# Quaternionic analysis

(Redirected from Quaternion variable)

In mathematics, quaternionic analysis is the study of functions with quaternions as the domain and/or range. Such functions can be called functions of a quaternion variable just as functions of a real variable or functions of a complex variable are called.

As with complex and real analysis, it is possible to study the concepts of analyticity, holomorphy, harmonicity and conformality in the context of quaternions. It is known that for the complex numbers, these four notions coincide; however, for the quaternions, and also the real numbers, not all of the notions are the same.

## Discussion

The projections of a quaternion onto its scalar part or onto its vector part, as well as the modulus and versor functions, are examples that are basic to understanding quaternion structure. An important example of a function of a quaternion variable is

${\displaystyle f(q)=uqu^{-1}}$

which rotates the vector part of q by twice the angle of u.

The quaternion inversion ${\displaystyle f(q)=q^{-1}}$ is another fundamental function, but it introduces questions f(0) = ? and "Solve f(q) = 0." Affine transformations of quaternions have the form

${\displaystyle f(q)=aq+b,\quad a,b\in \mathbb {H} .}$

Linear fractional transformations of quaternions can be represented by elements of the matrix ring M2(H) operating on the projective line over H. For instance, the mappings ${\displaystyle q\mapsto uqv,}$ where u and v are fixed versors serve to produce the motions of elliptic space.

Quaternion variable theory differs in some respects from complex variable theory as in this instance: The complex conjugate mapping of the complex plane is a central tool but requires the introduction of a non-arithmetic operation. Indeed, conjugation changes the orientation of plane figures, something that arithmetic functions do not change. In contrast, the quaternion conjugation can be expressed arithmetically:

Proposition: The function ${\displaystyle f(q)=-{\tfrac {1}{2}}(q+iqi+jqj+kqk)}$ is equivalent to quaternion conjugation.

Proof: For the basis elements we have

${\displaystyle f(1)=-{\tfrac {1}{2}}(1-1-1-1)=1,\quad f(i)=-{\tfrac {1}{2}}(i-i+i+i)=-i,\quad f(j)=-j,\quad f(k)=-k}$.

Consequently, since f is a linear function,

${\displaystyle f(q)\ =f(w+xi+yj+zk)\ =wf(1)+xf(i)+yf(j)+zf(k)\ =w-xi-yj-zk\ =q^{*}.}$

The success of complex analysis in providing a rich family of holomorphic functions for scientific work has engaged some workers in efforts to extend the planar theory, based on complex numbers, to a 4-space study with functions of a quaternion variable. These efforts were summarized in 1973 by C.A. Deavours.[1] He recalls a 1935 issue of Commentarii Mathematici Helvetici where an alternative theory of "regular functions" was initiated by R. Fueter through the idea of Morera's theorem:[2] quaternion function F is "left regular at q" when the integral of F vanishes over any sufficiently small hypersurface containing q. Then the analogue of Liouville's theorem holds: the only quaternion function regular with bounded norm in E4 is a constant. One approach to construct regular functions is to use power series with real coefficients. Deavours also gives analogues for the Poisson integral, the Cauchy integral formula, and the presentation of Maxwell’s equations of electromagnetism with quaternion functions.

Though H appears as a union of complex planes, the following proposition shows that extending complex functions requires special care:

Proposition: Let ${\displaystyle f(z)=u(x,y)+iv(x,y)}$ be a function of a complex variable, ${\displaystyle z=x+iy}$. Suppose also that u is an even function of y and that v is an odd function of y. Then ${\displaystyle f(q)=u(x,y)+r\ v(x,y)}$ is an extension of  f  to a quaternion variable ${\displaystyle q=x+yr,\quad r^{2}=-1,\quad r\in \mathbb {H} }$.

Proof: Let r* be the conjugate of r so that q = xy r*. The extension to H will be complete when it is shown that f(q) = f(xy r*). Indeed, by hypothesis

${\displaystyle u(x,y)=u(x,-y),\quad v(x,y)=-v(x,-y)\quad }$ so that one obtains
${\displaystyle f(x-yr^{*})=u(x,-y)+r^{*}v(x,-y)=u(x,y)+r\ v(x,y)=f(q).}$

## Homographies

The rotation about axis r is a classical application of quaternions to space mapping.[3] In terms of a homography, the rotation is expressed

${\displaystyle U(q,1){\begin{pmatrix}u&0\\0&u\end{pmatrix}}=U(qu,u)\thicksim U(u^{-1}qu,1),}$

where ${\displaystyle u=\exp(\theta r)=\cos \theta +r\sin \theta }$ is a versor. If p * = −p, then the translation ${\displaystyle q\mapsto q+p}$ is expressed by

