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I have just started to study differential forms. I don't yet fully understand the definition of what a differential form is (it's a $p$-times covariant tensor field) but I know that if $U$ is an open subset of a manifold $M$ a differential form $\varphi$ on $M$ can be represented (locally on $U$) as

$$ \varphi = \sum_{(k_{1}, \dots, k_p )}^n f_{(k_1, \dots, k_p)} dx^{k_1} \wedge \dots \wedge dx^{k_p}$$

I suppose that a "smooth" differential $p$-form is one where the functions $f_{(k_1, \dots, k_p)}: U \to \mathbb R$ are smooth so let them be smooth. I think of the wedge product in the sum above a "infinitesimal volume elements".

Ok. So far so good. To simplify things consider the example of a function $f: U \to \mathbb R$ where $U \subseteq \mathbb R^n$.

Now to my problem: Wikipedia states that

"since ${\partial x^i \over \partial x^j}=\delta_{ij}$ it follows that $\displaystyle df = \sum_{i=1}^n {\partial f \over \partial x^i} dx^i$"

I compare this expression to the expression for the change of variable in the partial derivative given shortly before this claim:

$$ {\partial f \over \partial x^j} = \sum_{i=1}^n {\partial y^i \over \partial x^j} {\partial f \over \partial y^i}$$

so somehow the meaning of $dx^i$ is ${\partial x^i \over \partial x^j}$ except... there is no partial derivative ${1 \over \partial x^j}$ in the left hand side expression $df$. So I am confused.

What is $dx^i$? An example would really help. A simple one in two dimensions or so.

self-learner
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  • I think you can interpret $dx_i$ as the differential of the $i^{\text{th}}$ projection along the $i$-th coordinate. That is, $\text{pr}_i: x\in\mathbb{R}^n\mapsto x^i.$ Then we have $\text{pr}_i(x+h)=x^i+h^i=\text{pr}_i(x)+h^i.$ If write $dx^i: h\mapsto h^i,$ then we see $\text{pr}_i(x+h)=\text{pr}_i(x)+dx^i(h).$ – azc Oct 20 '14 at 05:55

1 Answers1

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If you fix a point $p \in U \subseteq \mathbb{R}^n$, then a differential $1$-form is a linear map from $T_pU$ (the tangent space to $U$ at $p$) to $\mathbb{R}$. The tangent space is spanned by the vectors $\frac{\partial}{\partial x_i}$, where $\frac{\partial}{\partial x_i}$ can be thought of as a vector ``pointing in the $x_i$ direction." Then you can express any linear map $T_p U \to \mathbb{R}$ in terms of the linear maps $dx^i$, where $dx^i$ is the map that sends the vector $a_1 \frac{\partial}{\partial x_1} + \cdots + a_n \frac{\partial}{\partial x_n}$ to $a_i$ (i.e. it picks out the $\frac{\partial}{\partial x_i}$ component). That is, the basis $dx^i$ for the linear maps from $T_p U$ to $\mathbb{R}$ (the dual space of $T_p U$) is the dual basis to $\frac{\partial}{\partial x_i}$.

As the point $p$ varies, the linear map given by the differential form changes smoothly (which is defined as you describe, by requiring each of its coefficient functions to be smooth). So you can think about $dx^i$ on all of $U$ as the differential $1$-form that takes in a vector field and picks out its component in the $\frac{\partial}{\partial x_i}$-direction, yielding a real-valued function on $U$.

cws
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  • Thank you. Your answer has been incredibly helpful to me. – self-learner Oct 21 '14 at 01:53
  • I just have one more question: you call $dx^i$ a basis for the linear maps -- would it be correct to call it a local basis? (because the $x^i$ are local coordinates?) – self-learner Oct 23 '14 at 05:59
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    Absolutely. The $dx^i$ form a basis for the tangent space at the point $p$, a single vector space. As $p$ varies, the tangent space varies, so you get a different basis at each point. Also, choosing different coordinates near $p$ leads to a different basis at $p$, so the meaning of $dx^i$ depends on the coordinates. – cws Oct 23 '14 at 07:30