Mathematics - 2019-06-13 (page 2 of 2)

I knew the def of UFD that included units and irreducibles. But, now I saw this def which wikipedia claims is more useful: A unique factorization domain is an integral domain R in which every non-zero element can be written as a product of a unit and prime elements of R. See this.

Semiclassical

i'm not sure I can visually tell if that's really truly a regular dodecahedron

but it looks really plausible

think I know a reasonable way to check tho

Silent

The second version seemed rephrasing but, then i recalled that prime=irreducible guaranteed in PID only. So, how are two defs of UFD given are equal?

ÍgjøgnumMeg

@Silent no prime = irreducible in UFD

and all PIDs are UFDs

Akiva Weinberger

How are you specifying the cutting plane?

Coordinates of normal vector?

Semiclassical

4:25 PM

Right

and then applying the rotation matrices to see what they rotate to

Silent

@ÍgjøgnumMeg Really? Wow!

Semiclassical

my way of checking is to take the matrix M of these rotated cutting plane normals and compute M^T M

since that computes all the mutual dot products between these (unit vectors)

hmm, okay, not quite

I think this approach may be useful more generally tho

ÍgjøgnumMeg

@Silent :)

also hi chat

Akiva Weinberger

So how are you finding the plane that's a ratio of $1:\phi$ between them

That's a spherical geometry problem, no?

Semiclassical

hmm

i was just adding unit vectors, with components 1,phi, and normalizing the result

but that doesn't really make sense

(maybe approximately so but not exactly)

so anyways. at this point, I think it's a very plausible conjecture that the regular dodecahedron occurs

but I can't say I've actually verified it yet

bolbteppa

4:47 PM

Cayley–Dickson construction

In mathematics, the Cayley–Dickson construction, named after Arthur Cayley and Leonard Eugene Dickson, produces a sequence of algebras over the field of real numbers, each with twice the dimension of the previous one. The algebras produced by this process are known as Cayley–Dickson algebras, for example complex numbers, quaternions, and octonions. These examples are useful composition algebras frequently applied in mathematical physics. The Cayley–Dickson construction defines a new algebra similar to the direct sum of an algebra with itself, with multiplication defined in a specific way (different...

Anybody see this as obvious?

In a sense you're just copying the transition from real to complex but why the conjugates in e.g. the quaternion product (yeah it's implicit in the real to complex product but still)

Silent

@ÍgjøgnumMeg, I am a little confused: wikipedia says 'The uniqueness part is usually hard to verify, which is why the following equivalent definition is useful: A unique factorization domain is an integral domain R in which every non-zero element can be written as a product of a unit and prime elements of R.' But I don't see how this formulation of UFD def addresses uniqueness issue!

ÍgjøgnumMeg

@Silent because then you can use essentially the same proof as the fundamental theorem of arithmetic

i.e. write down two prime factorisations and set them equal and then use primality to prove uniqueness

Semiclassical

proof is in the pudding, i think. start with a power-associative algebra and show that the cayley-dickson construction always gives you power-associative algebra whose dimension is twice as large

whether or not the construction seems 'obvious' is immaterial to whether it works

Silent

@ÍgjøgnumMeg Oh! until we prove that 'every non-zero element can be written as a product of a unit and prime', we can't say it is UFD and we still have possibility that irreducible$\ne$prime. That's where two definitions differ! Thanks a lot.

bolbteppa

Think it's to preserve the norm hmm

Semiclassical

5:00 PM

@akiva playing with my latest approach (the dot product one) I'm growing less confident that the regular dodecahedron actually occurs

Silent

@ÍgjøgnumMeg Doesn't this only prove that there is only one prime factorization? Since we are not sure if we are in UFD yet, there can be 'irreducible factorization' different from that prime factorization, right?

