groups-rings-fields

[fossil widget]

if a + bi is unit length then by definition a^2 + b^2 = 1

because a^2 + b^2 is the squared length of the vector, and 1^2 = 1

[indigo surge]

ok hold on

is a^2 + b^2 unit length, or is it literally equal to 1?

[fossil widget]

when you say a complex number is unit length you are saying that a^2 + b^2 = 1

unit length means its length is 1. a^2 + b^2 is how you compute its squared length

do you know what a vector is? if not i don't think any of this will make any sense

[indigo surge]

no no ik what a vector is

I'm familiar with linear algebra

and I've known about complex numbers for a while, it's just that them being used as rotations is new

ok so this might be a dumb question

but even tho a + bi is unit length, it still affects z?

[fossil widget]

yes, it doesn't affect the length of z but it will rotate it

[indigo surge]

got it

and multiplying by the conjugate of of a + bi (a - bi), neither affects the length nor rotates z?

[fossil widget]

it will rotate it in the opposite direction

[indigo surge]

which cancels out?

[fossil widget]

so if you apply both a + bi AND a - bi then the rotations will cancel out

but if you only apply a - bi then it will rotate, just in the opposite direction of a + bi

[indigo surge]

I see

now what if u do a similar thing as qvq^(-1)?

[fossil widget]

(a + bi)z = rotates it
(a - bi)z = rotates it in the opposite direction
(a + bi)(a - bi)z = (a^2 + b^2)z = 1z = z = doesn't rotate it at all because the rotations cancelled

[indigo surge]

so u do something like (a+bi)z(a-bi)?

[fossil widget]

yea, something like that

[indigo surge]

what would that do?

would it also cancel the rotation?

[fossil widget]

it undoes rotation in one of the planes but doubles it in the other

[indigo surge]

cuz ik in the context of quaternions it reinforces one but cancels out another

[fossil widget]

yes

[indigo surge]

but what about in complex numbers?

[fossil widget]

in complex numbers it just undoes it

[indigo surge]

oh so same as multiplying them right next to each other?

[fossil widget]

(a + bi)z(a - bi) = z will also undo the rotation

[fossil widget]

correct

[indigo surge]

interesting

[fossil widget]

quaternions work the same way as complex numbers in 1 of the planes so it will cancel in that plane

but in the other plane, by moving it to the other side it will flip the rotation again

so it will double it

[indigo surge]

I see

and, ik that 4D rotations can have either simple rotations or double rotations

quaternions always represent double rotations?

or can they represent single rotations as well?

[fossil widget]

yea, you can think of the quaternion as breaking the 4D space into two perpendicular 2D planes and executing the rotation in both planes

[fossil widget]

you can do a single rotation using the sandwitch trick

[indigo surge]

right right

but a quaternion on its own

is always a double?

[fossil widget]

yea

[indigo surge]

gotchu gotchu

damn this is a lot of new concepts 😅

going back to the complex numbers real quick

is multiplying by something like (a - bi) or (a + bi) considered a rotation?

or multiplying just by i?

or both?

[fossil widget]

i is equal to 0 + 1i which is also a unit length complex number

it will rotate by 90 degrees

[indigo surge]

oh ok

[kind temple]

multiplication by any number with unit norm corresponds to a rotation

[fossil widget]

imagine the complex number as a 2d vector, whatever angle it makes with the x-axis (real axis) will be the angle it rotates by

[indigo surge]

and anything other than "i" will just rotate by a different angle?

I see

[indigo surge]

and u can figure out that angle using the dot product formula?

[indigo surge]

oh ok

[fossil widget]

you can use arctan

[indigo surge]

oh?

[fossil widget]

but yes you can find the angle between two complex numbers with the dot product

[indigo surge]

how would u use arctan?

just plug in the two complex numbers into it?

[fossil widget]

no you'd use arctan to find the angle of a single complex number with the real axis (x-axis)

[kind temple]

https://en.wikipedia.org/wiki/Atan2

the atan2 function

or some slight variation of it

[indigo surge]

oh

ok I'd have to look into how to do that then

but dot product also works right?

to figure out the angle between complex number and x axis

also, how do u represent x-axis?

is it just 1 + 0i or something?

cuz just that feels like a point

[kind temple]

you may discover later you can write any non-zero complex number as a positive real number multiplied by a complex number on the unit circle.

this explains why complex multiplication corresponds to a dilation + a rotation

[fossil widget]

1 + 0i

[indigo surge]

is 1 + 0i not just a point?

[fossil widget]

you have to think of them as vectors

a complex number is just a 2d vector. a is just the x-component, b is just the y-component

[indigo surge]

Ok so is 2 + 0i also the x-axis?

just not normalized?

[fossil widget]

1 is the unit vector along the x-axis, i is the unit vector along the y-axis

[fossil widget]

yes

[indigo surge]

wdym by "non-zero complex number"?

0 + 5i?

something like that?

[kind temple]

sure

[indigo surge]

oh

[indigo surge]

got it

[kind temple]

just anything that doesn’t have it’s real and imaginary parts both zero

a + bi with a != 0 and b != 0

[indigo surge]

oh

ok so

complex numbers are often literally just thought of as 2D vectors?

the real part is the x-axis and imaginary part is y-axis?

[fossil widget]

yes

historically the concept of a vector was derived from the complex numbers (and the quaternions)

[indigo surge]

and (unit-length) complex numbers can be used as rotations

oh

[fossil widget]

you can think of complex numbers as outdated notation for 2d vectors, if you want

[kind temple]

that’s often how the complex numbers are defined. R^2 with a multiplication

[indigo surge]

wdym by "R^2 with a multiplication"?

[kind temple]

R^2 is a vector space. its elements are ordered pairs of real numbers (x,y). in particular, you can add these pairs of real numbers component-wise.

to “upgrade” it to C, you define
(x,y) • (u,v) = (xu - yv, xv + yu)

from there you can redefine 1 and (1,0) and i as (0,1) and write each ordered pair (x,y) as

(x,y) = (x,0) + (0,y) = x(1,0) + y(0,1) = x + yi

and you can check that the multiplication works how it should work

so C is R^2 with a multiplication

[indigo surge]

so wait

how is applying a rotation using a vector different from applying a rotation using a matrix?

cuz that's what I'm used to seeing

[kind temple]

well, each vector (x,y) such that x^2 + y^2 = 1 determines a rotation matrix

[x -y]
[y x]

and vice versa

if you have a 2d rotation matrix, you can get a 2d vector with unit norm

[kind temple]

but the “rotation using a vector” is encoded in the definition of complex multiplication

[indigo surge]

oh what

so if u have a unit vector

u can make an equivalent rotation matrix?

or something like that?

[kind temple]

yea

[indigo surge]

oh

ok I need to read up on all of that

I didn't know about that

ok now, back to quaternions real quick

so, earlier, mc said that one of the rotations a quaternion causes works the same was as the rotation caused by a complex number?

e.g. multiplying the quaternion by its inverse cancels it out

[fossil widget]

yea, what's your question

[indigo surge]

sorry I'm just trying to think all this thru

[fossil widget]

q*qv will cancel the rotations in both planes, but moving q* to the right will cause only 1 to cancel

[indigo surge]

interesting, is there a name for that? like a property or something?

or is that just something that happens?

cuz yeah it's interesting how in complex numbers both versions completely cancel the rotation

[fossil widget]

it just follows from the fact that rotations commute in the first plane but anticommute in the second

[indigo surge]

but in quaternions, their results differ

[fossil widget]

yea, that's the main difference

[indigo surge]

I see

idk if there even is an answer to this question but, which plane is which? if that makes sense?

as in, how do u know which plane's gets doubled, and which plane's rotation gets canceled out?

is it possible to determine that, or is that just some background magic?

[fossil widget]

the plane spanned by your quaternion and 1 is the first plane, the one that behaves the same as with a complex number

and then whichever plane is perpendicular to that is the second one

which gets doubled

[indigo surge]

by "1" do u mean the identity quaternion?

(1, 0, 0, 0)?

[fossil widget]

yea

that might help

[indigo surge]

ok so

that axis pointing up in that picture

is the axis that points in the direction of ur vector?

[fossil widget]

its the axis you want to rotate around

[indigo surge]

mhm

so that's the one that gets doubled?

[fossil widget]

the plane perpendicular to that axis (which will also be perpendicular to the whole plane)

is the one the rotation gets doubled in

let me explain it like this

just like you can split a complex number z into a component along 1 (real component) and a component perpendicular to 1 (imaginary component)

you can split a quaternion into a component along 1 (real component) and a component perpendicular to 1 (in that picture it's a)

the plane spanned by 1 and a (or alternatively 1 and q, it's the same plane) is the one that behaves like the ordinary complex plane

[indigo surge]

so the plane that's spanned by q and 1?

