# Generalization of Gauss-Stokes theorem [closed]

The Gauss-Stokes theorem defined for a vector field is as follows: $$\int_V A^\alpha_{;\alpha}\sqrt{-g}\,d^4x = \oint_{\partial V} A^\alpha\,d \Sigma_\alpha\, .$$

But how do we generalize it to do integration of Tensor fields over manifolds? What if we have a tensor of rank $(m,n)$? What form does the theorem take?

• Adding extra indices to A is equivalent to just adding more equations each of which is true by the one index case. I.e, you can just write however many extra indices you like on A on both sides of your equation and it will still be true. – jacob1729 Jun 3 '18 at 10:41
• Would Mathematics be a better home for this question? – Qmechanic Jun 3 '18 at 13:40

You don't. Or it depends on what do you mean on integration of tensor fields.

Generally speaking if you are working on $\mathbb R^n$ with all the standard structures this implies, then a tensor field of type (k,l) is really just a map $\mathbb R^n\rightarrow\mathbb{R}^{n^{k+l}}$, and you can perform the integration $$I^{a_1...a_k}_{b_1...b_l}=\int d^nx\ T^{a_1...a_k}_{b_1...b_l}(x),$$ which will result in an algebraic tensor of type (k,l) (eg. not a tensor field).

On a manifold, on the other hand, you cannot have tensor-valued integrals. The reason for that, sticking with the local coordinate approach, is that, if we consider a vector field $Y^a(x)$ (we can consider a general type (k,l) tensor field but considering only a vector field saves typing time), the field values at different points transform differently. Eg. if we have the coordinate transformation $x\mapsto x'$, then we have $$Y^{a'}(x')=\frac{\partial x^{a'}}{\partial x^a}(x)Y^a(x),\ \text{but}\ Y^{a'}(y')=\frac{\partial x^{a'}}{\partial x^a}(y)Y^a(y),$$ with the two transformation matrices obviously not being the same in general. This means that it is not a meaningful operation to add $Y^a(x)$ and $Y^a(y)$ for $x\neq y$.

But an integral is essentially the limit of sums of the type $$\Delta x^1\cdot...\cdot\Delta x^n Y^a(x)+\Delta y^1\cdot...\cdot\Delta y^n Y^a(y)+...$$ for all points on the manifold, which, as it was just explained, is not meaningful.

This problem is not fixed by multiplying by $\sqrt{|g|}$ or anything like that.

On the other hand, without invoking differential forms directly, we can say that over an $n$ dimensional manifold, we can integrate totally antisymmetric tensors of type $(0,n)$, because every such tensor field $\rho_{a_1...a_n}$ is of the form $$\rho_{a_1...a_n}=\rho\pi_{a_1...a_n},$$ where $\rho$ is a scalar density, and $\pi_{a_1...a_n}$ is the Levi-Civita symbol. Actually this is only true for oriented manifolds, in positive coordinates, but I'd prefer to handwave this away now.

We know that scalar densities can be integrated in a coordinate-free manner, so we define a volume element $d\mu^{a_1...a_n}$ such that $$\int d^nx\ \rho(x)\equiv\int d\mu^{a_1...a_n}\rho_{a_1...a_n}(x).$$

Because $\pi^{a_1...a_n}\pi_{a_1...a_n}=n!$, clearly $$d\mu^{a_1...a_n}=d^nx\frac{1}{n!}\pi^{a_1...a_n}.$$

So far this seems superflous compared to simply integrating a density, but the point is that if $\Sigma$ is some $k$ dimensional surface in the "bulk" manifold $M$, then there is no simple relationship between densities of $M$ and densities of $\Sigma$. However there is a simple relationship between antisymmetric tensor fields of the two manifolds. Namely, if the $k$-surface $\Sigma$ is given by the embedding maps $$x^a=x^a(\xi^1,...,\xi^k),$$ and we use the notation $$e^a_i=\frac{\partial x^a}{\partial \xi^i},$$ then if $\lambda_{a_1...a_k}$ is a completely antisymmetric $k$-tensor field in $M$, its pullback to $\Sigma$ is $$\lambda_{i_1...i_k}=\lambda_{a_1...a_k}e^{a_1}_{i_1}...e^{a_k}_{i_k},$$ where on $\Sigma$, $\lambda_{i_1...i_k}$ is of the form $$\lambda_{i_1...i_k}=\lambda\pi_{i_1...i_k},$$ where $\lambda$ is a scalar density on $\Sigma$.

This also means that $$\int_\Sigma d^k\xi\ \lambda(\xi)=\int_\Sigma d\mu^{i_1 ... i_k}\lambda_{i_1 ... i_k}(\xi)=\int_\Sigma d\mu^{i_1 ... i_k} e^{a_1} _{i_1} ... e^{a_k} _{i_k} \lambda_{a_1 ... a_k}(\xi)=\int_\Sigma d\Sigma^{a_1...a_k}\lambda_{a_1...a_k}(\xi),$$ so as it turns out you can integrate completely antisymmetric covariant $k$-tensor fields over $k$-dimensional surfaces.

Now, assume that $\Sigma$ is a $k+1$ dimensional surface, with a $k$ dimensional boundary $\partial\Sigma$. Then the integral theorem is $$\int_{\partial\Sigma} d(\partial\Sigma)^{a_1...a_k}\lambda_{a_1...a_k}=(k+1)\int_\Sigma d\Sigma^{a_1...a_{k+1}}\partial_{a_1}\lambda_{a_2...a_{k+1}},$$ where I think, but not sure, that the factor of k+1 is needed.