serious-discussion

[rocky shuttle]

!xy

[quasi jetty]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[rocky shuttle]

what it means to apply xy

[scarlet mango]

thats not even an equation

[torpid bay]

or is it gprod wraith?

[scarlet mango]

and how tf do we wedge a wedge?

head explosion syndrome incoming

[zealous garden]

It's all wrong don't do this

[torpid bay]

$x\wedge x = 0, x\wedge y = -y\wedge x$

[fathom swallow]

Yeatte

[scarlet mango]

that makes sense

[zealous garden]

$xy$ transforms $v$ as $(xy)^{-1}v(xy)$

[fathom swallow]

wraithlord_kojima

[torpid bay]

oh that thing yeah

[scarlet mango]

that looks like linear algebra

[zealous garden]

Let me finish this job and I'll tell you what's what kids

[scarlet mango]

I'm actually so invested and I am not even sure if I know where this is all going

[torpid bay]

man I forgot too

[scarlet mango]

I didnt even know GA was a thing till a few hours ago

[zealous garden]

We'll hopefully achieve a derivation of the sandwich product between arbitrary versors in arbitrary geometric algebras over arbitrary Fields

[scarlet mango]

theres a sandwich product???

[zealous garden]

Yummy innit

[scarlet mango]

in the next episode

[zealous garden]

It's going straight to the heart of physics

[scarlet mango]

please ping me if you have the time to explain all of that

[zealous garden]

Unironically

[zealous garden]

I'm hoping to free up in like 4 minutes here

[scarlet mango]

thats a really cool cliffhanger

[zealous garden]

Shouldn't take long to explain the mirrors

[scarlet mango]

dont underestimate me

[torpid bay]

while wraith is gone I'll go back to quaternions >:)

[zealous garden]

Trust, we start with relatively few definitions and like 2 intuitions

[scarlet mango]

my powers of doubting till I understand everything are immense

[scarlet mango]

okeokeoke

start from beginning

[torpid bay]

I forget the beginning tho

[scarlet mango]

hahahah

pick somewhere and start there

any new knowledge is fine

ill prolly backtrack with questions abt everything

[torpid bay]

really tho, wraith will do a much better job of explaining Geometric Algebra

[scarlet mango]

It's fine, you can try your way

[torpid bay]

I only have a small bit of knowledge about it because I needed it for cursed physics euqations

[scarlet mango]

thats really cool to me

there isnt even a channel in the mathcord for it

[torpid bay]

do you know cross product? it's where I needed to end up using it in 3D

[scarlet mango]

yessss

like i×j= k and stuff

I learnt abt it few days ago. relearned actually but ye

[torpid bay]

the cross product works in 3D, as it takes two vectors and gives back the vector that is perpendicular to the 2

however, in 4D, you don't have just 1 vecctor perpendicular to i and j vectors, you have many more

[scarlet mango]

and there is always one unique vector that is perpendicular to any two vectors?

[torpid bay]

yep

[scarlet mango]

ahhh I see

[torpid bay]

being any combination of k and w

4D*

[rocky shuttle]

unique up to scaling

[torpid bay]

natrually, if we want to work in 4D using things similar enough to the cross pdocut, we need osmthing new

this occured when I was working with derivatives using time, x,y and z isntead of just x,y,z

[scarlet mango]

something that gives out a unique result

[torpid bay]

yes

[scarlet mango]

I see I see

And what is the answer to that?

do we just one up the conditions to be met?

[torpid bay]

the wedge product acts quite similar to the cross product

[scarlet mango]

but it stays in the same plane when wedging two vectors, no?

[torpid bay]

yes

[scarlet mango]

how is it any similar then?

[torpid bay]

$i\times j = -j\times i = k \ x\wedge y = -y\wedge x$

[fathom swallow]

Yeatte

[scarlet mango]

hmm

[torpid bay]

in this case rather than another thing k, we just leave it as and think of the thing as a 2-blade on its own

[zealous garden]

Alright I am present

[torpid bay]

[rocky shuttle]

Hi present, I'm zan.

[scarlet mango]

I don't get that

[torpid bay]

let's let wraith do it

[scarlet mango]

okeokeoke

[torpid bay]

the wege is not entirely the same as the cross, as we ignored the existsnce of 1-blades vs 2-blades tho

[zealous garden]

Let us begin with a vectorspace V over a field F, for motivation we will look at examples from R^3 and R^4. We equip this space with a symmetric bilinear form, examples will use the Euclidean dot product and perhaps a minkowski interval

[torpid bay]

^

[zealous garden]

Any questions so far

[scarlet mango]

yes

lemme think of them

what does it mean to equip a space with a bilinear form?

and a symmetric at that

[zealous garden]

Excellent question, it means that we have a Bilinear form, a map that takes two vectors and gives a scalar, and if you make either argument constant it is linear in the other

Symmetric means that it is commutative

So B(u,v)=B(v,u)

This is essentially what defines lengths and angles

Two vectors are perpendicular or orthogonal if B(u,v)=0

The squared length of a vector is B(v,v)

[scarlet mango]

huhhhh

[zealous garden]

This is a Definition

[rocky shuttle]

you can think of it as the dot product

[zealous garden]

That's what orthogonal means

[scarlet mango]

is it because they end up in the same place?

[scarlet mango]

yes, is that the same thing?

[zealous garden]

The dot product is an example of what were talking about

The easy ubiquitous example

[rocky shuttle]

the dot product is a particular symmetric bilinear form, yeah

[zealous garden]

Everything else can be derived by say flipping the sign on some terms of the dot product, or ignoring some terms

[scarlet mango]

hmm I see

[zealous garden]

x^2-y^2, throw away z^2 for example

For a less trivial one

So yes the dot product is your motivating example

Think about dot products in 3D

[rocky shuttle]

oh, you don't even assume non-degeneracy

[zealous garden]

No need

[scarlet mango]

thats not so easy for me

[zealous garden]

Let $v$ be a vector s.t. $B(v,v)\neq 0$, there is a subspace $U<V$ s.t. $u\in U \iff B(u,v)=0$

[fathom swallow]

wraithlord_kojima

[scarlet mango]

so anything that takes two vectors, and spits out a scalar has a bilinear form?

