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I have recently encountered Stokes theorem on embedded submanifolds of $\mathbb{R}^n$, and I didn't manage to find a proof for $C^1$ vector fields over $C^1$ manifolds, infact I have only seen that proof for $C^1$ vector fields on $C^2$ manifolds.

Now, it seems to me that lots of the special cases of stokes theorem (such as the divergence theorem) can be generalized to $C^1$ manifolds and I'm starting to wondering if the general case works also for such manifolds.

Does anyone have a reference, or maybe is so kind to write a proof himself, for such a generalization of the theorem, or, conversely, knows a counterexample?

Thanks in advance and excuse me for my english, I hope the question is clear enough.

Lorenzo Vanni
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    On a $C^k$ manifold, a vector field is at best $C^{k-1}$. Perhaps you mean a $C^1$ vector field on an ambient $C^2$ manifold? See Michael Taylor’s Measure Theory and Integration, Appendix I, for the divergence theorem for $C^1$ vector fields on Lipschitz domains $\Omega\subset\Bbb{R}^n$. – peek-a-boo Dec 05 '23 at 23:15
  • You are right, I confused a couple of traductions. I guess now my question is: is there a generalization of Stokes theorem that implies generalized versions of the 'Stokes type theorems', such as the divergence theorem? Thank you for the reference btw. – Lorenzo Vanni Dec 05 '23 at 23:50
  • I’m sure they exist, I just don’t know precise references (except for the $C^1$ forms on $C^2$ manifolds case for which I wrote an answer once… but you already know this case). Afterall, in analysis, we have tons of ways of lowering regularity, and approximating results (weak derivatives, distributional derivatives, convolutions, Sobolev spaces etc), so you need to systematically extend these notions to manifolds then prove a general result there (balancing regularity of the boundary and regularity of the forms). – peek-a-boo Dec 06 '23 at 00:04
  • @peek-a-boo I think we can define $C^1$ vector fields (and forms) on $C^1$ manifolds by requiring them to be restrictions of $C^1$ vector fields defined on the ambient euclidean space. This extends and agrees with the usual definition of $C^1$ vector fields on $C^2$ manifolds. Then the proof of Stokes' theorem should work with no problems. – Martin Frenzel Aug 08 '24 at 11:13
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    @MartinFrenzel if a manifold is $C^1$, then the tangent bundle is only $C^0$, so we can only have continuous vector field on the manifold. Next, you’re not talking about manifolds in general, but submanifolds. However, what you’re suggesting is having a $C^1$ vector field on an ambient $\Bbb{R}^n$, which is a $C^{\infty}$ manifold (well it’s also real-analytic, and even better a vector space) so you actually made the situation easier :) – peek-a-boo Aug 08 '24 at 11:49
  • @peek-a-boo, the question started out talking about embedded $C^2$ submanifolds, so I thought it was fair to keep this assumption for the $C^1$ case. Otherwise I agree there is no consistent definition of $C^1$ forms. But do you agree that in this case Stokes' theorem holds? I'm currently writing it up to check that I'm not missing something. – Martin Frenzel Aug 08 '24 at 12:00
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    I’d suggest you not write it up yet. There’s a strong version of the theorem I referenced above. Also, I once wrote an answer here, but I should assume there $\partial U$ is $C^2$ otherwise my proof as written there doesn’t work. If you want to write an answer just make sure (so you don’t end up wasting time) you actually prove something more general (and not make the same mistake I did). – peek-a-boo Aug 08 '24 at 12:21
  • @peek-a-boo Thank you for the link to your post and to the book (which probably has the most interesting appendix list I have ever seen in a maths book). I think that your proposed theorem is correct with $C^1$ boundaries, by the fix that you suggest in the discussion below: In each boundary chart, we can "retract" the boundary a little bit into the $C^2$ interior of the manifold and perform the analysis there, and then apply standard integral convergence theorems to get what we want. – Martin Frenzel Aug 08 '24 at 13:17
  • @peek-a-boo I also think that the case for embedded manifolds doesn't really have any of these difficulties, since we can slightly redefine what the integral of a form is in a way that recovers the $C^2$-manifold case. I'll put up the details as an anwser shortly. – Martin Frenzel Aug 08 '24 at 13:21

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I'm sure there is a much better way to do this. Anyway here is my Claim:

