Let $M$ be a Riemannian smooth compact manifold. It is known that $M$ has a triangulation, for any dimension. But do we know if there exists a triangulation such that all simplices have same volume ?

I've been looking for some references, but I can't find any dealing with this problem. I would appreciate if someone had an answer - even partial.


I don't know a reference, but I think that we can use a theorem of Moser to get what you want.

Start with any triangulation $\mathcal{T}$ of $M$, whose cardinal is denoted by $k$ and denote by $\omega$ the Riemannian volume form. There is a positive smooth function $f$ such that $\int_M f\omega=\mathrm{vol}(M)$ (i.e. $f$ has average $1$) and such that for all $T\in\mathcal{T}$: $$\int_T f\omega = \frac{\mathrm{vol}(M)}{k}$$ (simply adjust $f$ in the interior of each simplex).

Now, Moser proved in 1965 (answering a question of a MO user by the way) that, since $f\omega$ and $\omega$ have the same total volume, there is a diffeomorphism $\phi:M\to M$ such that $\phi_*(f\omega)=\omega$. Then by the change of variable formula, $$\int_{\phi(T)} 1\,\omega= \int_T f\,\omega = \frac{\mathrm{vol}(M)}{k}$$ so that the collection $\{\phi(T)\}$ is a triangulation of $M$ with all simplices of the same volume.


It may be useful to contrast triangulations with equidissections: partitions into simplices of the same volume, such that the simplices have pairwise disjoint interiors (but might not meet face-to-face, edge-to-edge, etc.). Equipartitions are difficult to achieve even just triangulating a polygon in the plane. Most polygons have no such equidissection. The first paper below shows "that for almost all polygons no equidissection exists." The second paper shows nonexistence holds if the vertex coordinates are algebraically independent.

Kasimatis, E. A., and S. K. Stein. "Equidissections of polygons." Discrete Mathematics 85.3 (1990): 281-294. (Elsevier link.)

Su, Zhanjun, and Ren Ding. "Dissections of polygons into triangles of equal areas." Journal of Applied Mathematics and Computing 13.1-2 (2003): 29-36. (Springer link.)

          (Image from Wikipedia.)
I suspect you are not finding literature because the triangulations you seek rarely exist.

  • 4
    $\begingroup$ I don't think the question is about triangulation by geodesic simplices. Otherwise, there would be a simpler answer as soon as $n\ge3$: most manifolds don't have totally geodesic submanifolds, so we don't have any geodesic simplex. $\endgroup$ – Benoît Kloeckner Apr 28 '16 at 8:26
  • $\begingroup$ @BenoîtKloeckner: Thanks, and apologies for misinterpreting. $\endgroup$ – Joseph O'Rourke Apr 28 '16 at 14:33

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