[fossil widget]

and the plane perpendicular to that is the one that behaves differently, in which the angle is doubled

[indigo surge]

wait what?

they are pointing in different directions in the diagram

and are orthogonal to each other, no?

[fossil widget]

the identity quat (1) and your quat (q) span a plane

call it plane 1

the rotation axis a is also in plane 1

plane 2 is perpendicular to plane 1, it's not in that drawing

so plane 2 will be perpendicular to a since a is in plane 1

does that make sense?

[indigo surge]

oh ok so everything in that diagram is meant to be on the same plane?

[fossil widget]

yes

it's comparing plane 1 of the quaternion with a regular complex plane and showing the similarity

here's a final thing that might help you finally put it all together

purple arrow is your quat (q).
blue arrow is the identity quat (1)
red arrow is the rotation axis.
purple ring is plane 1
red ring is plane 2, perpendicular to plane 1

and the animation is showing the rotation being applied in both planes

[indigo surge]

ok I kinda get htat

is "a" a pure quaternion then?

[fossil widget]

yea

its perpendicular to the identity quat so it has no real component

its pure 3d, you could say

and plane 2, which is the plane you want to rotate in, is perpendicular to it

you dont want to rotate in plane 1, so you would use the sandwitch trick to undo that purple rotation

and youd double the red rotation

and then you can cut the angle in half to componsate for the doubling

[indigo surge]

ok so u don't want to rotate in the plane that is spanned by q and 1?

[fossil widget]

correct

[indigo surge]

oh wait

hold on I think that makes sense

real quick so, a is the axis of rotation?

[fossil widget]

yes

[indigo surge]

oh ok and u want to rotate around the axis of rotation

that's why u don't wanna rotate the plane that it's in?

[fossil widget]

yes

[indigo surge]

ok ok

[fossil widget]

in a regular 3d rotation, the part of your vector that lies along the axis of rotation is supposed to stay the same

imagine rotating a globe, all the points on the line of rotation through the globe should not move

only the points perpendicular to it should rotate

so you definitely dont want the component of your vector lying along the axis of rotation to rotate into the 4th dimension

[indigo surge]

and does that happen when the "w" becomes non-zero?

[fossil widget]

yea, the part of your vector that's supposed to stay the same will get shorter because it's rotating in that plane

part of it "transfers" to the w-component

anyway if you're trying to do a regular rotation in 3d space you need to only rotate in plane 2

[indigo surge]

I see

ok this is sorta coming together

now, I saw this about double rotations on the wikipedia page:

"For each rotation R of 4-space (fixing the origin), there is at least one pair of orthogonal 2-planes A and B each of which is invariant and whose direct sum A ⊕ B is all of 4-space."

that makes it sound like planes A and B don't rotate at all

but as we just discussed, they do get rotated

am I just misunderstanding what it's saying there?

[fossil widget]

yes

it just means the planes are mapped to themselves

so like a vector lying in one of the planes will still lie in that plane after the rotation

which is what you would expect

it will be rotated, but still lie in that plane

if you just read the next sentence it makes it clear: "Hence R operating on either of these planes produces an ordinary rotation of that plane."

[indigo surge]

ok so it just means that the vectors that lie on those planes will all still lie on those planes after the rotation is done?

[fossil widget]

yes

[indigo surge]

ok

that's weird, cuz qv moves the vector off of the plane

by giving it a "w" component

but oh well

[fossil widget]

no it's still in that plane

it moves it off the a axis

but it still lies in plane 1 (spanned by 1 and q (or a))

[indigo surge]

oh

so even tho w becomes non-zero, it's still in the plane?

[fossil widget]

yea, it's in plane 1 still

[indigo surge]

I see

[fossil widget]

i think you may be overthinking it. it's really not that complicated. the quat q breaks 4d space into 2 perpendicular planes and executes a rotation in both of those planes

and that's it

some other quat v can be broken into 2 components, one in each of those planes, and those components will rotate

[indigo surge]

yeah ur probably right lol

it's always 2 perpendicular planes in a double rotation in 4D right?

[fossil widget]

yea, which planes in particular will depend on how q is pointing relative to 1

but plane 1 is always the plane spanned by q and 1 and plane 2 is whatever plane is perpendicular to plane 1

[indigo surge]

got it

ok I'ma leave all the complicated behind-the-scenes stuff alone for now

quick question about the single vs double rotation tho

is there a simple way to tell whether a quaternion (or a 4x4 matrix for that matter) will produce a simple (1 plane) rotation or double (2 plane) rotation?

or not really?

[fossil widget]

a quaternion always breaks 4d space into 2 planes and rotates in both of those planes

if you want a rotation in a single plane you need to use the sandwitch product trick

to cancel one of them out

so the answer is a single quat q always does a double rotation

it can never do a single rotation by itself

[indigo surge]

I see

ok wow this was a lot of great info

thank u very much for ur time

[fossil widget]

you're welcome, good luck

[indigo surge]

thanks

now just to generalize, higher-dimension rotations can rotate more than 2 planes at once right?

e.g. 6D, 10D and whatnot?

[fossil widget]

yep

[indigo surge]

thanks

oh wait by the way

regarding that simple/double rotation stuff

that can only be done by unit quaternions right?

or 4x4 matrices with det +/- 1?

cuz everything else also scales, shears, etc.?

[fossil widget]

yea if the quaternion isnt a unit quaternion it will also scale

same goes for complex numbers

[indigo surge]

I see

but does the one/two plane thing hold?

like, a 4D linear transformation will either affect one or two planes at the same time?

it's just that it won't be a pure rotation?

[wicked patio]

Nonunit quaternions should do the same thing as unit quaternions but just also scale by the magnitude right

scaling leaves all planes invariant

[indigo surge]

can't they also shear?

[wicked patio]

It's just a unit quaternion times a positive real number, right

The positive real should just scale, if I'm understanding the representation H → R^4×4 we're using correctly

[indigo surge]

ye but a non-unit quaternion

[wicked patio]

Is a unit quaternion (rotation) times a positive real (scaling)

[fossil widget]

no, just scale and rotate. a quaternion isnt equivalent to a general 4d linear transformation

[dire wren]

Is every torsion free abelian group isomorphic to a subgroup of Q^n under addition for some cardinal n?

[lusty marlin]

No, ℚ^ℕ isn't a subgroup of ℚ^n for any n.

[dire wren]

n doesn't have to a natural

n is a cardinal

[lusty marlin]

Oh ok

[indigo surge]

oh ok, thanks!

[dire wren]

I mean I am reasonably certain this is true, but i am unable to prove it

[indigo surge]

oh yeah and also

is there a reason that for 2D u can extend the real number system by just one parameter (a + bi) for a total of 2 parameters to allow for 2D rotations

but for 3D u can't extend by 2 parameters and get something like (a + bi + cj), but instead u have to extend by 3 parameters to get (a + bi + cj + dk)?

is it something to do with "division algebra"?

and how (a + bi + cj) are not gauranteed to have a multiplicative inverse?

plz ping, thanks!

[dire wren]

Rotations in 3d space are parameterized by 3 values, if you also want scaling that comes upto 4 values

[indigo surge]

yeah but I was referring to extending the real number system

like how complex numbers for 2D are 2 parameters but how quaternions for 3D are 4 parameters

[dire wren]

well it depends on what properties you want

[indigo surge]

?

[dire wren]

If you don't care about division, then (a,b,c)*(d,e,f) = (ad,be,cf) works

[indigo surge]

ok, thanks

[dire wren]

But if you want to garantee invertibility
And let us say you are extending C

so you want to assign some values to i*j and j*j

but now if i*j = a+bi+cj

then (i-c)*j = a + bi
but (i-c)*((a+bi)/(i-c)) = a+bi

so (i-c)(j-(a+bi)/(i-c)) = 0

But neither of those is 0

[frail shoal]

yes: let A ∈ Ab be torsion-free. for every nonzero a ∈ A, we have <a> ⊆ A and <a> ≅ Z embeds into Q. Q is an injective object in Ab, so we get a map from A to Q extending this embedding: in particular, a is not in the kernel of this map. using the universal property of Q^(A \ {0}), we have a map from A to Q^(A \ {0}), and by construction, no nonzero a ∈ A can be in this map's kernel

[dire wren]

What is an injective object?

[mint seal]

I haven't seen this argument, nice one

[frail shoal]

in general, given an injective R-module I, an R-module M embeds into some I^n if every cyclic submodule of M embeds into I, although actually im not sure if the converse is true

dw ill respond to what you said

im just saying this to say it

[lusty marlin]

Nice

[frail shoal]

I is injective if, for all A ⊆ B, for all f: A -> I, we have a map g: B -> I such that g restricts to f on A

so i guess im implicitly assuming the proof that Q is injective

[dire wren]

Yes

Do you have any hints on the proof of that?

