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We were working on a smoothing problem and ran across the apparently simple following question: X is a triangulated smooth manifold of dimension $n$, and $\alpha$ is an exact differential form of degree $n$. Assume moreover that its integral vanishes on each $n$-simplex.

Is it possible to find a primitive $\beta$ of $\alpha$ (that is, $d\beta=\alpha$) such that $\beta$ vanishes tangentially to the $(n-1)$-skeleton?


The last sentence means that, for any vectors $v_1,...,v_{n-1}$ based on a same point and tangent to an $(n-1)$-simplex, we have $\beta(v_1,...,v_{n-1})=0$.


What we already know : take any primitive $\beta$ of $\alpha$, then, by stokes its integral on the boundary of each $n$-simplex is zero. However, its integral on each $(n-1)$-simplex is a priori not zero. Lemma 2 p.165 of Singer-Thorpe (Springer edition) implies that $\beta$ can be found to have integral 0 on each $(n-1)$-simplex. But can we then find $\beta$ to be uniformly zero tangentially to those simplices?

We know how to prove it if we replace the triangulation by a "sweet" cubic complexe.

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1 Answer 1

I think, the answer can depend on how you interpret the question. Let me show that the answer is negative for one of the interpretations already in the case of $2$-dimensional manifolds. We study the question locally in a neighbourhood of a vertex of a triangulation, so the condition on the integral over $n$-simplexes does not play any role. The obstruction for the existence of $\beta$ comes form the local behaviour of curves at a vertex.

Lemma. Let $\alpha=dx\wedge dy$ on $\mathbb R^2$. For $n\ge 8$ there exist $\gamma_1,...,\gamma_n$, smooth rays on $\mathbb R^2$ that meet at $0$ with different tangent vectors and such that there is no $\beta$ defined in any neighbourhood of 0 with $d\beta=\alpha$ and vanishing been restricted to $\gamma_i$.

It is clear that this lemma implies the negative answer to a version of the question, when we are not allowed to deform the triangulation.

Proof of Lemma. Suppose by contradiction that $\beta$ exists. Then $\beta_1=\beta-\frac{1}{2}(xdy-ydx)$ is a closed 1-form. So we can write $\beta_1=dF$, where $F$ is a function defined in a neighbourhood of $0$, $F(0)=0$. Since the number $n$ of the rays is more than $2$, $dF$ should vanish at zero. Moreover, it is not hard to see, that since the number of rays is more than $4$, the quadratic term of $F$ vanishes at zero too.

Now, since $\beta$ vanishes on $\gamma_i$, the restiction of $\beta_1$ to $\gamma_i$ equals $\frac{1}{2}(ydx-xdy)$. So we get the formula for $F$, resticted to $\gamma_i$ $$F=\frac{1}{2}\int_{\gamma_i}ydx-xdy.$$ Now, we will chose the rays $\gamma_1,...,\gamma_8$. Namely $\gamma_1(t)=(t,t^2)$, $\gamma_2(t)=(t,t-t^2)$, and take $\gamma_3,...,\gamma_8$ by consecutively rotating $\gamma_1,\gamma_2$ by $\pi/2$, $\pi$, $3\pi/2$.

It is not hard to see, that $F$ is cubic modulo higher terms in $t$ when it is restricted to $\gamma_i$. At the same time $F$ is positive on $\gamma_{1},\gamma_3,\gamma_5, \gamma_7$ and negative restricted to other rays. So it changes its sign at lest $8$ times on a little circle surrounding $0$. This is impossible for a cubic Function (in a little neighbourhood the cubic term of $F$ should be dominating). Contradiction.

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Leodnid, the condition on integral don't play a role for my reasoning. You can change this form $dx\wedge dy$ outside a little neighborhood of 0 , so that all integrals are zero. The obstruction happen just at 0 , the problem will stay. –  Dmitri Feb 15 '10 at 23:48
    
Thanks a lot! Your counter-example is very clear. Unfortunately deformations were not possible. It is true in the case of cubic complexes (in particular, more than $2^n$ cubes joining at one are not allowed) –  user4039 Feb 16 '10 at 18:04

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