I've learnt about Stokes' Theorem and the divergence theorem that relate integrals of functions over manifolds to integrals of related functions around the boundary of the manifolds but all in 3-dimensional vector calculus.

Can anyone recommend references for generalisations to 4 dimensions or higher? Cheers!


There is indeed a generalisation, although it requires some more advanced language. As @Artem Alexandrov mentioned this requires the use of differential forms. Note: Mathematicians should see the caveat at the end of post.

I'll give the result first, because I think that's what OP wants to see, and then go on to explain the concepts used. The general statement of Stokes' theorem relates the integral over the boundary, $\partial{M}$, of a manifold, $M$, of a form, $\omega$, to the integral over the boundary of the (exterior) derivative of the form: $$ \int_{\partial M} \omega = \int_{M} d\omega.$$ In this language, Stokes' area theorem is recovered if $M = A$, is some two dimensional space with one dimensional boundary $S$ and $\omega = f_{i}(\mathbf{x})dx ^{i}$ is some vector function with the differential factors $dx^{i}$ (see below for more details). Then the action of the exterior derivative, $d$, is to differentiate and anti-symmetrise as follows: $$d \omega = d \left(f_{j} dx^{j} \right) = \epsilon_{ij} \partial_{i} f_{j} \,dx^{i} dx^{j}.$$ If $f_{i} = (f_{x}, f_{y})$ then $$d\omega = \partial_{x}f_{y} dy dx - \partial_{y}f_{x} dx dy = (\partial_{x} f_{y} - \partial_{y}f_{x})dx dy$$ which we note takes the form $(\nabla \wedge f) \cdot \mathbf{n}\, dx dy = (\nabla \wedge f) \cdot \mathbf{n} \, dA$ where $n$ is normal to the surface and so $$\int_{S} \mathbf{f}(x, y) \cdot d\mathbf{x} = \int_{A} (\nabla \wedge \mathbf{f}(x, y)) \cdot d \mathbf{A}$$ which is Stokes' theorem. In a similar way the divergence theorem can be recovered.

Now, some more information about forms!

You're probably used to the idea of forming a scalar product from two vectors, $\langle u, v \rangle \in \mathbb{R}$ maps vectors in the tangent space of the manifold to a real number (I will deal with only real valued functions here). We can make say this in fancier language by considering "covectors" which I'll define as $\omega_{u}$ such that $\omega_{u}(v) := \langle u, v \rangle$. We can think, then, of $\omega_{u}$ as a map $\omega_{u} : T_{M} \rightarrow \mathbb{R}$ taking vectors (in the tangent space $T_{M}$) to real numbers. Mathematicians say that $\omega_{u}$ lives in the cotangent space $T^{\star}_{M}$, which is the dual space to $T_{M}$.

What has this got to do with calculus? Well it turns out that in a chosen basis we can use $\frac{\partial}{\partial x^{\mu}}$ as a basis for $T_{M}$ so that an abstract vector $v = v^{\mu} \frac{\partial}{\partial x^{\mu}}$ with $v^{\mu}$ the (contravariant) components of the vector. Also we have for covectors that $\omega = \omega_{\mu}dx^{\mu}$ where the $dx^{\mu}$ form a basis in $T_{M}^{\star}$ and $\omega_{\mu}$ are the components of the one form. To see this makes sense, we note that $\frac{\partial}{\partial x^{\mu}} x^{\nu} = \delta_{\mu}^{\nu}$ so that defining $dx^{\nu} (\frac{\partial}{\partial x^{\mu}}) = \frac{\partial x^{\nu}}{\partial x^{\mu}} = \delta^{\nu}_{\mu}$ we get the result $\omega(v) = \omega_{\nu}v^{\mu}dx^{\nu}(\frac{\partial}{\partial x^{\mu}}) = \omega_{\nu}v^{\mu}\delta^{\nu}_{\mu} = \omega_{\mu}v^{\mu}$ giving the scalar product between components of the one form and the vector. The map $\omega$ from the tangent space to its dual is therefore achieved via the metric, $\omega_{u,\, \mu} = u_{\mu} = g_{\mu\nu}u^{\nu}$ (we call $u_{\mu}$ the covariant components) so that $\omega_{u}(v) = u_{\mu}v^{\mu} = \langle u, v \rangle$.