MatheinBoulomenos

@Silent start with one prime factorization and show that it equals every irreducible factorization

ÍgjøgnumMeg

Hey @Mathein

MatheinBoulomenos

Hey @ÍgjøgnumMeg

Semiclassical

@AkivaWeinberger ok, scratch my last comment. pretty sure I've got it now

Akiva Weinberger

5:15 PM

Sorry I was gone for a while

Semiclassical

np

The direction is actually simpler than I thought

Take the top diamond to be in the direction (1,0,0) and the bottom diamond to be in the direction (0,1,0)

then the regular dodecahedron occurs in the direction of $(1,\phi,0)$

Akiva Weinberger

I think I said that :P

Semiclassical

lol, you did anticipate it pretty well

Akiva Weinberger

59 mins ago, by Akiva Weinberger

$(0,1,\phi)$?

The coordinates of the vertices of the icosahedron that I got from Wikipedia

Semiclassical

yeah

it's the same as those, so long as the axes of reflection symmetry for the tetrahedral group are taken to be the Cartesian axies

so that's neat

back later

Akiva Weinberger

5:24 PM

If the Cayley–Dickson construction were obvious to me I would remember what it was

(Both of these are subject to change though)

I mean I guess I could try to derive it

If $(1,0)=1$

then I guess $(i,0)(0,1)=(0,i)$?

Hm

ÍgjøgnumMeg

@Mathein alles roger? :)

Akiva Weinberger

Do we want $(0,1)(i,0)=(0,-i)$?

MatheinBoulomenos

@ÍgjøgnumMeg Ja, soweit alles gut, danke!

ÍgjøgnumMeg

Guuuut :P

MatheinBoulomenos

Und bei dir?

Akiva Weinberger

5:28 PM

I think we want perpendicular roots of -1 to anticommute

ÍgjøgnumMeg

Ja geht schon :) muss a kle motivation finden was zum lernen

Semiclassical

“Applying the Cayley-Dickson construction to the sedenions yields a 32-dimensional algebra, sometimes called the 32-ions or Trigintaduonions.”

bolbteppa

Trigintaduonions...

Octooctonions vs trigintadunions

Semiclassical

Sounds like an alien species from Dr Who

7

Q: What comes after the ducentiquinquagintasexions?

Hypercomplex numbers that use the Cayley-Dickson construction seem to follow a Latin naming convention related to the size of the algebra (which is always a power of two). As an English.SE question, I'm interested in the larger names that I haven't been able to find on the web. Here is what I kno...

bolbteppa

So $\overline{(a,b)} = (\overline{a},-b)$ is the general conjugation rule again a tiny bit random if you just think of going from real to complex

Not so random if you accept quaternions, but if you want to 'get' them from Cayley-Dickson from first principles without any pre/propter-hoc unjustified steps hmm

Semiclassical

5:35 PM

Meh. Like I said earlier, what “justifies” the C-D construction is that it works ie it always gives a new algebra that’s twice as big as the old

Akiva Weinberger

@bolbteppa What's the norm of $(i+j)(i-j)$

or in general $(e_0+e_1)(e_0-e_1)$ where $e_0$ and $e_1$ are any two perpendicular numbers satisfying ${e_0}^2={e_1}^2=-1$

$2$, right?

But if you multiply it out you get $i^2-ij+ji-j^2=-1-ij+ji+1=ji-ij$

We know that $ij$ and $ji$ both have a norm of $1$

The only way for the sum of two numbers of norm 1 to have norm 2 is if they're equal

With me so far?

So $ji=-ij$

or $i$ and $j$ anticommute

bolbteppa

Using properties of quaternions it's $(i + j)(i - j) = i^2 + ji - ij - j^2 = - 1 - 2 i j + 1 = - 2 k$ so $2$ yeah

Akiva Weinberger

Even if we weren't in the quaternions

Even if it were $e_0$ and $e_1$

and we didn't know what $e_0e_1$ was

this proves that $e_0e_1$ must equal $-e_1e_0$

Any pair of perpendicular roots of $-1$ must anticommute

bolbteppa

You mean - we will need $ij$ and $ji$ to have norm $1$

However we define the norm eventually

Akiva Weinberger

That's 'cause $i$ and $j$ have norm 1 and we want the product to preserve the norm

bolbteppa

5:49 PM

Right

Akiva Weinberger

(the same reason we wanted $(i+j)(i-j)$ to have norm $2$ - each of the factors have norm $\sqrt2$)