[frail shoal]

the issue is that i never studied the proof too well, it's in aluffi, but i uhhh skimmed it

[dire wren]

I see, i don't need the proof though, just a hint

[frail shoal]

you might need zorn's lemma?

unsure, i should reference a proof

actually, you might not need it for Q specifically? unsure

this criterion requires zorn to prove

i shall actually study the proof for the first time in my life and get back to you

[verbal valley]

the proof is not too bad

[dire wren]

Oh, can't I just do,

if a --> 1

and b^k = a^j , f(b) = j/k
and if this is not true for any j or k, f(b) = 0

That doesn't prove injectivity, but that's enuf for the qn i gave

I think

[frail shoal]

that might work

if so it'd be a better solution cuz it's constructive

[dire wren]

ig, thx

Also i haven't rlly studied modules yet

[verbal valley]

(injective => extending from ideals is trivial, and in the other direction, you form a poset of submodules with an extension of f, hit this with zorn, and then using the condition for ideals you can actually extend the homomorphism further than the upper bound unless it really is extended all the way)

but yea choice is required

[frail shoal]

ah yea fair. you can replace R-module with abelian group, R-linear map with group homomorphism, R with (Z, +), and I with subgroups of (Z, +)

if you're curious

it turns out to be equivalent to an abelian group being "divisible", i.e. for every a ∈ A and natural number n there exists b ∈ A such that nb = a

[verbal valley]

(this happens for R any PID)

[dire wren]

That makes some amount of sense

So no cyclic abelian group is injective?

[frail shoal]

yea

[dire wren]

Weird, i would expect like C2 to be injective.

[verbal valley]

yeah standard examples of injective abelian groups are Q/Z and Q

[frail shoal]

yea, the issue comes trying to extend the map from 2Z to Z/2 given by sending 2 to 1

[dire wren]

Yeh, i see

That was stupid

[dire wren]

Makes sense

[verbal valley]

tbf Z/2 is an injective module over itself

but this is very special behaviour

[dire wren]

Z/p should be ig

[frail shoal]

yea this concept becomes a lot more rich once you start learning modules, and it's already kinda hairy as it is

[dire wren]

Ok thx alot

[verbal valley]

(yeah trying to pin down what injective modules look like for arbitrary rings is uh kinda terrible lol -- in a certain sense they usually have to be very big)

[dire wren]

Can you "injectivize" an abelian group by introducing formal divisors?

[verbal valley]

A localization style construction like that only works if the module is torsion free (exercise: why?) but the idea of taking the smallest injective group containing a given one is a thing you can do

this is called taking the injective hull

[dire wren]

I mean like that would take Z/2Z to Q/Z

[verbal valley]

no it would take it to zero

you can inject Z/2Z into Q/Z but that's not done by localizing

[dire wren]

Idk what localising is

[frail shoal]

ok the issue is that this doesn't work: consider the group Z^2 and the subgroup generated by (1, 0). then we would send (1, 0) to 1 and (0, 1) to 0 and also (1, 1) to 0, which isn't linear

[verbal valley]

it's exactly that kind of "introduce formal divisors" idea you were talking about -- like how we construct Q by just adding an inverse for every nonzero integer

you can look up the construction on your own time tho i need to get out of bed lol

[fading acorn]

tfw $\varinjlim_{s\in S}R[1/s]$

[verbal valley]

i prefer the description where you take a colimit over this diagram personally :3

[frail shoal]

the proof might necessarily be very tricky and zorn-y: how would we prove that we can get a map from Z^N -> Q that extends (1, 0, 0, ...) -> 1?

[dire wren]

Ah, ok, can I just take the quotient group

for the subgroup of things which share a power with a

[frail shoal]

yea that group is a subgroup

hm, but the issue is that sends everything to 0

[dire wren]

How?

Oh yeh, I use choice here

I take a representative element from every coset

Such that the representatives form a subgroup

I can do that by choice non trivially but pretty easily

Like i first well order all the cosets, and for each coset, if I am forced to take an element i take it, else i choose anything

And by abelianness I am guaranteed that i will only be forced to choose one thing

Then i send the reps to 0 and everything else to the obvious thing

@frail shoal does this work?

[frail shoal]

ohhh ok i didn't realize you meant something more like this

ok i will think about this

i feel I'd need more detail to be able to say

[sonic coral]

most of the time injectives are infinitely generated

[frail shoal]

but the general shape of the proof seems to be what you said

[frail shoal]

more precisely it'll look like this

[dire wren]

Icy

[next obsidian]

You can say things much more precisely, if there’s a finitely generated (nonzero) one then your ring is Artinian

So like if your ring isn’t Artinian, there just straight up isn’t any

Altho maybe you know that, so it’s more for @verbal valley

Also, if you want to know the structure of injectives, I think really the only time it’s tractable is to look at minimal injectives

There’s a whole structure theory for these due to Matlis, and this is somewhere in the middle of Matsumura

But it comes down to them being a direct sum of injective hulls of various residue fields of the ring

And the amount of each one can be read off of Exts and stuff, and this gives rise to the notion of Bass numbers

[sonic coral]

we just covered all of what you said in my commutative algebra class last semester so the reminder was indeed helpful

[next obsidian]

Oh I forget, you might need a Noetherian hypothesis to make what I said true about Artinian

It might possible be true without Noetherian in the local case, but the statement I’m thinking about is about a local ring, and I think to go from general to local you need Noetherian to ensure that the fg injective module stays injective?

[verbal valley]

oh neat

this was fact i didn't know

[rocky cloak]

You'll need some connectedness assumption at least. Otherwise you can just take a product of an artinian ring and a non-artinian one.

Alternatively I guess you could change to the injective envelope of any fg module being fg

[next obsidian]

Well I’m not sure, cuz for a Noetherian local ring, fg injective => Artinian right?

So if you’re Noetherian and have a fg injective I, then I_p is a fg injective for any p, and thus A_p is Artinian thus its dim 0, thus p is maximal

So now every prime ideal is maximal, and so the ring is dim 0 and Noetherian => Artinian

[rocky cloak]

I_p can be 0

Like take k a field and R and ring, then
0xk is an injective Rxk-module

Tells you nothing about R

[next obsidian]

Oh yeah fair fair fair

[rocky cloak]

But say you're Noetherian and I is an fg injective. Then there exists a maximal ideal m with I_m non-zero.

Then A_m is dim 0, so m is a minimal prime. If A is Noetherian you have finitely many minimal primes. Let I be the intersection of all except m. Then I is not contained in m, so I+m = A and InM is the nilradical.

Then A/nilradical is a product of rings, means A is a product of rings.

So if we throw in the connected assumption we should be good

[next obsidian]

Yeah ok

[next obsidian]

I think even more simply V(m) and V(intersection of other minimal primes) is a disconnection right?

So this is basically saying if you have a minimal, maximal prime, you can rip apart that part from the rest of Spec

[rocky cloak]

Yes.

Even more complicated geometry-y*

[next obsidian]

Which makes sense cuz such a point is like a floating point off in space by itself

[rocky cloak]

You need Noetherian for it to work though. But maybe that's par for the course when it comes to geometric intuition

[next obsidian]

Yeah

[indigo surge]

hey guys

so, yesterday I asked about quaternions and I ended off by asking why 3D space had to be represented using a 4D object (a + bi + cj + dk) instead of a 3D object (a + bi + cj)

and, I did a bit digging on my own and I just want to make sure I understand the reason

is it basically because closure breaks?

because u if u define ij to be something else (e.g. "k"), then u leave 3D space and enter 4D space, but if u define ij to be something like a constant (e.g. 1 or 2), then u enter 2D space after multiplying

did I understand that right?

plz ping, thanks

[rocky cloak]

Yes, if you try to define ij as some combination of 1, i and j you won't be able to define a well defined multiplication. So it will have to be a 4th thing.

A separate, less formal, reason why the quaternions are 4d can be imagined like this: Instead of thinking of quaternions as 3D space with a multiplication you can think of it encapsulating something about rotations of 3d space. Just like a complex number can be thought of as a length and a rotation of 2d space you can think of a quaternion as a length and a rotation of 3d space (this is true except that q and -q describe the same rotation, which is in fact related to spin on Physics)

As 3d space has 3 planes of rotation, length + rotation should be 4 dimensions.

[indigo surge]

I see, thanks

and this might be a dumb question too, but why does a 3D object like (a + bi + cj) not work, whereas a 3D vector like (1, 2, 3) works just fine?

cuz can't a + bi + cj also just be represented as a 3D vector like (a, b, c)?

