Mathematics - 2018-11-15 (page 1 of 2)

user330477

00:00

@TedShifrin What is equivalent to what you are talking about?

Ted Shifrin

@user330477: I'm waiting for you to tell me why a matrix of rank $>k$ cannot be in the closure of $M_k$.

Ugh, the site is still acting up.

MatheinBoulomenos

@LeakyNun I have an example

Leaky Nun

@MatheinBoulomenos :D

Ted Shifrin

heya @Mathein

MatheinBoulomenos

hey @Ted

still writing on the details though

Leaky Nun

00:03

> Genetic algorithm designed to find low-energy configurations of carbon exploits edge case in the physics model and superimposes all the carbon atoms

@Semiclassical this is funny

user330477

@TedShifrin In that case, there would exist a $l$ by $l$ minor which is zero. Then what do we do?

Ted Shifrin

I don't follow ... what is $l$ here?

user330477

@TedShifrin I am just using the theorem you talked about earlier. Here, $\max{n,m}\leq l>k$

Ted Shifrin

Aren't we working with square matrices to start with?

Or maybe not. I have no idea.

All I care about is one particular matrix of rank $>k$. Why is it not in the closure of $M_k$?

So, yes, use what we talked about. If it has rank $l>k$, then ...

user330477

@TedShifrin Do we have to use some density argument here as well?

Ted Shifrin

00:09

Huh?

The point is: how did you show the matrices $M_k$ formed a Zariski-closed subset?

user330477

@TedShifrin I think you mean closure of $M_{k} \setminus M_{k-1}=M_k$ earlier. For your question, right now, we do something like this math.stackexchange.com/questions/43192/…

Ted Shifrin

$M_k$ is matrices of rank $\le k$, right? So I meant what I said.

MatheinBoulomenos

@LeakyNun Let $F$ be any non-archimedean local field with absolute value $| . |$, let $\mu$ be a Haar measure on $F$ normalized such that $\mathcal{O}_F$, the valuation ring, has measure $1$, then consider the two-dimensional mirabolic group $G:=\left\{\begin{pmatrix} x & y \\0 & 1\end{pmatrix}, x \in F^\times, y \in F \right\}$, we can regard a function $f:G \to \Bbb{R}$ as a function in two variables on $F^\times \times F$, now we can define a left Haar integral (which gives a Haar measure by Riesz) by $f \mapsto \int \frac{f(x,y)}{|x|^2} \mathrm{d}\mu(x) \mathrm{d}\mu(y)$

Ted Shifrin

OMG @Mathein.

non-archimedean local field faints

MatheinBoulomenos

it was just a natural example for me :P

Ted Shifrin

00:16

So can you describe what you linked in your own words, @user330477? That's what I've been trying to get you to say for an hour.

@Mathein: In general, something like the Heisenberg group seems reasonable to look at.

MatheinBoulomenos

Haar measures here are really easy to compute here: they turn into sums

because our function is locally constant

Ted Shifrin

In general, turning left into right or vice versa requires looking at the adjoint representation, I guess, @Mathein.

user330477

@TedShifrin The only problem, i have is how do I show that $MA^{mn}$ is Hausdorff?

Ted Shifrin

The Zariski topology is never Hausdorff.

I have no idea where this is coming from.

user330477

@TedShifrin I am highly confused.

MatheinBoulomenos

00:21

@LeakyNun the function should actually be the square root of what I wrote, I forgot the "square" part about square-integrable

Ted Shifrin

@user330477. Me too. Your original question was very straightforward, and Zev's answer you linked shows you how to do it. So I don't understand ...

@Mathein: Are you using $f$ for different functions? I'm confuzled.

MatheinBoulomenos

no, I tried a characteristic function first but it didn't work

Ted Shifrin

Oh, so your "we can just consider the characteristic function $f$, then consider ..." is a typo.

MatheinBoulomenos

yeah

Ted Shifrin

OK. Serves me right for looking :D

MatheinBoulomenos

00:24

thanks

Ted Shifrin

But how're you seeing left versus right there?

Oh, I see, different formulas for the Haar integral.