As a consequence, $e_0e_1$ is also a root of $-1$ since $(e_0e_1)^2=e_0e_1e_0e_1=-e_0e_0e_1e_1=-(-1)(-1)=-1$

Now we have a complete multiplication table for the space spanned by $\{1,e_0,e_1,e_0e_1\}$

and we can see that it's isomorphic to the quaternions

akhil krishnan

@Akiva Weinberger what does we expect about the dynamics of a dynamical system when we encounter a baby Mandelbrot?

bolbteppa

Why does $i+j$ have norm $\sqrt{2}$

Akiva Weinberger

Pythagorean theorem

$i$ and $j$ are perpendicular (by hypothesis)

(This doesn't work if we don't assume they're perpendicular)

bolbteppa

But maybe we have a Lorentzian metric :p

Akiva Weinberger

5:52 PM

Hm, didn't think of that :P

But if it's a Lorentzian metric then we have some things with norm 0

and I'm pretty sure one of the axioms of the norm is $|x|=0\iff x=0$

Besides

Take $\Bbb C\oplus\Bbb C$

with the multiplication axiom $(a_0,a_1)(b_0,b_1)=(a_0b_0,a_1b_1)$

Define $i:=(i,i)$

and $j:=(i,-i)$

bolbteppa

This gets into the division algebra question but only some of the outcomes of CD are division algebras so seems a bit iffy to use something special like this in a general construction

Akiva Weinberger

Note that $(i+j)(i-j)=0$

with this

akhil krishnan

@N.Maneesh what do we expect from the dynamics when we encounter a baby Mandelbrot ???

Akiva Weinberger

and I think this space has a natural Lorentzian norm?

So such algebras certainly do exist

but let's assume the thing we want to construct doesn't have that

In any case

What about the space spanned by$$\{1,e_0,e_1,e_2,e_0e_1,e_0e_2,e_1e_2,e_0e_1e_2\}?$$

With the rules that ${e_i}^2=-1$ and $e_je_i=-e_ie_j$

We have a complete multiplication table for this now

I claim

(1) this is isomorphic to the octonians

(2) this approach is equivalent to the Cayley–Dickson approach

(3) multiplication commutes with the norm

Uh, that last one is probably the easiest to check

I have no idea how to prove any of these three things by the way

but I'm 90% sure they're true

and that we can figure this out

I'm mostly this confident because we essentially proved that this is the only way for this to look like

I'm less confident now

I don't actually think the one after this is the sedenions

bolbteppa

Yeah the argument for $ij = - ji$ is interesting

It may fail because you're exploiting division algebra properties

Akiva Weinberger

6:01 PM

The sedenions aren't even normed, are they?

bolbteppa

They are not a normed division algebra i.e. zero divisors start existing

Akiva Weinberger

Wait

The thing I just defined can't be the octonions

'cause it's associative

Derp

(…so then what is the thing I just defined?)

Hm

What's $(e_0e_1e_2)^2$

if we assume associativity and also the things from before

Mike Miller

@AkivaWeinberger Suppose the convex polyhedron $C_G$ is noncompact. Then it contains some line $\ell$. To see this, choose a sequence of points $x_n \in C_G$ with $\|x_n\| = n$; passing to a subsequence we may arrange so that $x_n/n \to x\in S^{n-1}$. Because $C_G$ contains zero, it also contains every vector of the form $\sum \lambda_i x_i$ where $\sum |\lambda_i| \leq 1$. In particular, because $tx_n/n \to tx$, and $|t/n| \leq 1$ for large enough $n$, we see that $tx \in C_G$ for all $t$.

This vector $x$ has $x_n \leq 1$, because $x \in C_G$, but in fact because $\ell \subset C_G$ we have $tx_n \leq 1$ for all $t \in \Bbb R$. This is only possible if $x_n = 0$. Furthermore, $(gx)$ has the same property for all $g$; that is, $(gx)_n = 0$ for all $g$. So the subspace spanned by $gx$, as $g$ varies over all of $G$, is a $G$-invariant subspace contained in $\Bbb R^{n-1}$.

