Mathematics - 2017-11-01 (page 1 of 7)

12:27 AM

Oh I see, I agree. My point was $v^1 \wedge \cdots \wedge v^n$ literally eats a parallelopiped, because it's an element of $\Lambda^n V^* = \hom(\Lambda^n V, \Bbb R)$

@LeakyNun It need not be, I don't think.

If $n$ is the dimension of $V$, the exterior power becomes one dimensional, but that's a separate fact.

@BalarkaSen it needs to be, because the dimension of $\Lambda^k V$ is $\dbinom n k$

it wouldn't be a scalar multiple of $e_1 \wedge e_2 \wedge \cdots \wedge e_n$

Ah of course it would be a linear combination of exterior products of the basis elements. Yeah that fails.

As I said, top exterior power of $V$ is one dimensional

yes

omg does this give yet another definition/intuition of binomial coefficient lol

Hi

What's cooking

hi

?

12:32 AM

Like, what's up

wedge product

Is there a way to use $\sum_k\binom nk=2^n$ in this in some way?

Like $\bigoplus\lambda^kV$ or something

@AkivaWeinberger that's the dimension of the exterior algebra, no?

would have dimension $2^n$, don't know if it's useful or anything

What you get is that $v_{i_1} \wedge \cdots \wedge v_{i_k}$ is linear combinations of $e_I$'s where coefficients of $e_I$ are volume of projections of the parallelpiped spanned by $v_{i_j}$ to the $I$-plane

I think

It stores volumes of all the various projections to the elementary $k$-subspaces

12:35 AM

Does the fact that every set is uniquely determined by its members follow from axiom of extensionality?

They're equivalent?

Extensionality says that $x\in A\iff x\in B$ implies $A=B$, right?

Then yeah.

@AkivaWeinberger is the dimension ever useful?

That's why $\{a,a\}=\{a\}$, for example.

is it useful to know that $\Bbb C : \Bbb R = 2$?

@AkivaWeinberger yeah to necessity or equivalence?

the latter I assume

12:36 AM

Equivalence

hmm thanks

The full exterior algebra is only philosophically useful; it's a graded algebra

Hm so that's what you call it

Right and you can wedge arbitrary elements in it

ayup

If $V$ is $\Bbb R^3$ then you get an eight-dimensional thing and part of me wonders if you can throw the octonions at it and get stuff to happen

12:38 AM

yay i can safely say the former statement as written is itself an axiom

(Cont'd) Probably not, I dunno

Shrug man

You're the dude with weird but cool ideas

I've been kinda dry with those lately

@AkivaWeinberger hmm, might be useful

I know you can get a seven-dimensional "cross product" on $\Bbb R^7$

12:40 AM

why do we only have powers of 2 as the dimension of our extensions?

like, C:R = 2, H:R = 4, and then 8 and 16

the proof requires algebraic topology

There's a theorem that the only division algebras (IIRC) over $\Bbb R$ have dimensions 1,2,4,8

@BalarkaSen actually what's the statement?

oh, division algebras

what do we lose in 16?

and beyond that you can have weird stuff that aren't division algebras

@Leaky Associativity

12:41 AM

and in 32?

@LeakyNun 16 has zero divisors

We lose associativity at 8

No, that's octonions, nevermind

Sniped

(16=Sedenions)

Sadonions

sadist onions

what do we have left in 16 lol

12:43 AM

Happy onions?

It's still an algebra, I guess @LeakyNun

@LeakyNun There's this neat map from $S^3\to S^2$ called the Hopf map where the preimages of points are all circles.

It's quite easy to prove R and C are the only commutative division algebras, I believe

And it's not nullhomotopic (which means it's "really nice" in this context)

then what do we lose in 32

There's also Hopf map from $S^7\to S^4$ and $S^{15}\to S^8$. (IIRC)

Talking about what?

12:45 AM

But no more after that. Hopf maps end there.

There are multiple Hopf maps S^7 --> S^4

'Cause one way to construct them uses $\Bbb C$, $\Bbb H$, and $\Bbb O$.

@BalarkaSen Really? Hm.

Eg the exotic 7-sphere fibers over S^4 with S^3 fibers

Milnor to the rescue

@AkivaWeinberger At least topologically. Smoothly that's not S^7 anymore so shrug emoji

Oh I guess also $S^1\to S^1$ but that's just doubling the circle over itself

The point is it looks like there's this $S^{2^{n+1}-1}\to S^{2^n}$ pattern but no

Stops at $n=3$

@AkivaWeinberger just like the division algebra

12:48 AM

So there's something fundamentally different about the topology of spaces at these dimensions

gasps

And indeed there's a connection!

