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I want to understand what it actually means to integrate a differential form on a manifold. Being a mathematician, the explanation I always get is that they simply follow the right transformation rule. This seems like a very inelegant reason since one also needs partitions of unity to make a global definition. Therefore I have tried to come up with explanations myself.

Firstly I asked myself why can I not simply integrate a function. I have found out that is because there is no intrinsic definition of volume on a manifold. In the differential form $f \mathrm{d}x$ on $\mathbb{R}$ the $\mathrm{d}x$ keeps track of length measurement. However it does so on the tangent space and not on a manifold. I have tried to map the tangent space at a point to the manifold via the flow of a vector field but my attempt was unsuccessful.

Hence here I am, asking physicists what is their interpretation of integration of differential forms, perhaps using some physical examples.

Thank you in advance for your time.

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    $\begingroup$ what it actually means For a physicist "what mathematics means" will be related (where intuitively possible) to what is happening physically. That's going to be specific to a problem. I'm not sure you are really asking this and that you are seeking some other kind of understanding in an "intuitive mathematical sense". $\endgroup$ – StephenG Aug 23 '18 at 21:51
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The crucial concept which is necessary for the integration of differential forms on manifolds is the parametrization of the manifold. Actually, this also needed for "normal integration" over a set $M \in \mathbf R^k$, furthermore vectors tangent to the parameterized manifold are needed. Let's go step by step.

  1. Math on manifolds need charts and in general one chart only covers part of the manifold, we actually need a covering of the manifold $\{ \cal U_i\}_{i\in \mathbf N}$, then we can write for a n-form $\omega$:

$$\omega = \sum_{i\in \mathbf N} h_i \omega$$ with $$\sum_i h_i =1$$ where $h_{i\in \mathbf N}$ is a partition of 1. The partition is chosen in a way that for each $i$ a chart can be chosen that covers completely $\cal U_i$. In the following we will always assume that only one chart is needed by arguing that if more charts are needed instead of $\omega$ $h_i\omega$ is considered. This was just the prelude.

  1. Now we assume a parametrization $g: \mathbf{R}^n\rightarrow \cal U_i \in \mathbf{R}^k$ respectively $\in M\,$ (n<k). At each point of the grid on $\cal U_i$ smooth tangent vectors $(\xi_{1,i_1},\ldots, \xi_{n,i_n})$ can be easily obtained by ($\Delta x_{i_1}$ and $\Delta x_{i_n}$ are the $x_{1}$- respectively $x_{n}$-intervals of the grid in $\mathbf{R}^n$ where $i_1, \ldots, i_n$ the indices which enumerate the grid points)

$$ \xi_{1,{i_1}} = (\frac{\partial g_{x_1}}{\partial x_1}, \ldots, \frac{\partial g_{x_k}}{\partial x_1})|_{g(x_1,\ldots ,x_{n})}\,\,\Delta x_{i_1}$$

$$ \ldots $$

$$ \xi_{n,{i_n}} = (\frac{\partial g_{x_1}}{\partial x_n}, \ldots, \frac{\partial g_{x_k}}{\partial x_n})|_{g(x_1,\ldots ,x_{n})}\,\,\Delta x_{i_n}$$

The differential form $\omega$ res. $h_i\omega$ is evaluated on these tangent vectors in the following form:

The n-form is evaluated on the tangent vectors $\xi_{1,i_1},\ldots,\xi_{n,i_n}$ at all grid points $(g(x_{i_1},\ldots, x_{i_n}))$ of $\cal U_i$ and for the evaluation of the integral

$$\int_{\cal U_i} h_i\omega \approx \sum_{i_1} \ldots \sum_{i_n}h_i\omega(\xi_{1,i_1},\ldots \xi_{n,i_n})$$

Riemann sums have to built up from the resulting values $\sum_{i_1} \ldots \sum_{i_n}\omega(\xi_{1,i_1},\ldots \xi_{n,i_n})$. If everything is done correctly, one ends up with an approximate integral where even the Jacobi determinants (which are needed for integration in k-dimensional space) are included automatically (n-forms are made like to produce the Jacobi determinant automatically!) and the only missing step is to take the limit $ \Delta x_{i_1} ,\ldots , \Delta {x_{i_n}} \rightarrow 0$ to obtain the value of the integral. A nice property of this definition is that the integral is independent of the parametrization $g$ which is chosen. So actually the form can be evaluated on any tangent vectors $\xi$ as long they can be associated to a smooth parametrization $g$ (or $f$ etc.). In this explanation I tried to be rather general, it is more intuitive to imagine for instance a 2-form over a surface $M$, in that case $n=2$ and $k=3$. This is a very common application in electrodynamics.