Therefore the $G$-action is reducible to an action on $V \oplus W$, where $W \subset \Bbb R^n$ contains the basis vector $e_n$. Write $C_{G,W} \subset W$ for the polyhedron determined by the action on $W$; then the polyhedron $C_G = V \times C_{G,W}$.

Akiva Weinberger

@bolbteppa Did I make a mistake or is it $1$

Mike Miller

It's clear from this description that the converse holds: $C_G$ is noncompact if and only if $G \subset O(n)$ lies in some $O(n-1) \times O(1)$ subgroup corresponding to a subspace $\Bbb R^{n-1} \subset \Bbb R^n$ containing $e_n$. Your conjecture was more or less correct.

Akiva Weinberger

6:06 PM

@Semiclassical and I found a lot of interesting things about this

@MikeMiller

Mike Miller

nice, not sure if you already got this property or not

Akiva Weinberger

Did you see our stuff with bipyramids and trapezohedra?

bolbteppa

$\{1,i,j,k,ij,ik,jk,ijk \} = \{1,i,j,k,k,-j,i,-1 \}$ right? So $(e_0 e_1 e_2)^2 = (ijk)^2 = (k^2)^2 = (-1)^2 = 1$ it's just the quaternions

Mike Miller

It looked like you were cooking up examples, and in particular coming up with some less regular shapes

Akiva Weinberger

@bolbteppa Yeah

Hm

and that satisfies the same multiplication rules

Hmmmm

OK hold on what's $(e_0e_1+e_2)(e_0e_1-e_2)$

Mike Miller

6:08 PM

My statement above is wrong. $C_G$ is noncompact if and only if $G$ is reducible, aka, does not lie in any $O(k) \times O(n-k)$ corresponding to $\langle e_n\rangle \subset \Bbb R^k \subset \Bbb R^n$

Akiva Weinberger

$-e_0e_1e_2+e_2e_0e_1$

…equals $0$

Wow

OK so if we want to avoid zero divisors

bolbteppa

$(e_0 e_1 + e_2)(e_0 e_1 - e_2) = (ij + k)(ij - k) = (k + k)(k - k) = 0$

Akiva Weinberger

@bolbteppa We don't know that $e_0e_1=e_2$ yet

But we do know that $e_0e_2=-e_2e_0$

And I think that forces us to have $(e_0e_1+e_2)(e_0e_1-e_2)=0$

meaning, if we want to avoid zero divisors,

that $e_0e_1$ has to equal $\pm e_2$

@MikeMiller Makes sense

Alright so hold on I want to get the octonions out of this

so let's break associativity

I know that $e_0e_1+e_2$ should have norm $\sqrt2$

(We're keeping the norm thing)

and $e_0e_1-e_2$ should also have norm $\sqrt2$

and also $e_0e_1$ should be a root of $-1$

Dunno if that's needed but it seems like it should happen

meaning $(e_0e_1+e_2)(e_0e_1-e_2)=e_2(e_0e_1)-(e_0e_1)e_2$ should have norm $2$

meaning $(e_0e_1)e_2=-e_2(e_0e_1)$

OK so what now

I'm now a lot less confident that this will lead us to the C–D construction

Akiva Weinberger

6:35 PM

OK maybe there's some way to prove $e_0(e_1e_2)=-(e_0e_1)e_2$

Nah I dunno

@bolbteppa What even is the claim with Cayley–Dickson algebras?

They're not normed.

Did they just go "Hey look this is a series of algebras, and the first four ($\Bbb R,\Bbb C,\Bbb H,\Bbb O$) are familiar"?

anakhro

Movement to use \mathfrak instead of \mathbb

Akiva Weinberger

Like, the Cayley–Dickson algebra of dimension $2^n$ is the unique algebra of that dimension such that… what?

I mean technically the matrix group $\Bbb R^{n\times n}$ is an algebra

as is the direct sum of any pair of algebras

bolbteppa

So basically, it makes sense to copy what you do going from $\mathbb{R}$ to $\mathbb{C}$ in combining pairs of real numbers into $a + ib$ where we added a square root of $-1$, $i^2 = - 1$, and used this to find a new multiplication and call the whole thing $\mathbb{C}$. We can find the conjugate $\overline{a+ib}$ by wanting the norm $(a+ib)(\overline{a+ib}) = a^2 + b^2$ to exist in $\mathbb{C}$.

