Mathematics - 2016-12-05 (page 2 of 6)

Null

9:01 AM

i really don't understand $f\in\mathbb{R}^2/U$ then. $U=\{\frac{x}{2}|x\in\mathbb{R}\}$

arctic tern

what?

Null

what is a generic element of a quotient space? :s

arctic tern

Perhaps you want to understand what the elements of the quotient space $\Bbb R^2/U$ look like, where $U$ is the line $y=x/2$, i.e. the subspace $\{(x,x/2):x\in\Bbb R\}$, i.e. the span of $(1,1/2)$.

An element of a quotient space $V/U$ looks like $v+U$.

Tobias Kildetoft

@Null First, you need an actual subspace. The $U$ there is not a subset

Null

@TobiasKildetoft with pointwise addition and scalarmultiplication it is.

arctic tern

9:06 AM

If $U$ is a line through the origin in Euclidean space, then $v+U$ is the image of $U$ under the map $f(x)=x+v$ which is just translation in the direction of $v$. But translating a line just yields a parallel line. Indeed, the cosets of $U$ are precisely the collection of all lines parallel to it.

Tobias Kildetoft

@Null How you define addition and multiplication does not change whether it is a subset

arctic tern

@Null You wrote $\{x/2:x\in\Bbb R\}$, which equals $\Bbb R$ itself, which is not a subset of $\Bbb R^2$

Null

i see

this is defintly not what I meant^^

@arctictern so the span of a non-0-element of the quotient space is simply the quotient space?

arctic tern

In your case, $U$ is one-dimensional and $\Bbb R^2$ is two-dimensional, so the quotient space $\Bbb R^2/U$ is also one-dimensional. In any one-dimensional vector space, the span of any nonzero vector is the whole thing.

Null

mmh, so a quotient space is $n$-dimensional, if the subspace we modulo by is $n$-dimensional?

arctic tern

9:11 AM

no

If $W$ is a subspace of $V$ then $\dim(V/W)+\dim(W)=\dim(V)$.

(-:

Tobias Kildetoft

Heh, nice

This formulation is also nice because it is the "shadow" of a stronger statement (about bases)

Null

is a quotient space a special vectorspace?

Tobias Kildetoft

Hmm, so the official abbreviation for Algebras and Representation Theory is Algebr. Represent. Theory. That just does not look very good to me

arctic tern

@Null what do you mean?

Null

@arctictern that a quotient space has to satisfie the same axioms as a vectorspace, i.e. addition has to be closed

arctic tern

9:22 AM

a quotient space $V/W$, where $W$ is a vector subspace of $V$, is also a vector space, yes

Null

@arctictern but 1+1 doesn't have to be 2 in every vectorspace or? simplest example would be $\mathbb{Z}/ 2\mathbb{Z}$. But 2 still makes sense if we say $2\equiv 0$?

arctic tern

what does the symbol 2 refer to if not 1+1?

also the elements of vector spaces are not generally numbers

Null

@arctictern but a vectorspace over R has numbers as scalars or?

arctic tern

yes a vector space over R has real numbers as scalars

Null

i think i have big knowledge holes already...

Sophie

9:39 AM

you're a topologist?

Fargle

rimshot

Null

11:18 AM

@NaCl i now understand your problems with math haha

DHMO

Hi

Null

@DHMO hi, how are you?

DHMO

fine

Null

that is good

Null

11:42 AM

there exists a vectorspace over $\mathbb{R}$ with exactly 13 elements: false, any vectorspace over $\mathbb{R}$ has infinite elements?

(or 1, with 0 as the only elements)

Tobias Kildetoft

11:54 AM

@Null Right

meow-mix

12:08 PM

i don't know why, but i always find $0/0$ questions pretty nonsensical from a field standpoint; $0$ has no [multiplicative] inverse, and infinity isnt even an element of the field, so how do you suppose you will take it's inverse?

Null

@meow-mix what about non-Fields?

Lozansky

12:26 PM

How is $[a,b] \times[\alpha, \beta]$ a rectangle?

Tobias Kildetoft

@Null Even in rings, it makes no sense

DHMO

@Lozansky well, because it is a rectangle?

or else what would it be

basically $a\le x \le b$ and $\alpha \le y \le beta$ defines a rectangle

in the 2D plane

@TobiasKildetoft why?

Lozansky

@DHMO Yeah I figured $[a,b]$ and $[\alpha, \beta]$ were vectors

DHMO

@Lozansky I thought they are intervals

Lozansky

They are, I've been matlabbing too much

DHMO

12:38 PM

in matlab vector = interval?

