Mathematics - 2019-09-05 (page 1 of 2)

Akiva Weinberger

00:00

Also after learning Japanese the /k/ in /ki/ has gotten really fronted

Almost /kçi/

MatheinBoulomenos

fun fact: due to mispronouncing double consonants, beginners in Italian frequently accidentally ask "how much anuses do you have?" instead of "how old are you?"

Akiva Weinberger

Oh also /hi/ is sometimes [çi]?

Leaky Nun

yeah think of "huge" (/hj/)

Akiva Weinberger

Like "here" became [çiɚ] (instead of [hiɚ]) for me 'cause of Japanese

Leaky Nun

@MatheinBoulomenos same with Spanish (ano vs año)

Akiva Weinberger

00:02

Double consonants are fun

Shame Modern Hebrew lost them

Leaky Nun

modern Hebrew also got rid of uvulars

Akiva Weinberger

? The chet is uvular

Leaky Nun

and those strong plosives what were they called

Akiva Weinberger

It lost pharyngials

"Yalla" got borrowed from Arabic and the double l lost its gemination

And the h in "ahlan" became silent

So an informal "hi" in Hebrew is "alan"

(Still spelled though)

@LeakyNun Emphatics?

Leaky Nun

yeah

Akiva Weinberger

00:05

I like implosives. English should use them

/ˈɪŋɠɪʃ/

MatheinBoulomenos

I think it's weird that there are voiceless nasals

I can't imagine pronouncing a voiceless nasal

Akiva Weinberger

Mhaybe

At some point I ought to learn how to pronounce Vietnamese

I only worry that I would go mad in the effort

Leaky Nun

it's mainly the tones right

Akiva Weinberger

Also the consonants

Written Vietnamese is quite something as well

Fun fact: instead of a space between each word, it has a space between each syllable

Leaky Nun

what's wrong with the consonants

retroflex?

Akiva Weinberger

00:07

/c/ for one

Can't remember the others

Leaky Nun

it ain't that hard

Akiva Weinberger

I i ma gine a space be tween each sy lla ble makes sense for a lan guage that used to have no spaces and a cha rac ter for each sy lla ble

Pinyin should do that

or maybe it shouldn't I dunno

O ma e wa mo o shi n de i ru

MatheinBoulomenos

have you ever thought of making a conlang?

Akiva Weinberger

Too much effort

Also too much effort to learn other conlangs

so it's an art form that I can't even really consume

Also I would pronounce it with an American accent

(I can't do accents)

(I mean, I can't do accents well; I can definitely do them confidently)

At one point I tried to learn how Turkish pronunciation works

Apparently the 'r' is really weird

It's almost like an 'sh'

Teshukula"sh"

"shrzhr"

Luckily Japanese pronunciation is about as easy as they come

There are nuances but they're not too hard

If I was raised Japanese and had to learn English I would have such a hard time

My Spanish pronunciation is atrocious

mainly 'cause I wasn't conscious of it going in

MatheinBoulomenos

my Italian pronuncation is terrible

Leaky Nun

00:16

I hate initial consonant clusters in the wrong order

Akiva Weinberger

and so I thought that the only thing I needed to do was learn to roll my 'r's

MatheinBoulomenos

mainly because I have difficulties with the trilled r

user193319

During my class on Lie Groups, my professor parenthetically remarked that the (complete?) classification of finite dimensional algebras over $\Bbb{R}$ was obtained via topological k-theory and that no algebraic proof is known (note, I might have misquoted him). Does this sound familiar to anyone? Does anyone know of any resources on this? I'd like to read more about it.

Akiva Weinberger

"Tres tristes tigres..."

Leaky Nun

@MatheinBoulomenos who can pronounce /ks-/

MatheinBoulomenos

00:17

@user193319 I think you mean division algebras

not general algebras

Akiva Weinberger

@user193319 Something like, the reals ($\Bbb R$), complex numbers ($\Bbb C$), quaternions ($\Bbb H$), and octonions ($\Bbb O$)?

(Complexes?)

MatheinBoulomenos

general finite dimensional algebras sounds hopeless to me

Akiva Weinberger

(Complices?)

Leaky Nun

complexes

user193319

Possibly.

And the proof uses topological k-theory?

