# Analytic continuation of Riemann zeta $\zeta(s)$ from the complex $\mathbb{C}$ to quaternion $\mathbb{H}$?

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 $$\mathbb{H}/\{1\}$$?

• thanks ++ -- change accordingly – annie heart Jul 11 at 22:00
• The Wikipedia page on Quaternion Analysis has some good starting points, as does this Reddit link – Brevan Ellefsen Jul 11 at 22:07
• 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... – DonAntonio Jul 11 at 22:08
• 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 – Conrad Jul 11 at 22:10
• 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 $\mathbb{H}/\{1\}$? – annie heart Jul 11 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!

• 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, $\mathbb{C}$. – The_Sympathizer Jul 12 at 8:29
• 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 – reuns Jul 12 at 9:43
• thanks very much voted up. – annie heart Jul 12 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 $$q = P DP^{-1}, \qquad \zeta(q) =P \zeta(D) P^{-1}$$

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