Is there a determinant that in the limit becomes the Riemann zeta function?

Question

Let the matrix A be: $$A(n,k): \text{If } n=k \; \text{ then } 1+\frac{1}{n^{\frac{1}{4} (3+2 i t)} k^{\frac{1}{4} (3+2 i t)}}\text{ else }\frac{1}{n^{\frac{1}{4} (3+2 i t)} k^{\frac{1}{4} (3+2 i t)}}$$

which is equivalent to: $$A(n,k): \text{If } n=k \; \text{ then } 1+\frac{1}{n^{\frac{1}{2} (3+2 i t)}}\text{ else }\frac{1}{n^{\frac{1}{4} (3+2 i t)} k^{\frac{1}{4} (3+2 i t)}}$$

which is equivalent to $A=M+I$ where $M$ equals:

$M(n,k)=\frac{1}{n^{\frac{1}{4} (3+2 i t)} k^{\frac{1}{4} (3+2 i t)}}$

and $I$ is the identity matrix in linear algebra:

$$I(n,k): \text{If } n=k \; \text{ then } 1 \text{ else } 0$$

The first few determinants starting at the upper left corner of the infinite matrix $A$ are:

The $1$x$1$ matrix: $$2$$ The $2$x$2$ matrix: $$2^{1+\frac{1}{2} (-3-2 i t)}-2^{\frac{1}{2} (-3-2 i t)}+2$$ The $3$x$3$ matrix: $$2^{1+\frac{1}{2} (-3-2 i t)}-2^{\frac{1}{2} (-3-2 i t)}+3^{\frac{1}{2} (-3-2 i t)}+2$$ ...

The plot of the determinant of the $10$ x $10$ matrix looks like this:

where the vertical lines are at the Riemann zeta zeros on the critical line.

Compare this to the actual Riemann zeta function on the critical line:

Is there a infinite sized matrix, which is not totally degenerate, such that the determinant is the Riemann zeta function?

Associated Mathematica program:

(*start*)
Clear[n, k, nn, t, x];
nn = 10;
A = Table[
   Table[If[n == k, 1 + 1/n^(1/4 (3 + 2 I t))/k^(1/4 (3 + 2 I t)), 
     1/n^(1/4 (3 + 2 I t))/k^(1/4 (3 + 2 I t))], {k, 1, nn}], {n, 1, 
    nn}];
pol = Det[A];
Show[Plot[Re[pol], {t, 0, 60}],
 Graphics[
  Table[Line[{{Im[ZetaZero[n]], -30}, {Im[ZetaZero[n]], 30}}], {n, 1, 
    12}]], ImageSize -> Large]
Plot[Re[Zeta[1/2 + I*t]], {t, 0, 60}, ImageSize -> Large]
(*end*)

