# Higher dimensional version of Stoke's Theorem / Divergence theorem

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.

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.

• 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 – user239743 Jan 12 at 0:29
• Are those products just much copies of one vector or form? – user239743 Jan 12 at 0:30
• 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. – lux Jan 12 at 0:36
• That's great - a really clear answer. Thanks to you very much – user239743 Jan 12 at 1:16