UPDATE This version is substantially improved from the one posted at 8 AM.
I now think I can achieve mathbbZ/p using O(logp) vertices. I'm not trying to optimize constants at this time.
Let B be a simplicial complex on the vertices a, b, c, a′, b′, c′ and z1, z2, ..., zk−3, containing the edges (a,b), (b,c), (c,a), (a′,b′), (b′,c′) and (c′,a′) and such that H1(B)congmathbbZ with generator (a,b)+(b,c)+(c,a) and relation
2large((a,b)+(b,c)+(c,a)large)equiv(a′,b′)+(b′,c′)+(c′,a′).
I think I can do this with k=6 by taking damiano's construction with p=2 and adding three simplices to make the hexagon (h1,h2,ldots,h6) homologous to the triangle (h1,h3,h5).
Let Bn be a simplicial complex with 3+nk vertices ai, bi, ci, with 0leqileqn, and zij with 0leqileqn−1 and 1leqjleqk−3. Namely, we build n copies of B, the r-th copy on the vertices ar, br, cr, ar+1, br+1, cr+1 and zr1, zr2, ..., zrk−3. Let gammai be the cycle (ai,bi)+(bi,ci)+(ci,ai).
Then H1(Bn)=mathbbZ with generator gamma0 and relations
gammanequiv2gamman−1equivcdotsequiv2ngamma0
Let p=2n1+2n2+cdots+2ns.
Glue in an oriented surface Sigma with boundary gamman1sqcupgamman2sqcupcdotssqcupgammans, genus 0, and no internal vertices.
In the resulting space, sumgammaniequiv0 so pgamma0equiv0, and no smaller multiple of gamma0 is zero. We have use 3+klog2p vertices. This is the same order of magnitude as Gabber's bound.
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