[zealous garden]

It has to be linear in either argument

Or in other words, distributive property

(Plus commutes with scalars)

[rocky shuttle]

sure (this is where non-degeneracy comes in )

[torpid bay]

ill be back later

[zealous garden]

This is the orthogonal complement to $v$

[fathom swallow]

wraithlord_kojima

[scarlet mango]

okeoke have fun whatever youre doing

[zealous garden]

Taking inspiration from the Euclidean plane, and using that B(v,v) is the squared length of v, it may be easy to see that

$P_v(u)=B(u,v)\frac{v}{B(v,v)}$ is the projection of $u$ onto $v$

[fathom swallow]

wraithlord_kojima

[zealous garden]

Do you see this?

[scarlet mango]

not really

I'm too much of a rookie to keep up with your definitions sorry

[zealous garden]

One may use that the dot product B(a,b) is equal to |a||b| cosθ

Sad, we'll pare it back then

So if I want to project one vector onto another, do you know how the dot product does it?

[zealous garden]

This

[scarlet mango]

uhm

[zealous garden]

Then the angle between them is 0, its just the product of vector lengths

When the angle between them is 90 degrees, its 0

[scarlet mango]

what does it mean to project a vector unto another?

[zealous garden]

Obie

[scarlet mango]

yes I understand that

[zealous garden]

Uh

Orthogonal projection

[scarlet mango]

this too

[scarlet mango]

like we do in geometry?

[zealous garden]

Yes

The very same

[scarlet mango]

okay sure

[zealous garden]

So like P_x(x+y)=x

[scarlet mango]

so the projection of i to j, dotted with j is 0

No

sorry

bad examlle

take i and j to be non perpendicular

lol

[zealous garden]

So check it out the dot product between two vectors u,v; projects the vector onto the other and multiplies their lengths

So we need to get rid of the length of the vector we want to project onto, so it doesn't stretch it out, right?

[scarlet mango]

I think

[zealous garden]

So we want to say $P_v(u)=(u\cdot v) v$

Where $\norm{v} = 1$ right?

[fathom swallow]

wraithlord_kojima

[scarlet mango]

uhm

I think that makes sense

[zealous garden]

Good

So this works fine in euclidean space

[scarlet mango]

how do we achieve that

[zealous garden]

Wdym

[scarlet mango]

to make |v|=1?

[zealous garden]

AHAHAHAHAA

[scarlet mango]

HEYYYY

[zealous garden]

That's the catch!

[scarlet mango]

I am still learning linear algebra intro stuff

[zealous garden]

If we have a good nice well behaved metric vectorspace, like the dot product we were using earlier

And the reals

We can just say v=v/||v||

[scarlet mango]

that looks like the vector unit

[zealous garden]

And ||v||=sqrt(B(v,v))

[scarlet mango]

unit vector or something*

[zealous garden]

Yeah

That's what it is

The issue is you need to take a square root

And you need B(v,v) to be positive

[scarlet mango]

why can we just say that tho lol

[zealous garden]

You take some x that doesn't satisfy it, then make v=x/||x||

Then v satisfies it cus it's already normalized

But you can't do these normalization shenanigans in an arbitrary metric vectorspace over arbitrary fields so we need a slight change

$(u\cdot v)v$

whenever we can't normalize $v$, ends up with TWO extra factors of $\norm{v}$ which means $\norm{v}^2$

[fathom swallow]

wraithlord_kojima

[zealous garden]

Which we CAN get without taking roots or caring about signs

[scarlet mango]

how do we normalize a vector

i told you not to underestimate me lmfao

[zealous garden]

https://tenor.com/view/you-don't-invincible-omni-man-jk-simmons-that's-the-neat-thing-gif-4390372980992221432

[scarlet mango]

HUH

[zealous garden]

Generally speaking, a unit vector might not even exist

Like in a rational vectorspace

But

So we have $P_v(u)= B(u,v)\frac{v}{B(v,v)}$

As $B(u,v)=u\cdot v$ and $B(v,v)=\norm{v}^2$

[fathom swallow]

wraithlord_kojima

[zealous garden]

Because $B(u,v)v$ is $B(v,v)$ too big

[fathom swallow]

wraithlord_kojima

[zealous garden]

.

This also works when B(v,v)<0, as happens in spacetime

[scarlet mango]

this is so fucked up

[zealous garden]

Isn't it

[scarlet mango]

I have another question!!!!

[zealous garden]

But you can project now!

[zealous garden]

Yeah?

[scarlet mango]

no I can't

[zealous garden]

We're moving towards the quaternions slowly but surely

[zealous garden]

Do you not get the formula?

[scarlet mango]

how tf do we go from not being able do normalize v, to ending up with TWO factors of |v|]?

[zealous garden]

So v=x/|x| right

[scarlet mango]

If you see what books I'm reading you'll be surprised my head hasnt exploded yet

[scarlet mango]

in terms of difference of skill

[scarlet mango]

yes

[zealous garden]

So x= |x| v

Then B(u, x) x = B(u, |x|v)|x|v= |x|^2 B(u,v)v

Divide by |x|^2

Relabel x to v

Add a hat to the artist formerly known as v

[scarlet mango]

how did we just pull out the scalar?

[zealous garden]

B is Bilinear

Scalars can be pulled out because of that

[scarlet mango]

alright

I actually understand

[zealous garden]

Excellent, next step we define a Clifford Algebra, or Geometric Algebra

[scarlet mango]

[zealous garden]

You need a vectorspace V, and a symmetric bilinear form B, like we were just talking about

Then you say, we can multiply vectors

[scarlet mango]

true

[zealous garden]

BUT, whenever I see v^2, replace it with B(v,v)

So let's see some consequences of this

Let x,y be orthogonal vectors, so B(x,y)=0. Then (x+y)^2=x^2 + xy + yx + y^2 = x^2 + y^2 + xy + yx

[scarlet mango]

lemme decipher this

[scarlet mango]

how did you just multiply x and y without a dot or a cross or whatever?

[zealous garden]

That's the geometric product, the Clifford product

The multiplication where x^2=B(x,x) and we say little to nothing else

[scarlet mango]

but

but

[zealous garden]

By Pythagorean theorem, this is equal to x^2+y^2

[scarlet mango]

we had the length of x, squared in our previous proof

[zealous garden]

Which implies xy+yx=0->xy=-yx

[zealous garden]

Correct

|x|^2=x^2=B(x,x)

This is going to be key to our simplification

[zealous garden]

I was going to make a point here about the wedge product and how it falls out but we dont need it actually

[scarlet mango]

huhhhhh

[zealous garden]

This is by definition

Both of those are defined to be B(x,x)

[scarlet mango]

yes that, I understand thank goddd

[zealous garden]

Here's what's up

B(x,x)=x^2, so x/B(x,x) = x/x^2, which is x^-1

So we can rewrite our projection formula again!

