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In Gromov's famous book, it says "simplical volume of every oriented surface of genus $ \ge 1$ satisfies ${\left\| {\left[ S \right]} \right\|_\Delta } = 4g - 4 = - 2\chi \left( S \right) = - 2({k_0} - {k_1} + {k_2})$ (Euler characteristic in this case is negative). One proof is to use proportionality theorem for compact locally homogeneous space V, ${\left\| {\left[ V \right]} \right\|_\Delta } = cvol\left( V \right)$, where the constant c depends on the local geometry of V. But Gromov says it's elementary! Since the simplicial volume is the infimum of the number of simplices over all homotopy triangulations of V. It's easy to see that "we need at least 4g-4 simplices to triangulate S for a surface with negative Euler characteristic, ${k_2} \ge - 2\chi \left( S \right)$? I can't see it. Please give an explanation.

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  • $\begingroup$ You did not define your notations, but I guess that $k_0, k_1$ and $k_2$ are respectively the number of vertices, edges and faces. Then, in a triangulation of $S$, every edge is adjacent to two faces and every face to three edges, so we have $k_1=3/2k_2$. You plug it into Euler's identity : $k_0-k_1+k_2=\chi(S)$ and it gives you $k_2 \geq -2 \chi(S)$. $\endgroup$
    – Arnaud
    May 22, 2013 at 10:21

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Well, the simplicial volume is not really the minimal number of simplices in a homotopy triangulation but the minimal (more precisely: the infimum) sum of (modulus of) coefficients in a fundamental class.

In any case, the inequality $\parallel S\parallel\ge 4g-4$ follows easily from Gauss-Bonnet: take a hyperbolic metric on S, by Gauss-Bonnet its area is $\pi(4g-4)$. But any fundamental cycle can be straigtened such that the simplices become straight geodesic simplices, which then have area $\pi-\alpha-\beta-\gamma\le\pi$. ($\alpha,\beta,\gamma$ are the interior angles.) So the number of simplices or the sum of coefficients must be at least 4g-4.

To see that actually equality holds, observe that the surface of genus g can be realised by gluing the edges of a 4g-gon (whose interior angles add up to $2\pi$). Obviously this can be decomposed into 4g-2 triangles, which already shows $\parallel S\parallel \le 4g-2$. To get the sharp bound, you can use that the surface of genus $h(g-1)+1$ is an h-fold covering of the surface of genus g. Thus $h\parallel S_g\parallel\le \parallel S_{h(g-1)+1}\parallel\le 4h(g-1)+2$ and thus $\parallel S_g\parallel \le 4g-4+\frac{2}{h}$ which approaches 4g-4 when h goes to infinity.

The paper of Gromov may be hard to read for beginners, so you may prefer to look at the textbooks Benedetti-Petronio:"Lectures on hyperbolic geometry" or Ratcliffe "Hyperbolic manifolds". Other comprehensible online sources are http://arxiv.org/pdf/math/0504106v1.pdf or http://homepages-nw.uni-regensburg.de/~pac16580/Pagliantini_PhDthesis.pdf and of course the classic http://library.msri.org/books/gt3m/PDF/6.pdf

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  • $\begingroup$ This argument works for genus $\ge 2$. For the torus you can easily see $\parallel T\parallel=0$ because there are self-maps of arbitrarily high degree. $\endgroup$
    – ThiKu
    May 23, 2013 at 4:47

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