[mint seal]

a + bi + cj works fine with addition; the problem is trying to define multiplication of them in such a way that makes a division algebra

that turns out to not be possible

but yes, a + bi + cj can be identified with (a,b,c), and with the natural addition and scaling operations, this is isomorphic to the vector space R^3

but doing this we're no longer claiming i^2 = j^2 = -1

[dull ginkgo]

Well it’s about how 3D space is represented. Any 3 dimensional vector space can “represent” positions and translation in 3D space, however another important facet of 3D space is rotation and translation, I.e Euclidean transformations.

(Special) Euclidean transformations in 2D and 3D space are rotations about an axis and translations. Algebraicly, 2D rotations and translations are encoded in complex multiplication, where
ux + b, |u| = 1 represents such transformations and can easily be composed.

To “encode” 3D rotations , we want multiplication of some algebra in some way to encode those invertible rotations (I.e be division).

To do this, the quaternions do this by conjugation by the exponential of a purely imaginary, normalized element

But this is quite different from the complex case.

Notice how for complex numbers, we use “the whole” space. As in, a + bi are our components…

But for quaternions to be used as such, we only use the “imaginary” part bi + cj + dk. The real part for this purpose is kinda like an “intermediate” thing.

But that additional “real” part needed in the intermediate steps is why we need 4D

This is especially worsened by the fact that rotations in 3D are noncommutative

[fossil widget]

if you can rotate a vector by 90 degrees in some plane, then as long as you can also scale it, you can rotate it by any angle in that plane. so in some sense a 90 degree rotation is the most fundamental. in 2d you have only one plane you can rotate in and i is your "90 degree rotator" for that plane. but in 3d you have 3 planes you can rotate in so you need 3 "90 degree rotators" (i, j, and k), and you still need the real part for the scaling.

if you want to make a 3d unit vector represent a 90 degree rotator, you can immediately see that you cant do it consistently with only 3 dimensions

like if you imagine that the unit vector will rotate any vector perpendicular to it by 90 degrees, well that's fine

but what if the vector is parallel to it

then the first problem is you have an infinite number of planes you could potentially rotate in, but side stepping that issue, if you just pick one

that plane will intersect the plane perpendicular to it

but then you wont have two applications of the rotator resulting in a flipping of your original vector: u(uv) = -v, which is a feature you definitely want. and this is resolved by moving to 4d because in 4d you can have two planes that don't intersect

[dull ginkgo]

It just took me a while to realize that the quaternion form of rotations is mechanically way different than the complex one

[fossil widget]

they function pretty similarly

[dull ginkgo]

Similarly yeah

Just what took me the longest time to grapple is the fact that we use both the real and imaginary for positions in the complex but only the imaginary ones for quaternions

[fossil widget]

ah yea that's true

[dull ginkgo]

Also a degree of freedom problem,

Linear-conformal maps in 2D are a rotation and scaling, 2 degrees of freedom.

Linear-conformal maps in 3D are an axis-rotation and a scaling, 1 for the scaling, 2 for the axis, and 1 for the rotation, you need 4 degrees of freedom, so 3 degrees of freedom are not sufficient

[frail shoal]

just realized that every 3D rotation has an axis

that's kinda cursed

[fossil widget]

wow, i hope that coincidence doesn't lead to anything confusing

[dull ginkgo]

Yep, one real eigenvalue

that eigenspan is the axis

[frail shoal]

oh yea, if a rotation has a real eigenvalue, it must be 1 or -1

[dull ginkgo]

Must be 1

Otherwise it’d be a reflection

[frail shoal]

what about x, y -> -x, -y

wait

yea

[dull ginkgo]

In 2D that’s a rotation

In 3D making all the coordinates negative is a reflection

[frail shoal]

yea

ok what about x, y, z -> -x, -y, z

[dull ginkgo]

That’s a rotation of 180 degrees about z

[frail shoal]

yes

[dull ginkgo]

Oh shit yeah

180 degree rotations are a bit pathological in that description

But yeah

@frail shoal speaking of coolness

Rodriguez rotation formula is:

$M = \cos(\theta) \mathbf{I} + \frac{1 - \cos(\theta)}{u^T u} u u^T + \frac{\sin(\theta)}{\Vert u \Vert}[u]^\cross$, where $M$ is the rotation matrix, and $[u]^\cross$ is the antisymmetric tensor form of the vector. The symmetric part is the first two terms and the antisymmetric part is the latter two.

[cloud walrus]

Tits metric
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[dull ginkgo]

So you can actually "find" the axis and the angle from a rotation matrix in most cases

Noting that $tr(M) = 1 + 2\cos(\theta)$, you can then determine sine, and thus if sine is nonzero (i.e $\cos(\theta) \neq \pm 1$, you can divide the antisymmetric part, $A = \frac{1}{2} (M - M^\mathsf{T})$, by sine to get the antisymmetric tensor associated with $u$,

[cloud walrus]

Tits metric

[dull ginkgo]

[[(x,y,z)]^\times= \begin{bmatrix} 0 & -z & y \ z & 0 & -x \ -y & x & 0 \end{bmatrix}]

[cloud walrus]

Tits metric

[frail shoal]

is it like a 1-form

not that I'd find it particularly intuitive cuz of that

it's just all that my analysis class has been talking about

wait shit there's also a SYMMETRIC part?

[dull ginkgo]

of course

(M + M^T)/2

[dull ginkgo]

Sorta

Those are another thing lumped under the umbrella of the "exterior" / "antisymmetric" side of tensor nonsense

[frail shoal]

oh wait duh

i was bamboozled all mysterious

[dull ginkgo]

Differential forms are in a sense exterior algebra objects stretched across a whole manifold, i.e they are locally antisymmetric forms

[dull ginkgo]

This is just an isomorphism for a specific case of the exterior algebra

where antisymmetric bilinear forms in 3D are isomorphic to 3D due to a convenient overlap in dimension

[dull ginkgo]

Do you know what a 2-form etc are

[frail shoal]

yea

[dull ginkgo]

2-forms locally are antisymmetric bilinear forms, which for 3D euclidean space you can identify with vectors due to that isomorphism

same thing with curl

[frail shoal]

oh yea sounds reasonable

[dull ginkgo]

Though this is just dealing with the algebraic side, not gluing together the algebraic things into a smooth section of the overall bundle :p

[frail shoal]

anyway im thinking about the eigenvalues of a rotation. they always have to be complex units, and the product of conjugate units is always 1. additionally, the determinant of a rotation is always 1. so that means, in odd dimensions, rotations must have axes, cuz we always have an odd number of real eigenvalues, so they can't all be -1

[dull ginkgo]

Yes

4D rotations are bastards

I.e they can have two axes of rotation, for two eigenvalue pairs

Actually just two rotations, no axes

[frail shoal]

yea

kinda wacky

[dull ginkgo]

Yep lmfao

[dull ginkgo]

This is Euler’s theorem iirc

4D anything just sucks

[indigo surge]

so is it a division algebra that we want, or just closure?

[indigo surge]

oh ok

[indigo surge]

ah that makes sense

[indigo surge]

well yeah u use a 3x3 matrix for that right?

[indigo surge]

?

[dull ginkgo]

well, a rotation in quaternions of a purely imaginary vector is
$$v = q u q^{-1},\quad \Re[u] = \Re[v] = 0$$

[cloud walrus]

Tits metric

[dull ginkgo]

Notice how the real part isn’t used to describe position… but for the 2D complex one it is used.

[mint seal]

historically Hamilton was trying to find what we would today call a 3D real normed division algebra. He eventually found the quaternions, which is one of those but 4D. Later it was proven no 3D one exists, and the only ones that do exist are 1D (R), 2D (C), 4D (H), and 8D (O)

If we define (a,b,c) + (x,y,z) = (a+x, b+y, c+z) and (a,b,c) * (x,y,z) = (ax, by, cz) we get a commutative ring, but it fails to be a division algebra

[indigo surge]

oh I see what u mean

[indigo surge]

division algebra just means that every element in the set has a multiplicative inverse, right?

[mint seal]

every element except 0, but yeah

[dull ginkgo]

Note again that rotations are not just quaternion multiplication

[indigo surge]

right they can be represented by 3x3 matrices

[indigo surge]

but yeah could I plz get some clarification on this?

ManifoldCuriosity said that "we're no longer claiming i^2 = j^2 = -1"

could I get some clarification about why that matters?

[dull ginkgo]

At a high level this isn’t enough degrees of freedom to describe all linear conformal maps

[mint seal]

have you learned what a vector space is?

[indigo surge]

a set of vectors that is closed under vector addition and homogeneity (scalar multiplication)

right?

