Algebra 1 and pre-algebra worksheets are something that all would love to have been able to dominate in high school. Sometimes they do not dominate it there and that's when it could be a bit difficult.
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Loops and their multiplication groups
A thread in 15 parts
(0/15)
Recall that a quasigroup (Q,*) is a set Q with a binary operation * such that for each a,b in Q, the equations a*x=b and y*a=b have unique solutions x,y. Groups are quasigroups and this property is usually one of the first things proved in elementary group theory.
(1/15)
Note that we don't assume associativity of *!
A loop is a quasigroup with an identity element. The story of why they are called loops is an interesting one and may even be true, but I will save it for another day. I am going to focus on loops in this thread.
(2/15)
Natural examples of nonassociative loops:
- The nonzero octonions under multiplication
- The sphere S^7 under octonion multiplication
- I have discussed other examples
For each x in a loop Q, define the left & right translations L_x, R_x : Q->Q by L_x(y)=xy and R_x(y)=yx. These mappings are permutations of Q. The composition L_x L_y of two left translations is not necessarily a left translation because Q is not necessarily associative.
(4/15)
A thread in 15 parts
(0/15)
Recall that a quasigroup (Q,*) is a set Q with a binary operation * such that for each a,b in Q, the equations a*x=b and y*a=b have unique solutions x,y. Groups are quasigroups and this property is usually one of the first things proved in elementary group theory.
(1/15)
Note that we don't assume associativity of *!
A loop is a quasigroup with an identity element. The story of why they are called loops is an interesting one and may even be true, but I will save it for another day. I am going to focus on loops in this thread.
(2/15)
Natural examples of nonassociative loops:
- The nonzero octonions under multiplication
- The sphere S^7 under octonion multiplication
- I have discussed other examples
Rethinking Vector Addition
— Michael Kinyon (@ProfKinyon) December 1, 2020
or
How I Learned to Stop Worrying and Love Nonassociativity
A thread in 29 tweets
(0/28)
For each x in a loop Q, define the left & right translations L_x, R_x : Q->Q by L_x(y)=xy and R_x(y)=yx. These mappings are permutations of Q. The composition L_x L_y of two left translations is not necessarily a left translation because Q is not necessarily associative.
(4/15)
In light of my tweet thread about the category of finite sets and commutative monoids (https://t.co/jnY0wZZbxq), I thought I might try to say what the analogue is for braided monoidal things (although much of this is still somewhat hypothetical).
It's also just kind of a cool combinatorial structure! I've been talking to @CreeepyJoe about this lately, as well as @grassmannian.
The first thing you have to know is that, in a braided monoidal category you can still have commutative monoids. Since a braided monoidal category C has a "twist" map for every object β(x):x⊗x→x⊗x, if x is a monoid you can ask for the following diagram to commute:
Remember that being symmetric monoidal just means that if you take the twist map above and do it twice, you get the identity map, but braided monoidal doesn't mean that. But it's okay! You can still define commutative monoids here.
But so anyway, we can talk about commutative monoids in braided monoidal categories.
So okay, here's a thread on the category of finite sets and a way in which it controls algebraic structure in symmetric monoidal categories. I think it's some really pretty stuff.
— Jonathan Beardsley (@JBeardsleyMath) December 6, 2020
It's also just kind of a cool combinatorial structure! I've been talking to @CreeepyJoe about this lately, as well as @grassmannian.
The first thing you have to know is that, in a braided monoidal category you can still have commutative monoids. Since a braided monoidal category C has a "twist" map for every object β(x):x⊗x→x⊗x, if x is a monoid you can ask for the following diagram to commute:
Remember that being symmetric monoidal just means that if you take the twist map above and do it twice, you get the identity map, but braided monoidal doesn't mean that. But it's okay! You can still define commutative monoids here.
But so anyway, we can talk about commutative monoids in braided monoidal categories.
OK, I may be guilty of a DoS attack attempt on mathematicians' brains here, so lest anyone waste too much precious brain time decoding this deliberately cryptic statement, let me do it for you. •1/15
First, as some asked, it is to be parenthesized as: “∀x.∀y.((∀z.((z∈x) ⇒ (((∀t.((t∈x) ⇒ ((t∈z) ⇒ (t∈y))))) ⇒ (z∈y)))) ⇒ (∀z.((z∈x) ⇒ (z∈y))))” (the convention is that ‘⇒’ is right-associative: “P⇒Q⇒R” means “P⇒(Q⇒R)”), but this doesn't clarify much. •2/15
Maybe we can make it a tad less abstruse by using guarded quantifiers (“∀u∈x.(…)” stands for “∀u.((u∈x)⇒(…))”): it is then “∀x.∀y.((∀z∈x.(((∀t∈x.((t∈z) ⇒ (t∈y)))) ⇒ (z∈y))) ⇒ (∀z∈x.(z∈y)))”. •3/15
Maybe a tad clearer again by writing “P(u)” for “u∈y” and leaving out the quantifier on y, viꝫ: “∀x.((∀z∈x.(((∀t∈x.((t∈z) ⇒ P(t)))) ⇒ P(z))) ⇒ (∀z∈x.P(z)))” [✯]. Now it appears as an induction principle: namely, … •4/15
… “in order to prove P(z) for all z∈x, we can assume, when proving P(z), that P(t) is already known for all t∈z∩x” (n.b.: “(∀z.(Q(z)⇒P(z)))⇒(∀z.P(z))” can be read “in order to prove P(z) for all z, we can assume Q(z) known when proving P(z)”). •5/15
\u2200x.\u2200y.((\u2200z.((z\u2208x) \u21d2 ((\u2200t.((t\u2208x) \u21d2 (t\u2208z) \u21d2 (t\u2208y)))) \u21d2 (z\u2208y))) \u21d2 (\u2200z.((z\u2208x) \u21d2 (z\u2208y))))
— Gro-Tsen (@gro_tsen) February 12, 2021
First, as some asked, it is to be parenthesized as: “∀x.∀y.((∀z.((z∈x) ⇒ (((∀t.((t∈x) ⇒ ((t∈z) ⇒ (t∈y))))) ⇒ (z∈y)))) ⇒ (∀z.((z∈x) ⇒ (z∈y))))” (the convention is that ‘⇒’ is right-associative: “P⇒Q⇒R” means “P⇒(Q⇒R)”), but this doesn't clarify much. •2/15
Maybe we can make it a tad less abstruse by using guarded quantifiers (“∀u∈x.(…)” stands for “∀u.((u∈x)⇒(…))”): it is then “∀x.∀y.((∀z∈x.(((∀t∈x.((t∈z) ⇒ (t∈y)))) ⇒ (z∈y))) ⇒ (∀z∈x.(z∈y)))”. •3/15
Maybe a tad clearer again by writing “P(u)” for “u∈y” and leaving out the quantifier on y, viꝫ: “∀x.((∀z∈x.(((∀t∈x.((t∈z) ⇒ P(t)))) ⇒ P(z))) ⇒ (∀z∈x.P(z)))” [✯]. Now it appears as an induction principle: namely, … •4/15
… “in order to prove P(z) for all z∈x, we can assume, when proving P(z), that P(t) is already known for all t∈z∩x” (n.b.: “(∀z.(Q(z)⇒P(z)))⇒(∀z.P(z))” can be read “in order to prove P(z) for all z, we can assume Q(z) known when proving P(z)”). •5/15