Akiva Weinberger

00:18

I mean technically the set of $n\times n$ matrices is an algebra

but not a division algebra

@user193319 Dunno

MatheinBoulomenos

yes, I heard this before

iirc it's due to Milnor

Akiva Weinberger

What's the property they satisfy?

I've always heard it as "normed division algebras"

MatheinBoulomenos

simple algebra

Akiva Weinberger

Can we weaken that to "division algebras with units"?

MatheinBoulomenos

@AkivaWeinberger you can drop the normed, but it makes the problem harder

user193319

00:20

13

Q: finite dimensional real division algebras

A celebrated theorem of Milnor and Kervaire asserts that any finite dimensional division algebra over the real numbers has dimension 1,2,4 or 8. This result is established using methods from algebraic topology, such as K-Theory. Now for any given natural number $n$ the existence of such an alge...

Akiva Weinberger

$\Bbb C$ with $z\times_Aw:=z\bar w$ technically works

but doesn't have a unit

MatheinBoulomenos

@AkivaWeinberger division algebra implies unital by definition I think

user193319

Very cool. I'd like to learn my about topological k-theory because I would eventually like to learn operator k-theory.

MatheinBoulomenos

or maybe not

I'm not sure

Akiva Weinberger

I thought it was just that left- and right-multiplication by a nonzero constant are bijections

so basically no zero divisors

(in the finite-dimensional case)

@user193319 Oh yeah we want finite dimensional or else you can do the set of rational functions I think

Polynomial divided by polynomial

Don't know if that was mentioned

Oh you said that

Never mind

user193319

00:23

Operator K-theory

In mathematics, operator K-theory is a noncommutative analogue of topological K-theory for Banach algebras with most applications used for C*-algebras. == Overview == Operator K-theory resembles topological K-theory more than algebraic K-theory. In particular, a Bott periodicity theorem holds. So there are only two K-groups, namely K0, which is equal to algebraic K0, and K1. As a consequence of the periodicity theorem, it satisfies excision. This means that it associates to an extension of C*-algebras to a long exact sequence, which, by Bott periodicity, reduces to an exact cyclic 6-term-sequence...

Operator algebras are so damn cool.

Leaky Nun

@AkivaWeinberger @MatheinBoulomenos til the half-laplacian operator $D$ with $DD\psi = -\Delta \psi$

MatheinBoulomenos

something more artihmetic about divison algebras: we have a complete classification of associative divison algebras over $\Bbb Q$, but it's quite complicated

genescuba

00:46

Every graph that has an induced subgraph $C_n$ where $n > 5$ has an induced subgraph $P_4$.

This is true because essentially you can just pick four vertices in the $C_n$ and the induced subgraph will give you a $P_4$ right?

Akiva Weinberger

Which one's $P_4$ again?

genescuba

a path is just a walk that does not resvisit vertices

so essentially $P_4$ is just visiting 4 vertices

Akiva Weinberger

Oh OK

so a line

Yeah makes sense then

genescuba

The contrapositive of this Let $G$ be a connected simple graph. If $G$ has an odd cycle then there exists an
induced subgraph $P_{4}$ or $C_{3}$ .

is Let $G$ be a connected simple graph. If $G$ has no induced subgraph $P_4$ or $C_3$ then it does not have any odd cycles

is this correct?

Akiva Weinberger

01:02

Yeah

I think the first form is more useful though

Either the odd cycle has length 3 or it has length at least 5

genescuba

Thank You @AkivaWeinberger

Akiva Weinberger

02:48

Almost all stories are post-apocalyptic stories from the dinosaurs' perspective

Jacksoja

that is so deep akiva

@AkivaWeinberger thank you for sharing that

Akiva Weinberger

03:01

It's about as deep as it deserved to be

krauser126

03:13

I have a set R that contains all numbers n = 4k + 1 given k in an integer > 0 and a set S that contains all numbers n = 8k + 1. Trying to prove that S is a subset of R. Could use a hint or two :)

Jacksoja

to show that S is a subset of R , is to pick an arbitrary element of S and show that it is also in R, for all elements of S

in other words, if n can be written as 8k+1 when k is an integer, it can also be written as m= 4k' +1

you can maybe start with all multiples of 4 are also multiples of 2, just to get an intuition @krauser126

