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We all know that Riemann Hypothesis (RH) has many equivalent statements.

There is one statement which expresses RH in term of Redheffer matrix, there is another equivalent statement of RH which involves the Farey sequences.

Are there similar equivalent statement for GRH (Generalized Riemann hypothesis ) ?

Thank you for your attention.

david
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1 Answers1

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The determinant of the Redheffer matrix is explained in this paper :

  • Start from the two $N\times N$ matrices $$D_{d,n} = 1_{d | n}, \qquad\qquad E_{m,d} =1_{m | d} \mu(d/m)$$ Then $$(ED)_{n,m} = \sum_d D_{n,d}E_{d,m} = \sum_d 1_{d | n}1_{m | d} \mu(d/m) = \sum_k 1_{mk | n} \mu(k)$$ $$ = 1_{m | n} \sum_{k | n/m} \mu(k)= 1_{m = n}$$

    i.e. $ED = I$.

    Then consider the matrix $C_{d,n} = 1$ if $n=1$ and $d \ne 1$, and the Redheffer matrix $A = C + D$ i.e. $A_{d,n} = 1$ if $d|n$ or $n=1$.

    Since $D,E$ are upper triangular matrices with only ones on the diagonal, you see that $\det(D) = \det(E) = 1$, and $$\det(A) = \det(C+D) = \det(E(C+D))=\det(EC+I)$$

    It is then easy to see that $EC+I$ is lower triangular with diagonal entries $(EC+I)_{1,1} = \sum_{n=1}^N \mu(n) = Mertens(N)$ and $(EC+I)_{n,n} = 1$, i.e. $\det(A) = Mertens(N)$.

    And the RH is that $\det(EC+I)= \mathcal{O}(N^{1/2+\epsilon})$. $$ $$

For the GRH it works exactly in the same way :

  • pick a Dirichlet character $\chi(n)$, and start from the two $N\times N$ matrices $$D^\chi_{d,n} = 1_{d | n}\chi(n/d), \qquad\qquad E^\chi_{m,d} =1_{m | d} \mu(d/m)\chi(d/m)$$ Then $$(E^\chi D^\chi)_{n,m} = \sum_d D^\chi_{n,d}E^\chi_{d,m} = \sum_d 1_{d | n}\chi(n/d)1_{m | d} \mu(d/m)\chi(d/m)$$ $$ = 1_{m | n} \sum_{k | n/m} \mu(k)\chi(n/m)= 1_{m = n}$$

    i.e. $E^\chi D^\chi = I$.

    Then consider the matrix $C^\chi_{d,n} = 1$ if $n=1$ and $d \ne 1$, and the $\chi$ Redheffer matrix $A^\chi = C^\chi + D^\chi$ i.e. $A^\chi_{d,n} = \chi(n/d)$ if $d|n$ and $A^\chi_{d,1} = 1$.

    Since $D^\chi,E^\chi$ are upper triangular matrices with only ones on the diagonal, you see that $\det(D^\chi) = \det(E^\chi) = 1$, and $$\det(A^\chi) = \det(C^\chi+D^\chi) = \det(E^\chi(C^\chi+D^\chi))=\det(E^\chi C^\chi+I)$$

    It is then easy to see that $E^\chi C^\chi+I$ is lower triangular with diagonal entries $(E^\chi C^\chi+I)_{1,1} = \sum_{n=1}^N \mu(n)\chi(n)$ and $(E^\chi C^\chi+I)_{n,n} = 1$, i.e. $\det(E^\chi C^\chi+I) = \sum_{n=1}^N \mu(n)\chi(n)$.

    And the GRH is that $\det(A^\chi)= \mathcal{O}(N^{1/2+\epsilon})$.

reuns
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  • @user952009, thank you for the quick answer. Is there an equivalent statement of GRH using Farey sequence ? – david Jul 07 '16 at 04:02
  • @david : did you try to understand why the Farey sequence are related to the RH ? for example that $|F_n| = |F_{n-1}| + \varphi(n)$ ? then how do you relate $\varphi(n)$ to $\zeta(s)$ ? and what is the equivalent statement in term of $L(s,\chi)$ ? – reuns Jul 07 '16 at 14:51
  • https://en.wikipedia.org/wiki/Farey_sequence#Riemann_hypothesis has a entry on the relation between Farey sequence with RH, but I had not seen the similar relation between Farey sequence with GRH. – david Jul 07 '16 at 16:47
  • @david : It is similar to what I wrote for the Redheffer matrix. you have to write all the steps for relating $F_n$ with RH, and replace $\mu(n)$ by $\mu(n) \chi(n)$ for obtaining a relation between $F^\chi_n$ and GRH – reuns Jul 07 '16 at 16:51