P_v(u)=B(u,v)/v

[scarlet mango]

What even was our projection formula

[zealous garden]

And this one works everywhere

[zealous garden]

For now it's this

[scarlet mango]

ohhh

[zealous garden]

We can also define (a•b)=B(a,b) to simplify stuff

But do you already see how its getting easier?

[scarlet mango]

tbh

no

[zealous garden]

HAHAHA

[scarlet mango]

brother

[zealous garden]

You don't think (u•v)/v is easy?

[scarlet mango]

like

what does it mean????

ya know

you have no idea what a headache this shit gives me

[scarlet mango]

like, why is this our projection formula to begin with?

[zealous garden]

Well we started with B(u,v)v with |v|=1 right

[scarlet mango]

yes

[zealous garden]

We then showed that |v| can't be 1 sometimes, so we had to figure out a slightly different form for longer or shorter v, right

[scarlet mango]

never noticed that but ok lol

[zealous garden]

Well I didn't show you a counter example just said they exist

So take the rational plane, Q^2, and look at the subspace spanned by (1,1) so pairs of rational numbers of the form (q,q)

Notice there is no vector of length 1

In particular such a vector would have q=sqrt(1/2) which is not rational

So it's not in the rational vectorspace we were looking at

[scarlet mango]

how am I supposed to visualize this

[zealous garden]

The rational points of the plane

This is standard geometry

[scarlet mango]

wait so this was a counter example that you didnt provide earlier?

[zealous garden]

It means that there are triangles with rational sidelengths that are similar to triangles with irrational side lengths and longest side length=1

[zealous garden]

Yes

An example of why we can't normalize

We might not have the number we need to normalize with!

[scarlet mango]

how is that the result omg lol

[zealous garden]

Another example is imagine B(v,v)=-1, we have no imaginary unit in the reals to "normalize" with

[scarlet mango]

actually, a better question would be; how do we fix this problem? but you can answer both

[zealous garden]

Well consider the special case of an orthogonal basis of Q^2, then we can form right triangles with rational sidelengths. The hypotenuse is the longest side and the norm of the vector. So if we scale such that the hypotenuse is unit length, the sidelengths are no longer rational

[zealous garden]

By not normalizing, or more specifically not taking square roots which begins the mess

And divide by the squared length

[scarlet mango]

now thats true af

[scarlet mango]

how doth that help?

[zealous garden]

It corrects for the stretching done by not getting to normalize v

[scarlet mango]

I see

[zealous garden]

Since we use the stretched v twice, it will stretch by the square of its magnitude

So we cancel this out

Now it turns out, that v divided norm v^2, is equal to v^2 by our little definitions up there

x^2=B(x,x) if you remember

[scarlet mango]

why do we use it twice?

[zealous garden]

So x * x/B(x,x)=B(x,x)/B(x,x)=1

[zealous garden]

Look at the projection

B(u,v)v

[scarlet mango]

what does v divided norm v^2 even mean?

[zealous garden]

x/B(x,x)

We divide it by the square of its length

So shrink it if its big, expand it if its small

That way if it was shrinking vectors before, it stretches them out now

[scarlet mango]

I get that part

[zealous garden]

So we say B(u,v)v^-1, so the two v's stretch in opposite ways but still represent the same direction on the same line

And the stretches correct eachother, the same as if you had a unit v and used our first formula

[scarlet mango]

how how how how

I'm so sorry for the brain rot

I'm not crashing out, just sayin

[zealous garden]

Lmfao

[scarlet mango]

I understand how it works in a similar function. I just dont get that previous chat

[zealous garden]

So do you get it now? I see the thumb

Is it the direction and line part?

[scarlet mango]

I dont get why this is true

[zealous garden]

If so ignore

[scarlet mango]

uhm

alright

so we have a projection formula for any case now

[zealous garden]

Yes

So now earlier I told you about the orthogonal complement of a vector do you remember

The set of vectors that have a dot product of 0 with your chosen vector, the orthogonal complement

[scarlet mango]

[scarlet mango]

so the orthogonal complement is a set of vectors?

[zealous garden]

This is the same thing as the hyperplane orthogonal to the vector. Now I pause for example

[zealous garden]

Yes

Example, the y axis is the orthogonal complement of (any vector on) the x axis

The X,Y plane is the orthogonal complement to the Z axis

The present moment is the orthogonal complement to the forward arrow of time

Get it?

[scarlet mango]

HUH

[zealous garden]

Okay forget it if you don't get it

It's just an example

[scarlet mango]

lmao

[zealous garden]

It's to help you understand so if you dont get it, move on preferably

As it is not the point, merely beside it

Anyways

[scarlet mango]

I understood the other two

[zealous garden]

So let's say you look at a mirror, you are looking along the vector that is the orthogonal complement of the mirror, a normal vector right

We dont even care if its unit, just orthogonal

How do you reflect a vector u across the mirror represented with normal n?

Can you see it?

[scarlet mango]

what do you mean "reflect"?

[zealous garden]

I mean that every point on the mirror should be unchanged, distances should be preserved, and if you were on one side of the mirror before, you're on the other now

Flipping across a line

Right?

[scarlet mango]

so like

--, \ , | , /, --

[zealous garden]

Yes

The first two reflect across the middle and become the latter two

[scarlet mango]

yes(?)