Tobias Kildetoft

@DHMO Because division by $0$ is not defined in any (non-zero) ring

DHMO

@TobiasKildetoft division itself is not defined in any ring either

Lozansky

No, but $[x_1, x_2....]$ denotes a vector

DHMO

so your statement is a vacuous truth

@Lozansky I see

Tobias Kildetoft

@DHMO Division can however sometimes be defined for some elements, but never for $0$

DHMO

12:39 PM

@TobiasKildetoft why not?

Tobias Kildetoft

@DHMO Because that would result in the $0$-ring

DHMO

@TobiasKildetoft could you elaborate?

Tobias Kildetoft

@DHMO This is an easy consequence of the definition. If $0$ has an inverse then $1 = 0$ and the ring has just that one element

DHMO

@TobiasKildetoft ok, thanks.

Vrouvrou

1:12 PM

can someone help me for this : math.stackexchange.com/questions/2041655/bounded-in-l1

Null

could someone help me? If a line $f$ is viewed as a "mere" subset of $\mathbb{R}^2$. Then it doesn't make sense to talk about $\langle f\rangle _{\mathbb{R}}$ or?

or does it indeed make sense, because a line is a collection of vectors?

Antonio Vargas

@Vrouvrou I don't understand your question. The first line is the definition of "convergence to $0$ in $L^1$". What are you asking?

Semiclassical

1:44 PM

Hey @AntonioVargas

Antonio Vargas

@Semiclassical Oi

Vrouvrou

@AntonioVargas I'm asking if $\Phi(|u_n-u|)\leq H\in L^1$ or a subsequence of it ?

Antonio Vargas

@Vrouvrou, ah, the answer is no. See here: math.stackexchange.com/q/1166871/5531

Vrouvrou

@AntonioVargas so only for a subsequence

Antonio Vargas

@Vrouvrou I don't think any subsequence of the example given in the top answer is dominated

Than again I only looked at it for ~3 seconds, so I may be wrong

Vrouvrou

1:55 PM

i found a theorem which that there exist a subsequence which dominated by an L^1 function

Antonio Vargas

Ok I looked at it for 3 more seconds and I stand by my original assertion.

@Vrouvrou there is clearly no subsequence of Crostul's example which is dominated.

Vrouvrou

You understand french ?

Antonio Vargas

Nope

DHMO

@Vrouvrou Pourquoi tout le monde est francophone?

Vrouvrou

@AntonioVargas

Antonio Vargas

2:03 PM

@Vrouvrou I told you I don't understand French

Vrouvrou

there two words in french the rest are symboles

Antonio Vargas

Ok, I thought about it for 10 seconds this time, and I take it back. There is a subsequence of Crostul's example which is dominated.

(If you know the answer, why ask me?)

DHMO

2:36 PM

Why is $\displaystyle \left\{\sin\left(\frac{1}{n}\right)\;\Big|\;n \in \mathbb{N} \right\}$ not open?

Alessandro

not open in R with the standard topology, right?

DHMO

@Alessandro I'm failing to understand why it is not open

Alessandro

Hm, can an open set in R be countable?

DHMO

@Alessandro $\Bbb N$

I thought discrete sets are open

Alessandro

That's not open

DHMO

2:50 PM

why not

Alessandro

Pick a poin in N, you can find an open ball around not entirely contained in N

So N is not open in R

(With the standard topology)

DHMO

oh, thanks

let's explore the discrete topology of $\Bbb R$ lol

discrete topology of uncountable set lol

Alessandro

Everything is open in the discrete topology

DHMO

but does its uncountability add any special effects to it?

Alessandro

It depends what you mean, it's neither separable nor second countable since R it's uncountable

DHMO

2:55 PM

@Alessandro what do "separable" and "second countable" mean lol

Alessandro

A topological space is second countable if its topology has a countable basis and separable if it has a dense countable subset

R with the standard topology is both for example

DHMO

@Alessandro never mind

you're explaining two terms with 4 terms i don't know lol

what is the basis of R?

Danu

@TedShifrin Actually, I'd like to return to this---it's clear that the first Chern class distinguishes a $\Bbb Z$'s worth of isomorphism classes and that there are $\Bbb Z$'s worth of isomorphism classes... But how do I prove that it really classifies? A priori, that's not obvious, is it? It might distinguish only a subgroup (still iso to $\Bbb Z$). By additivity it is enough to prove that it vanishes iff the bundle is trivial, I guess.

DHMO

@Alessandro Sei invitato a questa sala

$Mathematics$ General topology

For any discussions about general topology. For instructions h...

