Editing Quaternion (section)

== Square roots ==

=== Square roots of −1 ===
In the complex numbers, <math>\mathbb C,</math> there are exactly two numbers, {{mvar|i}} and {{math|−''i''}}, that give −1 when squared. In <math>\mathbb H</math> there are infinitely many square roots of minus one: the quaternion solution for the square root of −1 is the unit [[sphere]] in <math>\mathbb R^3.</math> To see this, let {{nowrap|{{math|''q'' {{=}} ''a'' + ''b'' '''i''' + ''c'' '''j''' + ''d'' '''k'''}} }} be a quaternion, and assume that its square is −1. In terms of {{mvar|a}}, {{mvar|b}}, {{mvar|c}}, and {{mvar|d}}, this means

<math display=block>\begin{align}
a^2 - b^2 - c^2 - d^2 &= -1, \vphantom{x^|} \\[3mu]
2ab &= 0, \\[3mu]
2ac &= 0, \\[3mu]
2ad &= 0.
\end{align}</math>

To satisfy the last three equations, either {{nowrap|{{math|''a'' {{=}} 0}}}} or {{mvar|b}}, {{mvar|c}}, and {{mvar|d}} are all 0. The latter is impossible because ''a'' is a real number and the first equation would imply that {{nowrap|{{math|''a''<sup>2</sup> {{=}} −1}}.}} Therefore, {{nowrap|{{math|''a'' {{=}} 0}}}} and {{nowrap|{{math|''b''<sup>2</sup> + ''c''<sup>2</sup> + ''d''<sup>2</sup> {{=}} 1}}.}} In other words: A quaternion squares to −1 if and only if it is a vector quaternion with norm 1. By definition, the set of all such vectors forms the unit sphere.

Only negative real quaternions have infinitely many square roots. All others have just two (or one in the case of 0).{{citation needed|date=September 2017}}{{efn|The identification of the square roots of minus one in <math>\mathbb H</math> was given by Hamilton<ref>{{cite book |author=Hamilton, W.R. |year=1899 |title=Elements of Quaternions |edition=2nd |page=244 |publisher=Cambridge University Press |isbn=1-108-00171-8}}</ref> but was frequently omitted in other texts. By 1971 the sphere was included by Sam Perlis in his three-page exposition included in ''Historical Topics in Algebra'' published by the [[National Council of Teachers of Mathematics]].<ref>{{Cite book |last=Perlis |first=Sam |chapter=Capsule 77: Quaternions |title=Historical Topics in Algebra |chapter-url=https://archive.org/details/historicaltopics0000nati/page/38/mode/2up |chapter-url-access=registration |publisher=[[National Council of Teachers of Mathematics]] |location=Reston, VA |series=Historical Topics for the Mathematical Classroom |volume=31 |year=1971 |page=39 |isbn=9780873530583 |oclc=195566 }}</ref> More recently, the sphere of square roots of minus one is described in [[Ian R. Porteous]]'s book ''Clifford Algebras and the Classical Groups'' (Cambridge, 1995) in proposition 8.13.<ref>{{Cite book |last=Porteous |first=Ian R. |author-link=Ian R. Porteous |chapter=Chapter 8: Quaternions |url=https://www.maths.ed.ac.uk/~v1ranick/papers/porteous3.pdf |title=Clifford Algebras and the Classical Groups |series=Cambridge Studies in Advanced Mathematics |publisher=[[Cambridge University Press]] |location=Cambridge |volume=50 |pages=60 |year=1995 |doi=10.1017/CBO9780511470912.009 |isbn=9780521551779 |oclc=32348823 |mr=1369094 }}</ref>}}

==== As a union of complex planes ====
Each [[antipodal points|antipodal pair]] of square roots of −1 creates a distinct copy of the complex numbers inside the quaternions. If {{nowrap|{{math|''q''<sup>2</sup> {{=}} −1}},}} then the copy is the [[image (mathematics)|image]] of the function

<math display=block>a + bi \mapsto a + b q.</math>

This is an [[injective function|injective]] [[ring homomorphism]] from <math>\mathbb C</math> to <math>\mathbb H,</math> which defines a field [[isomorphism]] from <math>\Complex</math> onto its [[image (mathematics)|image]]. The images of the embeddings corresponding to {{mvar|q}} and −{{mvar|q}} are identical.