MatheinBoulomenos

yeah

Ted Shifrin

I would do that in the smooth category with Maurer-Cartan forms.

Érico Melo Silva

hlo

Ted Shifrin

heya @Eric

MatheinBoulomenos

00:27

the idea is that in the formula $\begin{pmatrix} x & y \\0 & 1\end{pmatrix} \cdot \begin{pmatrix} x' & y' \\0 & 1\end{pmatrix}=\begin{pmatrix} xx' & xy'+y \\0 & 1\end{pmatrix}$ you can ignore the $+y$ if you're integrating over the whole space, so you get the left action scales both entries and the right action scales only one entry

Ted Shifrin

You get more telegraphic every week, @Eric.

MatheinBoulomenos

hi @ÉricoMeloSilva

Érico Melo Silva

what do u mean

Ted Shifrin

fewer letters in your words

Érico Melo Silva

o ya

Ted Shifrin

00:27

LOL

It's like you're paying a premium for each letter.

Érico Melo Silva

yup

i didnt write down anything for the $\partial \bar{\partial}$ lemma thing and now i dont see at all why you need d exactness

Ted Shifrin

Oh, so, as Demonark would say, RIP that exercise?

Érico Melo Silva

ya i also cant find where it is in GH cuz it's not in the index lol

ik it's in voisin i think

Ted Shifrin

Oh, I thought I gave the page number.

Érico Melo Silva

nah at least if u did i mightve already been gonezo

Daminark

00:33

I have been summoned

Ted Shifrin

Oh, maybe I said the page number in my notes.

Érico Melo Silva

oh maybe, ill look

Ted Shifrin

I'll find it in the book.

Érico Melo Silva

i assumed it'd be where u talk about harmonic theory for kahler mflds

user193319

Does an isometry of a compact metric space always have a fixed point?

Daminark

00:35

Rotate the circle

user193319

Shoot!

Érico Melo Silva

rekt

Daminark

Hopefully you didn't need that face for something

Érico Melo Silva

yeah @Ted i dont see it but i might be a big fool

Ted Shifrin

Eric: p. 149.

Érico Melo Silva

00:39

ok word ill look at bok

oh duh how did i miss this

Ted Shifrin

In the case of that proof, type considerations will get you what you need. This is like p. 17 (cont) of my notes.

I don't remember the counterexamples. We should work on that if you don't get 'em.

Érico Melo Silva

type?

Ted Shifrin

$(p,q)$-type.

Érico Melo Silva

oh ok the grading gotcha

Ted Shifrin

Right.

Érico Melo Silva

00:43

hmmm ok i will think

Ted Shifrin

OK, then I won't.

Érico Melo Silva

ill be back w thoughts but probably not tonight

i have the general test tmr

Ted Shifrin

What general test?

I won't be around tonight much anyhow.

Érico Melo Silva

GRE

Math geek

This is an example given in the textbook C.W Patty Foundation of Topology, Let $\leq$ denote the dictionary order relation on $I\times I$ determined by less than or equal to on $I=[a,b], a,b\in \mathbb R$, Let $\mathscr T$ denote the order topology on $I\times I$. Then $(I\times I,\mathscr T)$ is locally connected but not locally pathwise connected. Let $p\in I \times I$. Let $U_p$ neighbourhood of $p$.

How do I prove that that there does not exist any pathwise connected neighbourhood of $p$ that sits in $U$?

Ted Shifrin

00:44

Oh.

Érico Melo Silva

at like 8am

Ted Shifrin

For some points $p$ there's a little path, @Mathgeek. You have to pick particular points.

Érico Melo Silva

so ill be sleeping v early

Ted Shifrin

No problem, @Eric. I am disappearing shortly and won't be back until sometime tomorrow.

Math geek

@TedShifrin okay

Érico Melo Silva

00:46

aight ill see ya tmr then, im outta here to think complex geo

Ted Shifrin

Night, and good luck on GRE (not that you need it).

Érico Melo Silva

ya i just crushed a practice test w/o knowing the format so i am not worried

Ted Shifrin

@Mathgeek: You should draw pictures.

