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In [1] on page 118 the authors introduce differential k-forms $\omega$ on $U \subset \mathbb{R}^n$ by \begin{equation} \omega : U \subset \mathbb{R}^n \to\Lambda^k\mathbb{R}^n \end{equation} where $\Lambda^k$ is the space of k-vectors and $(\mathbb{R}^n)^\ast$ is the dual of $\mathbb{R}^n$.

Now they define the 1-form $dx^i$ by \begin{equation} \tag{1} dx^i : x \in \mathbb{R}^n \mapsto e^i \in \Lambda^1 \mathbb{R}^n \simeq (\mathbb{R}^n)^\ast \end{equation} where $e^1,\ldots,e^n$ denotes the dual basis of $\mathbb{R}^n$.

They conclude \begin{equation} dx^i(x) = x^i \tag{2}. \end{equation}

It seems to me that by $x^i$ they don't mean the $i$-th component of $x$, but the coordinate-map \begin{equation} x^i : v \in \mathbb{R^n} \mapsto v_i \in \mathbb{R} \end{equation} for this is the only way that (1) and (2) don't contradict each other.

Question 1: correct so far?

Now for any smooth function $f : U \to \mathbb{R}$ they define \begin{equation} df : x \in U \mapsto \sum_{i=1}^n f_{x^i}(x) dx^i \in \Lambda^1 \mathbb{R}^n \simeq (\mathbb{R}^n)^\ast \end{equation} where I suppose $f_{x^i}(x)$ is the $i$-th partial derivative of $f$ (the notation is never clarified).

Now $df(x)$ is an element of $(\mathbb{R}^n)^\ast$ and we can apply it to some $v \in \mathbb{R}^n$. This yields \begin{equation} df(x)(v) = \left(\sum_{i=1}^n f_{x^i}(x)dx^i \right) (v) = \sum_{i=1}^n f_{x^i}(x)dx^i(v) = \sum_{i=1}^n f_{x^i}(x) e^i = Df(x). \end{equation}

Question 2: isn't $df(x)(v)$ supposed to be $D_v f(x)$ (the directional derivative?) Where is my mistake?

Any help is much appreciated!

[1] Giaquinta, M., G. Modica und J. Souek: Cartesian currents in the calculus of variations. I, Nr. 37 in A Series of Modern Surveys in Mathematics [Results in Mathematics and Related Areas. 3rd Series. A Series of Modern Surveys in Mathematics]. Springer, Berlin, 1998, ISBN 3-540-64009-6. Cartesian currents.

mjb
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  • @ChristianBlatter Are you sure about that? But then $dx^i$ does not map to $(\mathbb{R}^n)^\ast$ as in the definition of differential forms. Isn't the mistake rather that $df : x \mapsto \sum f_{x^i}(x)d x^i$ instead of $df : x \mapsto \sum f_{x^i}(x)d x^i(x)$? – mjb Jun 26 '13 at 16:12
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    The line "$\omega:\ U \subset \mathbb{R}^n \to\Lambda^k\mathbb{R}^n$" should be replaced by "$\omega\ : U \subset \mathbb{R}^n \to\bigl(\Lambda^k\mathbb{R}^n\bigr)^$". Formula (1) should be replaced by $$dx^i : x \in \mathbb{R}^n \mapsto \langle e^i,x\rangle\in{\mathbb R}\ ,$$ so $dx^i\in \bigl( \Lambda^1 \mathbb{R}^n\bigr)^ \simeq (\mathbb{R}^n)^* $. In this way $dx^i(x)=x^i$, i.e. $dx^i$ computes the ith coordinate of x with respect to the basis $(e_1,\ldots,e_n)$. – Christian Blatter Jun 26 '13 at 18:41
  • @ChristianBlatter Thank you! I am so sorry, but I still don't get it. I agree with your replacement of the first line. But then $\omega(x) \in (\Lambda^k \mathbb{R}^n)^\ast$ and not $\omega \in (\Lambda^k \mathbb{R}^n)^\ast$. This somehow seems to contradict your definition of $dx^i$ as $\omega = dx^i \in (\Lambda^1\mathbb{R}^n)^\ast$ instead of $\omega = dx^i(x) \in (\Lambda^1\mathbb{R}^n)^\ast$. Doesn't it? – mjb Jun 27 '13 at 07:43
  • Has anyone considered what I said in my first comment that instead of $df : x \mapsto \sum f_{x^i}(x) dx^i$ the authors just could have said $df : x \mapsto \sum f_{x^i}(x) dx^i(x) \in (\mathbb{R}^n)^\ast$, because then everything works out: $df(x)(v) = (\sum f_{x^i}(x) dx^i(x))(v) = (\sum f_{x^i}(x) e^i)(v) = \sum f_{x^i}(x)v_i=D_vf(x)$? Then this would be the only thing one would have to change. Does that make sense? – mjb Jun 27 '13 at 07:50

2 Answers2

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Yeah, this seems to be very sloppy. They should've said $dx^i = e^i$ (this is used in your final equation), and it should be a directional derivative. When they simplify $dx^i(v)$, they should get $e^i(v)$, and instead they leave off the $v$.

Yes, $f_{x^i}$ is meant to denote a partial derivative with respect to $x^i$.

Muphrid
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  • Same question as for ChristianBlatter above... – mjb Jun 26 '13 at 16:14
  • I don't see how your question to Christian applies to me. Further, the $dx^i$ may depend on position, so they must be functions of $x$. – Muphrid Jun 26 '13 at 16:24
  • Maybe I misunderstood. You mean they should have said $dx^i=e^i$ in (2), right? This makes sense for the last equation, but if $dx^i=e^i$ then $dx^i(x)=\langle e^i,x \rangle \in \mathbb{R}$. That means $dx^i$ does not map to $(\mathbb{R}^n)^\ast$ contrary to the definition of 1-forms. Is the definition wrong? – mjb Jun 26 '13 at 18:23
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    I see; we read the $(x)$ in $dx^i(x)$ to mean two different things. I read it to mean $dx^i$ at the position $x$, which is by definition $e^i$. You read it to mean $dx^i$ acting upon the vector $x$, which would be $e^i \cdot x = x^i$. It may be the latter is actually meant after all. I don't think this violates what is written. Really, $dx^i$ takes two things in: (1) a position to be evaluated at and (2) an input vector that is a member of the tangent space. – Muphrid Jun 26 '13 at 18:33
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You are correct. From what you've put here, this seems to be a sloppily written/edited text.

Ted Shifrin
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