A question on Farey Sequences 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.
 A: 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 $\det(A^\chi)= \mathcal{O}(N^{1/2+\epsilon})$.
