Today I was toying around with reflecting points over other points, which I expanded to reflecting lists of points over other lists of points. In the process, I found an interesting yet somewhat bizarre algebraic structure to describe what I was doing, and I would like to know if there exists a name for it or if similar structures have been studied previously.

What follows is a description of the "observed properties" of this algebraic structure. I'm not currently certain which properties should be considered axioms and which are derived from others.

Let $(R, +, \cdot )$ be a set $R$ with two binary operations. Furthermore, each element of this set belongs to one of two possible "types." Each element of the first type (to be indicated with a subscript $_0$ such as $a_0$) represents a "translation," while each element of the second type ($a_1$) represents a "reflection." If a subscript is not provided then the relation should hold regardless of the type.

Addition acts as a "list-like" operator, based on how it interacts with multiplication (described later). It is associative but not commutative.

$$(a + b) + c = a + (b + c)$$

When two elements are added, their types xor together.

$$a_x + b_y = c_{x \oplus y}$$

Any element added to its negation (additive inverse) results in the identity.

$$a + -a = I_0$$ $$a + I_0 = I_0 + a = a$$

Any type-$1$ element is its own negation, as is the identity, but all other type-$0$ elements do not equal their negation.

$$a_1 = -a_1$$ $$I_0 = -I_0$$

Swapping the order of operands require that you negate both operands and the expression. The negation does not "distribute" over addition.

$$a + b = -(-b + -a)$$

I believe, but am not entirely certain, that $(R, +)$ qualifies as a group.

The multiplication operation represents the translation/reflection of one element by/over another:

  1. reflection of $a_1$ over $b_1$

$$a_1 \cdot b_1 = c_1$$

$$a_1 \cdot a_1 = a_1$$

  1. translation of $a_1$ by $b_0$

$$a_1 \cdot b_0 = c_1$$

  1. reflection of $a_0$ over $b_1$

$$a_0 \cdot b_1 = -a_0$$

  1. translation of $a_0$ by $b_0$

$$a_0 \cdot b_0 = a_0$$

The identity plays a special role here as well:

$$a \cdot I_0 = a$$ $$I_0 \cdot a = I_0$$

Multiplication is a non-associative, non-commutative, yet flexible operation, meaning that the following always holds:

$$(a \cdot b) \cdot a = a \cdot (b \cdot a)$$


$$(a \cdot b) \cdot -a = (a \cdot b) \cdot a = a \cdot (b \cdot a) = a \cdot (b \cdot -a)$$

My somewhat bizarre addition operation was constructed to give multiplication of property of "folding over" addition from the left. That is,

$$a \cdot (b + c) = (a \cdot b) \cdot c$$ $$a \cdot (b + c + d + \ldots + y + z) = a \cdot b \cdot c \cdot d \cdot \ldots \cdot y \cdot z$$

I call this "folding" based on programming terminology, and I think this property is one of the more interesting ones. It follows from the above that there is no concept of left(?) division:

$$a \cdot (b + a) = (a \cdot b) \cdot a = a \cdot (b \cdot a)$$ $$(b + a) \ne (b \cdot a)$$ $$a_0 \cdot b_0 = a_0 = a_0 \cdot c_0$$ $$b_0 \ne c_0$$

Multiplication also distributes right, similar to the concept of "mapping" in programming.

$$(b + c) \cdot a = (b \cdot a) + (c \cdot a)$$

Negating the first argument of a multiplication expression is equivalent to negating the whole expression.

$$-a \cdot b = -(a \cdot b)$$

There also also some other useful equations to relate addition, multiplication, and negation:

$$-a + b + a = b \cdot a$$ $$a \cdot b = c \leftrightarrow c \cdot -b = a$$ $$a_0 \cdot b = a_0 \cdot -b$$

Again, I am not entirely confident on the self-consistency of these relationships. I don't have a real axiomatization yet. With the properties above I've been able to "prove" some simple equivalences such as

$$c_x \cdot (c_x \cdot a_1) \cdot b_x = c_x \cdot (b_x \cdot a_1 \cdot c_x)$$

which corresponds to this diagram (which uses "r" for "reflection" instead of "$\cdot$"):

enter image description here

Have similar algebraic structures been named/studied before? I am particularly interested in structures that obey $a \cdot (b + c) = (a \cdot b) \cdot c$. If possible, I would also like to find ways to reduce the role of "typing" in the properties listed above.

