Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $\Delta^k$ the standard $k$-dimensional simplex, $\Delta^k=\{(x^0,\dots,x^k)\in \mathbb{R}^{k+1}|\sum_{i=0}^kx^i=1\}$, and let $\Omega^\bullet(\Delta^k)$ be the de Rham complex of smooth differential forms on $\Delta^k$ (as usual a differential form is smooth on $\Delta^k$ if it admits a smooth extension to an neighborhood of $\Delta^k$). It is well known, dating back at least to Poincare', that $\Omega^\bullet(\Delta^k)$ is an acyclic complex (in strictly positive degree).

Now consider the following graded subspace $\Omega^\bullet_{ctb}(\Delta^k)$ of $\Omega^\bullet(\Delta^k)$: for any point $p$ in $\Delta^k$, let $\Delta_p$ be the lowest dimensional simplicial subset of $\Delta^k$ the point $p$ belongs to and let $\pi_p$ the orthogonal projection on the linear subspace determined by the $\Delta_p$; and say that a differential form $\omega$ is "constant towards the boundary" if for any point $p$ in $\Delta^k$ there is a neighborhood $U$ of $p$ such that $\omega=\pi_p^*(\omega\vert_{\Delta_p})$ on $U$ (these differential forms are called "differential forms with sitting instants" in Cech cocycles for differential characteristic classes).

Since the de Rham differential commutes with pullback (and restriction is a special case of pullback), it is immediate to see that $\Omega^\bullet_{ctb}(\Delta^k)$ is actually a subcomplex of $\Omega^\bullet(\Delta^k)$, and my question is: is $\Omega^\bullet_{ctb}(\Delta^k)\hookrightarrow \Omega^\bullet(\Delta^k)$ a quasiisomorphism? (equivalently, is $\Omega^\bullet_{ctb}(\Delta^k)$ acyclic in positive degree?). Or, in completely explicit terms, is it true that if $\omega$ is a closed differential form on $\Delta^k$ which is constant towards the boundary, one can find a primitive $\eta$ which is constant towards the boundary?

I've worked out the lowest $k$ cases by hand, and the answer is yes. For $k=1$ this is almost trivial and for $k=2$ it is just a bit more tricky, working inductively. Since no change in the homotopy typ happens as dimension of the simplex increases, I guess acyclicity of $\Omega^\bullet_{ctb}(\Delta^k)$ will be true for any $k$, but the inductive argument I have in mind gets more and more involved at every step, so that I rapidly loose control over it. On the other hand I am convinced that a clever application of Poincare' lemma (i.e. by choosing a contraction of the simplex to a point along a vector field which is orthogonal to the boundary and with a sink at the center of the simplex), but I'm not completely sure of this argument.

(do not hesitate to close this question if it is too localized)

share|improve this question
1  
How about using sheaves? Show that the sheaf of such forms is acyclic (no higher sheaf cohomology) and note that the usual Poincare lemma applies stalkwise. –  Tom Goodwillie Jun 15 '11 at 15:07
1  
Sure! proving stalkwise Poincare' lemma should be pretty immediate, and then the usual primary school spectral sequence does the job! Thanks a lot, Tom, I will blame myself for ages for having not thought of sheaves before asking! :) –  domenico fiorenza Jun 15 '11 at 15:45
1  
(and proving stalkwise Poincare' is not pretty immediate: it is completely and totally immediate. thanks again) –  domenico fiorenza Jun 15 '11 at 15:50
    
Finally, to show that the sheaf of such forms is acyclic it would suffice to prove that there exist partitions of the unity constant toward the boundary of the simplex (subordinate to a given open cover), which should not be hard. –  domenico fiorenza Jun 15 '11 at 17:23
    
(and indeed it is not hard at all: one can mimic step by step the usual construction of partitions of unity on smooth manifolds, but with bump functions which are constant toward the boundary of the simplex) –  domenico fiorenza Jun 15 '11 at 18:12

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.