[zealous garden]

Alright so can you guess how to reflect a vector u across the mirror normal to n, or v, or whatever you wanna call it

If I tell you that you use the projections we just discussed

[scarlet mango]

first of all

we are implying that a mirror has actually something on the other side

like mirror dimension

[zealous garden]

https://tenor.com/view/spirit-in-the-sky-doctor-and-the-medics-goin-on-up-80s-music-glam-rock-gif-18013888

[scarlet mango]

celeste and shi

[zealous garden]

Lmfao no

Funny though

[scarlet mango]

i was being serious

[zealous garden]

We are just using the reflectional symmetries of space to talk about transformations on space

[scarlet mango]

I prefer the mirror world

[zealous garden]

It turns out that EVERY distance preserving transformation is made up of reflections

Rotations, translations, screw motions, it's all about composing mirrors on mirrors

So this is the first step

Reflecting through 1 mirror

You can get philosophical if you really wish

But no such statements are concretely implied unfortunately

[scarlet mango]

no man I wanna get mathematical

[zealous garden]

Good

So what I just told you is a major theorem in GA

Cartan Dieudonne

[scarlet mango]

how tf is that a theorem lmao

[zealous garden]

It's true

Try it out

Play with your left hand and right hand

Imagine mirrors between them

Reflect them around

[scarlet mango]

how do i even do that lmfao

[zealous garden]

Notice how reflections through intersecting mirrors rotates your hand around the axis

Just do it lol

Put your hand out

[scarlet mango]

alright

[zealous garden]

Put your arm out as a mirror

[scarlet mango]

WHAT DOES THAT EVEN MEANNNNNN

said I calmly

[zealous garden]

Move your opposite hand to the opposite side, as your first hand takes the place of the arm

[vivid halo]

[zealous garden]

I'm grabbing the gif

[scarlet mango]

pause man pause

how do I put my hand out while using my arm as a mirror

like

what?

as you can see I'm not a very bright individual

[zealous garden]

So put your hand out, somewhere

Now choose where you want a mirror to be, place your other arm there as a bar to mark it

[scarlet mango]

palm down, up, sideways?

[zealous garden]

Any way you want

[scarlet mango]

alright

[zealous garden]

.

[scarlet mango]

I have chosen my mirror

[zealous garden]

Now in one motion you

• put the opposite hand (the one on the arm acting as a mirror), on the other side of the mirror, facing the first hand like a reflection would
• make the first hand take the place of the mirror, to mark your spot still

Can you do this?

It's not the easiest but if you can strongly visualize the mirrors you're using as you do it, it works out

[scarlet mango]

So i mirror one hand, then the other

[zealous garden]

Yes

Two reflections

Notice how your hand just rotates around where the mirrors would meet

Or if they would never meet (parallel), your hand translates in a straight line

[scarlet mango]

how the?

when did we have two mirrors

[zealous garden]

Is that amazement or confusion

[scarlet mango]

latter

[zealous garden]

I said do it twice

[scarlet mango]

and do I visualize all 4 things?

the two hands, and the two reflections

like

Idk man

I feel stupid lmao

[zealous garden]

Do you see

[scarlet mango]

I know I am, its just not cool to feel it too

[zealous garden]

Not perfect illustration

[scarlet mango]

THATS NOT WHAT YOU TOLD ME TO DO

Im calm btw

so

whats the black and white line?

[restive salmon]

Wordle 1522 4/6*

⬜⬜🟨⬜⬜
🟩⬜⬜⬜⬜
🟩⬜⬜🟩🟩
🟩🟩🟩🟩🟩
ah @heady pagoda

[zealous garden]

Yes it is

[zealous garden]

Two different mirrors

[scarlet mango]

okay, and could you explain what each other curve is?

and the two spots ofc

I'm sorry. I've been at this for 5 hours now, and visualizing things has never been my strong suit

I did warn you of my incompetence

[zealous garden]

So given the blue point as a vector, and a vector for the black line, what is the yellow point

[zealous garden]

Color coded curves to show you step by step

Anyways the answer is, you subtract twice the projection onto the normal

[scarlet mango]

if we entered the magical realm of mirror world it would be the reflection of the blue and the green point (?)

wtf

So far out

[scarlet mango]

blue and yellow are obviously reflections of each other

[zealous garden]

You see Y=B+G, G=-2 P_N(B), Y = B - 2 P_N(B) is the reflection of B through N

Right?

[scarlet mango]

so

I get Y=B+G

G is equal to -2(projection of B)??

thats def wrong

by me

[zealous garden]

Twice projection of B onto N

[scarlet mango]

what is N?

[zealous garden]

The black normal vector to the black mirror

So you get it?

[scarlet mango]

man

im sorry

im trying my best

[zealous garden]

$R_v(u)=u-2 P_v(u)$

[fathom swallow]

wraithlord_kojima

[scarlet mango]

the reflection of u onto v is u -2(projection of u onto v)

what is u and v?

[zealous garden]

u is a vector, v is a vector. Specifically we take v to represent its orthogonal complement

We can actually take both of them to represent their orthogonal complements and get the same result

[scarlet mango]

can we stick to the simpler one

[zealous garden]

So the reflection of u THROUGH v, is u minus twice the projection of u ONTO v

Wdym

This is the same thing

[scarlet mango]

oh lol

waitwait

[zealous garden]

Yeah we're summarizing and writing it up before we move on to the next piece

Questions shoot

[scarlet mango]

I am so stuck

Can you show me which vectors are u and v in our example?

[zealous garden]

u is the blue arrow, v is the black arrow

[scarlet mango]

yes thank you

[zealous garden]

R_v(u) is the yellow arrow

[scarlet mango]

waitttt are we in 3D?

[zealous garden]

Shhh, doesn't matter actually

[scarlet mango]

because u and v arent perp

[zealous garden]

I'm drawing the picture in 2D

[scarlet mango]

so we must be in 3D

[zealous garden]

But this equation we're working on holds in any dimension

No this example is in 2D

u and v dont need to be perp

This is general

[scarlet mango]

oh true sorry

[heady pagoda]

sorry my internet is getting hammered out of shape give me a moment

[scarlet mango]

so the projection of u thru v is the blue arrow? or is it the green one

I love how you expected me to be able to sniff this out using my hands lmfao

[zealous garden]

The projection of u onto v is HALF the green arrow

[restive salmon]

Ahh its alright

[scarlet mango]

I seeeee lemme check it out again i think i get it lol

I SEE IT NOW

I see it I see it I see it

[zealous garden]

Based!!

[scarlet mango]

hahah its basic subtraction lmao

[zealous garden]

Ready for the next step?

[scarlet mango]

no but lets go

[zealous garden]

Lmfao love the spirit

So armed with this formula for a reflection through v, we're missing one crucial identity for the geometric product uv

[scarlet mango]

the geometric product is based on the fact that x² = |x|² right?