Let $M^m\subset\mathbb R^{m+n}$ be an embedded $C^1$ manifold with $C^1$ boundary, and let $\omega\in\Omega^{m-1}(M)$ be $C^1$ with compact support, then $$\int_M d\omega=\int_{\partial M}\omega $$

Here, as usual, a $C^1$ manifold with $C^1$ boundary is defined to be a subset that is locally the image of $\mathbb R^m$ the upper half space $\mathbb H^m\subset \mathbb R^m$ under a $C^1$-embedding $\phi:\mathbb R^m\to \mathbb R^{m+n}$. And a $C^1$ form $\omega\in\Omega^k(M)$ is just the restriction of a $C^1$ form in $\Omega^{k}(\mathbb R^{m+n})$ to $M$. And we define the integral of a continuous, compactly supported form $\eta\in\Omega^m(\mathbb R^{m+n})$ over $M=\phi(\mathbb H^m)$ to be

$$\int_M \eta =\int_{\mathbb H^m}\phi^*\eta,$$

and we use ($C^1$) partitions of unity to define $\int_M \eta$ over manifolds with more than one chart. Here, $\phi^*\eta$ is defined as usual by

$$\phi^*\eta(v_1,\cdots,v_m)=\eta(d\phi(v_1),\cdots,d\phi(v_{m+n}))$$

(i.e. here we consider $\phi$ not as a smooth map from $\mathbb H^m$ to $M$, but from $\mathbb H^m$ to $\mathbb R^{m+n}$). The theorem then follows if we can prove that

$$\int_{\partial\mathbb H^m}\phi^*\omega=\int_{\mathbb H^m}\phi^*(d\omega).$$

Unfortunately, the forms we are integrating on the left and right hand side are both generally only continuous. And applying the fundamental theorem of calculus requires that the form on the left be continuously differentiable.

My argument consists of splitting the form into pieces, and constructing charts for each piece such that the form on the left becomes differentiable, and applying Stokes' theorem on each piece.

Step 0

I start by quickly showing why this the claim is true when $n=0$. In this case we can find an open cover consisting of cubes (products of open intervals, $U_i=(a^{(1)}_i,b^{(1)}_i)\times\cdots\times(a^{(m)}_i,b^{(m)}_i)$ ) $U_i\subset \mathbb R^n$ such that $U_i$ is (1) either entirely contained in the interior of $M$, or (2) that we have (up to reordering of coordinates) $$U_i\cap M=\{x\in U_i:x_m\leq q(x_1,\cdots,x_{m-1})\}$$ or (3) $$U_i\cap M=\{x\in U_i:x_m\geq q(x_1,\cdots,x_{m-1})\}$$ for some $C^1$ function $$q:(a^{(1)}_i,b^{(1)}_i)\times\cdots\times(a^{(m-1)}_i,b^{(m-1)}_i)\to(a^{(m)}_i,b^{(m)}_i).$$ Now we may use a partition of unity to reduce to the case that $\omega$ is supported in one of these cubes. Let's say that we're in the case (2). Then write $\omega=\sum_ia_idx_1\wedge\cdots\wedge \widehat{dx_i}\wedge\cdots\wedge dx_m$ we get that $$\int_{U_i}d\omega=\int_{a^{(1)}_i}^{b^{(1)}_i}\cdots\int_{a^{(m-1)}_i}^{b^{(m-1)}_i}\int_{a^{(m)}_i}^{q(x_1,\cdots,x_{m-1})}\sum_i\frac{\partial a_i}{\partial x_i} dx_{m}\cdots dx_1.$$ Taking out the sum and applying the fundamental theorem of calculus to each $\int \frac{\partial a_i}{\partial x_i}dx_i$, we get $$=\int_{a^{(1)}_i}^{b^{(1)}_i}\cdots\int_{a^{(m-1)}_i}^{b^{(m-1)}_i}a_m(x_1,\cdots,x_{m-1},{q(x_1,\cdots,x_{m-1})})dx_{m-1}\cdots dx_1$$ $$=\int_{\partial U_i}\omega.$$And so we're done. (I'm assuming that the reader has seen the proof of the ordinary Stokes' theorem before)

Step 1

Let $A:\mathbb R^{m+n}\to\mathbb R^{m+n}$ be linear and nonsingular, then let $A(M)$ be the image of $M$ under $A$. Then $A(M)$ is still a $C^1$-manifold, and for every continuous $(m-1)$-form $\eta$ we have $$\int_{A(M)}(A^{-1})^*\eta=\int_{M}\eta.$$ Now for every $x\in M$, let $U_x\subset M$ be a neighbourhood of $x$ and $A_x:\mathbb R^{n+m}\to\mathbb R^{n+m}$ be linear such that every projection of $A_x(U_x)$ onto a subspace of $\mathbb R^{n+m}$ spanned by any $m$ of the $n+m$ standard basis vectors is a diffeomorphism onto its image. This is doable using inverse function theorems, a bit of linear algebra ...