${\displaystyle U(q,1){\begin{pmatrix}1&0\\p&1\end{pmatrix}}=U(q+p,1).}$

Rotation and translation xr along the axis of rotation is given by

${\displaystyle U(q,1){\begin{pmatrix}u&0\\uxr&u\end{pmatrix}}=U(qu+uxr,u)\thicksim U(u^{-1}qu+xr,1).}$

Such a mapping is called a screw displacement. In classical kinematics, Chasles' theorem states that any rigid body motion can be displayed as a screw displacement. Just as the representation of a Euclidean plane isometry as a rotation is a matter of complex number arithmetic, so Chasles' theorem, and the screw axis required, is a matter of quaternion arithmetic with homographies: Let s be a right versor, or square root of minus one, perpendicular to r, with t = rs. Rotation about the axis parallel to r and passing through s is expressed[4] by the homography composition

${\displaystyle {\begin{pmatrix}1&0\\-s&1\end{pmatrix}}{\begin{pmatrix}u&0\\0&u\end{pmatrix}}{\begin{pmatrix}1&0\\s&1\end{pmatrix}}={\begin{pmatrix}u&0\\z&u\end{pmatrix}},}$

where ${\displaystyle z=us-su=\sin \theta (rs-sr)=2t\sin \theta .}$ Now in the (s,t)-plane the parameter θ traces out a circle

${\displaystyle u^{-1}z=u^{-1}(2t\sin \theta )=2\sin \theta (t\cos \theta -s\sin \theta ).}$

in the half-plane ${\displaystyle \lbrace wt+xs:x>0\rbrace .}$ Any p in this half-plane lies on a ray from the origin through the circle

${\displaystyle \lbrace u^{-1}z:0<\theta <\pi \rbrace }$ and can be written
${\displaystyle p=au^{-1}z,\ \ a>0.}$

Then up = az, with

${\displaystyle {\begin{pmatrix}u&0\\az&u\end{pmatrix}}}$

as the homography expressing conjugation of a rotation by a translation p.

## The Gâteaux derivative for quaternions

Since the time of Hamilton, it has been realized that requiring the independence of the derivative from the path that a differential follows toward zero is too restrictive: it excludes even ${\displaystyle f(q)=q^{2}}$ from differentiability. Therefore a direction-dependent derivative is necessary for functions of a quaternion variable.[5][6]

The Gâteaux derivative of a quaternionic function f(x) is given by

${\displaystyle \partial f(x)(h)=\lim _{t\to 0\in R}(t^{-1}(f(x+th)-f(x)))}$

where h is a quaternion indicating the direction in which the derivative is to be taken. On the quaternions, the Gateaux derivative will always be linear in h, so it may be expressed as

${\displaystyle \partial f(x)(h)=\sum _{s}{\frac {{}_{(s)0}\partial f(x)}{\partial x}}h{\frac {{}_{(s)1}\partial f(x)}{\partial x}}.\,\!}$

The number of terms in the sum will depend on the function f. The expressions

${\displaystyle {\frac {{}_{(s)p}\partial f(x)}{\partial x}},\ p=0,1\,\!}$

are called components of the Gateaux derivative.

For the function f(x) = axb, the derivative is

${\displaystyle \partial f(x)(h)=ahb\,\!}$

and so the components are:

 ${\displaystyle {\frac {{}_{(1)0}\partial axb}{\partial x}}=a\,\!}$ ${\displaystyle {\frac {{}_{(1)1}\partial axb}{\partial x}}=b\,\!}$

Similarly, for the function f(x) = x2, the derivative is

${\displaystyle \partial f(x)(h)=xh+hx}$

and the components are:

 ${\displaystyle {\frac {{}_{(1)0}\partial x^{2}}{\partial x}}=x\,\!}$ ${\displaystyle {\frac {{}_{(1)1}\partial x^{2}}{\partial x}}=1\,\!}$ ${\displaystyle {\frac {{}_{(2)0}\partial x^{2}}{\partial x}}=1\,\!}$ ${\displaystyle {\frac {{}_{(2)1}\partial x^{2}}{\partial x}}=x\,\!}$

Finally, for the function f(x) = x−1, the derivative is

${\displaystyle \partial f(x)(h)=-x^{-1}hx^{-1}}$

and the components are:

 ${\displaystyle {\frac {{}_{(1)0}\partial x^{-1}}{\partial x}}=-x^{-1}\,\!}$ ${\displaystyle {\frac {{}_{(1)1}\partial x^{-1}}{\partial x}}=x^{-1}\,\!}$