Akiva Weinberger

Are the Cayley–Dickson algebras unique for not being matrix algebras or direct sums?

bolbteppa

Iterating this procedure for pairs of complex numbers after adding a second 'independent' root of $-1$ $j$, $j^2 = - 1$, where from
\begin{align}
(a,b) &= a + b j \\
&= (a_1 + i a_2) + (b_1 + i b_2) j \\
&= a_1 + a_2 i + b_1 j + b_2 (ij) \\
&= a_1 + a_2 i + b_1 j + b_2 k
\end{align}
we see we are forced to attach meaning to the product $ij$ which we just call $k$ for now, noting $ij = k$ implies $ik = - j$ and $k j = - i$.

Akiva Weinberger

6:49 PM

And then to make the norm thing work you make $ji=-ij$

bolbteppa

Actually writing out
\begin{align}
a_1^2 + a_2^2 + b_1^2 + b_2^2 &= (a_1 + a_2 i + b_1 j + b_2 k) (a_1 \pm a_2 i \pm b_1 j \pm b_2 k) \\
&= a_1^2 + a_1 (\pm a_2 i \pm b_1 j \pm b_2 k) \pm a_2^2 i^2 + a_2 i (a_1 \pm b_1 j \pm b_2 k) \\
&\pm b_1^2 j^2 + b_1 j (\pm a_1 \pm a_2 i \pm b_2 k) \pm b_2^2 k^2 + b_2 k (a_1 \pm a_2 i \pm b_1 j)
\end{align}
explicitly has given me all the other quaternion relations e.g. $ji = - k$ etc and all the signs

In this Baez says you can define CD as basically just adding a new root of minus one to the algebra you have, along with the relations in equation (1) on that page, but appreciating (1) from first principles idk yet

There's like a paragraph at the end motivating why it ends up non-associative for the octonions which looks cool trying to figure it out, but this approach above of adding a new $\sqrt{-1}$ and using the norm to constrian things isn't so bad

Akiva Weinberger

Why is Baez everywhere

bolbteppa

He turns things into children's pictures :p

Quaternions, octonions, projective planes, Hurwitz, Clifford algebras, projective planes, exceptional lie algebras, all unified, that's miraculous in less than 50 pages all with good explanations

user76284

7:07 PM

Does anyone know why ChatJax is not enabled by default?

It's annoying to have to manually activate it every time I enter chat. Am I doing it wrong?

MatheinBoulomenos

@AkivaWeinberger there are other algebras, e.g. the algebra of upper triangular matrices which is 3-dimensional non-commutative and not a direct sum

Oh I missed the dimension $2^n$ thing

Tobias Kildetoft

So we are asking about dimension $2^n$ indecomposable algebras which are not matrix algebras?

there should be plenty of those I think

Rithaniel

So, I know Quaternions are non-commutative and Octonions are not even associative, but do you continually lose properties as you progress upward?

Tobias Kildetoft

@Rithaniel What do you mean by "progress upwards"?

Ryan Unger

7:23 PM

@bolbteppa hot take, if your math can be turned into children's pcitures it's bound to be boring :P

Rithaniel

Informally, the "$2^n-$ions" as $n$ increases.

Akiva Weinberger

@MatheinBoulomenos @TobiasKildetoft Hence my question: what makes the Cayley–Dickson algebras special

@Rithaniel Yeah

Tobias Kildetoft

@AkivaWeinberger I am not even sure what those are

MatheinBoulomenos

@TobiasKildetoft can you classify indecomposable 4-d algebras over $\Bbb C$ with quivers?

