I'm reading a book on differential forms, and it says:

A basis for $(\mathbb{R}_p^3)^*$ is obtained by taking $(dx_i)_p, i=1,2,3$, where $x_i:\mathbb{R^3}\to\mathbb{R}$ is the map which assigns to each points its $i$th coordinate.

But everywhere else I'm reading that "the map which assigns to each points its $i$th coordinate" is written as $dx_i$, not just $x_i$? So shouldn't the quote have read (with the bold d indicating my change):

A basis for $(\mathbb{R}_p^3)^*$ is obtained by taking $(dx_i)_p, i=1,2,3$, where d$x_i:\mathbb{R^3}\to\mathbb{R}$ is the map which assigns to each points its $i$th coordinate.

If not, then what is $dx_i$? if $x_i$ are the coordinate functions, then what is $dx_i$ defined to mean?

Edit: Since the notation tag asks users to give the source of the notation: the book is "Differential forms and Applications" by Manfredo P. do Carmo.

edit2: Compare the above quote to the quote from "Calculus a complete course" By Adems and Essex:

Differentials as basis for 1-forms For $1\leq i\leq n$, let $dx_i$ be the 1-form that assigns to $v\in\mathbb{R}^n$ its $i$th component $v_i$:$$dx_i(v)=v_i$$

So now I'm confused again... How do these 2 definitions work together? Is Adems just neglecting to mention that we are actually working in a tangent space?

edit3: If you are also confused about differential forms, I would recommend reading "An introduction to Manifolds" by "Loring W. Tu". I'm reading it right now and it gives an a lot more detailed and elementary introduction to this stuff. More suitable for someone like me who has never before heard anything about manifolds/differential forms and is self-studying.

  • $\begingroup$ Yes, Adems is neglecting to mention that you are actually working in a tangent space. $\endgroup$ – Qiaochu Yuan Mar 6 '15 at 17:36
  • $\begingroup$ @QiaochuYuan Thanks again for all your help. $\endgroup$ – user2520938 Mar 6 '15 at 17:52

$dx_i$ is not a function; it is a differential form. In general, if $f : \mathbb{R}^n \to \mathbb{R}$ is a smooth function, then $df$ is the differential form which in coordinates is given by

$$df = \sum \frac{\partial f}{\partial x_i} dx_i.$$

Admittedly I now have to tell you what $dx_i$ is. One way of describing what $dx_i$ does is that it assigns numbers to tangent vectors: if $v \in T_p(\mathbb{R}^n)$ is a tangent vector $(v_1, \dots v_n)$ at a point $p$, then

$$dx_i(v) = v_i.$$

That is, it assigns to a tangent vector its $i^{th}$ coordinate. The confusing thing about this situation is that each tangent space at each point of $\mathbb{R}^n$ can be canonically identified with $\mathbb{R}^n$ itself; this stops being true on more general smooth manifolds, and here it's easier to tell the difference.

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  • $\begingroup$ Hm, so is it correct to say that if $f:\mathbb{R}^n\to\mathbb{R}$ then $df_p(x)=f_1(p)x_1+f_2(p)x_2+...+f_n(p)x_n$. So that $d$ basically just creates from a functional $f$ on $\mathbb{R}^n$ a new linear functional on $\mathbb{R}^n$ with coefficients given by the partial derivatives of $f$? $\endgroup$ – user2520938 Mar 5 '15 at 21:43
  • $\begingroup$ @user2520938: if $x$ refers to a tangent vector and $f_i$ refers to the partial derivatives, then yes, but you'll confuse yourself down the line if you don't build a habit of distinguishing points in $\mathbb{R}^n$ from tangent vectors to points in $\mathbb{R}^n$. $\endgroup$ – Qiaochu Yuan Mar 5 '15 at 21:44
  • $\begingroup$ Oke I think I get it now. It's just a bit confusing that in $\mathbb{R}^n$ the tangent space at a $p$ is (or at least is very similiar to) just $\mathbb{R}^n$ again (right?). But in general the function $df_p(x)$ is a linear functional on the tangent space of $p$ right? Again with the coefficients given the partial derivates of $f$ evaluated at $p$. $\endgroup$ – user2520938 Mar 6 '15 at 6:39
  • $\begingroup$ @user2520938: right, keeping in mind that "coefficients" and "partial derivatives" in general both depend on a choice of local coordinates at $p$. $\endgroup$ – Qiaochu Yuan Mar 6 '15 at 8:10
  • $\begingroup$ Oke thanks for the clarification. Now I feel like I can actually read the book and know what they are talking about. $\endgroup$ – user2520938 Mar 6 '15 at 8:54

Rather than dealing with $\mathbb{R}^3$, I think it is more illuminating to look at an abstract $V$ finite dimensional vector space, because that one does not have a natural basis.