MatheinBoulomenos

@Daminark did you figure out the thing with the corollary of Jordan's theorem about characters? I think I got something but it might be way too complicated

Daminark

Not quite

Math geek

00:54

Ted Shifrin

Draw neighborhoods of points in your picture, @Mathgeek. I don't know what that circle is supposed to be.

Math geek

yellow path won't go left right?since relation greater than satisfy points right. in the line $y=1$

neighbourhood of (0,1)

Ted Shifrin

So what is a neighborhood of $(0,1)$ in the dictionary order? I assume that's a point at the top? You're doing $[0,1]\times [0,1]$?

MatheinBoulomenos

So the idea is this: for a group $G$, let $R(G)$ be the $\Bbb{Z}$-submodule of the class functions on $G$ $F_C(G)$ (using the notation from Serre here) generated by all irreducible characters, then for a subgroup $H$, a restriction of a character is again a character so we get that $\mathrm{Res}_H^G:R(G) \to R(H)$ is well-defined, note that the irreducible characters form a $\Bbb Z$-basis for $R(G)$, so using Jordan's theorem,
we can show that the map $\mathrm{Res}_H^G:F_C(G) \to F_C(H)$ is not injective

Ted Shifrin

Forget paths for now. What are neighborhoods of points?

MatheinBoulomenos

00:57

but since the characters form an orthogonal basis for $R(G)$, we get that $\Bbb C \otimes_{\Bbb Z} R(G) \cong F_C(G)$

Leaky Nun

thanks @MatheinBoulomenos

MatheinBoulomenos

now if we assume that $\mathrm{Res}_H^G R(G) \to R(H)$ is injective, then tensoring with $\Bbb C$ would give us that $\mathrm{Res}_H^G:F_C(G) \to F_C(H)$ is injective ($\Bbb C$ is a flat $\Bbb Z$-module), so the map $\mathrm{Res}_H^G:R(G) \to R(H)$ has nontrivial kernel

Ted Shifrin

What is a basis element around $(1/2,1/2)$? What is a basis element around $(1/2,1)$ or $(1/2,0)$?

Math geek

@TedShifrin sorry it will be like a open rectangles. right?

Ted Shifrin

NOOOOO.

That's product topology.

This is what you have to visualize correctly.

Zee

01:01

@TedShifrin there is a tough question that I saw , I don’t want any help solving it but I wanna understand

Can a complete manifold with non negative ricci curvature have finite volume

MatheinBoulomenos

let $f \in R(G)$ be an element in the kernel of $\mathrm{Res}_H^G$, then $f=\sum_{i=1} a_i \chi_i$, where the $a_i$ are integers and the $\chi_i$ are the irreducible characters of $G$, so now because the $a_i$ are integers, we can just consider $g=\sum_{i=1} 2|a_i| \chi_i$, then both $g$ and $g+f$ have positive integer coefficients, so they are actually characters

Zee

How is volume defined here ?

MatheinBoulomenos

but since $f$ is in the kernel of $\mathrm{Res}_H^G$, these restrict to the same character

Ted Shifrin

@Zee: What about a sphere?

Volume is defined by the Riemannian metric.

Math geek

these types of line. right?

Ted Shifrin

01:08

@Mathgeek: Totally not.

Remember this is an order topology. So basis elements are open intervals in the order. So a neighborhood of $(1/2,1/2)$ is an open interval from one point less than that to another point greater than that. How could that look?

Math geek

okay

eigenvalue

Can anybody help me with the following? -> Consider $(\mathbb{R}^{4},-tdt^{2}+dx^{2}+dy^{2}+dz^{2})$ This is a flat pseudo Riemannian manifold. With the coordinate transformation $X=x,Y=y,Z=z,T=sign(t)\frac{2}{3}\sqrt{sign(t)t}^{3}$ the metric has the form g=-sign(t)dT^{2}+dX^{2}+dY^{2}+dZ^{2}.

What are the geodesics? For some reason I get the following... which is not the geodesics as shown in the paper: $T(X)=\pm\frac{X}{\sqrt{sign(t)}}+c$.