[mint seal]

yeah

it requires addition and scalar multiplication, but not multiplication of one vector by another

the set of all a + bi + cj can be identified with the set of all (a,b,c)

that is R^3

[indigo surge]

oh is that why they are able to represent R^3?

cuz they don't require the multiplication?

[mint seal]

what is "they"?

[indigo surge]

3D vectors

[mint seal]

why 3D vectors are able to represent R^3?

[dull ginkgo]

What do you mean represent

[indigo surge]

yes

cuz they don't require multiplication

[mint seal]

what else would represent R^3 but a 3D vector?

that's what R^3 is made of

[dull ginkgo]

I am confused what about Euclidean space you mean by represent

[mint seal]

I'm also confused

[indigo surge]

sorry I'm still just asking about 3D vectors vs (a + bi + cj)

[mint seal]

if you want to speak of some kind of space of numbers of the form a + bi + cj, you have to state what the rules are with them

[dull ginkgo]

That’s just a 3D vector space with basis 1, i and j

[indigo surge]

right yeah, and that's why I'm just confused as to why something like a 3D vector (a, b, c) can represent R^3 just fine but (a + bi + cj) can't

[mint seal]

and yeah, one thing you can do is say addition and scalar multiplication are defined in the obvious ways, and forget about multiplying those points together

then what you have is isomorphic to the vector space R^3

[dull ginkgo]

It can represent position and translation

As can any 3D vector space

However, adding an additional multiplicative structure that somehow encodes rotations and scalings is a different story

[mint seal]

it can

[indigo surge]

?

but it can't

cuz it lacks closure

[mint seal]

what closure?

[indigo surge]

of multiplication

[mint seal]

I said we don't do that

R^3 as a vector space doesn't have that

[frail shoal]

i think in the context of this discussion, the reason complex numbers and quaternions are so special is that they encode 2D / 3D rotations, which are an entirely different kind of thing than vectors in a vector space

[indigo surge]

oh wait

[dull ginkgo]

And again, the way it encodes rotations isn’t JUST multiplication, it’s conjugation on a 3D subspace

[mint seal]

yeah, and further, there's more to the quaternions that that

[indigo surge]

3D vectors and (a + bi + cj) can both encode position and translation, as u guys said

but they can't encode rotations?

[frail shoal]

oh damn, never knew that part specifically

[mint seal]

you can just multiply quaternions as any two points of R^4

[dull ginkgo]

Rotating a purely imaginary quaternion is conjugation by the exponential of theta/2 * the unit imaginary vector of the axis

[indigo surge]

is this right?

I think I might be getting it

and that is right regarding both 3D vectors and (a + bi + cj)?

[dull ginkgo]

That’s not just an algebraic fact, once again it’s the fact that it’s not enough degrees of freedom

[indigo surge]

but u can technically multiply 2 3D vectors tho, can't u?

it's just it'll have to be a row vector and a column vector

[velvet hull]

you can multiply them, but you will not get what you want

for instance if you take the component wise product then [1 0 0] * [0 1 0] = [0 0 0], so the product of two nonzero vectors is zero, which doesn't really make sense geometrically

[tribal moss]

There are several different concepts of multiplying R^3 vectors together; one of them may be what you want, depending on what you want it for.

[indigo surge]

I see

[dull ginkgo]

All sorts of ways. But they all are structurally different

Most are meaningless for this context

[indigo surge]

gotchu

I see

so, sorry if I'm going a bit in circles, but is the main problem with (a + bi + cj) the fact that it's not closed under multiplication, or the fact that it doesn't form a "division algebra"?

or are those related?

[dull ginkgo]

What is a + bi + cj

[indigo surge]

like a quaternion without the fourth dimension

[mint seal]

they are related. The fact is there's no way to define multiplying them so that you get a division algebra

[indigo surge]

I sorta get why it doesn't work

because u can't set ij equal to anything

[indigo surge]

another way to restate "division algebra" is that there are no zero divisors, right?

[mint seal]

yup, together with commutativity and etc

[tribal moss]

No, that is not enogh.

[dull ginkgo]

Well then what is i * j?

[indigo surge]

right yeah that's what my current understanding is

u can't rlly define ij, so it doesn't work

[indigo surge]

division algebra requires commutativity as well?

I thought it just requires no zero divisors/every element other than 0 has multiplicative inverse

[tribal moss]

R[X] has no zero divisors, but it is not a division algebra.

[dull ginkgo]

Normed division algebra

[mint seal]

my bad, you're right.

The proper way to say it is for any fixed a and b, with a nonzero, the equation ax = b has a unique solution in x

[indigo surge]

R[X]?

is that just real numbers in the X dimension?

[tribal moss]

R[X] is the ring of all polynomials (in one variable) with real coefficients.

[dull ginkgo]

And xa = b

Left inverses that aren’t right inverses happen

[indigo surge]

ok

[mint seal]

actually no. I was tripping. A commutative division algebra is a field

[indigo surge]

oh ok

is the difference between an "algebra" and a "vector space" the fact that an algebra also requires multiplication to work?

whereas vector space is just additivity and scalar multiplication?

[dull ginkgo]

An algebra is a vector space with multiplication

[tribal moss]

Yes.

[indigo surge]

got it

[dull ginkgo]

I understand this stuff is fascinating but you need to build up a proper background in algebra to do this stuff

Make it a goal post to understand the background to then get back to this :3

[indigo surge]

yeah mb

it's just that I was trying to understand quaternions

and then all this stuff popped up 😅

but yeah, thanks for ur time everyone

I think this makes a lot more sense

[mint seal]

sure, happy studies! quaternions are super cool

[indigo surge]

thanks!

and yeah, I can see that

[tall igloo]

@mystic ether continuing from #point-set-topology

[mystic ether]

The image isn’t normal in general as tits metric pointed out

You quotient not by the image but by closure of the smallest normal subgroup containing it

[tall igloo]

right but it is if everything in sight is abelian

[mystic ether]

However if you meant that forgetful from Ab preserves

Then yeah I agree I think

[tall igloo]

okay cool

yeah im just saying that if everything already was abelian, then you can compute coker in Grp or in Ab and get the same thing

[mystic ether]

I think this kinda makes sense though

Because morally you can’t tell if a group is abelian by maps into it

So if you have forgetful functor, colimit properties defined by maps out of it and into other groups

Although actually this reasoning doesn’t fully work

Cause it doesn’t work with coproducts

[tall igloo]

yeah so here's a question that im too tired to think about

theres no coincidences in category theory so why can you compute coequalizers of abelian groups in Grp or Ab? like i said you can compute limits in either category, so something is going on

maybe this has to do with that one axiom of abelian cats where coker(ker) = ker(coker) idk

but any argument that is provided cannot work for coproducts, because finite coproducts and products agree in Ab but disagree when computed in Grp

[rocky cloak]

Same thing would be true in any algebraic category and a subcategory given by adding a property.

Any such subcategory should be closed under both subobjects and quotient objects

[tall igloo]

here your algebraic category is Grp and youre adding commutativity?

[tall igloo]

wdym by algebraic category exactly? i think i have heard this used to describe categories of the form Fun^{\Pi} (C^op, Set) where C has finite products and \Pi means pick out the product-preserving functors. this includes Ab, CRing, and Set for example but not Grp afaik

[rocky cloak]

I meant, like universal algebra.

[tall igloo]

ah rip ik nothing about universal algebra lol

[rocky cloak]

So a set with some operations satisfying some equations

[frail shoal]

did someone say

[frail shoal]

yea that's how I'd understand it to some degree

i think often one only considers when C consists of precisely x^n for all finite n and some fixed object x

then it's a "lawvere theory"

but it doesn't change much if you drop this restriction, cuz it just lets you have more "underlying sets"

anyway Grp is algebraic

we take C here to be the full subcategory of f.g. free groups

to recover a group from such a functor F: C^op -> Set, note that there is a map from <a> to <x, y> given by sending a to xy, and so F induces a map in the opposite direction, F(<x, y>) -> F(<a>), and because F is product-preserving, we get F(<x, y>) ≅ F(<a>)^2, other operations and properties are recovered similarly

[frail shoal]

universal algebra takes a more syntactic approach to arrive at essentially the same thing, what jagr said, we look at sets X equipped with functions X^n -> X that satisfy universally quantified equations, i.e. each variable is quantified by ∀, thus the name "universal algebra"

[swift root]

and arbitrary products

(valued in the original category)

[rocky cloak]

