# Derivation of one-form/vector equation in Carroll confusion

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 $\mathrm{\mathrm{d}x^{i}}\left(\frac{\partial}{\partial x^{j}}\right)=\delta_{j}^{i}$?

• 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. Oct 25 '15 at 18:00
• I've edited my question to try to make my confusion clearer. Oct 25 '15 at 18:18
• So to speak.... Oct 25 '15 at 18:30
• If you can see it, here's the link to p56 of the Schutz book: books.google.co.uk/… Oct 25 '15 at 19:34
• 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.
– Danu
Oct 25 '15 at 19:43

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 $\frac{df}{d\lambda} = df \circ \frac{d}{d\lambda} = df\left( \frac{d}{d\lambda}\right)$.

• 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. Oct 26 '15 at 10:47
• $\frac{df}{d\lambda}$ 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.
– udrv
Oct 26 '15 at 11:54
• 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 $df=\frac{\partial f}{\partial x^{\mu}}dx^{\mu}$? Oct 26 '15 at 13:42
• 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.
– udrv
Oct 26 '15 at 14:04
• 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? Oct 26 '15 at 18:32