Math geek

means p lies some where in the open parallel to x axis or open ray parallel to y axis. right

?

Ted Shifrin

No, @MathGeek.

You really need to sit down and figure this out and stop guessing.

The axes are very different. An interval like $(1/2,1/4)$ to $(1/2,3/4)$ is a neighborhood of that point.

You need to think about how the dictionary order works.

Math geek

okay

Ted Shifrin

01:12

Then think about a point on the top or bottom.

@eigenvalue: I'm not going to take the time to figure this out, but think about what the general solution will be. That certainly isn't it. I don't know about what happens around $t=0$.

OK, I'm gone.

Zee

I have misheard the question in class since the prof said it’s not easy

eigenvalue

@TedShifrin Is there a way to find out (see from he equations) whether the author is talking about null geodesics or timeline geodesics?

Zee

Take care ted

Daminark

01:30

@Mathein damn that is quite the argument

MatheinBoulomenos

@Daminark yeah but maybe there's something easier if Serre doesn't mention at all how you get from Jordan's theorem to the statement of the characters

Daminark

Possibly

Rithaniel

01:59

So, trying to prove that path connectedness implies connectedness. I'm trying to go for contradiction, but I think I'm missing a detail. Assuming that you have a space $X$ which is not connected, but is path connected. So, there exist open $A,B\subseteq X$ s.t. $A\cup B=X, A\cap B=\emptyset$, and I take $a\in A$, $b\in B$. Since $X$ is path connected, it must be possible to construct a continuous function $f:[0,1]\rightarrow X$ s.t. $f(0)=a$ and $f(1)=b$.

I think I should be showing that, since $[0,1]$ cannot be written as the union of two disjoint, open sets (in the standard metric topology), that somehow implies that such a continuous function is impossible. However, how to show that is eluding me.

nitsua60

02:37

I just invalidated that flag on a seemingly-inoffensive comment from an hour and a half ago. If the flagger thinks I've made an error feel free to ping me or invite me to a private room to explain--otherwise I'm assuming it was a mis-click.

Ted Shifrin

03:08

@Rithaniel You have it set up perfectly. Look at $f^{-1}(A)$ and $f^{-1}(B)$.

CaptainAmerica16

Good evening.

Ted Shifrin

Hello, @CaptainAmerica.

CaptainAmerica16

I learned the true meaning of an ordered field today. It wasn't as complicated as I expected. @TedShifrin

Ted Shifrin

No, CaptainAmerica, you already know.

CaptainAmerica16

Well, I learned that I know what an ordered field is. After "learning" what it was, I realized it's pretty spelled out in Spivak. I'm just more certain now.

Ted Shifrin

03:16

OK.

CaptainAmerica16

That's pretty much it, I guess.

@TedShifrin Good night then.

Ted Shifrin

Well, OK, good night. :)

CaptainAmerica16

Ok, well actually not goodnight. What shape is that on your avatar thingy? I keep thinking it's the part you hold on a kite, which it obviously isn't. @Ted

Ted Shifrin

I think the kite thing is pretty much a standard cylinder.

I never answer that question, however. Lots have asked.

CaptainAmerica16

:-/ I don't know if that's you admitting it's a thing or you stating it's a cylinder.

I'll add that to my list of things to find out if it's the last thing I do.

Ted Shifrin

03:25

Huh?

CaptainAmerica16

Huh?

Ted Shifrin

I'm not sure what you're talking about with kites. But I thought that would be a regular cylinder. My shape certainly isn't.

CaptainAmerica16

Oh, ok. Yeah.

That's what I was trying to say.

I mean when I look at your thing it reminds me a kite handle, but I know that it isn't - which is why I asked. Then you said you'll never say, so now I'll have to find out if it's the last thing I do.

I should have just gone to bed.

Ted Shifrin

I'm just a meanie.