I guess it should be all limits and all filtered colimits as well

[frail shoal]

universal algebra does essentially everything the product preserving functors pov can do, at the cost of less categorical flexibility, but with the benefit of being a lot more concrete and easier to understand at first brush

well ok also it's older, dating back to the early 1900's

[frail shoal]

so bearing in mind that this can be formulated in categorical language, one can also precisely define what it means to "add an property"

well in this context we often call it an equation

here it's "ab = ba", by default we drop the ∀ symbols cuz we assume them

but it's short for "∀a, ∀b, ab = ba"

if you're curious, here's the syntactical presentation of groups typically given in universal algebra:

• we have 3 operations, a binary operation written with concatenation, an identity "e" (i.e. a nullary operation), and a unary inverse operator written as ^-1
• a(bc) = (ab)c
• ae = a
• ea = a
• aa^-1 = e
• a^-1a = e

[frail shoal]

one can explicitly use these syntactical presentations to recover "free" structures over them, and then use those to recover the categorical perspective, similar to what i described for groups

[swift root]

those can all be expressed as a quotient of a subalgebra of a product

[frail shoal]

ok i think im yapped out

[frail shoal]

HSP jumpscare

[swift root]

the goated theorem

[frail shoal]

oh yea i should mention that one of the gems of universal algebra is the HSP theorem, which says that an "equational class" (i.e. classes of algebraic structures described by equational properties) is precisely those subcategories closed under product, then subalgebra, then quotient, in that order, with the "correct" definitions of each that it would take a bit to explain

[lime badge]

I have a very silly question, but let R be an integral domain and suppose I have a map F: R^l ---> R^m

Consider coker F = R^m/im F. This is certainly a torsion module isn't it?

[velvet hull]

not necessarily, you can just take the embedding of R into R^2

then coker is just R

[lime badge]

Hmm yeah, and then the torsion submodule (as an R-module) is 0 in this example.

I'm in a practical situation where R = k[x,y], k is a field, and I need to compute the k-dimension of Tor(coker F).

To do this, does it suffice to determine a monomial basis for coker F and check which of them have non-zero annihilators?

[velvet hull]

on second thought, I'm not sure if that is enough to compute Tor(coker F) - it's possible for a linear sum of non-torsion elements to be torsion so I'm not convinced it is sufficient to check this on a basis

do you have a better handle on what the structure of coker F looks like?

[twilit wraith]

alright notation is kinda tripping me up here again

so previously in this text it was implied that when you see S^-1 A for A a ring that S is a subset of A

but when im looking at S^-1 (M1 x ... x Mn) that makes it seem like S is in that product of modules, and for S^-1 M1 x ... x S^-1 Mn that each S is in each Mi

but the notation is the same so i thought its just that S is a subset of the ring which each Mi is over

im fairly convinced its the latter but now im struggling to think about (c) properly

so like usually when i see (1, 1/2, ...) as a counterexample its that its not possible for it to be in the image

[velvet hull]

it is this

[twilit wraith]

but if S is simply a multiplicative subset of Q couldnt i just pick s^-1 (s/1, s/2, ...) to map into it

[wraith cargo]

I mean this is how localization of modules is defined

[twilit wraith]

oh i suppose the issue is that this map is not well-defined

because for any s i can have this

sorry the backwards map wouldnt be well defined bc the forwards map isnt injective

[twilit wraith]

yeah i mean this makes sense

i guess now im just not sure how exactly the counterexample on the bottom works

[twilit wraith]

bc naively this seems to me to be a problem even in the finite case

[twilit wraith]

maybe the problem is instead this idk

i said that the obvious map is $s^{-1}(m_1, \cdots) \mapsto (s^{-1} m_1, \cdots)$

[cloud walrus]

hiidostuff

[twilit wraith]

nvm im a goober

the infinite product of Q is supposed to be S^-1 Z so that Q is the field of fractions

ts makes sense now

[lime badge]

Hmm, good point.

I do have a specific matrix I want to get the cokernel of, but its not obvious from looking at it what it should be. Maybe its more helpful to know the relationship between dim_k Tor_S(coker M') and dim_k Tor_R(coker M)

[dire wren]

Am a bit confused on the definition of cofinal, is it supposed to mean a subfamily? Because otherwise I am thinking what they are asking to prove is wrong? I mean can't I take F to be the family consisting of only the trivial subgroup, and I can take H_i to be something more interesting?

[rocky cloak]

For a general poset you would talk about a subset being cofinal.

In this paragraph it seems F is specifically the set of all finite index normal subgroups.

You might note they say the completion with respect to {H_i} is the same as for "the full family F" implying that {H_i} is a part of F

[dire wren]

Ah, that makes sense

Thx

[indigo surge]

hi

so yesterday I've more or less cleared up my uncertainty about quaternions and algebras

I just have one little question:

why is the cross product considered to be an "algebra"?

isn't the cross product just an operation?

plz ping, thanks

[tidal schooner]

yes, what they mean is that the cross product is an operation that makes R^3 into an R-algebra

[rocky cloak]

The definition of what an algebra is varies a bit, but the thing everyone agrees is that an algebra should have an addition and a bilinear multiplication.

And R^3 with the cross product as multiplication is exactly that

Other people have more restrictions on what the multiplication operation should satisfy

[tribal moss]

The trouble in the table is that there isn't a commonly used short word for "R^3 with the cross product" which is how the algebra would be described, so instead the author just put a description of the product operation in the "which algebra are we talking about" column.

[indigo surge]

ok so what does "bilinear" mean exactly?

like I've seen the definition but idk what it means

also can't R^3 not be an algebra?

cuz it just doesn't work cuz u have nothing to assign ij to?

[stone fulcrum]

Linear but in a bi way

Jk. Linear over both inputs

[rocky cloak]

Bilinear means linear in both arguments, so
(rx + sy) * z = r(x*z) + s(y*z)
and similarly in the second argument.

R^3 cannot be a division algebra (i.e. where we can also divide). It's also not possible to make an algebra spanned by 1, i, j with i^2 = -1 (even if we don't require division)

[indigo surge]

that's it?

[rocky cloak]

It's isomorphic to so(3) in a pretty natural way. But I think introducing that would create more confusion than clarity

[tribal moss]

Hmm, point.

[indigo surge]

right that's why I was confused about the inclusion of the cross-product there

[mint seal]

the cross product is bilinear

it lacks an identity element and inverses, though

[indigo surge]

right so, it can't be considered an "algebra" then, can it?

[mint seal]

it is an algebra but not a division algebra

[elfin wraith]

The term algebra isn’t 100% settled, but I think most people would agree that’s an algebra over a field

[mint seal]

more specifically it's a non-unital, non-associative algebra

[indigo surge]

really?

but wait

I thought algebra was vector space + multiplication

cross product isn't necessarily a vector space, is it?

it's just an operation?

[cursive spindle]

Another word I hate in math

[velvet hull]

wow i didnt know you hated algebra

[elfin wraith]

It’s a bilinear map, R^3 has a natural vector space structure on it coming from R

[indigo surge]

ok

[indigo surge]

is that even possible btw? cuz I thought R^3 can't be an algebra

[elfin wraith]

Can’t be a division algebra

[indigo surge]

but u still can't multiply them together

[elfin wraith]

You can with a bit of effort show that only R^{2^n} can be a division algebra, and with a lot of effort show that only n=0,1,2,3 work

[elfin wraith]

What do you mean?

[indigo surge]

like u can't multiply (a + bi + cj) by (d + ei + fj)

[elfin wraith]

Ok so you have that R^3 is a vector space over R right, you know how that works. Then we turn it into an R-algebra using the cross product, that gives a reasonable notion of multiplication of vectors, and by reasonable I specifically mean it’s a bilinear map

[indigo surge]

oh

hm

well alright ig

it's just weird that all the other examples were number systems but for R^3 it was an operation

[quiet pelican]

How do you do the first one?

[elfin wraith]

You consider like the map on RP^n xRP^n induced by multiplication and look at the cohomology rings

Theres then a fiddly bit of counting to be done, which is kinda easier if you restrict to binary, but yeah there’s an argument in Hatcher lol

Deducing that you only have R C H and O is then quite a bit harder and you need some homotopy theory, I don’t know how that argument goes so maybe it’s not so bad but it does take more, it’s possibly an Adam’s spectral sequence argument if I remember correctly but I’m not particularly confident in saying that

[somber goblet]

wow

[dull ginkgo]

Please don’t deal with quaternions and the like until you have a proper understanding of the underlying algebra

You are going in circles

[delicate orchid]

(x,y,z)(a,b,c) = (xa,yb,zc)

[cursive spindle]

Since you seem to be interested in quaternions why not read https://kconrad.math.uconn.edu/blurbs/ringtheory/quaternionalg.pdf

[dense mica]

Atheism

[mint seal]

I'm not sure what you mean by this - the examples mentioned in that table are complex numbers, R^3, quaternions, polynomials, and square matrices. Of those, only the complex numbers and quaternions are commonly called "number systems".

