct 25 '15 at 19:34
  • $\begingroup$ Perhaps it's a good idea for you to pick up a book on differential geometry if you're looking for a more thorough exposition of this issue. I recommend Lee's "Introduction to Smooth Manifolds"; he has a chapter about this stuff somewhere early in the book. Note, however, that this will be significantly time consuming. $\endgroup$
    – Danu
    Oct 25 '15 at 19:43
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You don't need $dx^i \left( \frac{\partial}{\partial x^j}\right) = \delta^i_j$ to derive the relation.

The easiest way to see this is working backwards from the regular expression for $\frac{df}{d\lambda}(x)$. If we denote the tangent vector as $t^\mu(x) = \frac{dx^\mu}{d\lambda}$, we have: $$ \frac{df}{d\lambda}(x) = t^\mu(x) \partial_\mu f(x) = df(t(x)) = df\left(\frac{dx^\mu}{d\lambda}\right) = \left(df \frac{d}{d\lambda} \right)(x) $$

where in the last equality above the tangent vector $t(x)$ is regarded as a vector application $t = \frac{d}{d\lambda}$. So technically $\frac{df}{d\lambda} = df \circ \frac{d}{d\lambda} = df\left( \frac{d}{d\lambda}\right)$.

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  • $\begingroup$ Apologies for my limited (high school) maths, but what's the difference between my $\frac{df}{d\lambda}$ and your $\frac{df}{d\lambda}\left(x\right)$? Thanks. $\endgroup$
    – Peter4075
    Oct 26 '15 at 10:47
  • $\begingroup$ $\frac{df}{d\lambda}$ is a function, as in "member of a function space", while $\frac{df}{d\lambda}(x)$ is its value at $x$. Also, notice that the text uses $\frac{d}{d\lambda} = t^\mu\partial_\mu$ as an application on functions $f$ that produces a function $\frac{df}{d\lambda}$, while $t=\frac{d}{d\lambda}$ acts on coordinate vectors $x$ and produces a tangent vector t(x). I guess this slight abuse of notation is the main source of confusion: just make sure to keep track of the correct domain in each case. $\endgroup$
    – udrv
    Oct 26 '15 at 11:54
  • $\begingroup$ So I need to think of everything in your derivation being evaluated at a point $x$? Are you saying $\partial_{\mu}f(x)=df$? I can't see why that is. Doesn't $df=\frac{\partial f}{\partial x^{\mu}}dx^{\mu}$? $\endgroup$
    – Peter4075
    Oct 26 '15 at 13:42
  • $\begingroup$ Yes, everything is evaluated at x, but the conclusion is independent of x. And no, when I wrote $t^\mu\partial_\mu f(x) = df(t(x))$ I meant df as a 1-form ( en.wikipedia.org/wiki/Differential_form#Concept): $d_x f(v) = \partial_v f(x) = v^\mu\partial_\mu f(x)$. In our case $v = t(x)$ and I omitted the x in $d_x$ for simplicity since it already shows in $t(x)$. The rest is just substituting the explicit form of $t(x)$ and casting it as a vector application on x. $\endgroup$
    – udrv
    Oct 26 '15 at 14:04
  • $\begingroup$ But isn't $\partial_{v}f(x)=d_{x}f(v)$ the equivalent of $\frac{df}{d\lambda}\left(x\right)= \textrm{d}f\left(\frac{dx^{\mu}}{d\lambda}\right)$, which is equivalent to $\frac{df}{d\lambda}=\textrm{d}f\left(\frac{d}{d\lambda}\right)$, in other words, the very equation we are trying to derive? $\endgroup$
    – Peter4075
    Oct 26 '15 at 18:32

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