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Has the $\zeta(s)$ function, $\sum_n 1/n^s$, been generalized to quaternions, so $\zeta(q)$ for $q$ a quaternion?

Euler defined it for $s$ integers, Chebyshev for $s$ real, Riemann for $s$ complex. So it is natural to explore $s$ a quaternion. But perhaps this does not lead to an interesting Quaternion Hypothesis?

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You can define $\zeta(q)$ for any quaternion with real part $\ne 1$ by plugging it into the Taylor series for $\zeta(x)$ around a real point far enough from $1$ in the appropriate direction.

This is not very exciting, though, because every quaternion $q$ decomposes as $a+b\ell$ where $a$ and $b$ are real and $\ell$ is a quaternion satisfying $\ell^2=-1$. So for every fixed $q$, all terms in the Taylor series will lie in the complex plane spanned by $1$ and $\ell$, and therefore no really new phenomena show up.

This shows that it doesn't matter which center for the Taylor series we choose, as long as it is closer to $q$ than it is to $1$. It also shows that we can extend the function continuously to quaternions with real part $1$ (other that $1$ itself).

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  • $\begingroup$ Thanks for this response. If $q=2+0i+0j+0k$, will $\zeta(q)$ will still equal $\pi^2/6 =\zeta(2)$? In other words, would the extension agree with standard $\zeta$ for real $q$, and for complex $q$? $\endgroup$
    – Joseph O'Rourke
    Commented Aug 25, 2018 at 12:18
  • $\begingroup$ @JosephO'Rourke: Yes, because all the terms in the Taylor series would be unchanged, and convergence works the same. (You can also use the Laurent series around $1$ to get values for all of $\mathbb H\setminus\{1\}$ in one go). $\endgroup$
    – hmakholm left over Monica
    Commented Aug 25, 2018 at 12:26

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