[zealous garden]

Specifically, we have uv + vu = 2B(u,v)

[zealous garden]

Yep

Some call this the contraction axiom

Vectors square to scalars

[scarlet mango]

I mean i dont understand the reasoning behind this but ok

[scarlet mango]

so uv=vu

[zealous garden]

It's one of those "because magic happens" things for now

[zealous garden]

Nope!

[heady pagoda]

Wordle 1522 3/6*

⬛🟩⬛🟨⬛
⬛🟩🟨⬛🟩
🟩🟩🟩🟩🟩

[zealous garden]

So we can split uv into two terms

[restive salmon]

I liked it alot better when ur internet was hammered

[zealous garden]

[u,v]=uv-vu and {u,v}=uv+vu

Divide by half if you want

[scarlet mango]

what does this even mean? like, what is the difference between {} and []. and what does the comma in between u and v mean?

[zealous garden]

Swap u,v; you'll see that [u,v]=-[v,u]
{u,v}={v,u}

[zealous garden]

It's a function thing, those are definitions

[scarlet mango]

so, do both these things just hold true? and how do we even use them here?

[zealous garden]

This is just for argument sake

If its confusing let go

The point is thay uv≠vu

[scarlet mango]

okeoke

[zealous garden]

You can eventually break it down into uv=u•v+u \wedge v

[scarlet mango]

and we magically derive that uv+vu = 2B(u,v)

[zealous garden]

Yep, because the dot product commutes, and the wedge doesn't

So we end up with twice the dot product

[scarlet mango]

interesting

[zealous garden]

I forget how to derive it exactly but I'm sure it's stupid simple and comes from squaring (u+v) or something

[scarlet mango]

so the reason uv is not vu is the wedge?

[zealous garden]

Essentially yes

[scarlet mango]

whoaaaa i get it

[zealous garden]

So the wedge product falls out of the geometric product

Where the dot product is 0

Look at this

If u^2=0 for all vectors, then

(u+v)^2=u^2+uv+vu+v^2=uv+vu=0

But we only have that to simplify

We can't say anything else about uv and vu except, uv=-vu

This is the wedge product

So u^v = - v^u

This encodes span!

[scarlet mango]

why did u^2 and v² just vanish both?

[torpid bay]

u^u = 0

[zealous garden]

We take the dot product where everything measures to 0

Remember how I said sometimes we can throw away components from pythagoras?

This is then

So B(u,v)=0 for all u,v

This turns the geometric product into the exterior product

[scarlet mango]

wtf man

[zealous garden]

This is a detour

Not straight to the goal

[scarlet mango]

this is serious manipulation

[zealous garden]

But the point is that it is hiding inside of the algebra we are trying to figure out

Sometimes we use it to make things easier

But uv+vu=2B(u,v) okay

Are we okay with this

Can we move on

[scarlet mango]

youve defined this B function to be both a dot product and a GP if im not mistaken. Which one is it now lol

[zealous garden]

No, no it is not

[scarlet mango]

I'm ok with this, yes

[zealous garden]

But in the special special case that you multiply a vector with itself, they look the same

[scarlet mango]

lmao alright

[zealous garden]

The dot product and B are the same, for vectors

The GP and dot product are the same if you SQUARE a vector

[scarlet mango]

yup

[zealous garden]

So we can actually plug this in

Look at our reflections from before

[scarlet mango]

where do I plug it in at?

[zealous garden]

$R_v(u)=u-2P_v(u)=u-2B(u,v)/v=u-(uv+vu)/v=u-uv/v - vu/v= -vu/v$

[fathom swallow]

wraithlord_kojima

[zealous garden]

Or in other words, $R_v(u)=-vuv^{-1}$

[fathom swallow]

wraithlord_kojima

[zealous garden]

This is the reflection through a mirror right

[scarlet mango]

yes I agree

[zealous garden]

Now witness as I reflect through TWO mirrors, to rotate around their intersection like I showed you earlier

[scarlet mango]

I didnt even know the mirrors were to intersect

I had them parallel

[zealous garden]

This works for that too dont worry

Parallel mirrors intersect "at infinity"

Translations are rotations around an axis infinitely far away

It works out

[scarlet mango]

thats so cursed

[zealous garden]

$R_v(R_u(w))=-v(-uwu^{-1})v^{-1}=vuwu^{-1}v^{-1}=(vu)w(vu)^{-1}=:R_{uv}(u)$

[fathom swallow]

wraithlord_kojima

[scarlet mango]

OMG BRO

hold on

gotta validate

[zealous garden]

So this is a rotation,

$R_{uv}(x)=(vu)x(vu)^{-1}$

[fathom swallow]

wraithlord_kojima

[scarlet mango]

hmmmm I see

I mean i have no idea why that means to rotate

[zealous garden]

Yes you do

[scarlet mango]

they'd just go up and down?

[zealous garden]

Reflecting twice is rotating remember

We drew that one

[scarlet mango]

[zealous garden]

So we now took our math for reflecting, did it twice

[scarlet mango]

the reflection of the reflection is a rotation

[zealous garden]

This is math for rotating

Yes

Soo simple

[scarlet mango]

for you brother

not everyone is megamind

[zealous garden]

It is simple yes? Come come

[scarlet mango]

my brain is like

bro

[zealous garden]

What questions do you have

[scarlet mango]

reflection of reflection

what

[zealous garden]

Or are you just fried for the second

[scarlet mango]

wut

[zealous garden]

Yeah man, reflect twice and you rotate

[scarlet mango]

i just dont know how to logically define this

[zealous garden]

.

[torpid bay]

reflection of reflection comes up in other areas as well, i hadn't seen this projection stuff before

[zealous garden]

Call it a bireflection

[scarlet mango]

thats just a few lines and curves to me lmfao

[zealous garden]

Or 2-reflection

[scarlet mango]

wait

[zealous garden]

Can you see it yet fijo

I believe in you

[scarlet mango]

did we just do this: --, \ , | , /, --

[zealous garden]

I'm not sure what you meant by that exactly lol

[scarlet mango]

like

we pass thru one mirror

and then we pass thru the other

yeah my shape doesnt make sense

but like

we reflect thru the first

and the reflection is reflected thru the second

isnt just one reflection a rotation?

[torpid bay]

nah

[scarlet mango]

yeah im wrong

[torpid bay]

its two

[torpid bay]

two reflections always gives a rotation

[scarlet mango]

I just dont get why

[torpid bay]

it may be helpful to draw arrows on those dots then

[zealous garden]

It's kinda definitional

Like two reflections can technically give things you wouldnt initially call a rotation

Like a lorentz boost

Or a translation

[scarlet mango]

i mean doesnt that mean that R is reflected thru G and then Y is reflected thru Blue?

