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Let $M$ be an n dimensional manifold and $\omega$ a $k<n$ form on $M$. Prove that if for every submanifold $S$ in $M$ diffeomorphic to the $k$ dimension ball we have that $$\int_S \omega=0 $$ then $d\omega=0$.

I wanted to used contradiction trying to find a open set $B$ of $M$ diffeomorphic to the $k+1$ dimensional ball such that $$\int_B d\omega>0 $$ But I could not find it.

Any hint?

However If you suppose that $d\omega_p\not=0$ then you can find a neighborhood $U$ of $p$ diffeomorphic to the $n$-dimensional ball and $X_1,\dots X_k$ vector fields on $U$ such that $$\omega_q(X_1(q),\dots,X_n(q))>0 $$ but I don't how to do the next steps.

EQJ
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The problem is local so you can suppose you work in euclidean coordinates. If $\textrm d \omega \neq 0$ at some $p$, then you can find some diffeomorphism $\varphi$ such that $\varphi^*\textrm d\omega(p) = \Omega = \textrm d x_1 \wedge \textrm d x_2 \wedge \cdots \wedge\textrm d x_{k+1}$. By continuity, for some ball $B$, $$ \left\| \varphi^*\textrm d\omega(q) - \Omega\right\| < \epsilon, \text{ for all } q \in B$$ Then estimate $$ 0= \int_{\varphi(\partial B)}\omega = \int_B \varphi^* \textrm d\omega = \int_B (\varphi^* \textrm d\omega - \Omega) + \int_B \Omega \geq -\int_B \left\|\varphi^* \textrm d\omega - \Omega\right\| + \int_B \Omega = (1 -\epsilon) |B|$$ the integral with the norm, is the usuall riemann integral

yaqa
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  • I don't understand how to interpreted the nor the contuinuity nor $\phi* d\omega(p)=dx_1\wedge dx_2 \wedge \cdots \wedge dx_{k+1}$. I mean in the first you are saying that they are equal at a point and by continuity they must be very near to each other in a neighborhood of $p$, but that norm is a norm in the forms? – EQJ Oct 27 '17 at 19:27
  • $\omega$ is a section of the bundle of the k-linear alternating maps. Hence for any $p \in M$ , $\omega(p)$ is a linear mapping from $\bigwedge_1^k \textrm T_p M$ to $\mathbb R$. The idea is to apply a local diffeomorphism so that $\varphi^d\omega$ has a special type at $p$. Now, $\varphi^\omega$ is still a $k$--form, differentiable, a fortiori continuous. This means that if we move close to $p$, the form $\varphi^\omega$ will still be like $\varphi^\omega(p) = \Omega$. Now $\varphi^*\omega(p)(X_{1p}, X_{2p},\ldots, X_{kp}) = \omega(\varphi(p))(T\varphi(X_{1p}), \ldots, T\varphi(X_{kp}))$ – yaqa Oct 27 '17 at 19:40
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    And any form in an euclidean space, being a linear map, has a norm, the operator norm: $|\omega| = \sup |\omega(X)|, |X| \leq 1$ – yaqa Oct 27 '17 at 19:46
  • How do you find the diffeomorphism? – EQJ Oct 27 '17 at 21:33
  • i had a linear map in my mind. Actually it is just a morphism. Say for example that the $k$--form $\theta$ is not $0$. Hence there is a vector $X = u_1 \wedge u_2 \wedge \cdots \wedge u_k \neq 0$ such that $\theta\cdot X \neq 0$. All the $u_i$ should be linearly independent and i am able to create a linear map $T$ such that $T e_i = u_i$, $i = 1,2, \ldots, k$. – yaqa Oct 27 '17 at 22:30
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    To be honest i am not even sure if such a transformation is necessary, or even possible in its full strength (aka making a form so specific). The important thing is that the integral with the norm must be small and the integral of the constant form on the right should be positive and this was my initial intention when i was writing about $\Omega$. I ll check it out soon and be back with a better answer – yaqa Oct 27 '17 at 22:58