The primary motivation for this question is the following: I would like to extract some topological statistics which capture how arithmetic progressions of prime numbers "fit together" in a manner that will be made precise below.


Consider a nested family of simplicial complexes $K(p)$ indexed by prime $p \in \mathbb N$ defined as follows:

  1. the vertices are all odd primes less than or equal to $p$, and
  2. insert a $d$-simplex ($d \geq 2$) spanning $d+1$ vertices if and only if they constitute an arithmetic progression. Of course, one must also insert all faces, and faces-of-faces etc. so that the defining property of a simplicial complex is preserved.

For instance, $K(7)$ has the vertices $3,5,7$ and a single $2$-simplex $(3,5,7)$ along with all its faces. $K(11)$ has all this, plus the vertex $11$ and the simplex $(3,7,11)$. The edge $(3,7)$ already exists so only the other two need to be added. Thus, the fact that $(3,7)$ occurs in two arithmetic progressions bounded by $11$ is encoded by placing the corresponding edge in the boundary of two 2-simplices.


Has someone already defined and studied this complex? What I am mostly interested in is

How does the homology of $K(p)$ change with $p$?

If it helps, here are -- according to home-brew computations -- the statistics for the first few primes (Betti 0 and 1 over $\mathbb{Z}_2$). I've already confirmed that the sequence of Betti-1's is not in Sloane's online encyclopedia of integer sequences. If an intermediate K[p] is missing in the list, that means that the homology is the same as that for the previous prime.

K [3]: 1 0
K [5]: 2 0
K [7]: 1 0
K [13]: 2 0
K [17]: 2 1
K [19]: 1 2
K [23]: 1 4
K [31]: 1 6
K [37]: 2 6
K [43]: 1 7
K [53]: 1 8
K [59]: 1 9
K [61]: 1 10
K [67]: 1 12
K [71]: 1 17
K [73]: 1 20
K [79]: 1 23
K [83]: 1 26
K [89]: 1 31
K [97]: 1 32
K [101]: 1 35
K [103]: 1 41
K [107]: 1 43
K [109]: 1 47
K [113]: 1 53
K [127]: 1 58
K [131]: 1 62
K [137]: 1 67
K [139]: 1 73
K [149]: 1 78

Here's a more concrete question:

Is it true that the $d$-dimensional homology groups of $K(p)$ for $d > 1$ are always trivial?

  • 4
    $\begingroup$ "The primary motivation for this question...." Was that a pun? $\endgroup$ Aug 5, 2012 at 22:44
  • 13
    $\begingroup$ Gerry: sadly, yes. I fear that my sense of humor is progressively deteriorating... $\endgroup$ Aug 5, 2012 at 22:49
  • 2
    $\begingroup$ This way of constructing complexes must be interesting for sequences of integers other than primes, too! $\endgroup$ Aug 5, 2012 at 22:59
  • 12
    $\begingroup$ @David: you mean like this? arxiv.org/abs/1101.5704 $\endgroup$ Aug 5, 2012 at 23:01
  • 5
    $\begingroup$ Regarding connectedness of the inf. complex: if I did not overlook something stupid, it is 'clear' (in the sense that 'everybody' believes it but nobody can prove it) that for any tow odd primes q1,q2 there will be a prime p (indeed infinitely many) such that q1,p and q2,p can each be completed to a 3-AP of primes. In other words for q1,q2 fixed there will be a prime p such that 2p -q1 and 2p -q2 are both prime. More generally any system of linear equations is believed to be solvable 'in the primes', except if there is a 'local' obstruction (congr. or size), and for some this can be shown. $\endgroup$
    – user9072
    Aug 6, 2012 at 11:32

1 Answer 1


No. In fact, for $p = 435052917615787$, this will absolutely be false, as the second homology group will not vanish.

Note that a 2-dimensional "hole" in your complex is a 3-simplex all of whose faces are 2-simplices in $K(p)$, i.e. 4 primes such that each 3 of them lie in some arithmetic progression.

Of course, you would suspect that such a thing is plausible, and with some effort construct an example. Here's why this number works:

Consider $p = 398936189798617$. Then, if $d = 2124513401010$, one sees that $p+d$ is not a prime, while $p+2d,p+3d,\ldots,p+15d$ are all primes. Here $p+15d = 435052917615787$ is the prime for which we consider the complex.

In fact, the numbers were taken from the AP-k records at http://primerecords.dk/aprecords.htm#minimal, and $p+2d$ is just a beginning of an AP-23.

Now, one sees immediately that $p, p+6d, p+10d$ form a simplex, as it is a subsequence of the arithmetic progression of primes $p, p+2d, p+4d, p+6d, p+10d$.

Similarly, $p, p+6d, p+15d$ form a simplex, as do $p, p+10d, p+15d$ and $p+6d, p+10d, p+15d$. These come from sequences with differences $3d$, $5d$, and $d$ respectively.

However, $p,p+6d,p+10d,p+15d$ can only form a simplex if they all lie in some arithmetic progression of primes. But then its difference must divide $d$, contradicting the fact that $p+d$ is not prime.

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    $\begingroup$ Thanks! The numbers involved are huge. So maybe this is also a good candidate answer to the big list question somewhere on this site asking about "eventual counterexamples" $\endgroup$ Mar 5, 2016 at 2:14
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    $\begingroup$ This of course leads to the (harder) question: what is the smallest $p$ for which the second homology group of $K(p)$ is nontrivial? $\endgroup$ Mar 5, 2016 at 7:50

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