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Let $[n]:=\{1,\cdots,n\}$. It is known that $\{\log(p) \mid p \text{ is prime }\}$ is linearly independent over $\mathbb{Q}$. For a subset $A \subset [n]$ we can consider the matrix $L(A):=(\log(x) \mid x \in A)$ with $\operatorname{rank}_{\mathbb{Q}}(L(A)) = \omega(\prod_{x \in A} x )$, where $\omega(n)$ counts the distinct prime divisors of $n$. Then we have:

$$\sum_{ A \subset [n]} (-1)^{|A|} \omega\left(\prod_{x \in A} x\right) = 0$$

Which denoting $(n;k) := \#\{A \subset [n]\mid \omega(\prod_{x \in A} x) = k\}$ for $k=0,\cdots,n$ might be rewritten as:

$$\sum_{k=0}^n (-1)^k (n;k)=0$$

My question is, if this last quantity can be interpreted as one Euler characteristic [https://en.wikipedia.org/wiki/Euler_characteristic#Topological_definition]?

Here are some numbers $(n;k)$ computed:

1 [2]
2 [2, 2]
3 [2, 4, 2]
4 [2, 8, 6, 0]
5 [2, 10, 14, 6, 0]
6 [2, 10, 30, 22, 0, 0]
7 [2, 12, 40, 52, 22, 0, 0]
8 [2, 20, 80, 108, 46, 0, 0, 0]
9 [2, 24, 148, 232, 106, 0, 0, 0, 0]
10 [2, 24, 180, 488, 330, 0, 0, 0, 0, 0]
11 [2, 26, 204, 668, 818, 330, 0, 0, 0, 0, 0]
12 [2, 26, 332, 1308, 1714, 714, 0, 0, 0, 0, 0, 0]
13 [2, 28, 358, 1640, 3022, 2428, 714, 0, 0, 0, 0, 0, 0]
14 [2, 28, 390, 2280, 5646, 5884, 2154, 0, 0, 0, 0, 0, 0, 0]

Thanks for your help.

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    $\begingroup$ +1, I had never seen this identity pretty cool it if I understand correctly it gives 0 because it is an inclusion-exclusion formula, I read this question and came back to see if someone answered but it was deleted so is nice too see it back $\endgroup$
    – Dabed
    Mar 8 '20 at 20:04
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    $\begingroup$ @DanielD. thanks for your comment $\endgroup$
    – user6671
    Mar 8 '20 at 20:08
  • $\begingroup$ we can write $\omega( \prod_{x\in A} x ) $ as the sum over $p<n$ of $1$ if $p$ divides $x$ for some $x$ in $A$ and $0$ otherwise. Exchanging the order of summation proves your identity. This makes me think, if it's an Euler characteristic, this will just be a space that is a disjoint union over the primes. $\endgroup$
    – Will Sawin
    Mar 9 '20 at 1:58
  • $\begingroup$ Euler characteristic is based on the equation between the alternative sums of ranks of two kinds of groups (chains vs homology). Do you have two seemingly unrelated kinds of objects, etc. ? $\endgroup$
    – Wlod AA
    Mar 9 '20 at 10:50
  • $\begingroup$ $\sum_{k=0}^n(-1)^k\cdot\binom nk = 0\ $ because $(n-1)$-simplex is acyclic. Also, $\ \sum_{k=0}^n (-1)^k\cdot 2^{n-k}\cdot\binom nk = 1\ $ because $n$-cube is acyclic, etc. $\endgroup$
    – Wlod AA
    Mar 9 '20 at 11:24
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Ok, for now here there is a six-degree kind of connection or just plain numerology but lets hope someone can give a real explanation after

For manifolds by the Gauss-Bonnet is true that

$$\int_M K\;dA+\int_{\partial M}k_g\;ds=2\pi\chi(M)$$

And we can also make sense of the expression $\zeta'(\Delta, 0) = \frac{1}{12}\int_M K \, dA$

Now for the Riemann function $\zeta(s)$ and the Euler function $\phi(q)$ is true that

$$\zeta(s)=\prod_p(1-p^{-s})^{-1}=\exp \left( \sum_{p,n}\frac{p^{-ns}}{n}\right) = \exp\left(-\sum_n\frac{\Lambda(n)}{\log(n)} n^{-s}\right)$$

$$\phi(q)=\prod_k (1-q^k) = \exp\left(\sum_{k,n}\frac{q^{nk}}{n} \right) = \exp \left( \sum_n\frac{\sigma(n)}{n}q^{n} \right)$$

and taking the logarithm derivative and evaluating $s=0$ and $q=1$ we get that

$$\ln(\zeta(s))'|_{s=0}=-\sum_n\Lambda(n)$$

$$\ln(\phi(q))'|_{q=1}=\sum_n\sigma(n)$$

where $\Lambda(n)$ is the Von Mangoldt function and $\sigma(n)$ is the divisor function

So if we asume that $\zeta(s)$ and $\phi(q)$ play more or less analogous roles (and all this makes sense) that would make some kind of connection.

But of course more than the divisor function $\sigma(n)=\sum_{d\mid n}d$ we want the "prime" divisor function $\omega(n)=\sum_{d\text{ prime}\mid n}d$ and the alternating sum at that $\sum(-1)^n\omega(n)$ instead of $\sum_n\sigma(n)$

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  • $\begingroup$ I refer now to the comment added by Will Sawin which surely makes more sense. $\endgroup$
    – Dabed
    Mar 9 '20 at 3:34
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    $\begingroup$ thanks for your interesting answer $\endgroup$
    – user6671
    Mar 9 '20 at 5:20

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