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

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

• It depends entirely on context. It is not uncommon for $\otimes$ to be "multiplication but different", the fact that there is the circle surrounding the usual operation symbol emphasizing that. For quaternions, it could just be the usual multiplication of quaternions... noting that the multiplication of quaternions is already different enough than multiplication of reals. Commented Jun 5, 2023 at 11:56
• If you were to provide more context... a link to where you have seen it used or a page number reference, etc... then maybe we could be more certain. Absent such a reference, "different authors, different notational choices" is about all that can be said. Commented Jun 5, 2023 at 11:59
• The answer is necessarily context-dependent.. Can you give a specific reference for the case in question, namely the quaternions. Commented Jun 5, 2023 at 19:49
• @RobArthan the example (in the edit) was for quaternions. I added it between brackets. The context was differentiating functions for quaternions, as the dx suggests.
– mick
Commented Jun 5, 2023 at 19:51
• I meant that you should give a specific reference, to the book or paper or lecture notes or wherever you saw the example. Commented Jun 5, 2023 at 19:53

$$\newcommand\H{\mathbb{H}}$$ 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 $$\frac{dx^2}{dx} = I\otimes x + x \otimes I.$$

• Interesting. But how did you do the first step with the $d/dt$ part ?
– mick
Commented Jun 5, 2023 at 21:10
• I added one more line to the calculation Commented Jun 5, 2023 at 22:09

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*}

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.

• Thank you for your contribution
– mick
Commented Jun 7, 2023 at 19:48

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