The point 2) actually can be formulated more elegantly by the concept of a pullback, where integral of $\omega$ over $M$ is written like:

$$\int_M \omega := \int_D g^{\ast} \omega$$

where D represents the area in $\mathbb{R}^n$ which is mapped by the parametrization $g: D \rightarrow M$ (or one of the partitions $\cal U_i$) and $g^{\ast}\omega$ is the pullback of $\omega$. This implies also mapping (the differential of $g$) of the tangential vectors of $D$ to $\xi_{m,{i_m}}$ with $m=1,\ldots, n$. (in other words the tangent vectors $\xi_{m,{i_m}}$ are the push forwarded tangent vectors of $D$) $D$ is a (flat)subset of $\mathbb{R}^n$, so its tangential vectors lie inside $\mathbb{R}^n$ which solves one of your doubts. The definition of integrals of differential forms on manifolds $M$ via the pullback of the form on a subset $D$ of $\mathbb{R}^n$ is the mathematically correct definition. It also should be familiar to you.

Last, but not least: An important prerequisite for this to work is that the manifold $M$ is orientable. In physics this is (practically) always the case.

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  • $\begingroup$ Awesome answer! Can I ask a little question? The parametritzation of $M$, the function $g$, can we think of it as giving a coordinate system to the manifold? So that the independence of the parametritzation can be understood as independence repect to the coordinates one uses for the manifold? I supose that yes but just wanted to confirm. Thank you! $\endgroup$ – vin92 Jun 13 '20 at 10:39
  • $\begingroup$ @vin92 As $M$ is n-dimensional embedded in a k-dimensional space (n<k) actually n-dimensional coordinates are needed whereas g is k-dimensional (k>n). So $(x_1,\ldots,x_n)$ already serve as coordinates. For instance on a sphere 2 coordinates are needed, whereas the parametrisation is 3-dimensional. BTW: I've added the important detail (k>n and $M \in \mathbf R^k$) to my post to make it clearer. Otherwise it can easily lead to confusion. $\endgroup$ – Frederic Thomas Jun 13 '20 at 20:55
  • $\begingroup$ Thanks for the reply, I was/am indeed a little confused about $g$. In the case of the Sphere you mention, could be $\theta$ and $\varphi$ (typical spherical coordinates) be a possible choice for the coordinates? And if so, which is a possible explicit form for the parametritzation $g:(\theta,\varphi)\rightarrow g(\theta,\varphi)$? Since $g(\theta,\varphi)\in R^3$, I expect it to be something like $g(\theta,\varphi)=(g_1(\theta,\varphi),g_2(\theta,\varphi),g_3(\theta,\varphi))$. Could you provide a link with an example? Thank you very much in advance $\endgroup$ – vin92 Jun 14 '20 at 10:57
  • $\begingroup$ In case of spherical coordinates the parametrisation is $g(\theta,\varphi) = (\cos(\theta)\sin(\varphi), \sin(\theta)\sin(\varphi), \cos(\varphi))$. Another parametrisation can be chosen, for instance in cartesian coordinates $h(x,y)=(x,y,\sqrt{1-x^2-y^2})$. In the first case $(\theta,\varphi)$ are the coordinates, in the second $(x,y)$. So indeed change of parametrisation leads to a change of coordinates. BTW both do not cover the whole sphere. The parts not be covered by the given parametrisations, others have to be chosen (the same apart fro a sign change at the appropriate place). $\endgroup$ – Frederic Thomas Jun 14 '20 at 11:43
  • $\begingroup$ Oh, it was simpler than I thought! Sometimes is hard to see through the mathematical fromalism. Thank you very much! $\endgroup$ – vin92 Jun 14 '20 at 20:55

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