Thus you can see the metric as a map $g: T_{M} \rightarrow T^{\star}_{M}$ as mapping between tangent space and dual. The inner product is also a map $T^{\star}_{M} \times T_{M} \rightarrow \mathbb{R}$. Note that it makes perfect sense to integrate a one-form over a one-dimensional manifold, $\int_{M_{1}} \omega = \int_{M_{1}} \omega(\mathbf{x})_{\mu} dx^{\mu}$ is well defined.

The game goes on. With vectors we often form their tensor product, $T_{u, v, \cdots r} = u^{\mu}v^{\nu}\cdots r^{\sigma} \partial_{\mu}\partial_{\nu}\cdots \partial_{\sigma} \in T_{M}\otimes T_{M} \cdots \otimes T_{M}$, where I've used the shorthand notation $\partial_{\mu} := \frac{\partial}{\partial x^{\mu}}$ etc. This transforms as the tensor product of $n$-vetors. Likewise we can define higher order forms, an $n$ form looking like $$\omega = \omega_{\mu \nu \cdots \sigma}dx^{\mu}dx^{\nu}\cdots dx^{\sigma} \in T^{\star}_{M} \otimes T^{\star}_{M} \cdots \otimes T^{\star}_{M}.$$ This is done by the generalisation of the cross produt (denoted $\wedge$) to N-forms and in particular means that the form's components have specific (anti-) symmetry properties.

Now crucially there is a "natural" operation we can do on forms, a natural map from $n$ forms to $n + 1$ forms, called the "exterior derivative," $d$. The map is such that it acts on the covariant components of forms by differentiation and anti-symmetrisation $$(d \omega)_{\mu_{1} \cdots \mu_{n} \mu_{n+1}} = (n+1)\partial_{[\mu_{n+1}}\omega_{\mu_{1}\cdots \mu_{n}]}$$ where the square brackets indicate anti-symmetrisation. Note if we start with a $1$ form and apply $d$ then the components are anti-symmetric; thus applying $d$ a second time will return $0$: $d^{2} = 0$. In any case, $$ d \omega = (n+1)\partial_{[\mu_{n+1}}\omega_{\mu_{1}\cdots \mu_{n}]} dx^{\mu_{1}} \cdots dx^{\mu_{n}}dx^{\mu_{n+1}}.$$ Let us note, then, that it makes sense to integrate an $n$ form over an $n$ dimensional manifold, and an $n+1$ form over an $n+1$ dimensional manifold.

Stokes' theorem is the spectacular result that integrating a $d-1$ form, $\omega_{d-1}$ over the $d-1$-dimensional boundary of a $d$-dimensional manifold is equal to integrating the $d$-form $\omega_{d} = d \omega_{d-1}$ over the $d$ dimensional boundary. Indeed, a $d$ form will have the form $\omega_{d\, \mu_{1} \cdots \mu_{d}} dx^{1}\cdots dx^{d} = \omega_{d\, \mu_{1} \cdots \mu_{d}} dV$ where $V$ is the volume measure; $d \omega_{d} = 0$ due to the anti-symmetrisation of $d + 1$ indices taking on only $d$ values.

Note to mathematicians

I am purposefully simplifying part of this explanation to avoid getting lost in details. I know the tangent space, $T_p(M)$ is defined at a point, $p$, where $p$ has local coordinates $x$ that are the arguments of the vectors at that point. I think it should be clear above that the vectors are really vector fields and we want to work in the tangent bundle.

  • $\begingroup$ Thanks - that's really helpful.It helps me see the bigger ideas or the more detailed case. I dont get all of the details but I see general idea of inventing generalizations of vectors. I read more about this now $\endgroup$ – user239743 Jan 12 at 0:29
  • $\begingroup$ Are those products just much copies of one vector or form? $\endgroup$ – user239743 Jan 12 at 0:30
  • $\begingroup$ The way I've constructed them, yes. Except it turns out not all tensors can be constructed in this way from vectors. But by definition a tensor transforms as of it were a product of such vectors. $\endgroup$ – lux Jan 12 at 0:36
  • $\begingroup$ That's great - a really clear answer. Thanks to you very much $\endgroup$ – user239743 Jan 12 at 1:16

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