[torpid bay]

two circular relfects then ig

[zealous garden]

But we can use all the math for rotations, nearly, for them as well. So we call them rotations too

[zealous garden]

Sure, I intended the other order but you can read it in either direction

[scarlet mango]

so if we leave it at the first step isnt it at least a half rotation?

[zealous garden]

No just a reflection

To cut the rotation in half, cut the angle of the mirrors in half

[scarlet mango]

wait lemme as this a lil differently

does rotation mean exactly what springs to mind in a pleb like me?

[zealous garden]

Yes

[scarlet mango]

HUHHH

[zealous garden]

Yes this is how you rotate things with GA

Which, BTW, includes quaternions!

qxq^-1!!!

[scarlet mango]

I will never understand how one reflection isnt enough to rotate something

[zealous garden]

That's what your hand was meant to illustrate

[scarlet mango]

i dont think I'd survive if i asked what this means again lol

[zealous garden]

If you only reflect once, its the wrong hand

[scarlet mango]

yeah can you explain how to do it with my hands actually?

so, the two mirrors, i just imagine them without having to keep my arms in their place?

[zealous garden]

https://arxiv.org/pdf/2107.03771

[scarlet mango]

WAIT

ok

hear me out

[zealous garden]

Look at the pictures on there

[scarlet mango]

i will but lemme say my stupid theory first

so

[zealous garden]

Sure

[scarlet mango]

if I were in a room

with these two mirrors

and I stood exactly where our first vector is

and looked at the direction of it

would i see myself rotated?

[zealous garden]

The direction of what

[scarlet mango]

idk the mirrors direction

[zealous garden]

If you look at the mirror, you see yourself standing where you would be after a reflection through the mirror

[scarlet mango]

at any point except their intersection

[scarlet mango]

well, its not just one mirror. my reflection would be reflected thru the other mirror too

[zealous garden]

IRL, you wouldnt be able to see this because the mirror is blocking your vision

But, the mirror you would theoretically see his reflection through the second mirror

[scarlet mango]

but if the mirror doesnt block my vision then its not a mirror lullulululul

[zealous garden]

And this second reflection is a rotated you

[scarlet mango]

alright and then

i would see myself looking back at me?

how would it work?

so weird

lemme see the photos

I saw some weird shit

which one is our thing?

btw disregard my previous theory

wait actually

i might understand now

so

say im looking straight at a mirror

but of course its not our normal mirror

so I look at myself looking back at me

and that i guess doesnt count as me having rotated yet

[zealous garden]

Fig 1, Fig 2, Fig 7 are the most relevant so far

[scarlet mango]

alright

[zealous garden]

I'll let you digest all this, maybe another time I'll show you how to generalize this for any l-versor V and general multivector U

[scarlet mango]

WAIT

so

when we say mirror, do we mean a certain axis of symmetry?

[zealous garden]

Essentially

[scarlet mango]

yeah in a couple years

[zealous garden]

LMFAO

[scarlet mango]

but with even one axis of symmetry we can rotate shit

[torpid bay]

[zealous garden]

No you can't?

[scarlet mango]

brother i am still learning analytic geometry

[zealous garden]

Look at the circle, rotational symmetry implies infinite axes of symmetry

[scarlet mango]

but ive heard the term being uttered before

[zealous garden]

Rotational axis is different

Although, it is an axis, and rotations are symmetries.... but not in the way you were clearly just using the terminology

Alternatively we add an "axis" of symmetry at the center of the circle?

[scarlet mango]

so now we are strictly talking reflections

[zealous garden]

Yes so just reflections across lines, the usual elementary idea of axis of symmetry

[scarlet mango]

so in the figs shown

we start from the right and end up on the right again correct?

[zealous garden]

Yeah, they're numbered and labeled so you can see the order of operations

[scarlet mango]

ohhh didnt see that lol

hmmmm

something is clicking up there

i can see now why it wouldnt work with one mirror

[zealous garden]

So yeah, we get all of the symmetries of the infinite Euclidean plane out of this ordeal

[scarlet mango]

how tf do we infer that?

[zealous garden]

We know how to reflect the mirrors, which is that basic idea of axis of symmetry, and the other symmetries are built out of these reflections (Cartan-Dieudonne Theorem)

[scarlet mango]

this is so hardcore man

so

so

so

can you now

give me an example

where this knowledge is used?

because I honestly think I get it now

I feel like the whiplash guy

[zealous garden]

Hmmmm

What do you mean by an example of this knowledge being used exactly

Multiple ways I could interpret that

[scarlet mango]

Gimme whichever you think I will actually understand

youre the expert here

[zealous garden]

Applications tend to make just of some other facts I don't want to pull in too crazily... but I have an idea

[scarlet mango]

I'm listening

[zealous garden]

https://bivector.net/tools.html?p=2&q=0&r=0

https://bivector.net/tools.html?p=2&q=0&r=1

[scarlet mango]

oh god

[zealous garden]

This will let you sidestep not quite knowing some subtleties I do believe

It's a calculator

[scarlet mango]

how does it work?

[zealous garden]

My recommendation is simple, you should draw a coordinate plane and some triangles and lines on it, try reflecting or rotating some around, with equations

[zealous garden]

It explains the syntax right there, you put in your equation and it gives you the answer

This way I dont have to talk you through the general notion of the geometric product

As you end up with entities that you haven't met yet as intermediaries in your calculations

It's like how the imaginary numbers show up in the cubic formula even when all roots are real

[scarlet mango]

i feel like youre like 200 years older than me man

[zealous garden]

Once you do one or two you should have a feel for it, be convinced of it, and probably never want to do such things by hand again

[scarlet mango]

literally

can i ask; what level of maths is this?

like what kinda background doth one need?

[zealous garden]

Uhhhh

[torpid bay]

[zealous garden]

Well I took a somewhat circuitous route in effort of a level of generality and rigor that most sources actually won't go for

[scarlet mango]

youre a different kind of beast

[zealous garden]

There are introductions for high schoolers

[scarlet mango]

NAHHHH

[zealous garden]

Much simpler than what we covered

As well as introductions assuming you know LA, or tensors

[scarlet mango]

brother i struggle at deriving the ellipse equation and they doing reflections in high school?