In each case, the "multiplication" is a bilinear operation that takes in two elements of the algebra and outputs a third.

I would recommend not trying to work with a + bi + cj as representatives of R^3, and also forget about the "you can't multiply them" idea. A point of R^3 looks like (a,b,c) or (x,y,z), and you can multiply them in various ways, like componentwise or with the cross product. But, again, these do not result in division algebras.

[indigo surge]

mb

[south patrol]

Gatekeeping

[indigo surge]

oh ok

didn't think cross product was considered multiplying

I thought it was just its own thing

[mint seal]

It may also be helpful for you to think about how the complex numbers a + bi are essentially the same thing as R^2 with the multiplication of points defined by: (a,b)*(c,d) = (ac - bd, ad + bc).

in that setup, the imaginary unit i is identified with the point (0,1), and the equation i^2 = -1 becomes (0,1)*(0,1) = (-1,0)

[indigo surge]

mhm

[mint seal]

and there's a similar identification of quaternions a + bi + cj + dk with points of R^4, (a,b,c,d)

[indigo surge]

I'll read that, thanks

[mint seal]

but there's no good number system with only two imaginary units i and j that can be identified with R^3

[indigo surge]

right, because u can't define ij

[mint seal]

yup

[indigo surge]

ok so

from what people have said

R^3 can be made into an algebra

but not into a division algebra?

[south patrol]

Ye

(You can make it into an algebra over R in a boring way)

[indigo surge]

wdym?

[south patrol]

(a,b,c).(d,e,f)=(ad,be,cf)

[indigo surge]

oh so just component-wise multiplication?

[south patrol]

Yeah

[indigo surge]

ok

I thought it couldn't be an algebra cuz u can't multiply two terms like (a + bi + cj) together

but I guess that doesn't matter for an algebra

[indigo surge]

so this only prevents it from being a division algebra, right?

not being able to multiply terms like (a + bi + cj) together?

[south patrol]

Tbh ig I am unsure you far you understand what algebras are, like what is ur understanding of them

In an algebra you can multiply any two elements

[mint seal]

If we use (a,b,c)(x,y,z) = (ax,by,cz), we are essentially defining (a + bi + cj)(x + yi + zj) = ax + byi + czj. But that's clearly not how we usually multiply imaginary units

[south patrol]

Yeah the problem is defining a multiplication compatible with some ways you'd like i and j to behave

Namely you'd probably like to have i^2 = j^2 = -1

(In what I said above, i^2 = j^2 = 1)

[mint seal]

I know several of us here would really like to clear up the confusion, but this has gone on for a while and gotten fairly muddled up with different ideas. If you're still feeling confused, could you restate what you're wondering about?

(@indigo surge )

[velvet hull]

at this point start a thread

[indigo surge]

Vector space + multiplication of any two elements

That’s my understanding

[indigo surge]

Right, so that seems like it can’t be an algebra

That’s rlly my only point of confusion

[mint seal]

why not? It's a vector space with multiplication, as you just said

[indigo surge]

Well it’s like u said, “that’s not how we multiply imaginary units”

[mint seal]

an algebra doesn't have to have anything to do with imaginary units though

it's just incidental that R^2 and R^4 (and R^8) have algebra structures connected to imaginary units

[indigo surge]

Ohhh wait ok

I see

I think I might have been mixing the two up

Ok

So, in that case, is there a different name/grouping for regular R^n and the imaginary number format?

If that makes sense?

[mint seal]

well there's C instead of R^2 (complex numbers), H instead of R^4 (quaternions abbreviated in honor of Hamilton), and O instead of R^8 (octonions)

and we can say there's an isomorphism of algebras between C and R^2 with the relevant multiplication

and between H and R^4

[indigo surge]

And between O and R^8?

[mint seal]

yup

[cursive spindle]

yeah it's a great read imo

[indigo surge]

Okkkk, I see, thanks

Ok I’ll just have to read up on isomorphism more and what it is and whatnot

But that clears up my confusion on that

Thanks

[indigo surge]

I am away rn (typing on my phone) but when I get back I will fs

[indigo surge]

And only the dimensions that have an isomorphism with a complex number-like system can be division algebras?

[elfin wraith]

Over R anyway, yes. This is what I was saying earlier, you just get R, R^2, R^4, and R^8 but this is quite hard to actually show

[dull ginkgo]

@indigo surge Look up the Cayley-Dickson algebras

[indigo surge]

I see, thanks!

[indigo surge]

Interesting, will do

[oak helm]

Hi everyone, I just wanted to discuss stabalisers, centralisers and orbits

[near tapir]

In the fourth point, shouldn’t the primes be distinct? And wouldn’t be more useful to prove it for distinct prime powers?

[south patrol]

Yes for both your questions

[near tapir]

Thanks

[indigo surge]

hey guys

I think I've connected some dots and want to make sure I did so correctly

definitions like i^2 = -1 for complex numbers and i^2 = j^2 = k^2 = -1 for quaternions are not universal for all algebras in those dimensions, right?

those were just chosen to be that way for those specific cases in order for complex numbers and quats to be normed division algebras?

but u could also have something like i^2 = 0, j^2 = 2, and k^2 = -1

it's just that then, stuff would break and the algebra would not work out to be a clean, normed division algebra, right?

plz ping, thanks

[covert cliff]

There is an extension of the real numbers where we introduce an element epsilon for which epsilon^2 = 0. This is usually called the dual numbers

[indigo surge]

so, what I said was right?

[covert cliff]

An interesting application of adding these relations like this comes from physics. Dirac wanted to "take the square root of the laplacian" and what he ended up doing was defining an operator $D = \gamma^\mu\partial_\mu$ where ${\gamma^\mu, \gamma^\nu} = 2\eta^{\mu\nu}$. This is an example of a clifford algebra. Where you take a vector space and a bilinear form and then define a multiplication in manner relative to this bilinear form

[cloud walrus]

Khush

[covert cliff]

Yeah basically

[indigo surge]

interesting

well that makes everything come together a bit

that's weird tho, like, what is the point of being able to define it however u want?

if 95% of the time something is just gonna break

[covert cliff]

The thing that usually breaks is invertibility

So you still end up with an algebra, just not a division algebra

[indigo surge]

right

do those have any uses tho, or not really?

[foggy tartan]

there is division whenever the real part of the denominator is nonzero as per wiki, so you have division, just many zero divisors

[indigo surge]

but those can't be divided

cuz they can't have inverses

[foggy tartan]

hence why invertibility breaks

[indigo surge]

Right

But yeah that’s why I’m wondering

Do those regular, non-division algebras have a use?

Or not really?

[foggy tartan]

previous user can argue for dual numbers, but in a more general sense, there are lots of algebraic structures where division isn't a guarantee. famously rings

and there are many rings which are a useful language for dealing with some collection of objects where you can add and subtract and multiply. the ring C(X) of continuous functions on X eg

[covert cliff]

I gave an example in physics where its used

[covert cliff]

This one

[mint seal]

matrix algebras are highly useful and they're not division algebras

[cursive spindle]

Did you read what I sent?

They discuss this

[indigo surge]

Ohhh ok

[indigo surge]

Oh I see

[indigo surge]

No not yet, sorry I’ve been moving around a bit recently and have just been asking here and there when I had time

But I will tho

I will

[woeful sage]

Lol does atheist jesus not believe in his dad?

funny name

[quartz olive]

Hi got a questions about isomorphisms for fields

i thought i was being smart considering the invertible elements as a cyclic group and then finding generators for each, and then having my map send each generator to each other but i can seem to show the addition property. This is the question for reference

These are the fields F3[x]/f and F3[x]/g where f = X^3-x+1 and g = x^3-x-1

the generators i found were (x-1) and (x+1) how do i show for two arbitrary elements that phi(a+b) = phi(a) + phi(b)

[quiet pelican]

There’s a nice way to do that with first isomorphism theorem
Consider the map F_3[x] -> F_3[x]/g which sends x - 1 to x + 1
This is a well defined surjective homomorphism
What is the kernel of this map?

[quartz olive]

im lowkey dont know im gonna guess that the kernel is f so we get that they are isomorphic but i dont actually know why the ker is that

[quiet pelican]

Yeah
The kernel is the ideal generated by f

The way I’d reason it is that the kernel is the preimage of (g) under F[x] -> F[x] via x - 1 -> x + 1
And then just compute the preimage of g, and you’re done since it’s a bijection

[quartz olive]

the issue i have is not that i cant show that they are isomorphic its creating an actual map

i argued that they are isomorphic by F3 being a field the PID then f and g irreducible... then they both have same elements so they must be isomorphic

but if i need to create a map i need to show that addition, multiplication properties hold and thats its bijective right, theres no way to get around that is there?