Nicholas Roberts

@krauser126 If $n = 8k + 1$ for some positive integer $k$, do you see a way re-write the expression $8k + 1$ as $4m + 1$ where $m$ is another positive integer?

krauser126

03:31

I guess let m = 2k ? @NicholasRoberts And from there I can show that the expressions 4m+1 is equal to 4k + 1? Sorry I'm kind of new to proofs haha. I understand why it makes sense but just reasoning thru each step seems weird to me

Nicholas Roberts

@krauser126 exactly! Good job

To gain entrance into the set R, it needs to be of the form 4m + 1 for a positive integer m. Surely, 2k is a positive integer. So x = 8k + 1 = 4(2k) + 1 is an element of R

But 4m+1 is not equal to 4k + 1, 4m + 1 is an element of your set R with m = 2k

Kumar Nilesh

04:06

If a and b are independent vectors and ca+(1-c)b=0 then it implies c=0 and 1-c=0 \\or\\ c=o or (1-c)=0. There is confusion between and and *or * in the final solution. Please help.

genescuba

04:19

Let G be a simple graph in which the degree of each vertex is at least 3. Show that G has a cycle of even length. (Hint: Consider a maximal path.)

Can anyone help me with this problem, I am not sure how to get the cycle of even length

Ted Shifrin

04:47

@KumarNilesh Look at the definition of linear independence. If $c_1v_1+\dots+c_kv_k = 0$ and $v_1,\dots,v_k$ are linearly independent, what must be the case with $c_1,\dots,c_k$?

genescuba

@TedShifrin any idea on how to show the cycle of even length

krauser126

If A is a set consisting of m = 4 - x such that 3 divides x and B is a set consisting of m = 10 - 3x such that x >= 2 , how can I show that A = B ?

I cant seem to rewrite m so that both expressions look the same

Ted Shifrin

@genescuba: I have never studied or taught graph theory. Obviously it depends on the fact that the degree of each vertex is 3 or more. What does their hint tell you to do?

@krauser126: For starters, don't use the same letters.

How can you write every $m\in A$?

Jacksoja

@TedShifrin Hi Ted

Ted Shifrin

Hi @Jacksoja

Jacksoja

04:53

Long time no see , how are you?

Ted Shifrin

I'm doing pretty well, thanks, and you?

Jacksoja

that is good to hear, am doing also thanks

well*

krauser126

@TedShifrin I guess we could rewrite every element as 4 - 3p where p is any positive integers?

Ted Shifrin

Well, you never said the $x$ there was positive. Is it?

$p$ is any positive integer, not all.

krauser126

Yes my mistake.

Forgot to mention x is a natural number. So no negative integers.

Ted Shifrin

05:01

OK. So now how do you relate $4-3p$ and $10-3x$?

krauser126

One is 6 more than the other?

Ted Shifrin

Not so.

Note how important it is to have different letters $p$ and $x$ there.

The question is: If $m=4-3p$ with $p\ge 0$, can you choose $x\ge 2$ so that $m=10-3x$? And vice versa ... That's how you show two sets are equal.

krauser126

05:20

I get that part. I just dont get how I can justify that properly. I feel like that just entails going through every element starting with the base cases and showing that all elements are the same

Ted Shifrin

No, you answer my question "can you choose $x$?" very specifically. You don't need any induction or listing or anything. Showing that $m=4-3p$ can be written as $m=10-3x$ shows that $A\subset B$. Then you do the reverse argument.

Are you taking a beginning course on how to write proofs?

krauser126

Yea I am. Much of this is new to me

Ted Shifrin

Well, I just told you one of the fundamental things. The way you show $A\subset B$ is to show that any element $x\in A$ also lies in $B$. Here you're talking about $m\in A$. What algebra can you do with $4-3p$ to rewrite it as $10-3q$?

krauser126

Is it allowed to let q = p + 2 since if we look at both base cases where q = 2 and p = 0 , that holds true. And since p and q must be integers ≥ 0 and 2 respectively, we can rewrite 10 - 3q as 10 - 3(p + 2)

Ted Shifrin

I don't know why you're worrying about induction. You don't need an inductive proof for this problem. You just show $A\subset B$ .

