One way to define Riemann zeta function is by the analytic continuation of $$\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s} = \frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + \cdots$$ for the domain $Re(s)>1$ to the full complex plane in $\mathbb{C}$.

Thus, Riemann zeta function is defined for $s \in \mathbb{C}$ and $\zeta(s) \in \mathbb{C}$

My question is that do we gain anything new to do analytic continuation of Riemann zeta function such that a "modified Riemann zeta function" so

$s \in \mathbb{H}$ is in quaternion? and $\zeta(s) \in \mathbb{H}?$

Does this lead to any interesting result in the math literature?

Edit: more precisely, according to the comment, we seek for an analytic continuation of $\zeta(s)$ from the complex $\mathbb{C}/\{1\}$ to quaternion $\mathbb{H}/\{1\}$?

  • 1
    $\begingroup$ thanks ++ -- change accordingly $\endgroup$ – annie marie heart Jul 11 '19 at 22:00
  • 1
    $\begingroup$ The Wikipedia page on Quaternion Analysis has some good starting points, as does this Reddit link $\endgroup$ – Brevan Ellefsen Jul 11 '19 at 22:07
  • 1
    $\begingroup$ 1) The Zeta function is extended to $\;\Bbb C\setminus\{1\}\;$ , not the whole complex plane, (2) The quaternions are a non-commutative division ring. That could pose some problems to extend meaningfully the zeta function... $\endgroup$ – DonAntonio Jul 11 '19 at 22:08
  • 2
    $\begingroup$ I would say that there a ton more meaningful generalizations of the zeta functions (characters, ideals in algebraic number fields, Selberg zeta function) than this $\endgroup$ – Conrad Jul 11 '19 at 22:10
  • 1
    $\begingroup$ thanks -- Edit: more precisely, according to the comment, we seek for an analytic continuation of $\zeta(s)$ from the complex $\mathbb{C}/\{1\}$ to quaternion $\mathbb{H}/\{1\}$? $\endgroup$ – annie marie heart Jul 11 '19 at 22:10

You don't gain anything.

To extend a holomorphic function $f(z)$ of a complex variable $z=x+yi$ to a function of a quaternion variable, if its series' coefficients are real then it's just defined by

$$ f(x+yi)=u+vi \implies f(x+y\mathbf{t})=u+v\mathbf{t} \tag{$\circ$}$$

for unit vectors $\mathbf{t}$. (Every quaternion is expressible as $x+y\mathbf{t}$ for a unit vector $\mathbf{t}$, which is unique up to the signs of $y$ and choice of $\pm\mathbf{t}$). Equivalently, $f$ extends to quaternions by "rotating" the graph in $\mathbb{C}^2$ around to get a graph in $\mathbb{H}^2$. In other words, $f(pzp^{-1})=pf(z)p^{-1}$ for complex numbers $z$ and quaternions $p$ (note every quaternion is expressible as $pzp^{-1}$ for a complex number $z$ and quaternion $p$, but not uniquely).

The reason this happens is because the unit vectors (i.e. pure imaginary unit quaternions) $\mathbf{t}$ are precisely the square roots of $-1$ in $\mathbb{H}$, so algebraically they behave just like $i$ does in $\mathbb{C}$. If you look at the Dirichlet series definition of the zeta function $\zeta(s)$ for $\mathrm{Re}(s)>1$, they involve $1/n^s$ which is computed as $\exp(-\ln(n)s)$ Euler's formula $\exp(i\theta)=\cos\theta+\sin\theta\,i$ generalizes to quaternions since it follows entirely from $i$ being a square root of negative one. The same applies to the analytic continuation of $\zeta(s)$.

Same story for octonions.

In order to get something nontrivial, you would want to to start with a power series that has complex coefficients (so, isn't simply extended from a real variable function like $\zeta(s)$ is). There is extra freedom in how you define the monomials for a function of a quaternion variable, since each $a_nz^n$ may be replaced by

$$ \square z\square\cdots\square z\square $$

where the $\square$'s are complex numbers which multiply to $a_n$ and there are $n$ $z$s present.

