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Given a triangulation of a compact, orientable and differentiable manifold $M^m$, it is possible to give a formula (in terms of real coordinates of the ambient space) for a vector field with only non-degenerated zeros which is "good" in the sense that there is exactly $\beta_q$=($q$-Betti number of $M^m$) zeros, each one with index $(-1)^q$, over the barycentre of each $q$-cell of the triangulation, $q=0,\cdots, m$?

For example, if $P=(a_0,\cdots,a_k)$ is the barycentre of a $q$-cell, it seems to me that the gradient $v$ of the function $f:M^m\rightarrow\mathbb{R}$ given by $f=(x_0-a_0)^2-(x_1-a_1)^2+\cdots+(-1)^q(x_q-a_q)^2+(x_{q+1}-a_{q+1})^2+\cdots+(x_k-a_k)^2$ works, since in this case $sign (\ det \ d_Pv)=(-1)^q$.

But how we can "glue" these gradient fields to give a global vector field on $M^m$?

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joaohelder -- do you want the number of zeroes to be equal the number of the $q$-simplices or the $q$-th Betti number? – algori Jan 13 '11 at 16:37
At least in well-behaved situations (as the above considered) this is the same. Correct? – joaohelder Jan 13 '11 at 19:07
joaohelder -- any triangulation of the 2-sphere has 1-simplices, yet $H^1(S^2)=0$. In general there is no guarantee that a manifold $M$ admits a CW-structure such that the number of $q$-cells is the $q$-th Betti number. – algori Jan 13 '11 at 20:12

This is not quite what you asked for, but hopefully similar enough. If $M^m \subset \mathbb{R}^n$ is a closed submanifold, then for almost all $v \in \mathbb{R}^n$, the restriction of the function $x \mapsto \langle x,v\rangle$ is a Morse function $f:M \to \mathbb{R}$. The critical points of $f$ are all points $x \in M$ such that $T_x M$ is perpendicular to $v$. According to Morse theory, you get a CW decomposition of $M$ with a $p$-cell for each critical point of index $p$. The index of the gradient vector field of $f$ at a point of index $p$ is then $(-1)^p$.

Another approach which might be closer to what you want: see page 10 of this link

There is a picture of such a vector field on a triagulated manifold which is clearer than anything I could write in words (but no formula).

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In fact, this link was my motivation for the question. There is a similar justificative (2-dimensional) in the book "differential geometry in the large" (H. Hopf) but again it seems too much "intuitive". – joaohelder Jan 13 '11 at 19:16
I recommend to read Bill Thurston's answer to this question:…. What insight do you expect from an explicit formula? I believe that it is more interesting and important to convince yourself that the picture in the link is meaningful in dimensions larger than $2$. – Johannes Ebert Jan 16 '11 at 17:29

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