And then you show $B\subset A$. If $p\ge 0$, then $p+2\ge 2$, and so $4-3p=10-3(p+2)=10-3x$ for $x=p+2$ and $x\ge 2$.

Now show $B\subset A$. I'm disappearing. Have fun :)

Rithaniel

08:03

Suppose you have a ring $R$. Is there a name for "the largest subring of $R$ which is an integral domain?" Is there perhaps a rule for finding what that subring is?

Math Geek

09:27

online anyone??

Leaky Nun

10:39

@Rithaniel consider $R=k[X,Y]/(XY)$. Then the subring generated by $X$ and the subring generated by $Y$ are both integral domains. Which one of them is the largest?

Rithaniel

Hmmm, good point. So I guess it would be better to ask for "maximal" subrings which are integral domains.

By which I mean any subring which contains them would have to not be integral domains.

Leaky Nun

now can you prove that those exist?

Rithaniel

Maybe, but I don't have an approach off-hand.

I will put some thought into it.

Leaky Nun

10:57

@Rithaniel do you want a hint?

Rithaniel

I wouldn't necessarily object to one. It's not for an assignment or anything.

Leaky Nun

Zorn's lemma

Rithaniel

Well, looking up the actual statement of Zorn's lemma, yeah, it seems to follow almost immediately.

Just have to show that you have chains of subrings ordered by inclusion (which is trivial) and that these chains have upper bounds.

Though, perhaps you could construct a ring with a chain of subrings lacking an upper bound.

$F_2\times \bigcup_{n\in\mathbb{N}}F_{5^n}$, perhaps?

Though, that has an upper bound.

Okay, so perhaps you could take the union of all integral domain subrings in a chain of subrings of the parent ring, and have that be upper bound of the chain?

Since they're ordered by inclusion, the union of these subrings should still be rings.

Leaky Nun

11:29

@Rithaniel 1. what if the chain is empty? 2. why is the union an integral domain?

@MatheinBoulomenos hi

MatheinBoulomenos

@LeakyNun hi

Rithaniel

Well, the chain is definitely not empty, because the trivial ring is an integral domain.
The union is an integral domain because (I probably need to be most careful here, but in the next sentence, I'm not being very careful) if it weren't an integral domain, then one of the subrings being unioned would have to not be an integral domain, which is a contradiction.

genescuba

Can anybody help with a graph theory problem?

Let G be a simple graph in which the degree of each vertex is at least 3. Show that G has a cycle of even length. (Hint: Consider a maximal path.)

ÍgjøgnumMeg

12:16

Hey @Mathein, weißt du übrigens ab wann man immatrikulieren darf? :)

user193319

Okay. I see why $SL_n(\Bbb{R})$ is closed, because $SL_n(\Bbb{R}) = \ker \det = \det^{-1} (\{1\})$, where $\det$ is a continuous function from $GL_n(\Bbb{R}) \to \Bbb{R}^\times$. But my professor said it is also open. Why is that so?

Leaky Nun

12:30

@user193319 it isn't open

@AkivaWeinberger kapwing.com/videos/5d70ff1c2b9b4c001469a4ba

here's something I made

Akiva Weinberger

Neat

Skies burn

hi there. sanity check: math.stackexchange.com/a/396303/476584

wait

Nvm

Akiva Weinberger

@genescuba Maybe try to break the largest cycle into pieces

Like say you have a cycle that's $a_1-a_2-a_3-a_4-a_5-a_1$

genescuba

@AkivaWeinberger Basically my reasoning so far is that there is some maximal path from some start $v_1$ and end $v_k$.

$v_k$ has at least degree 3 so it has at least 3 neighbors. Since our path is maximal we have those 3 neighbors on the path somewhere

I'm not sure how to get even cycle from here tho

Akiva Weinberger

One of which is $v_{k-1}$

genescuba

12:44

Exactly, one of which is $v_{k-1}$

Akiva Weinberger

OK so hm

The other two are $v_i$ and $v_j$

with let's assume $i<j$

genescuba

Yup, makes sense

Maybe we do cases here then. Like assume i is odd/even?