[quiet pelican]

The first isomorphism theorem gives you an explicit map

[quartz olive]

my bad i thought it only said they were isomorphic thank you

[azure cairn]

how unclear is it to use <= to denote sub-* structure

like subspaces in linalg/topology, subgroup, subrings, you get the idea

[languid trellis]

I'm not quite sure what you mean by unclear. If you mean 'uncommon', then from personal experience I've had at least two lecturers use this notation. So it's not super uncommon.

In any case, as long as you make it clear what your notation means, and you are consistent in its usage, it shouldn't be an issue.

[azure cairn]

common is moreso what i meant

ty

[languid trellis]

You should be writing out what it means anyway. Like "Let H <= G be a subgroup of G" or something similar

[elfin wraith]

Agree with swiftee, pretty common in my experience, but theres never an issue doing these thing as long as youre clear about what you mean

[rocky cloak]

Using $\leq$ for subgroups and $\trianglelefteq$ for normal subgroup is the standard I would say

[cloud walrus]

jagr2808

[elfin wraith]

Of course its best to stick to a convention, but you dont have to

[rocky cloak]

Not so much for subrings, where I would probably think <= means ideal.

[foggy tartan]

i've seen \trianglelefteq for ideals, i thought that was nice

analogous to normal subgrps

[languid trellis]

Or <= for ideal, so that it is analogous to submodule xd

[foggy tartan]

use neither and just use fraktur for all your ideals to make it really obvious

[languid trellis]

that is way too far

[rocky cloak]

I'd go with \trianglefteq for two sided ideals, and then \leq for ideals

[tall igloo]

for a second I forgot there were non commutative rings and was like wdym two sided ideal 💀

[chilly radish]

Oh wow I never knew about this command I only knew about \unlhd

[foggy tartan]

what does unlhd stand for? always wondered

[quiet pelican]

$\unlhd$

[cloud walrus]

micoi the group objects (she/it)

[rocky cloak]

un is surely underline

$\lhd$

[cloud walrus]

jagr2808

[foggy tartan]

left head or something?

[rocky cloak]

Yeah, that was what I was gonna guess

It is the head of a left arrow I guess

Some confirmation from the documentation
https://tex.stackexchange.com/a/684365

[foggy tartan]

ah, nice

[next depot]

What’s up groupsss

[velvet tusk]

No

[fickle dirge]

This proof doesn't make any sense right? This is not a set... It won't even be a set if you take the set of all algebraic extensions up to isomorphism?

[velvet hull]

Well yes, it’s implicitly considering the elements of the set to be the same up to isomorphism and embedding

And under those assumptions it’s small enough to be a set

[swift root]

up to isomorphism as F-algebra is what youre prolly looking for

yes

[fickle dirge]

why under those conditions its small enough?

[swift root]

tbh I do much prefer the constructive definition

[velvet hull]

Very loosely you can identify algebraic extension of F with polynomials in F[x]

[rocky cloak]

I think you'll run into some trouble in making this poset if you consider things up to isomorphism.

But something you could do is just fix some set of cardinality max(|F|, |N|) and consider field structures on subsets of it

[velvet hull]

The only thing to account for is the same isomorphic extension may admit non isomorphic embeddings into an algebraic closure, but an extension of degree n can have at most n such embeddings

So it is reasonable to expect that this set would be close to being countable

[fickle dirge]

very weird this is the proof my logic professor gave..

would've thought logicians would be more rigorous 🤔

[fickle dirge]

I think that makes sense..

I got confused because the proof I found online was much longer

anyway thanks I will give this some more thought when I'm done with my exam

[velvet hull]

There’s an alternate non zorns lemma construction of Fbar that makes it immediately obviously a set

But poset techniques do show up in algebra and other areas so I don’t object to doing it this way

[fickle dirge]

I see

[fickle dirge]

yeah have seen that a couple times

[rocky cloak]

I guess the other proof I know would just be to take the polynomial ring where you adjoin a variable for each polynomial. Then quotient out so that the variable becomes a root of said polynomial and then take a maximal ideal to get a field.

[fickle dirge]

that is very cool

[languid trellis]

i think this still uses zorn implicitly because you need to find a maximal ideal containing ur poly?

not that you were claiming that it doesn't, i'm just curious

[rocky cloak]

It does depend on AoC. I don't think this is provable without

[languid trellis]

Has that been proven?

[rocky cloak]

Idk, but I would think so

[velvet hull]

you have to do that more than once, it's not guaranteed to terminate in one step

i.e. start with F_0 := F, do that to get F_1, and then take the union of all the F_i's and that is the algebraic closure

[rocky cloak]

It actually is guaranteed to terminate in one step. But that fact is much harder to prove then just doing the union trick

In characteristic 0 it's not so bad, you just need the primitive element theorem. For non-perfect fields you need some fideling

[wraith cargo]

you don't need AoC

you only need the ultrafilter lemma

which is known to be strictly weaker

[rocky cloak]

Yeah you don't need the full strength

So what I mean is not provable in ZF.

Couldn't really find a good source for it, but people on MSE sort of imply it is the case

Although it's apparently open if it is weaker than the ultrafilter lemma

[south patrol]

I think it's just there are also standard ways to get around these kinds of issues

But I am surprised you would give a proof like this in lecture notes or whatever lol

[rocky cloak]

Here's a paper citing the result to a German paper
https://metaphor.ethz.ch/x/2025/fs/401-2004-00L/sc/Banaschewski.pdf

[south patrol]

I imagine it's cause this is enough to get existence of prime ideals?

Yeah okay that is what that paper uses lol

[rocky cloak]

I think you can also use the compactness theorem

[south patrol]

Ah sure

[elfin wraith]

I think there’s an n cat cafe thread discussing this iirc

https://golem.ph.utexas.edu/category/2021/04/algebraic_closure.html

Not as directly as I remembered, but all the comments are very related

[fickle dirge]

Oh cool this was posted by Leinster, the course I'm taking next semester is following his category theory book

[elfin wraith]

Nice one! It’s a good book

[indigo surge]

I don’t believe in myself I guess 😉

[indigo surge]

Thanks

[fickle dirge]

how does the formal derivative for polynomial rings over a field relate to the limit derivative?

[somber goblet]

it agrees when you would expect it to, like in $\bR[x_1, \dots x_n]$

[cloud walrus]

lexi

[fickle dirge]

yeah I know, but is there some deeper kind of connection or is that all

[somber goblet]

in general, it gives you the ‘linear part’ of the polynomial at a point

so if you take the polynomial $f \in k[x]$, the formal derivative at $c \in k$ will correspond to the coefficient of the degree-1 term of $g(x) = f(x + c) - f(c)$

[cloud walrus]

lexi

[somber goblet]

what we get out of polynomials is that we can talk about ‘linear parts’ of things by breaking down the coefficients, even if there is no way to take limits

[fickle dirge]

ah that makes sense.. thank you!

[somber goblet]

what this is really nice for is that it lets us still talk about repeated roots

also

like we can check if the derivative at a root is 0 to see if it has multiplicity >1

and this works even in finite fields

[fickle dirge]

yeah pretty cool

[somber goblet]

formal derivatives are therefore quite handy for field theory and algebraic geometry

[fickle dirge]

cool! I basically only learned the definition and the repeated root thing but there wasn't any intuition so thought I'd ask

[somber goblet]

there’s very cool uses, like the dimension of the space of possible values of the formal derivative at a point telling you when a curve has a cusp

[fickle dirge]

what!!!

crazy

[somber goblet]

and it turns out if you’re working in multivar polynomials over ℂ, this agrees with the ‘usual’ notion of a cusp

very cool stuff

[fickle dirge]

very interesting! I'm not sure how one "should" approach algebraic geometry but I'm taking commutative algebra next semester so I think that's a step in the right direction

[next obsidian]

Having derivatives is important

When things have actions via other things it tells you stuff.

This is a really deep thing that won’t mean anything to you, but in commutative algebra (and from there algebraic geometry), the mixed characteristic case splits very distinctly into the ramified and unramified case

In the unramified case, for a regular ring at least, you take the form of a formal power series ring over a nice ring which means you admit derivatives

In the ramified case this isn’t the case

This means a bunch of cohomology modules and stuff don’t have an action by derivations which in the unramified case can be pretty crucial to proving things

So there’s a lot of stuff that’s known in all but ramified mixed characteristic, and I personally chalk up a lot of that to the fact you don’t have this power series ring presentation

Now the fact that means you don’t have derivatives is only part of why that sucks, but it’s certainly a big one

[somber goblet]

does ‘mixed characteristic’ mean like Spec ℤ

[next obsidian]

Yes and no