However, doing this will not give you differentiable functions. In fact, the limit definition

$$ f'(p)=\lim_{h\to0}\frac{f(p+h)-f(p)}{h} $$

generalizes to quaternions in two ways: a "left" derivative and a "right" derivative, depending on which side of $\Delta f$ you put $h^{-1}$ (note $h\to0$ within $\mathbb{H}$ now). This turns out to be extremely restrictive: the only left or right differentiable quaternion functions are affine functions $f(q)=qa+b$ or $f(q)=aq+b$ respectively. It's a small miracle complex differentiable yields such a rich theory.

Moreover, say you start with a holomorphic function $f$, pick two complex numbers $\alpha$ and $\beta$ such that disk of convergence of the Taylor series around $\alpha$ includes $\beta$ and vice-versa. This gives you two different series (one in $z-\alpha$ and one in $z-\beta$), and (I'm pretty sure) these almost never give you the same function of a quaternion variable!

| cite | improve this answer | |
  • $\begingroup$ I believe that the lack of holomorphic functions in the quaternions may be related to the fact of the great impoverishment of conformal maps in dimensions $D > 2$. The quaternions have $D = 4$. Conformal maps are uniquely rich at $D = 2$, and that is the dimension of the complex numbers, $\mathbb{C}$. $\endgroup$ – The_Sympathizer Jul 12 '19 at 8:29
  • 1
    $\begingroup$ For $c_k$ decreasing fast enough and $q \in M_2(\Bbb{C})$ then $f(q)=\sum_{k \ge 0} c_k q^k$ is analytic and holomorphic as a function of $4$-complex variables but not $M_2(\Bbb{C})$-holomorphic. The embedding $\Bbb{H} = \Bbb{C}+j\Bbb{C}\to M_2(\Bbb{C})$ contains some complex conjugates so it is not holomorphic and that adds one more problem. @The_Sympathizer $\endgroup$ – reuns Jul 12 '19 at 9:43
  • $\begingroup$ thanks very much voted up. $\endgroup$ – annie marie heart Jul 12 '19 at 19:04

$\Bbb{H}$ is just a sub-algebra of $M_2(\Bbb{C})$.

For $A \in M_n(\Bbb{C})$ use the Jordan normal form to obtain $A = P J P^{-1} = P (D+N)P^{-1}$ where $D$ is diagonal and $DN=ND$ and $N^n = 0$. Let $f(s) = (s-1)\zeta(s)= \sum_{k=0}^\infty c_k s^k$ which is entire then $$P^{-1} f(A)P =f(D+N)=\sum_{k=0}^\infty c_k (D+N)^k =\sum_{k=0}^\infty c_k \sum_{l=0}^{n-1} {k \choose l} D^{k-l}N^l= \sum_{k=0}^{n-1} \frac{N^k}{k!} f^{(k)}(D)$$ Note the obtained function $A \mapsto f(A)$ doesn't depend on the basepoint $s_0 = 0$ we chose to expand $f(s)$ in power series.

It is not hard to convince that something similar happens with a meromorphic function such as $\zeta(s)$ obtaining $$\zeta(A) = P \zeta(D+N)P^{-1}= P \sum_{k=0}^{n-1} \frac{N^k}{k!} \zeta^{(k)}(D)P^{-1}$$

where $\zeta^{(k)}(D)$ is the matrix of $k$-th derivatives $$\zeta^{(k)}(D) = \pmatrix{\zeta^{(k)}(D_{11}) & & \\ & \zeta^{(k)}(D_{22}) & \\ & & \ldots}$$

If $q \in \Bbb{H}\subset M_2(\Bbb{C})$ then $q q^* = q^* q = N(q) I$ so that $$q = P DP^{-1}, \qquad \zeta(q) =P \zeta(D) P^{-1}$$

| cite | improve this answer | |
  • $\begingroup$ thanks very much voted up. $\endgroup$ – annie marie heart Jul 12 '19 at 19:04
  • $\begingroup$ How is your answer comparing w/ @runway44 above? $\endgroup$ – annie marie heart Jul 12 '19 at 23:01
  • $\begingroup$ He starts from the diagonalization of the real algebra of quaternions whereas I start from applying power series to complex algebra of matrices $\endgroup$ – reuns Jul 12 '19 at 23:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.