Alessandro

I don't have time to explain all this stuff because I have to run for a lecture now, but it's found in any intro to topology book

Akiva Weinberger

3:14 PM

@TedShifrin I am ready to receive mumblings

s.harp

3:34 PM

How can I understand $\mathcal O(\mathbb P^n_k)$ where $k$ is a field and $\mathbb P_k^n$ is the affine scheme over the field?

Ramanujan

@DHMO hi!

DHMO

hi

s.harp

My exercise is to "determine" it $\mathcal O(\mathbb P^n_k)$, but I am completley lost somehow

I barely understand the point of what a scheme is

Ramanujan

What is d/dx(e^{-x} ?

DHMO

@Ramanujan use chain rule

Ramanujan

3:39 PM

D/dx(e^xl=e^x

DHMO

Let $u = -x$.
$\dfrac{\mathrm d}{\mathrm dx}e^{-x}$
$= \dfrac{\mathrm d}{\mathrm du}e^u \dfrac{\mathrm du}{\mathrm dx}$
$= e^u (-1)$
$= -e^{-x}$

sorry, have to go now

Ramanujan

Ok

But d/dx ( a^{x}) = a^x loga

If a =e and x=-x then?

D/dx(e^{-x}=e^{-x}?

s.harp

If $\mathfrak p\in A$ is a prime ideal in a ring, $A_{\mathfrak{p}}$ is defined to be the localisation $(A-\mathfrak p)^{-1}A$ or $\mathfrak p^{-1}A$?

Both $A-\mathfrak p$ and $\mathfrak p$ are multiplicative, so both things work somehow

ok since $\mathfrak p$ contains $0$ the first definition is stupid >_>

Null

3:58 PM

$\forall b\in\mathbb{R}:1+bi\not=0$ right? (i for imaginary unit)

anonymous2

4:14 PM

@MartinArgerami, check this out:

20

Canada

Proposed Q&A site for anyone interested in or learning about Canada.

Currently in definition.

TheGreatDuck

...

that's an unusual site

Danu

@TedShifrin I guess that, now I've proven that the Euler class is PD to the generic zero locus of a section, plus the fact that a nontrivial line bundle does not admit a globally nonvanishing section, to at least get that $e$ is PD to a homology class that is not a priori the homology class of the empty set. But how do I prove that the zero locus does not bound?

Is it true that if the zero locus is null-homologous, that I can deform the section to be globally nonvanishing?

I'm guessing not, since the zero locus may bound something else than a disk. (Sorry for the rambling---does it make sense?)

Aksel'sRose

4:48 PM

Trying to prove a transformation is linear but the transformation is defined as $A\mapsto det(A)$. If I would generally use $L(cx)=cL(x)$ and $L(u+v)=L(u)+L(v)$, do I just plug in $det(A), det(B)$ and scalar c?

(or disprove)

Akiva Weinberger

Suppose we have an $n\times n$ lattice of $n^2$ points. At each $2\times 2$ "square" of four adjacent points (that is, whose vertices are of the form $(i,j)$, $(i+1,j)$, $(i,j+1)$, and $(i+1,j+1)$), we either join the northeast and southwest vertices with an edge, or else we join the northwest and southeast vertices with an edge.

How many of the connected components of the resulting graph are trees?

So I came up with this problem just now, and I want to know if (a) my solution is correct and if (b) there's actually a blindingly obvious way to solve this that I didn't think of.

Evinda

Hello @quid !!! Can I ask you something about pdes?

quid

Hello @Evinda

That's not among my strength PDE, also I need to leave soon. But I can have a quick look.

arctic tern

@Aksel'sRose det(A) is not linear

Evinda

We consider the following Cauchy problem:

$$u_t- \Delta{u}+cu=f(t,x) \text{ in } (0,T) \times \mathbb{R}^n \\ u(0,x)=\phi(x) \text{ for } x \in \mathbb{R}^n$$
where $c \in \mathbb{R}$ is a constant. I want to find a formula that will give us the solution in closed form.

Can you give me a hint how can we find the formula? @quid

That is my question

quid

5:03 PM

I am afraid I have no good answer to this. Sorry @Evinda

Evinda

A ok. No problem @quid

Does anyone else have an idea?

Aksel'sRose

@arctictern How would I show this? Im horrible at counter-examples...

ill search the stacks and see what I can find also

arctic tern

stop being lazy and plug in some matrices to see if det(A+B) equals det(A)+det(B)

or, you could see if det(cA)=c det(A), even easier

Aksel'sRose

ha I actually didnt think of that at all! Blonde moment to the max....