[zealous garden]

Or introductions that teach the LA you need

[zealous garden]

LMFAOOO

[scarlet mango]

no fr tho this is some nice stuff

ive never learnt anything advanced like this

obv idk what their use is

but who cares

thank you for sacrificing your precious time

[zealous garden]

No problem

[scarlet mango]

ice been here 6 hrs now

[zealous garden]

Hey

[torpid bay]

[scarlet mango]

[torpid bay]

a little qustion btw

a few

[zealous garden]

https://youtube.com/shorts/DmWJBrFIwqQ?si=s_nxw9Bfu-z8fkFi

[zealous garden]

Check out these shorts bro

Lovely lovely stuff

[scarlet mango]

alright alright thanks

[torpid bay]

ah yeah that was my question

lel

[scarlet mango]

yeatte ask your questions its my turn to sit and eat popcorn watching you fry your brain

huhhh

not fair

this was such a humbling experience man

anyways

[zealous garden]

https://youtube.com/shorts/2d3A5x6Mwc4?si=TbQAWlyOUb3E_Snk

Last one

https://youtube.com/shorts/E49bk6yToLI?si=2sEjC2CWzn5FKyE_

[zealous garden]

What was the question

[zealous garden]

Ah I appreciate it

[torpid bay]

I was gonna ask about the spcifics of v^-1 and what this means specifically. and then i didn't quite look all the way up but

in the (uv+vu)/v, v is a vector or is it the magnitude?

mag right?

[zealous garden]

A vector

[warm gull]

wait so someone very helpful told me Calculus involves algebra, functions, coordinate geometry, indices and logarithms, sequences and series, limits, trigonometry. im not good with sequences, series, i havent studied limits yet, and in trig i only know the 6 trig functions, and unit circle am i good guys?

[torpid bay]

the bottom one too?

[zealous garden]

Yeah, v^2=B(v,v)->v*(v/B(v,v))=1

[zealous garden]

so v^-1=v/B(v,v)

[torpid bay]

and when you write v^-1 you also wrote it as 1/v right?

the y're the same

[zealous garden]

Yes

[torpid bay]

ahh ok ye, I dind't see that part hm

in sudgylacmoe vids I saw $e^{\frac{-I\theta}{2}}we^{\frac{I\theta}{2}}$, this is just another way of writing it yeah?

for rotation formula

[fathom swallow]

Yeatte

[zealous garden]

Yes

That uses different parameters from my approach, but better ones arguably

However I didn't want to explain bivector exponentials

[torpid bay]

you said two reflection didn't give a hyperbolic rotation, is there some other object when applied twice that give hyperbolic rotation?

[zealous garden]

I said it did

Lorentz boosts

[torpid bay]

oh u did oh cool

I read the negation of your statement ig

[zealous garden]

Next time we explain the above memes, heretofore referred to as Figure 1 and Figure A

[torpid bay]

it really is such a great meme

say

I want something like

$\partial _i \wedge \Psi_j$

that's valid, right?

[fathom swallow]

Yeatte

[zealous garden]

I'm really not sure

What is partial_i

What is psi_j

[torpid bay]

im looking to turn my 3 vector formulation of EM into a 4-vector one using geo vectors

it just refers to the index of a vector with coefficients of partial operators

[zealous garden]

"It"

Partial in the ith direction? Or Some sort of gradient

[torpid bay]

g radient

[zealous garden]

So partial wedge Ψ?

Curl?

[torpid bay]

yeah i want to extend the curl

,rotate -90

[torpid bay]

mm i used conflictin notation here

have psi_3 be the 3 vector formulation then

[undone pike]

is this some kind of quantum

electro

magnet stuff

[torpid bay]

elecromanteteong ye

[zealous garden]

Curl does generalize yes

It's the same thing as the exterior derivative tbf

Divergence also generalizes

Same thing as the interior derivative (codifferential, not interior product)

[undone pike]

interesting

[torpid bay]

ah so it is the same hm

I never got to post this when we were talking about reflections but

$r \circ r = r \ m \circ r = m \ r \circ m = m \ m \circ m = r$

I got this when figuring out symmetries of stuff

[torpid bay]

$\sum_{n=0}^{\infty}{n \choose -x}=\frac{\sin(x\pi)}{x\pi}\sum_{k=0}^{\infty}{\frac{k!x!}{(x+k)!}$ hm

[undone pike]

$r \circ r = r \ m \circ r = m \ r \circ m = m \ m \circ m = r$

[fathom swallow]

back to ginger

[chilly lynx]

how do i get better at understanding word problems?

[zealous garden]

Wtf is that

[fathom swallow]

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

[undone pike]

ror = rmor = mrom = mmom = r

[torpid bay]

$r \circ r = r \ m \circ r = m \ r \circ m = m \ m \circ m = r$

[fathom swallow]

Yeatte

[torpid bay]

composition of rotations and reflections(mirrors)

[burnt ledge]

mom = r

[undone pike]

correct

[torpid bay]

tho it's kinda a weird little thing I got with factorials

[undone pike]

oh interesting

why is mor = rom = m?

rotation + reflection = just reflection?

[torpid bay]

well, I just manually calculated a bunch of rotations and symmetries and thats what happeened

[undone pike]

oh that is true i guess

[torpid bay]

rotation o reflection = relfection yeah

just a different one

[undone pike]

cool

[torpid bay]

I think you can prove it by knowing r o r = r, and each operation unique (at least for nice cases), so r o m must not be any of the known symmetries, and all r are gotten to by r of r, r o m must get all the mirrors

I didn't account for something more general than a symmetry like a group, so this logic works fine for normal symmetries

[zealous garden]

Yes

[zealous garden]

Only in O(2)

[torpid bay]

$\sum_{n=0}^{\infty}{n \choose x}=\frac{\sin(x\pi)}{x\pi}\sum_{k=0}^{\infty}{ {{x-k} \choose {k}}^{-1}}$

[zealous garden]

Otherwise you get a trireflection

It's a reflection in the sense that a rotation is, though also improper (changes handedness)

Rotoreflections are the name

[undone pike]

rightt

[fathom swallow]

Yeatte

[zealous garden]

However it is a reflection in the sense of being an odd k-reflection

[undone pike]

interesting!