@LeakyNun There's a theorem which says only S^n for n = 1, 2, 4, 8 are H-spaces

I don't know much beyond this though

:o

maths is too amazing

12:49 AM

Balarka knows more about this than I do

H-space means you have a multiplicative structure, sorta

@Akiva I know epsilon more probably

but not beyond

but this is the Hopf invariant one problem

I meant n = 0, 1, 3, 7

in which you try to find maps between spheres for which the Hopf invariant equals 1.

I think Hopf invariant means exactly the linking number between two fibers

Hm

Hence the name "Hopf link"

@LeakyNun Yeah try to show that $S^3$ is the disjoint union of circles, each pair of which is a Hopf link

Probably the other way around :P

12:54 AM

@AkivaWeinberger nope :P

or perhaps that $\Bbb R^3$ is the disjoint union of (geometric) circles plus one line

First learn to foliate $S^3$ with torii

Then learn to foliate torii with circles

every pair of circles of which is a Hopf link, and every circle of which has the line going through it

Bob

I posted a DE question here: math.stackexchange.com/questions/2499033/…

It actually looks pretty

12:56 AM

hi

did I miss ted at all

For a starting point, put a torus in $\Bbb R^3$ and then find two circles on its surface that are linked to each other

1:07 AM

does anyone haave any challenging "find delta for each epsilon" limit problems?

all ive done is like rational functions and stuff

$\delta = 2$. Give me an $\epsilon$

quick

uhh'

the mitochondria are the powerhouses of the cell

?

you're disqualified, the answer is 42

disqualified from what??

Too big an epsilon

1:09 AM

@MeowMix the quickmafs exam

skrrrra

oh and ive done the classic $x^2 \sin x$ thing

is there anything interesting

there's no point doing epsilon-delta proofs to that extent

@Daminark did you watch the keemstar video i sent to Washington

alright whatevs

next chapter

(in case you're wondering, yes, I sent a keemstar meme to George Washington and he loved it.)

1:18 AM

ok fine

does anyone have interestng (real) limit problems

just throw out the $\delta$

preferably stuff with one-sided and limits to infnity cuz i havent practiced any of it

so an (n,m)-tensor is a function that eats m vectors and spits n vectors and is linear on all its inputs?

@BalarkaSen

Well, it's an element of $V^{\otimes n} \otimes (V^*)^{\otimes m}$

I'm still trying to understand tensor product

1:24 AM

Not much to understand, IMO

Multilinear maps = Linear maps from a tensor product. That's more or less the story

ugh wikipedia uses different terminology for (co)domain of relation

could you expand on it somehow? @BalarkaSen

I don't know what to expand on

What do you want to know?

like a bilinear map is a linear map from the tensor product of the vector space with itself?

what's the difference between $V \times V$ and $V \otimes V$ then?

Yes, a billinear map $f : V \times V' \to W$ gives rise to a linear map $g : V \otimes V' \to W$, and vice versa.

@LeakyNun $V \otimes W$ has multilinearity built into it. It's elements are of the form $v \otimes w$

And $(cv_1 + v_2) \otimes w = c (v_1 \otimes w) + v_2 \otimes w$

1:30 AM

can't the same be said for $\times$? $(cv_1+v_2,w) = c(v_1,w) + (v_2,w)$

$c(v_1, w) + (v_2, w)$ is not an element of $V \times W$

Writing addition of pairs like that does not formally make sense

$V \otimes W$ is a quotient of $V \times W$, is the point.

how else would you define $\Bbb R^2$?

I mean $(cv, w)$ is not $(v, cw)$ in $V \times W$

oh, right

That's the essential property of the tensor product space

1:33 AM

oh!

I see the difference now

you should have said that :P

4 mins ago, by Leaky Nun

can't the same be said for $\times$? $(cv_1+v_2,w) = c(v_1,w) + (v_2,w)$

no idea why I said that

You could read the definition from fricken wikipedia or something :P

its 7 am and i am on a loose mode so i dont care enough to go through the defn

alright

no problem

but yeah tensors are linear on simultaneously both coordinates is the point

right, $\Bbb R^2$ is definitely not bilinear

this is interesting

how is $V \otimes W$ a quotient of $V \times W$

@Balarka

1:35 AM

@EricSilva the other way round I think

$V \otimes W$ is bigger

right

wait

I don't think it works the other way

elements of $V \otimes W$ also don't look like $v \otimes w$

@EricSilva what do they look like?