Krijn

5:20 PM

I never really understand that with students, why they won't check the easiest counterexamples first

Then again, I might be doing that myself and some professor is writing this about me

Maybe god is writing in this way about that professor

And then of course Grothendieck is writing in this way about god

I should really start cooking dinner

arctic tern

no, grothendieck eschews examples

Krijn

I could imagine after trying to give examples of primes

Akiva Weinberger

@AkivaWeinberger I was wrong. This depends on the graph.

Semiclassical

hi chat

Evinda

Hi @Semiclassical

Semiclassical

5:34 PM

Here's a question, which I think has come up here before and which I'd like a better example of.

What's an example of a $\forall x (p(x))$ statement which is false but for which there is no constructible counterexample?

The one I came up with is the disproof of every irrational power of an irrational being irrational, by noting that $(\sqrt{2}^{\sqrt{2}})^{\sqrt{2}}=2$ is a counterexample iff $\sqrt{2}^{\sqrt{2}}$ isn't a counterexample.

But that doesn't quite satisfy me now.

Vrouvrou

Hello, can someone help me on measure theory

Brody

My instructor has $\langle \mathbf{u},\mathbf{v}+\mathbf{w}\rangle$ $=\langle\mathbf{u},\mathbf{v }\rangle+\langle\mathbf{u},\mathbf{w}\rangle$ as one of the axioms of an inner product.

Isn't this sort of "distributivity" usually stated for the first argument though? Or are the two equivalent?

Semiclassical

@brody they're equivalent. see here for discussion: en.wikipedia.org/wiki/…

You've got conjugate symmetry as well, after all.

So what's good for the first argument is good for the second as well.

Akiva Weinberger

@Brody Don't you also have $\langle\bf u,v\rangle=\langle v,u\rangle$? (For real vector spaces)

(For complex ones you conjugate a side)

Mahmoud

Hi.

Semiclassical

5:44 PM

Not sure how you define an inner product for a vector space whose field of scalars isn't R or C, come to think of it. In the former there's no need for conjugation, and in the latter it's obviously just complex conjugation.

Is there a notion of 'conjugation' for a generic field?

Brody

@AkivaWeinberger Yes, so it plays off from the commutativity axiom. I wasn't sure if there were more nuance to the fact one argument tends to be emphasized explicitly over the other.

Akiva Weinberger

@Semiclassical Probably some idea of automorphism

Semiclassical

Yeah. The bit I linked above has some remarks on that, but not a lot.

With the message seeming to be: The base field doesn't have to be R or C, but these really are the most natural options.

Brody

@Semiclassical I've read the section. In the class, we're just working with real numbers (field $\Bbb R$ and vectors strictly with components thereof), so no talk about conjugates as far as my instructor teaches. I guess in this case it's just linearity and not conjugate linearity, and which argument the axiom regards is a matter of preference.

Semiclassical

Yeah.

Brody

5:55 PM

well, the linearity is still conjugate linearity I suppose, but much simpler than the general case. although, it's odd that in our notes, the scalar associates to the first argument, but the inner product distributes over the second argument. doesn't really matter, but...

Semiclassical

In truth I'd probably have remembered it as being bilinear, with the symmetry axiom guaranteeing that $A\wedge B\equiv A\equiv B$.

i.e. technically it's just one or the other, but it's obvious enough that one just remembers the bilinearity.

(but then i am also pretty lazy about such things)

Akiva Weinberger

@Semiclassical What does that mean

koolman

Answer to this

0

Q: Determinant formed by cofactors

We are given delta(0) And let delta(1) denote the determinant formed by the cofactors of elements of delta(0) and delta(2) denote the determinant formed by cofactors of delta(1). Then we have to find value of delta(n) in term of delta(0). I tried it a lot . But does not got any start .

Semiclassical

@akiva Nothing terribly sophisticated. Just that if it's linear in the first argument, then the (conjugate) symmetry axiom guarantees that it's linear in the second argument and vice versa. So while it's not correct to say that the definition is that it's linear in both arguments, it really doesn't carry any harm.

Alessandro

@Semiclassical can't you pick a positive definite symmetric bilinear form and call that the inner product?

Semiclassical

6:05 PM

Sure, but then you've got the same notion of conjugacy as in the real case, i.e. <x,y>=<y,x>

Alessandro

You probably need an ordered field though

To make sense of the positive part

Semiclassical

Yup. Again, the bit on Wikipedia discusses that.

A pet peeve of mine: People using images instead of mathjax in posts.

Alessandro

Hm, maybe I should read it

Semiclassical

For diagrams, that's fine. For matrices, not fine.