Akiva Weinberger

Various weaker forms of associativity

Tobias Kildetoft

7:25 PM

@MatheinBoulomenos up to Morita-equivalence yes, I think

Ryan Unger

@AkivaWeinberger um, why

Akiva Weinberger

@Rithaniel For example: the octonions are alternative, which means $(xy)y=x(yy)$ and $(xx)y=x(yy)$

Alessandro Codenotti

Did you end up figuring the minimal conditions for the convex thing to work? @Ryan

Ryan Unger

no

Akiva Weinberger

Another definition of alternative is: the subalgebra generated by any two elements is associative

The 16-dimensional sedenions are not alternative

Ryan Unger

7:27 PM

I suggested to my prof to just change it to R^n

Rithaniel

What about $(xy)z$?

Ryan Unger

he wrote NVS in the draft

MatheinBoulomenos

one thing I like about alternative algebras is the Artin-Zorn theorem and that it is equivalent to a statement in projective geometry (as is Wedderburn's little theorem)

Akiva Weinberger

@Rithaniel $(xy)z=x(yz)$ is associativity, which the octonions don't have

Alessandro Codenotti

What are the standard examples of spaces which are Banach but not uniformly convex? $L^1$ and?

Akiva Weinberger

7:28 PM

Exercise: show that if $(xy)y=x(yy)$ and $(xx)y=x(xy)$ then $(xy)x=x(yx)$

Rithaniel

Yeah, but what does $(xy)z$ equal if not $x(yz)$?

Ryan Unger

@AlessandroCodenotti you've got me there

L^\infty

Akiva Weinberger

@Rithaniel I think, in the octonions, if $\{e_1,\dots,e_7\}$ are a basis for the imaginary octonions

meaning ${e_i}^2=-1$

Alessandro Codenotti

@RyanUnger fair enough, the dual of a space which isn't uniformly convex won't be either

Akiva Weinberger

then, if $e_ie_j\ne e_k$, you have $(e_ie_j)e_k=-e_i(e_je_k)$.

Ryan Unger

7:37 PM

ok people

the othe day something was said here

and now it showed up in my research

freaky

Alessandro Codenotti

Hmm so $c_{00}$ is convex in $\ell^1$, and $\left(1+\frac1{n^2}\right)_{n\in\mathbb N}$ should have distance $1$ from $c_{00}$, which is not attained

Ryan Unger

but there was no resolution

MatheinBoulomenos

define $[a,b,c]=(ab)c-a(bc)$ which is trilinear. Then $(xy)y=x(yy)$ means $[a,b,b]=0$ and $(xx)y=x(xy)$ means $[a,a,b]=0$. We get $[a,b,a]=[a,b,a]+[a,b,b]=[a,b,a+b]=[a,b,a+b]+[a,a,a+b]=[a,a+b,a+b]=0$

Ryan Unger

if $(X,d)$ is a compact metric space and $\varepsilon>0$, can $X$ contain infinitely many disjoint $\varepsilon$-balls

I am NOT asking for a cover

Rithaniel

Oh, this $[a,b,c]$. Is that what you call the "associator?"

Alessandro Codenotti

7:40 PM

Does my counterexample look good to you? @Ryan

Ryan Unger

I'm stuck in analysis counterexample hell

@AlessandroCodenotti $c_{00}$?

Alessandro Codenotti

No it doesn't work sorry, it has distance $0$ from $c_{00}$ and is indeed in the closure

Oh of course, $c_{00}$ is dense in $\ell^1$

MatheinBoulomenos

@RyanUnger choose an element from each ball, this has no convergent subsequence?

I must be missing something

Ryan Unger

hmm yeah that works

MatheinBoulomenos

@Rithaniel right

Ryan Unger

7:43 PM

thanks

Rithaniel

I've seen the term before. Now I know what it is.

MatheinBoulomenos

np

Alessandro Codenotti

Suppose that $X$ is a metric space which can be covered by finitely many $\varepsilon$-balls. Is it possible that there also are infinite covers by $\varepsilon$-balls with no finite subcovers?

