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I've noticed that it uses imaginary numbers. I know that sometimes when I have too few dimensions like (-1)^n, dots show where I might expect lines due to imaginary numbers. So perhaps there is a function lost to the other normed divisions? I know there is the upper limit of the octonions for normed divisons, so I was wondering if people have explored that aspect, and it would allow us to quickly rule it out.

Rashi Abramson
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    Relevant: http://math.stackexchange.com/q/703593/ – Sal Feb 02 '15 at 09:10
  • So your saying its absolutely possible? – Rashi Abramson Feb 02 '15 at 19:10
  • I don't understand your second sentence, but I think you should first ask: can you extend the Riemann zeta function to the quaternions? (This could be a good post.) I've never thought about it, but it's an interesting question. My guess is no--the Riemann zeta function is defined by a series in a half-plane, and extended to $\mathbb C$ by analytic continuation. You may still be able to make sense of the series for some region of quaternions, but I'm skeptical that it can be extended in a nice way to be a function of all quaternions. – Kimball May 12 '15 at 14:10
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    The second sentence was basically: can we connect the dots by going through a higher dimension? – Rashi Abramson Oct 21 '15 at 13:48
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    @Kimball Every octonion has the form $re^{\theta{\bf u}}$ for some real $r\ge0$, angle $\theta$ and pure imaginary $\bf u$ of unit norm. Algebraically, $\bf u$ will function exactly like $i$ (it's a square root of negative one), so we can extend any holomorphic function defined on a domain in $\Bbb C$ to a domain in $\Bbb O$ by simply rotating around the real axis. – anon Nov 02 '16 at 01:28

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Any holomorphic function which is analytic and real valued on $\mathbb{R}$ minus some discrete set of poles has a natural extension to quaternions by plugging quaternions into the Taylor series around any real value in its domain. These functions are symmetric across the real axis, i.e. $f(\overline{z})=\overline{f(z)}$. But extending to quaternions does not provide anything interesting: anytime $f(x+yi)=u+vi$, we have $f(x+y\mathbf{t})=u+v\mathbf{t}$ for any unit imaginary quaternion $\mathbf{t}$. (The unit imaginary quaternions are precisely the square roots of $-1$, so they behave algebraically and analytically the same as $i$. Indeed, the numbers of the form $x+y\mathbf{t}$ form an isomorphic copy of $\mathbb{C}$ in $\mathbb{H}$.) So the quaternion zeros of the Riemann zeta function are nothing more special than the complex zeros.

anon
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