Akiva Weinberger

I dunno man but let me think out loud

so $v_i,v_{i+1},\dots,v_{j-1},v_j,v_k,v_i$ is a cycle

and $v_j,v_{j+1},\dots,v_{k-1},v_k,v_j$ is a cycle

and $v_i,v_{i+1},\dots,v_{k-1},v_k,v_i$ is a cycle

Can those all be odd? Or must one of them be even

The first one's odd if $i$ and $j$ have different parity

genescuba

Well they should all be even according to the statement right?

what do you mean by different parity

Akiva Weinberger

@genescuba ? We just want to find at least one even one

so I'm trying to argue by contradiction, assuming they're all odd

genescuba

12:50

Sorry you are right

Akiva Weinberger

and deriving a contradiction

genescuba

I see

Akiva Weinberger

@genescuba One's even and one's odd

genescuba

makes sense

Akiva Weinberger

Or hold on let's draw something

1010011010

12:54

I have a question about residue / contour integration

I have a function which I'm integrating over a finite interval say [-a,a]

Akiva Weinberger

1010011010

The complete expression under the integral operator is say F(x)/(x-b)

Akiva Weinberger

@genescuba Something like that? ^

Not drawing all the vertices on that path

Skies burn

Why does he use irreducibility of V to deduce that the kernel of f_H is trivial? Surely this simply follows from <u,u>\geq 0 with equality only when u=0?

1010011010

The explicit form of F(x) is not known

Skies burn

12:55

2

A: Invariant hermitian forms and irreducible representations

I think it is most useful here to interpret any Hermitian form$~h$, which I'm supposing conjugate-linear in its first argument, as associated to a conjugate linear mapping $f_h:V\to V^*$ defined by $f_h(v)=h(v,\cdotp)\in V^*$. Now a first thing to check is that $H$, which is nonzero because $H(v,...

genescuba

Yup that looks correct @AkivaWeinberger

Akiva Weinberger

so let's say there's $a$ edges between $v_i$ and $v_j$ and $b$ edges between $v_j$ and $v_k$

1010011010

However some things can be said about it: for certain it is only defined between -a and a

Akiva Weinberger

(We can count edges because a cycle will have the same number of edges as vertices)

So one of our cycles has length $a+2$. Another has length $b+1$, and the third has length $a+b+1$

If $a+2$ is odd then $a$ is odd

If $b+1$ is odd then $b$ is even

1010011010

Now, b is between -a and a, variables a,b are real and F(x) has no additional poles

Akiva Weinberger

12:57

If $a+b+1$ is odd then $a$ and $b$ are either both odd or both even

These can't all be true at the same time

1010011010

Is there any hope of interpreting this integral in the principal value sense?

Akiva Weinberger

@1010011010 Maybe the limit of $\int_{-a}^{b-\epsilon}+\int_{b+\epsilon}^a$?

1010011010

@AkivaWeinberger Ok, but then what?

The function F(x) is not explicitly known, although I have a 4 page DIN A4 expansion of it lying around the office

Akiva Weinberger

Limit as $\epsilon$ goes to 0?

1010011010

Yes ok, but the individual integrals don't have a known value so there's no hope of solving it that way

Secretly i was hoping I could somehow perhaps take a contour along the real axis, then to $\pm \infty i$ or something

And hence I'm here investigating what steps and considerations I should take in order to find the value of this integral

Akiva Weinberger

13:01

I thought you said $F$ was only defined on that interval

1010011010

Mmh, good point

Akiva Weinberger

It's also defined on the complex plane?

1010011010

It's given by an integral equation over variables that I allow to be complex

Would that imply that there's some kind of complex extension of it?

genescuba

Ah I see, this is a very interesting approach @AkivaWeinberger

Akiva Weinberger

I dunno @1010011010

Sorry

1010011010

13:03

Np

AbhigyanC

13:23

Hey everyone

Need a little help

does epsilon delta give the value of the limit or does it prove the existence of a limit which we already know

As, I found a certain limit in multivariable calculus using polar coordinates, but I didn't prove its existence

Tobias Kildetoft

epsilon-delta is part of the definition

So what it "does" depends on what you do with it

AbhigyanC

@TobiasKildetoft In what sense?

Like, is it possible to find the limit using epsilon delta alone?

Tobias Kildetoft

Well, what do you mean by epsilon-delta doing something?

how do you "use" epsilon-delta?

