What is the meaning of this symbol $\otimes$, in particular for quaternions?

Question

What is the meaning of this symbol $\otimes$, in particular for quaternions ?

In other context it is

$\otimes$ sometimes refers specifically to the Kronecker product. In particular, we have $$ I_d \otimes B = \overbrace{B \oplus B \oplus \dots \oplus B}^d = \text{diag}(\overbrace{B,B, \dots, B}^d)\\ = \pmatrix{B\\&B\\&&\ddots\\&&&B} $$

Or it is some kind of Tensor product

https://en.wikipedia.org/wiki/Tensor_product

Or a partition of a space or set.

But here, for quaternions, I am confused.

edit

From the comments it seems it depends on the author and such. An example (for quaternions) is this :

$$\frac{d x^2}{dx} = 1\otimes x + x \otimes 1$$

I do not understand it.

For instance it was used here :

https://en.wikipedia.org/wiki/Quaternionic_analysis

Answer 1

$\newcommand\H{\mathbb{H}}$ Here's one possible way to interpret this equation, using directional derivatives: Let $\H$ denote the quaternions, and let $x, \dot{x} \in \H$. Then \begin{align*} \frac{dx^2}{dx}\dot{x} &= \left.\frac{d}{dt}\right|_{t=0} (x+t\dot{x})(x+t\dot{x}) \\ &=\left.\frac{d}{dt}\right|_{t=0} (x^2+t(x\dot{x}+\dot{x}x) + t^2\dot{x}^2)\\ &= \dot{x}x + x\dot{x}. \end{align*} Now let $I: H \rightarrow H$ be the identity map and define $$ I\otimes x: \H \rightarrow \H $$ by $$(I\otimes x)(y) = yx,$$ and $$ x\otimes I: \H \rightarrow \H $$ by $$(x\otimes I)(y) = xy.$$ Then we see that $$ \frac{dx^2}{dx} = I\otimes x + x \otimes I. $$

Answer 2

I've seen $\otimes$ in a few different meanings with quaternions so far.

For example, $q_{1} \otimes q_{2} \wedge \left\{ q_{1},\, q_{2} \right\} \in \mathbb{H}$ is sometimes used simply as the product or Hamilton Product of two quaternions. I liked to interpret it as a reminder that the product of quaternions is somewhat special in its non-commutativity. Also used as a short form for this where $v = q - \Re\left( q \right)$(aka $v$ is the vector part), if we consider quaternions to be four-dimensional vectors and assume their imaginary units are given by $i ~\hat{=} \left( 0,\, 1,\, 0,\, 0 \right)$, $j ~\hat{=} \left( 0,\, 0,\, 1,\, 0 \right)$ and $k ~\hat{=} \left( 0,\, 0,\, 0,\, 1 \right)$: $$ \begin{align*} q_{1} \otimes q_{2} &= \Re\left( q_{1} \right) \cdot \Re\left( q_{2} \right) - v_{1} \cdot v_{2} + \Re\left( q_{1} \right) \cdot v_{2} + \Re\left( q_{2} \right) \cdot v_{1} + v_{1} \times v_{2}\\ &\text{or}\\ q_{1} \otimes q_{2} &= \Re\left( q_{1} \right) \cdot \Re\left( q_{2} \right) + \Re\left( q_{1} \right) \cdot v_{2} + \Re\left( q_{2} \right) \cdot v_{1} + v_{1} \otimes v_{2}\\ \end{align*} $$

For more reference see this or here formula $12$ and $13$.

Another interpretation that I've only seen on Wikipedia so far is to define $\otimes$ as "Outer Product" when we interpret quaternions as $4$D vectors (see here): $$ \begin{align*} \left( q_{1} \otimes q_{2} \right)_{ij} &= \left( q_{1} \right)_{i} \cdot \left( q_{2} \right)_{j}\\ q_{1} \otimes q_{2} &= q_{1} \cdot q_{2}^{T}\\ \end{align*} $$

You also find it sometimes in the theory of algebras and fields, but there the meaning is not unique to quaternions.

Answer 3

You might be looking for the composition of Euler-Rodrigues symmetric parameters. They're defined in Shuster 1993, "A Survey of Attitude Representations' see equations (171), (185), (188) etc $$ \overline{\eta^{'}} \otimes \eta = \begin{bmatrix} \eta_{4}\eta^{'} + \eta_{4}{'}\eta - \eta^{'} x \eta \\ \eta_{4}{'}\eta_{4} - \eta^{'} \cdot \eta \end{bmatrix} (171) ...$$

Use of the composition as operators is shown in Barfoot et al. 2010 "Pose Estimation using Linearized Rotations and Quaternion Algebra"

Eqn (2) shows the right-hand and left-hand compound operators and their use in quaternion multiplication.

Barfoot et al form a matrix to implement the composition.

The quaternion lefthand compound operator + in use: $$q^{+} := \begin{bmatrix} \eta \bf{1} - \varepsilon^{x} && \varepsilon \\ -\varepsilon^{T} && \eta \end{bmatrix} $$

The quaternion righthand compound operator $\otimes$ in use: $$q^{\otimes} := \begin{bmatrix} \eta \bf{1} + \varepsilon^{x} && \varepsilon \\ -\varepsilon^{T} && \eta \end{bmatrix} $$

where $$\varepsilon^{x} = \begin{bmatrix}0 && -\varepsilon_3 && \varepsilon_2 \\ \varepsilon_3 && 0 && -\varepsilon_1 \\ -\varepsilon_2 && \varepsilon_1 && 0 \end{bmatrix} $$