@EricSilva Formally I should say $F(V \times W)$, the free vector space on $V \times W$. The tensor product space is $F(V \times W)/\sim$ where you define $\sim$ as $(v, w_1 + w_2) \sim (v, w_1) + (w, w_2)$ etc etc. I didn't mean vector space quotient

@Leaky Linear combos of those

1:37 AM

they're linear combinations of those

snippety snip

@Balarka what's the free vector space on $V \times W$

ah nvm

i see what you mean

good good etc

how large is the free vector space on $V \times W$

yeah its how the algebraists define tensor product

@BalarkaSen then how do you define it?

1:39 AM

i don't

...

tensors are things which transform like tensor in @Eric's word

that's the definition

@LeakyNun it's like, this big \('-')/

@BalarkaSen what?

w0t

1:40 AM

wat @_@

@Balarka do they not define it as a thing which makes a thing commute

"things which transform like a tensor"

it's the most clear definition conceivable

tensors transform like tensors duh

That sounds more circular than trig

hmm, as another example, $\Bbb R^3 \otimes \Bbb R^3$ has dimension $9$ while $\Bbb R^3 \times \Bbb R^3$ has dimension $6$

1:41 AM

only to normies

wait, then doesn't that mean $\Bbb R \otimes \Bbb R$ has dimension $1$

V otimes W has dimension dim(V) x dim(W) yes

@Leaky yes

quite

yes it does, its elements look like $v \otimes w$ (triggered)

1:42 AM

i am on a loose mode don't poke fun at me for my loose math

@Daminark bring it up w the physicists

this is underwear maths, the best math i can give you at 7 AM

how long have you slept?

@BalarkaSen quickmafs is only 2PM-2AM

i havent

@EricSilva i guess they define it as a universal object

1:43 AM

the great thing is that physicists don't even usually mean tensors, they usually are talking about tensor fields

@EricSilva now what's that?

Oh for fuck's sake...

a tensor at every point in your space

smoothly varying point-to-point

it makes the situation much tenser

@Daminark true

1:44 AM

it's when you stick a bunch of tensors in space and roll with it

Kek

@Balarka id define it via universal property to be maximally unclear

style points

very much

i wonder what nlab has to say about tensors

some symmetric monoidal grapefruits

"a tensor is an element of a tensor product"

The Gray tensor product of strict 2-categories is a tensor product in the multicategory of 2-categories and cubical functor?s. Likewise for Sjoerd Crans’ tensor product of Gray-categories.

1:46 AM

i checked

it aint even bad

ncatlab.org/nlab/show/tensor

yeah it's not living up to the n point of view

why is a linear transformation a tensor?

A linear transfo $V \to W$ is the same as a linear map $V \otimes W^* \to \Bbb R$

@Leaky $\text{Hom}(V, W) \cong V \otimes W^{*}$ canonically

@EricSilva hmm? really?

@BalarkaSen right

1:48 AM

if you believe me there should be no trouble believing Eric

or is it $V^{*} \otimes W$

I believe that Hom(V,W) monomorphs to that thing

$v^{\ast} \otimes w \mapsto v^{\ast}(\cdot)w$

that makes more sense

yeah V^* otimes W is the thing

yeah

1:51 AM

then isomorphism it is

it's a solid isomorphism

8/10

(let it go let it go x100 times] canonicalism never bothered me anyway

-10/10 brackets

{let it go let it go x360noscope times( braces never bothered me anyway

the confluence of memes has become supercritical

1:53 AM

so supercritical that it underwent total internal reflection

:O

memes are being superimposed like wavefunctions bro

@Balarka "A smooth manifold (see there for details) is a locally CartSp CartSp-representable object in the sheaf topos Sh(CartSp) Sh(CartSp)". I can't believe i never understood what a manifold was until now

Beautiful

indeed

1:55 AM

Ncatlab should be the textbook for every class

ncatlab is my city

Oh no, I think you know what this means

pls no

hahaha

end this now

marx has the swankest quotes

1:57 AM

"Marx debunked quotes yeaaaars ago" - Marx

“Hegel remarks somewhere that all great, world-historical facts and personages occur, as it were, twice. He has forgotten to add: the first time as tragedy, the second as farce.” - Karly boi

"It's ncatlab bro with the higher topos flow. 500 papers in a month, never done before. Passed all the competition man, Concise Course is next"

Yeah I have heard this one @Eric

it's a fucking good quote

Okay wait I'll need to think of this more later and I'll come up with a good quote

1:58 AM

Truly

/lyrics

i have to write up a presentation on the eighteenth brumaire for a class

and it's good fucking writing

@Daminark nlab is my city and if it weren't for j p may then the UC would be shitty