AbhigyanC

@TobiasKildetoft The definition, we use the definition

To prove the existence of an already known limit

Tobias Kildetoft

sure, that shows you that the limit is whatever you put into the definition

AbhigyanC

13:27

That's all I know to use epsilon delta definition for :(

@TobiasKildetoft So, my question is, is it possible to find the value of the limit L using the epsilon delta definition and the function alone?

Or, is it not possible to do this?

Tobias Kildetoft

sure, it is possible

but it may require some good ideas along the way

AbhigyanC

@TobiasKildetoft Could you link me to an example or share some content?

Tobias Kildetoft

I don't have any on hand

AbhigyanC

Could you try to do this one?

As x approaches zero?

correction

sin(sqrt(x^2+y^2))/(x^2+y^2)

Or just point me in the right way

Tobias Kildetoft

you mean as a function of $y$?

AbhigyanC

13:32

Nope, correction, (x,y)->(0,0)

My mistake

Semiclassical

13:44

Since that function is of the form f(sqrt(x^2+y^2)), you should only need to think of this as a one-variable limit

AbhigyanC

Cool... So, that means I'm using polar coordinates :D

As I wrote above

As, sqrt(x^2+y^2) is basically r. My point is not to just find the limit, but to find it using the epsilon-delta definition of the limit

Lelouch

Just to make sure I understood Galois Theory properly, let $P$ be an finite galois extension of $F$, and let $G = \operatorname{Gal}(P/F)$. For each subgroup $H$ of $G$, there exists one and only one field $P < Q < F$ such that $\operatorname{Gal}(P/Q) = H$ right ? (take the elements of $F$ which are invariant under the elements of $H$. This forms a field, call this $Q'$, so $H$ is a subgroup of $H' = \operatorname{Gal}(P/Q)$. Now, if $H \neq H'$, then pick an element $h \in H'$]

Semiclassical

13:59

This is probably elementary but: I’m looking at a text that defines a “Hilbert space with unitary element” as a pair $(H,\mathbf{1})$ where $H$ is a Hilbert space containing an element $\mathbf{1}$ such that $\langle \mathbf{1},\mathbf{1}\rangle=1$.

Is the point just that a given Hilbert space will contain multiple such unitary elements, so pick one for specificity?

Tobias Kildetoft

@Semiclassical I would guess so. And then you require all maps to preserve that unitary element

Semiclassical

hmm. They don’t have anything to the latter effect, at least not in that definition

Tobias Kildetoft

then I am not sure what the point is

Semiclassical

At that point in the book their immediate interest is in the map $E(\mathbf{x})=\langle \mathbf{1},\mathbf{x}\rangle$

As defining “expectation of x” in this Hilbert space. (They go on to demand that H have a lattice structure so that you can interpret the vectors as random variables)

So I guess it’s just to match with the fact that you always have 1 as a constant real r.v.

This approach seems rather boutique tho

RScrlli

15:36

someone could please explain me why $$\int u d\delta_y=\int u(x) \delta_y(dx)$$?

where $\delta_y$ is the dirac measure of $y$

I am curious about the part implying the dirac measure, I assume that $d\delta_y=\delta_y(dx)$ must hold, but I am unable to see why

Semiclassical

@RScrlli For comparison, Wikipedia has this in its page for Dirac measure (with y and x swapped to mimic yours): $$\int_X f(x)\,d\delta_y(x)=\int_X f(x)\delta_y(x)\,dx=f(y)$$

RScrlli

yes, I've seen this on Wikipedia, but I am trying to figure out why does this hold

it may be something very trivial though

Semiclassical

based on this answer, it may just be notational: math.stackexchange.com/a/2085858/137524

Pretty unhelpful notation if so tho

RScrlli

15:52

ohh I get it now, I was struggling to understand what was the meaning behind that operation, as you said it seems to be simply notational

thanks! @Semiclassical

Ted Shifrin

@RScrlli: Not that I'm adding anything, but if $\mu$ is a measure, then people write integrals $\int u d\mu$, so the LHS is doing that with the Dirac measure. On the right, they're writing this as an integral with the Dirac delta "function" $\delta_y$. Namely, you integrate $u\cdot \delta_y$ with respect to the usual $dx$.

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