@Ryan consider the subspace of $\ell^1$ made of sequences with zero even terms and finitely many nonzero terms. This is clearly convex. The sequence whose 0th term is $1$ and whose $2k+1$ term is $\frac1{(2k+1)^2}$ has distance $1$ from this set, but I don't think it is attained

cattt

8:15 PM

1

Q: Numbers which cannot be formed

We are given two numbers $a,b$ such that $a<b$. Now we have a set $\{a,a+1,a+2,\ldots, b\}$ (all number between a and b including them). Then, we have to find how many numbers cannot be formed from the above set. The only operation allowed on the set elements is addition. Note : We can add the...

number-theory algorithms greatest-common-divisor

Can Anyone help me in this

Alessandro Codenotti

@Ryan two questions: does the counterexample above work? And do you mind if I ask a question on MSE concerning the minimal adjectives to put in front of "normed vector space" to get existence (and uniqueness) of projections to closed convex sets?

bolbteppa

@RyanUnger children's pictures in the sense of tangent spaces to isometry groups projective spaces over quaternions/octonions etc..., rather than abstract root systems :p

@Rithaniel I think you get 'power associativity' for higher and higher powers as you go higher and higher i.e. lose power-associativity for lower powers as you iterate

Cayley–Dickson construction

In mathematics, the Cayley–Dickson construction, named after Arthur Cayley and Leonard Eugene Dickson, produces a sequence of algebras over the field of real numbers, each with twice the dimension of the previous one. The algebras produced by this process are known as Cayley–Dickson algebras, for example complex numbers, quaternions, and octonions. These examples are useful composition algebras frequently applied in mathematical physics. The Cayley–Dickson construction defines a new algebra similar to the direct sum of an algebra with itself, with multiplication defined in a specific way (different...

Ryan Unger

9:11 PM

@AlessandroCodenotti please ask

I’ll look at it when I get home

@ÉricoMeloSilva Rod is a very cool dude

I might have to do GR

ÍgjøgnumMeg

9:28 PM

@Alessandro was just reading something and the Schröder-Bernstein property popped up, weird that we were talking about it just the other day lol

Érico Melo Silva

9:44 PM

@RyanUnger was he Luk's advisor

Ryan Unger

@ÉricoMeloSilva Jonathan? Yes

Érico Melo Silva

oh cool, i met him at stanford

Ryan Unger

he told me to skip sinai's class haha

Érico Melo Silva

lol

Ryan Unger

He said Dafaermos might be teaching GR in the fall

Érico Melo Silva

9:47 PM

oh i heard good things about dafermos from soug

Ryan Unger

of course, he's Greek :P

Érico Melo Silva

i think souganidis said he was his babysitter or something weird

Ryan Unger

wot

Érico Melo Silva

idk if it was a joke or if he was serious

knowing souganidis it was probably him making a joke

Ryan Unger

Rod did make fun of me for specifying "hyperbolic GR"

he interrogated me on who told me there's another kind

Érico Melo Silva

9:52 PM

lol

idk shit about gr

Ryan Unger

I said Rick Schoen

@ÉricoMeloSilva did you take a hyperbolic PDE course from Schlag?

Érico Melo Silva

no he never gave one

oh wait maybe there was a minicourse like 4 years ago that i did

Ryan Unger

you knew enough analysis 4 years ago?

Érico Melo Silva

no i think i played video games on my laptop for it

in the back of the room

Ryan Unger

lmao

Érico Melo Silva

9:58 PM

i made sure i paid attention for precisely one lecture

more than that was a no go

bolbteppa

Those rules in (1) in the link above make perfect sense (I think), basically just summarizing $1l = l1, il = - li, jl = - lj, kl = - lk$ by saying $a l = l \overline{a}$

Cosinux

Hi. Given a graph, how can I find out whether there is a connected path in the graph, that visits each node only once?

Alessandro Codenotti

@ÍgjøgnumMeg in which context?

Cosinux

I've found an article about Eulerian path on Wikipedia, but I've read that visiting a node multiple times is allowed in Eulerian path, so it doesn't seem to be what I need.

ÍgjøgnumMeg

@Alessandro why $\Bbb C_p \cong \Bbb C$

Alessandro Codenotti

10:09 PM

Because algebraically closed fields are classified by cardinality and characteristic

By a result of Steinitz I think

ÍgjøgnumMeg

nise

Ryan Unger

@AlessandroCodenotti is that sequence in the closure of the convex set

Alessandro Codenotti

No