We may choose finitely many $U_1,\cdots,U_N$ of these $U_x$, such that the support of $\omega$ is in the union of the $U_i$. If $\rho_i$ is a subordinate partition of unity, then we have that $$\int_M d\omega=\sum_i\int_{A_i(U_i)} (A_i^{-1})^*(d\rho_i\omega)$$and $$\int_{\partial M} \omega=\sum_i\int_{A_i(\partial U_i)} (A_i^{-1})^*(\rho_i\omega)=\sum_i\int_{\partial A_i(U_i)} (A_i^{-1})^*(\rho_i\omega).$$ So we may assume that $M=A_i(U_i)$ for some $i$ and write $\omega$ instead of $\rho_i\omega$.

Step 2

Write$$\omega=\sum_I a_Idx_I,$$where $I$ ranges over tuples $I=(i_1<\cdots <i_{m-1})$, and $a_I$ is a $C^1$ function $\mathbb R^{n+m}\to\mathbb R$. We have$$\int_{\partial M}\omega=\sum_I\int_{\partial M}a_Idx_I.$$We will show that for each $I$ we have $\int_{\partial M}a_Idx_I=\int_{M}d(a_Idx_I)$. Hence may assume that $\omega=adx_1\wedge\cdots\wedge dx_{m-1}$, and so we get $$d\omega=\sum_{j>m-1}\frac{\partial a}{\partial x_j}dx_j\wedge dx_1\wedge\cdots\wedge dx_{m-1}.$$Now because of the work we did in step 2, the projection $\pi(x_1,\cdots,x_{n+m})=(x_1,\cdots,x_m)$ is a diffeomorphism onto its image. Equivalently, $M$ is given the graph $$\Phi(x)=(x,\phi(x)),\phi=(\phi_{m+1},\cdots,\phi_{m+n}),x\in U$$of some $\phi:\mathbb R^m\supset U\to\mathbb R^n$, for some $U\subset\mathbb R^m$ which is a $C^1$ manifold with $C^1$ boundary. We also have that $\partial M$ is the diffeomorphic image of $\Phi|_{\partial U}$.

Step 3

We have that $$\int_{\partial M}\omega=\int_{\partial U}\Phi^*\omega,$$$$\int_{M}d\omega=\int_{U}\Phi^*d\omega.$$And some calculus of forms shows that $$\Phi^*dx_i=dx_i$$ if $i\leq m$, and $$\Phi^*dx_i=\sum_{1\leq j\leq m}\frac{\partial \phi_i}{\partial x_j}dx_j$$ if $i>m$. So we get $$\Phi^*\omega=a\circ\Phi\cdot\Phi^*(dx_1)\wedge\cdots\wedge\Phi^*(dx_{m-1})$$ $$=a\circ\Phi\cdot dx_1\wedge\cdots\wedge dx_{m-1}$$ and this is a $C^1$-form! We also have$$\Phi^*d\omega=\sum_{i>m-1}\frac{\partial a}{\partial x_i}\circ\Phi\cdot \Phi^*dx_i\wedge \Phi^*dx_1\wedge\cdots\wedge \Phi^*dx_{m-1}$$ $$=\frac{\partial a}{\partial x_m}dx_m\wedge dx_1\wedge\cdots\wedge dx_{m-1}+\sum_{i>m}\frac{\partial a}{\partial x_i}\circ\Phi\cdot \frac{\partial \phi_i}{\partial x_m}dx_m\wedge dx_1\wedge\cdots\wedge dx_{m-1}$$And we see that $d\Phi^*\omega=\Phi^*d\omega$. So we've reduced the claim to showing that $$\int_{\partial U}\Phi^*\omega=\int_{U}d\Phi^*\omega,$$ for a $C^1$-form $\Phi^*\omega$, so we have reduced the claim to the case where the codimension $n=0$: As this follows from Step 0, we're done.

Martin Frenzel
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