# The equation $\zeta(q)=0$ for $q$ a quaternion

I know there have been several attempts to define a theory of functions of a quaternionic variable. I would like to know if a coherent and satisfying definition of the "Riemann" zeta function exists in this framework and if studying the equation $$\zeta(q)=0$$ for $$q$$ a quaternion could lead to some progress towards RH. If I'm not mistaken the principle of isolated zeros doesn't hold true anymore, but do we conjecture that the solutions of the above equation lie in a plane?

• I doubt that any "quaternionic" zeta function is really different from the usual complex zeta function. Also, which "theory of functions of a quaternionic variable" are you referring to? – Somos Apr 10 at 22:21
• Every quaternion $q$ is expressible as $a+b\mathbf{v}$ where $\mathbf{v}$ is a unit vector / purely imaginary quaternion (hence $\mathbf{v}^2=-1$). Then $f(a+bi)=c+di$ for specific complex numbers $a+bi$ and $c+di$ implies $f(a+b\mathbf{b})=c+d\mathbf{v}$. Ultimately, the "graph" of $f$ for quaternions is the exact same as the "graph" of $f$ for complex numbers, only rotated around the real axis (so any unit vector can behave as $i$ does). – arctic tern Apr 18 at 1:42

## 1 Answer

For $$M \in M_k(\Bbb{C})$$ and $$\|\exp(-M)\|< 1$$ and $$\zeta(M) = \sum_{n=1}^\infty \exp(-M \ln n)$$ and by analytic continuation for every $$M$$

• If $$M=PD P^{-1}$$ is diagonalizable then $$\zeta(M) = P \zeta(D)P^{-1},\qquad \zeta(D)= \pmatrix{\zeta(D_{11})& &\\ & \zeta(D_{22})& \\ & & \ddots}$$

• If $$M$$ is not diagonalizable then the Jordan normal form gives $$M = P (D+N) P^{-1}$$ where $$ND=DN, N^k=0$$ and $$\zeta(M) = P \sum_{m=0}^{k-1} \frac{N^m}{m!} \zeta^{(m)}(D) P^{-1}$$

• The quaternions are isomorphic to a diagonalizable subalgebra of $$M_2(\Bbb{C})$$

• For the convergence of $\sum_{n=1}^\infty \exp(-M \ln n)$ for $\|\exp(-M)\| < 1$ use the Jordan normal form again : $M = P(D+N) P^{-1}, \exp( -M \ln n) = P\sum_{m=0}^{k-1} N^m (-\ln n)^{k-m} \exp(-D\ln n) P^{-1}$ – reuns Apr 11 at 13:58
• Thanks reuns for your answer but there are several things I don't understand. First how is the norm of the matrix defined? Second, is the exponent $(k-m)$ a multiplicative power, an order of derivation, an iterate wrt composition? – Sylvain Julien Apr 11 at 15:49
• Any matrix norm compatible with the multiplication. Replace $\zeta(s)$ by $s \mapsto s^2$ to see where the $m$ comes – reuns Apr 11 at 16:25