(1/15)
Loops and their multiplication groups
A thread in 15 parts
(0/15)
(1/15)
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)
- The nonzero octonions under multiplication
- The sphere S^7 under octonion multiplication
- I have discussed other examples previously:
https://t.co/q5LjmxHEIF
https://t.co/UPHSMwQo75
(3/15)
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)
(4/15)
(5/15)
(6/15)
(7/15)
Now back to the general case where Q is any loop.
(8/15)
(9/15)
Loops can be studied via their multiplication and inner mapping groups. I will give one example of how this works.
(10/15)
(11/15)
1. N is the kernel of a homomorphism,
2. N is invariant under Inn(Q),
3. N is a block of Mlt(Q).
Such an N is called a normal subloop of Q. Q is simple if it has no nontrivial normal subloops.
(12/15)
Theorem (Albert 1941): A loop Q is simple if and only if Mlt(Q) is primitive.
(13/15)
More from Maths
\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
You May Also Like
Five billionaires share their top lessons on startups, life and entrepreneurship (1/10)
I interviewed 5 billionaires this week
— GREG ISENBERG (@gregisenberg) January 23, 2021
I asked them to share their lessons learned on startups, life and entrepreneurship:
Here's what they told me:
10 competitive advantages that will trump talent (2/10)
To outperform, you need serious competitive advantages.
— Sahil Bloom (@SahilBloom) March 20, 2021
But contrary to what you have been told, most of them don't require talent.
10 competitive advantages that you can start developing today:
Some harsh truths you probably don’t want to hear (3/10)
I\u2019ve gotten a lot of bad advice in my career and I see even more of it here on Twitter.
— Nick Huber (@sweatystartup) January 3, 2021
Time for a stiff drink and some truth you probably dont want to hear.
\U0001f447\U0001f447
10 significant lies you’re told about the world (4/10)
THREAD: 10 significant lies you're told about the world.
— Julian Shapiro (@Julian) January 9, 2021
On startups, writing, and your career:
BREAKING: President Donald Trump has submitted his answers to questions from special counsel Robert Mueller
— Ryan Saavedra (@RealSaavedra) November 20, 2018
Mueller's officially end his investigation all on his own and he's gonna say he found no evidence of Trump campaign/Russian collusion during the 2016 election.
Democrats & DNC Media are going to LITERALLY have nothing coherent to say in response to that.
Mueller's team was 100% partisan.
That's why it's brilliant. NOBODY will be able to claim this team of partisan Democrats didn't go the EXTRA 20 MILES looking for ANY evidence they could find of Trump campaign/Russian collusion during the 2016 election
They looked high.
They looked low.
They looked underneath every rock, behind every tree, into every bush.
And they found...NOTHING.
Those saying Mueller will file obstruction charges against Trump: laughable.
What documents did Trump tell the Mueller team it couldn't have? What witnesses were withheld and never interviewed?
THERE WEREN'T ANY.
Mueller got full 100% cooperation as the record will show.