  • $\begingroup$ Does $(a\cdot b)\cdot −a=(a\cdot b)\cdot a$ only hold for type $1$ numbers or also for type $0$ numbers? $\endgroup$ – vrugtehagel Mar 9 '17 at 8:28
  • $\begingroup$ @vrugtehagel Both types. It also holds when $a$ and $b$ are different types than each other. $\endgroup$ – PhiNotPi Mar 9 '17 at 8:36
  • $\begingroup$ Your $(R,+)$ looks mighty like a group of all translations and reflections of a real line with the operation of composition (which is a group indeed). Multiplication is another story; I don't quite understand what it "really" means. $\endgroup$ – Ivan Neretin Mar 9 '17 at 10:22
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    $\begingroup$ @IvanNeretin I think you're correct regarding $(R, +)$. Regarding multiplication, for type-$1$ numbers at least, it represents the reflection of points over other points. For example, $a_1 \cdot (b_1 + c_1)$ is reflection of $a_1$ over $b_1$ then again over $c_1$. The expression $(b_1 + c_1) \cdot a_1$ is the reflection of those points (individually) over $a_1$, giving $(b_1 \cdot a_1) + (c_1 \cdot a_1)$. It's also important to note that $b_1 + c_1 = d_0$, so $a_1 \cdot d_0$ is a translation of $a_1$ by $d_0$. $\endgroup$ – PhiNotPi Mar 9 '17 at 14:58
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    $\begingroup$ I didn't read the question, but +1 for the effort of posting such a question :-) $\endgroup$ – Surb Mar 10 '17 at 19:07

$(R, +)$ is a non-Abelian group of translations/reflections with composition

This part of the question was relatively straightforward (see comments). Type-$0$ elements are translations, with $I_0$ being the identity (no translation). Type-$1$ elements are reflections. Any two reflections are equivalent to a translation, a translation and reflection is still a reflection, etc. The most relevant feature of a group is that

$$a + b = -(-b + -a)$$

$(R, \cdot)$ is a "quandle" based on conjugation of $(R, +)$

This took me a while to find, because I had never heard of a "quandle" before. A quandle is a set with one operation $\triangleleft$. In our case, the $\triangleleft$ operation of the quandle is the $\cdot$ operation of our structure.

Although there different ways to create a quandle, one can be formed from any group using the operation of conjugation, that is

$$a \triangleleft b = b^{-1}ab$$

Or, in our case,

$$a \cdot b = -b + a + b$$.

The axioms of a quandle are:

  1. Self-distributive on one side: $$(a \triangleleft b) \triangleleft c = (a \triangleleft c) \triangleleft (b \triangleleft c)$$

    This self-distributive property is shown by

    $$(a\cdot b)\cdot c = a\cdot(b+c) = a\cdot(c + -c + b + c) = a\cdot(c + b \cdot c) = (a\cdot c)\cdot(b\cdot c)$$

    We can continue this to show $-(b\cdot c) = -b \cdot c$:

    $$(a\cdot b)\cdot c = (-b + a + b)\cdot c = -b\cdot c + a\cdot c + b\cdot c $$ $$(a\cdot c)\cdot (b\cdot c) = -(b\cdot c) + a\cdot c + b\cdot c $$

  2. For any pair $a$ and $b$, there is a unique $c$ such that

    $$c \triangleleft b = a$$

    (This is axiom equivalent to stating the existence of an inverse operation $\triangleright$.)

    In our case,

    $$c \cdot b = a$$ $$-b + c + b = a$$ $$c = b + a + -b$$ $$c = a \cdot -b$$

  3. The operation is idempotent:

    $$a \triangleleft a = a$$

    In our case,

    $$a \cdot a = -a + a + a = a$$

    Furthermore, in a result that lines up with our choice of inverse operation,

    $$a \cdot -a = a$$

    This third axiom is what separates "quandles" from the simpler structures of "racks".

Based on what I've read, quandles representing reflections were some of the first such studied (back in the 1940s). This particular quandle is slightly different than those because it deals with both reflections and translations.

An "involuntory" quandle is one in which

$$(a \triangleleft b) \triangleleft b = a$$

is always true, but in our case, this statement is generally only true when $b = -b$.

Sources: webpage by Sam Nelson and nLab and Wikipedia

Other properties arise from those listed above

The conjugation means that multiplication distributes-right over addition:

$$(a + b) \cdot c$$ $$-c + a + b + c$$ $$-c + a + c + -c + b + c$$ $$a \cdot c + b \cdot c$$

The folding-left property is a result of the underlying group property mentioned at the beginning of this answer:

$$c \cdot (a + b)$$ $$-(a + b) + c + (a + b)$$ $$-b + -a + c + a + b$$ $$(-a + c + a) \cdot b$$ $$(c \cdot a) \cdot b$$

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    $\begingroup$ I didn't read the answer, but +1 for posting such an answer :-) $\endgroup$ – Surb Mar 10 '17 at 19:07
  • $\begingroup$ Just noticed a typo. You typed "is" twice in the second bolded line. Otherwise great answer. +1 $\endgroup$ – Harnoor Lal Mar 10 '17 at 21:41
  • $\begingroup$ @HarnoorLal fixed, and thanks, $\endgroup$ – PhiNotPi Mar 10 '17 at 21:42

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