Your definition of $;A(n,k);$ is, imo, highly confusing: if $;n=k;$ , then it seems to be $;a_{n,n}=1+\cfrac1{n^{\frac12(3+2it)}};$ ...but then why to keep the $;n,,k;$ separatedly if they're they the same? — DonAntonio, Oct 22 '16 at 13:16
Reading your title I'd say this question is the holy grail of mathematics $>\ldots$ — Christian Blatter, Oct 22 '16 at 17:43
@ChristianBlatter not really. Construct two matrices $A,B$ such that $(AB){1,1} = \sum{n < N} n^{-s}, (AB){k,k} = 1$ and $AB$ upper triangular. Then $ \det(AB) = \sum{n < N} n^{-s}$. And note how this is the same idea as in https://en.wikipedia.org/wiki/Redheffer_matrix — reuns, Oct 22 '16 at 22:33
What is the definition of the determinant of a matrix of infinite size? http://mathoverflow.net/questions/1886/infinite-matrices — Did, Nov 23 '16 at 08:20
@Did here it works fine as the limit of the sequence of determinant — reuns, Nov 23 '16 at 08:26
@user1952009 What "works fine"? Even if some subdeterminants have a limit, what is the meaning of this limit and is it intrinsic in any way? — Did, Nov 23 '16 at 08:30
@Did The sequence converges absolutely, so the limit doesn't depend on a permutation of the basis. Now under a unitary change of basis, I don't know. And I'm not sure what you mean. — reuns, Nov 23 '16 at 08:32
@user1952009 Sorry but which sequence converges absolutely, to begin with? "Change of basis": this seems offtopic here. "Intrinsic": consider an increasing sequence $(S_n)$ of finite subsets of $\mathbb N$ such that $\bigcup S_n=\mathbb N$, and $D_n$ the determinant restricted to the lines and columns indexed by $S_n$, then is $(D_n)$ converging and is its limit independent of the choice of $(S_n)$? — Did, Nov 23 '16 at 08:39
@Did did you read my answer below ? You'll have the answer (the OP said $s=\frac14(3+2it)$ but any $Re(s) > 1/2$ will be fine) : the limit is $1+\zeta(2s)$ whatever sequence $S_n\subset S_{n+1}$ you choose — reuns, Nov 23 '16 at 08:40
@user1952009 Right, your trick (posted after my first comment) solves the definition problem. — Did, Nov 23 '16 at 08:45
@Did: I've looked into the given wikipedia-links on the problem of determinants of infinite-sized matrices. I've one doubt: if the infinite matrix has a LDU-decomposition where the entries of the diagonalmatrix D converge to 1 and if their product (the sum of their logarithm) converges - wouldn't this allow the notion of a generalized determinant? (The entries of D have a relatively simple analytical formula expressible in a form like $(1+a_n)$ with convergent sequence $a_n$) — Gottfried Helms, Nov 23 '16 at 13:25
@GottfriedHelms Maybe, maybe not. How is knowing the LDU decomposition of the infinite matrix sufficient to know the LDU decomposition of every finite minor? — Did, Nov 23 '16 at 13:28
@Did - well, now that's going offroad. I've never seen the problem of "every finite minor" before, and this is not really my expertise - so we better leave this idea away at this place... — Gottfried Helms, Nov 23 '16 at 13:31
@GottfriedHelms Just taking the notion of limit seriously, don't worry... — Did, Nov 23 '16 at 16:03
Mats - noticing that there are close-requests: my intuition is, that the core problem in your question is that of the possibility of finding arguments for the generalizability of the "determinant"-term from finite to infinite matrices of a certain type. The involving and the argumentative focus of the zeros-of-the-zeta-aspect seems a bit obfuscative for this attempt and this ambiguity is possibly the source for the "close"-requests. If I met your intention correctly then rewording/reducing your question would likely help to avoid closing and would improve that whole subject. What do you think? — Gottfried Helms, Nov 24 '16 at 08:38
@GottfriedHelms If you want to edit the question it is fine with me. I had forgotten that I had entered the matrix in my answer in the oeis a few years ago. The analytic continuation should be taken into account. My motivation for asking this question was that if you could define a determinant that is symmetric then maybe you could say something about the Hilbert-Polya conjecture. But I once proved to myself that eigenvalues cannot be arbitrary, and thereby include the zeta zeros. The symmetry of a possible determinant is my main hope here. — Mats Granvik, Nov 24 '16 at 13:00
Should have read: "But I once proved to myself that eigenvalues cannot be arbitrary, and thereby cannot include the zeta zeros." — Mats Granvik, Nov 24 '16 at 13:08

Gottfried Helms · Answer 1 · 2016-11-23T08:19:52.987

I construct the matrix, for instance size 4x4, symbolically, giving the entries at r'th row and c'th column symbolical names which we shall later expand into the actual formulae: $$ A_4= \small \begin{bmatrix} c_1 r_1+1 & c_2 r_1 & c_3 r_1 & c_4 r_1 \\ r_2 c_1 & c_2 r_2+1 & c_3 r_2 & c_4 r_2 \\ r_3 c_1 & c_2 r_3 & c_3 r_3+1 & c_4 r_3 \\ r_4 c_1 & c_2 r_4 & c_3 r_4 & c_4 r_4+1 \end{bmatrix} \tag 1 $$ The determinant of this matrix is $$ det(A_4) = 1+ c_1 r_1 + c_2 r_2 + c_3 r_3 + c_4 r_4 \tag 2$$ So this is a very simple expression and comparing with larger matrix-sizes we find the obvious generalization.
Now we replace the symbolic names by the actually intended terms $$r_k=c_k = k^{-1/4(3+2it)}$$ and have the determinant $$ \det(A_4) =1+ 1 \cdot 1 + 2^{{-1/4(3+2it)}} \cdot 2^{-1/4(3+2it)} + 3^{{-1/4(3+2it)}} \cdot 3^{-1/4(3+2it)} + 4^{{-1/4(3+2it)}} \cdot 4^{-1/4(3+2it)} \\ =2 + 2^{-1/2(3+2it)} + 3^{-1/2(3+2it)} + 4^{-1/2(3+2it)} $$ The function to plot, in additional dependance of the matrixsize $n$ is the real part of $$f_n(t) = 2 + \sum_{k=2}^n {1\over k^{ \, 3/2 + \, î \, t}} \tag 3$$ which is simply the partial-sum evaluation of the zeta at $s=3/2+ î t$ and can thus be written without all the matrix-overhead.

