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The square map $z\mapsto z^2$ from $\mathbb{C}\to\mathbb{C}$ has a straightforward interpretation as a double-cover with a singular point at $0$. More generally $z\mapsto z^n$ is an $n$-fold cover.

What is the map $z\mapsto z^2$ from $\mathbb{H}\to\mathbb{H}$ (quaternions)? From counting dimensions I expect the typical fiber to be finite. On the other hand, $(bi+cj+dk)^2=(-b^2-c^2-d^2)$ so each point on the negative real axis has a whole $S^2$ worth of quaternions squaring to it.

A possible answer I'd be after is that this map is some kind of blow-up of $\mathbb{R}^4$ along a half-line. Well, this cannot be right since the typical fiber has at least two points ($q^2=(-q)^2$).

Bruno Le Floch
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Any non-real quaternion $q$ generates an isomorphic copy of $\bf C$ inside $\bf H$. So $q^2$ is obtained by the usual complex recipe: square the absolute value and double the argument in the plane spanned by $1$ and $q$. Of course if $q$ is real then $q^2$ is its usual square in $\bf R$. Yes, if $q$ has zero real part (and $q \neq 0$) then squaring $q$ yields a negative real number (and thus forgets the plane spanned by $1$ and $q$).

Noam D. Elkies
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    It might be helpful to mention that any two distinct copies of $\mathbb C$ intersect exactly in $\mathbb R$, so the fiber of any non-real quaternion has size 2. – Kimball Aug 12 '16 at 07:13
  • Thanks. So restricted to quaternions of a given norm, this is a double cover of $S^3$ where the equator (purely imaginary quaternions) is all mapped to the same point (-(norm squared)). That's easier to visualize for any $S^2$ slice containing both poles (±(norm squared)). – Bruno Le Floch Aug 12 '16 at 07:18