A coordinate system, when abstracted from the "grid that spans space" is basically a machine that takes a point in some space (topological space, manifold, vector space, etc.) and turns it into an $n$-tuple of numbers, where $n$ is the dimensionality of the space, in your example, 3.

So, let $V$ be 3 dimensional so that it maches your example. Then a coordinate system, which I'll denote as $\Psi$, is a $V\rightarrow\mathbb{R}^3,\ V\ni\mathbf{p}\mapsto\Psi(\mathbf{p})=(x^1(\mathbf{p}),x^2(\mathbf{p}),x^3(\mathbf{p}))$ map. I denoted a point of $V$ as $\mathbf{p}$, which is also a vector obviously, since $V$ is a vector space. I denoted the components of $\Psi$ with $x^1,x^2,x^3$, since $\Psi$ has values in the real coordinate space $\mathbb{R}^3$, I can talk about components.

NOW, as such we can also view a coordinate system as the $x^i(\mathbf{p})$ maps separately, in this case, we can view $x^i$ as the map that assigns the $i$th coordinate to the point $\mathbf{p}$.

What is very important is, that there is absolutely no reason to assume that the coordinate system is a linear map. In linear algebra it is generally assumed, because linear algebra only deals with linear maps, but analyis doesn't. And in differential geometry, on a manifold, it cannot even be a linear map, because a manifold is generally not a vector space. If you use a coordinate system that is NOT linear, you will "ruin" the vector space structure of your space. Of course it will still be a vector space, but adding the coordinates of two points, will not give you the coordinates of the "real" sum of the two points.

SO, when using nonlinear coordinate systems, you basically have to use a manifold-like approach, where the only quantities that work in coordinate representations as vectors are fields evaluated at the same point, so there is a difference between vectors and points.

The conclusion from all this is that, $x^i$ is the map that assigns a point its $i$th coordinate. The exterior differential $\mathrm{d}$ on a scalar field, however, is the same as its Fréchet-differential, and we know from analysis that the Fréchet-differential of a $F:V\rightarrow\mathbb{R}$ scalar function (and the $x^i$ coordinate maps are scalar functions) at point $\mathbf{p}_0$ is a $\left.\mathrm{d}F\right|_{\mathbf{p}_0}:V\rightarrow\mathbb{R}$ linear map, and the $\mathbf{p}_0\mapsto\left.\mathrm{d}F\right|_{\mathbf{p}_0}$ assignment is a "linear map field" aka a dual vector field.

So, $x^i$ is a (not necessarily linear) map, that assigns a point its $i$th coordinate, and $\mathrm{d}x^i$ is a dual vector field that assigns a vector field its $i$th coordinate function.

One more thing to note, that if $F(\mathbf{p})=\omega(\mathbf{p})$ is a linear map, then its Fréchet derivative at all points is $\omega$ itself, therefore if the $x^i$ are linear maps, then basically $\left.\mathrm{d}x^i\right|_{\mathbf{p}}=x^i$.

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  • $\begingroup$ Thanks for the answer. It's pretty late hare, so I'll read it tomorrow. $\endgroup$ – user2520938 Mar 5 '15 at 21:52
  • $\begingroup$ Oke I think this is pretty clear. However, does the coordinate system map $\Psi$ not at least have to be a bijection? $\endgroup$ – user2520938 Mar 6 '15 at 6:44
  • $\begingroup$ @user2520938 It doesn't have to be a global bijection, however it has to be a local diffeomorphism. Basically, it has to only map a patch of $V$ to a patch of $\mathbb{R}^3$, but on those patches it has to be invertible and continuously differentiable for both the map and its inverse. $\endgroup$ – Bence Racskó Mar 6 '15 at 13:57

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