I wouldn't see, why the determinant-expression (2) should not be extended formally to infinite matrices as long it designs a convergent sum, or an expression which could be analytically continued for the infinite-size case.

(Remark: I do not yet understand why we have the real part of the exponent being 3/2 instead of 1/2, is this perhaps a typo in the question?)

reuns · Answer 2 · 2016-11-23T08:28:21.353

1

Fix $N,s$. Your $N \times N$ matrix is $A = I+u^Tu$

where $u = (1,2^{-s},3^{-s},\ldots,N^{-s})$ is a row vector.

Hence thanks to the determinant lemma $$\det(A) = (1+u u^T)\det(I) = 1+\sum_{n=1}^N n^{-2s}$$

and Sherman-Morrison gives $B = A^{-1} = I - \frac{ u^Tu}{1+uu^T}$ i.e. $B(n,k)= 1_{n=k}-\frac{(nk)^{-s}}{\det(A)}$

edited Nov 23 '16 at 08:28

answered Nov 23 '16 at 08:20

reuns

77,999

score 0 · Answer 3 · answered Nov 22 '16 at 17:41

0

Mathematica:

Table[M = 
  Table[Table[
    If[Mod[n, k] == 0, MoebiusMu[n/k]/(n/k)^s, If[n == 1, 1, 0]], {k, 
     1, nn}], {n, 1, nn}];
 Expand[FullSimplify[Det[M]]], {nn, 1, 12}]

Output:

{1, 1 + 2^-s, 1 + 2^-s + 3^-s, 1 + 2^-s + 3^-s + 4^-s, 1 + 2^-s + 3^-s + 4^-s + 5^-s, 1 + 2^-s + 3^-s + 4^-s + 5^-s + 6^-s, 1 + 2^-s + 3^-s + 4^-s + 5^-s + 6^-s + 7^-s, 1 + 2^-s + 3^-s + 4^-s + 5^-s + 6^-s + 7^-s + 8^-s, 1 + 2^-s + 3^-s + 4^-s + 5^-s + 6^-s + 7^-s + 8^-s + 9^-s, 1 + 2^-s + 3^-s + 4^-s + 5^-s + 6^-s + 7^-s + 8^-s + 9^-s + 10^-s, 1 + 2^-s + 3^-s + 4^-s + 5^-s + 6^-s + 7^-s + 8^-s + 9^-s + 10^-s + 11^-s, 1 + 2^-s + 3^-s + 4^-s + 5^-s + 6^-s + 7^-s + 8^-s + 9^-s + 10^-s + 11^-s + 12^-s}

https://oeis.org/A143142

answered Nov 22 '16 at 17:41

Mats Granvik

7,396

1

So that whole complicated construction reduces simply to the n'th partial sum of the zeta like $d(n,s) = sum(k=1,n,k^{-s}) $ ? Of course one would expect that this approximates the zeta function and its evaluations/plot over the imaginary part of the exponent should also show the approximations of the zeta zeros. But what is then the real question? – Gottfried Helms Nov 23 '16 at 07:23
Again this is a (sequence of) triangular matrix whose determinant is trivial. – reuns Nov 23 '16 at 08:35
and you should look in details at the $N \times N$ matrix $A_{i,j} = \lfloor \frac{N}{ij} \rfloor$ and at this paper https://arxiv.org/pdf/0807.4145.pdf on the $2\sqrt{N}\times 2\sqrt{N}$ matrix $\tilde{A}_{i,j} = \lfloor \frac{N}{s_is_j} \rfloor$ – reuns Nov 23 '16 at 09:01

Is there a determinant that in the limit becomes the Riemann zeta function?

3 Answers3

Linked