[zealous garden]

Rotations generalize to the Even k-reflections, forming a subgroup

So all reflection is O(p,q,r), all Even reflections is SO(p,q,r)

[undone pike]

now i'm not sure if i'm following :p

[zealous garden]

Orthogonal groups

You know them?

[torpid bay]

rotate o mirror = mirror, and rotate + mirror = rotoreflect in the generalization right?

[zealous garden]

Yes it's just addition of k

[torpid bay]

[undone pike]

yes

i think

i know the groups corresponding to orthogonal matrices

[zealous garden]

do you know what it means

what an orthogonal map is

nevermind matrices, see if you can do it without saying something about matrix representations

[velvet dagger]

If you're not sure, try out some of this problem set

[undone pike]

i'm not sure

i was gonna say a linear transformation

[velvet dagger]

Specifically 1, 2, 6, and 7

[undone pike]

a real one that just "reorients axes"

[zealous garden]

an orthogonal map is one that preserves a (the) dot product

[undone pike]

these are making sense, thank you 👍

yes

but i don't recognize the notation O/SO(p,q,r)

[zealous garden]

O is the Orthogonal Group, SO is the Special Orthogonal Group

specifically the ones preserving real symmetric bilinear forms of inertia p,q,r

the linear maps on R^p+q+r satisfying B(u,v)= B(Tu, Tv) where, given a """""nice""""" basis, B(v,v) is positive for p vectors, negative for q vectors, and 0 for r vectors

[undone pike]

interestingg

this is the part i didn't know

[zealous garden]

you can also say it's the isometries of R^p,q,r onto itself

though strictly speaking, with r neq 0, I've only covered the weak orthogonal groups when talking about reflections

[zealous garden]

as the weak orthogonal group is the subgroup of the orthogonal group which leaves null vectors (B(v,v)=0) invariant, as opposed to doing whatever it wants with them like the definition here implies

[undone pike]

i see

i can't think of an example of such a transformation

[zealous garden]

yes you can

[undone pike]

like say in minkowski space

[zealous garden]

O+(n,0,1)=E(n)

SO+(3,1) is the lorentz group yeah

SO+(3,1,1) is the poincare group!

[undone pike]

hmmm

does SO(3, 1) mean p = 3, q = 1, r =0?

ah

[zealous garden]

SO+(3,1) is also the group containing special conformal transformations of the plane (or sphere ig)

[undone pike]

that is interesting

[zealous garden]

the Orthogonal groups are quite an interesting family of Lie Groups

[undone pike]

i'm not sure how to view the poincare group as a linear transformation, since it includes translations

[latent edge]

I hate orthogonal groups ngl

[zealous garden]

many subfamilies as well. O(n,0,0)=O(0,n,0) for your Orthogonal Maps, O+(n,0,1), O(n+1,0,0) and O(n,1,0) can all do projective geometry

[zealous garden]

I hate number theory so we'll call it even

also looks like I need homology or number theory or some shit to finish my classification of metric vectorspaces wtf

is this something easily available? Does anyone know how to enumerate the isomorphism classes of metric vectorspaces over a Field?

[zealous garden]

O(n,1) and O(1,n) for spacetimes

O(p+1,q+1,r) for conformalization

[latent edge]

I like Lie groups/algebras

[velvet dagger]

Yup, once you go through these you get a good image about what orthogonal and unitary maps mean geometrically

[latent edge]

orthogonal ones tend to mess it up

[velvet dagger]

As opposed to just matrices

[zealous garden]

Why

What about the complex orthogonal maps dami

No love for O(n,C) huh

Tsk tsk tsk

[velvet dagger]

HAH I don't think much about O(n,C)

[zealous garden]

tbh I'd have been stunlocked if you did

[undone pike]

what class is this problem set from? :p

[velvet dagger]

Analysis, but a very esoteric analysis class

[limber thunder]

i love esoteric classes

[elfin smelt]

i love klasse orthogonal theory

[torpid bay]

$\sum_{n=0}^{\infty}{n \choose -x}=\frac{\sin(x\pi)}{x\pi}\sum_{k=0}^{\infty}{\frac{k!x!}{(x+k)!}$ hm

[zealous garden]

What?

[deft tide]

Whats ᵽᵉ − ᵽᵉ − ᵽ = ᵽ(ᵉ − ᵼ) ᵉ + ᵽ + ᵽ = ᵽ + ᵽᵉ − ᵉ − ᵽ = ᵉ − ᵽ − ᵽ − ᵽᵉ + ᵽ = − ᵽᵉ + ᵽ - ᵽ + ᵽ = ᵉ + ᵽ - ᵽᵉ − ᵽᵉ − ᵽ = ᵽ(ᵉ − ᵼ) ᵉ + ᵽ + ᵽ = ᵽ - ᵽ

[zealous garden]

<@&268886789983436800>

Oh my bad

[undone pike]

interesting

[zealous garden]

I thought I saw this appear in 5 different channels at the same time

My bad mods

[deft tide]

Because I said that in them

[zealous garden]

It's just two (channels)

Oh

[deft tide]

It's 2?

[zealous garden]

Still don't do that homie

[deft tide]

What is the answer

[zealous garden]

!help

[quasi jetty]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[zealous garden]

But tbf this looks like gibberish

[deft tide]

Do I just make a help channel or som

I don't wanna waste someones time

[elfin smelt]

KO-theory

the K in k theory comes from klasse

[zealous garden]

Oh?

Oh no the demons are trying to make me join them

nG is here isn't he

[nimble sparrow]

guys how long does it take for an average men

like say in mid 20s?

??

[soft nymph]

Wdym

[nimble sparrow]

??

[nimble sparrow]

are you answering my question with another question?

[soft nymph]

No
I just dont understand u question

What do u mean

How long does it tske average man for what

[nimble sparrow]

i mean say if a man is in mid 20s?

like around 24to 26 idk?

[calm vessel]

The chat reads like a stroke 😵‍💫

[real shell]

[calm wolf]

where can I discuss stuff on analytic number theory if I want to get insights? I'm working on a paper relating prime number sequences to roots of L functions using a method called spectroscopy where you construct a corresponding theta function (ie \theta(x)=\sum_{p_i\leq x} where p_i's have some property from the sequence) then finding the error term from that and taking the fourier transform, where the peaks correspond to the roots of L functions related to the sequence

[ocean robin]

maybe you could look in #advanced-number-theory ?