Every non-real quaternion generates a [[subalgebra]] of the quaternions that is isomorphic to <math>\mathbb C,</math> and is thus a planar subspace of <math>\mathbb H\colon</math> write {{mvar|q}} as the sum of its scalar part and its vector part:

<math display=block>q = q_s + \vec{q}_v.</math>

Decompose the vector part further as the product of its norm and its [[versor]]:

<math display=block>q = q_s + \lVert\vec{q}_v\rVert\cdot\mathbf{U}\vec{q}_v=q_s+\|\vec q_v\|\,\frac{\vec q_v}{\|\vec q_v\|}.</math>

(This is not the same as <math>q_s + \lVert q\rVert\cdot\mathbf{U}q</math>.) The versor of the vector part of {{mvar|q}}, <math>\mathbf{U}\vec{q}_v</math>, is a right versor with –1 as its square. A straightforward verification shows that
<math display=block>a + bi \mapsto a + b\mathbf{U}\vec{q}_v</math>
defines an injective [[algebra homomorphism|homomorphism]] of [[normed algebra]]s from <math>\mathbb C</math> into the quaternions. Under this homomorphism, {{mvar|q}} is the image of the complex number <math>q_s + \lVert\vec{q}_v\rVert i</math>.

As <math>\mathbb H</math> is the [[union (set theory)#Arbitrary unions|union]] of the images of all these homomorphisms, one can view the quaternions as a [[pencil of planes]] intersecting on the [[real line]]. Each of these [[complex plane]]s contains exactly one pair of [[antipodal points]] of the sphere of square roots of minus one.

==== Commutative subrings ====
The relationship of quaternions to each other within the complex subplanes of <math>\mathbb H</math> can also be identified and expressed in terms of commutative [[subring]]s. Specifically, since two quaternions {{mvar|p}} and {{mvar|q}} commute (i.e., {{math|''p q'' {{=}} ''q p''}}) only if they lie in the same complex subplane of <math>\mathbb H</math>, the profile of <math>\mathbb H</math> as a union of complex planes arises when one seeks to find all commutative subrings of the quaternion [[ring (mathematics)|ring]].

=== Square roots of arbitrary quaternions ===

Any quaternion <math>\mathbf q = (r,\, \vec{v})</math> (represented here in scalar–vector representation) has at least one square root <math>\sqrt{\mathbf q} = (x,\, \vec{y})</math> which solves the equation <math>\sqrt{\mathbf q}^{\,2} = (x,\, \vec{y})^2 = \mathbf q</math>. Looking at the scalar and vector parts in this equation separately yields two equations, which when solved gives the solutions

<math display=block>
\sqrt{\mathbf q}
= \sqrt{(r,\, \vec{v})}
= \pm\left(\sqrt{\tfrac12\bigl({\|\mathbf q\|+r}\bigr)},\ \frac{\vec{v}}{\|\vec{v}\|}\sqrt{\tfrac12\bigl({\|\mathbf q\|-r}\bigr)}\right),
</math>

where <math display="inline">\|\vec{v}\| = \sqrt{\vec{v}\cdot\vec{v}}=\sqrt{-\vec v\vphantom{v}^2}</math> is the norm of <math>\vec{v}</math> and <math display="inline">\|\mathbf q\| = \sqrt{\mathbf q^*\mathbf q} = \sqrt{r^2 + \|\vec{v}\|^2}</math> is the norm of <math>\mathbf q</math>. For any scalar quaternion <math>\mathbf q</math>, this equation provides the correct square roots if <math display="inline">\vec{v} / \|\vec{v}\|</math> is interpreted as an arbitrary unit vector.

Therefore, nonzero, non-scalar quaternions, or positive scalar quaternions, have exactly two roots, while 0 has exactly one root (0), and negative scalar quaternions have infinitely many roots, which are the vector quaternions located on <math>\{0\} \times S^2\bigl(\sqrt{-r}\bigr)</math>, i.e., where the scalar part is zero and the vector part is located on the [[n-sphere|2-sphere]] with radius <math>\sqrt{-r}</math>.