Akiva Weinberger

@Semiclassical I was referring to the statement with the $\wedge$ and $\equiv$

Semiclassical

6:10 PM

Sure. I was being lazy there: A="inner product is linear in first argument" and B="inner product is linear in second argument"

back later

alan2here

Does it ever make sense to define a value (N), thats not a complex (or real, integer etc...) number, where infinitesimal * N = 1?

Krijn

@alan2here Sort of, in $\mathbb P(\mathbb C)$, the projective line, there are some funny thing going on

Look it up on wikipedia, it's a pretty good article I think

alan2here

can you type its name without mathbb code?

Krijn

Might not be exactly what you are looking for

The projective line

alan2here

TY, might be interesting either way.

all lines ususally have ends that are infinitely far from any given point on the line

if two libes met at a point at the far infinite end of the end in one direction, then there angles must be the same, + or - infinitesimal, so they just point towards each other infinitesimaly slightly

Kasmir Khaan

6:22 PM

1

Q: Surface integral (divergence theorem )

Evaluate: $\int \int \bar{F}d\bar S $, $ \bar{F}=(2-x^{2}yz+y^{3},xy^{2}z+ye^{z},y^{2}+z+e^{z})$ $\gamma : y^{2}+z^{2}=x^{2} $ between the planes $x=1$ and $x=2$ , The normal pointing away from the $x$-axis. Thanks in advance.

multivariable-calculus

help with this please guys

alan2here

Is this the idea wirh projective lines?

not sure if that helps, but TY anyway

*with

I guess I might need to be using surreals for any number of infentesimal to be anything other than 0.

(1/inf) * N = 1

(1/inf)=1/N

inf=N

N=inf

there solved :)

I guess infinitesimal * infinite is undefined and therefore has a solution set of everything, being the compliment of the null set, and therefore "=1" is a perfectly good solution here but so would any other value.

Akiva Weinberger

6:49 PM

@Semiclassical Yeah, sorry — that occurred to me shortly after I wrote that and left the chat

Astyx

7:14 PM

Hello everyone ... except it seems I'm alone right now

Mahmoud

Nah. @Astyx

Astyx

How are you ?

Mahmoud

I am still alive in the middle of all the frustrating problems of Math and Life. What about you @Astyx ?

Astyx

Same

Mahmoud

This situation isn't just going to vanish away in time, it's seems like a fundamental rule of existence, so we probably should consider it normal to have problems.

GPhys

7:23 PM

What is typical length of US math PhD?

I only know the stats for physics

Mahmoud

Google says it's between 4 and 6 years.

GPhys

was mostly wondering if the median weighs heavily toward 4 years

Mahmoud

@GPhys

Check this disappointing article, cbsnews.com/news/12-reasons-not-to-get-a-phd :/

GPhys

I'm already in a PhD program, @Mahmoud

Mahmoud

@GPhys Ok. Best Of Luck $:)$

Aksel'sRose

7:43 PM

Im looking at the following question, and im not following why the answer is true. Can someone explain it in a bit more detail?

0

Q: If $\|U(x)\| = \|x\|$, and $x$ is in an orthonormal basis, must $U$ be unitary?

I had this question on my final yesterday. I still don't know the answer. Can someone please tell me the answer/explain it to me? Thanks. Ps: I wrote that yes it must be unitary.

linear-algebra

Astyx

@Aksel'sRose What part is not clear to you ?

Semiclassical

^ Not much help we can give if you don't say what you do understand.

Aksel'sRose

Is the second sentence of the answer a proposition/definition?

Astyx

A proposition

Aksel'sRose

hmm, I dont think I have seen that one.

Astyx

7:50 PM

You can show the first and second sentence are equivalent

Huh wait actually

Semiclassical

That $\Vert \beta_i \Vert = 1$ is just by virtue of it being an orthonormal basis.

Aksel'sRose

because its a normalized orthogonal basis, right?

Astyx

No this is not true actually

Take $f:z\mapsto \Re(z) + \Im(z)$

Then $f(i) = f(1) = 1$

But $f(1+i) = 2 \ne |1+i| = \sqrt 2$

(right?)

Semiclassical

no clue.

Vrouvrou

hello, is it right to say that : if $(f_n)$ do not converge to $f$ then ther exist a subsequence $(f_{n_k})$ such that $||f_{n_k}-f||>\varepsilon, \forall k $

Astyx

7:57 PM

Depends what $\epsilon$ is

Vrouvrou

$\varepsilon>0$

Astyx

Where do you define it ?

Vrouvrou

in \mathbb{R}

!

Astyx

Okay, re-ask you question with all terms defined please (otherwise I can't answer)