Night :)

CaptainAmerica16

Yes >:|

Ok, I'm really leaving now. Until next time Shifrin.

user131753

03:57

Is the following system of arithmetic, say $\mathsf{APA}$, incomplete where $\mathsf{APA}$ denotes the system obtained by removing the induction axiom schema from $\mathsf{PA}$ and adding the following axiom schema, ${\displaystyle \forall {\bar {y}}\neg(\varphi (0,{\bar {y}})\land \forall x(\varphi (x,{\bar {y}})\to \varphi (S(x),{\bar {y}}))\to \forall x\varphi (x,{\bar {y}}))}$?

user131753

I think that it is because its incompleteness would follow from the Robinson arithmetic. Am I right?

Sharath Zotis

04:22

Does the ring $R = \mathbb { Z } _ { 7 } \times \mathbb { Z } _ { 25 }$

have any zero divisors?

I say it doesn't but I am not extremely confident about my answer

hmm actually now that I think about it: (0,1) * (1,0) are both zero divisors

30 zero divisors!

Secret

04:49

Seeing today's horribly disorganised conversation, wishes this chat can be multiplied by its corresponding zero divisor

Anjani Gupta

05:08

Hi all! Its known that a locally compact group admits a Haar measure on itself. I wanted to know where does the sigma algebra of that Haar measure come from? I want to what the sigma algebra is, ans then carry out some integration over that. And I don't know what kind of integration would fit in there.

Gaurang Tandon

05:46

Hey all, can we not modify this proof by saying that there would exist an integer m such that $a^m=a$, then $a^{m-1}=1$, so $a^{m-2}$ is the inverse since $m>2$?

Daminark

How do you know that $ax = y$ and $bx = y$ implies $a = b$?

Oh nvm you have cancellation

Sharath Zotis

$\text { If } \operatorname { char } ( D ) = 0 , \text { then all nonzero elements of } D \text { have infinite additive order. }$ where $D$ is an integral domain

So to show this I basically assume $char(D) = 0$

which means for all $a \in D$ there is no such $m$ such that $m \cdot a = 0_R$

my question is there may be no such m for all $a$ such that $m \cdot a = 0_R$

Gaurang Tandon

06:03

@SharathZotis sorry not taught characteristic yet :/

Sharath Zotis

No worries @GaurangTandon

however how do I know that there isn't an m for some nonzero element, $x \in D$ such that $m*x = e = 0_R$

Right, so then do we even need char(D) = 0?

do all nonzero elements in an integral domain have infinite additive order?

Gaurang Tandon

@SharathZotis it's an integral domain, isn't it? then there can't be any such $m$ :/

no zero divisor property iirc

Secret

@SharathZotis Well, we have finite fields where every element does not have characteristic zero

Sharath Zotis

Right, but Im curious as to why then we need to have "if $char(D) = 0$"

Secret

and finite fields are examples of integral domains

Sharath Zotis

06:09

Wait Im a bit confused

m is not an element of the integral domain D

Basically what I know is that there is no such m such that $a + a \cdots + a = 0_R$ m times

for all $a \in D$

What I am trying to show is that for any element $x \in D$ there is no $n$ such that $a + a + \cdots + a = e = 0_R$

n times

Secret

By distributivity we have:

$a+a+a+a+ \cdots + a = 0_{R}$

$a (1+1+1+1+\cdots +1) = 0_R$

Rings are closed under $+$ and $\times$, thus the pile of ones has to be some element in the integral domain, and cannot be outside

Will Hunting

Hey guys thanks for the stars. I love stars, though I love money more.

Secret

actually screw that

104

Q: Example of infinite field of characteristic $p\neq 0$

Can you give me an example of infinite field of characteristic $p\neq0$? Thanks.

$\operatorname { char } ( D ) = 0 , \implies \text { all nonzero elements of } D \text { have infinite additive order. }$

Jacksoja

@WillHunting when are you going to man-up and make new video on Youtube?

You made me cry last time you bastard @WillHunting

Will Hunting

@Jacksoja Hiii, I have 5 now. Taking a break because I am not feeling well.

Rajesh Dachiraju

06:19

My recent arXiv paper that I am excited about : A Function Fitting Method

Sharath Zotis

Im confused @Secret isnt that what I am trying to show?

Rajesh Dachiraju

arxiv.org/abs/1811.01336

Jacksoja

@WillHunting Feel better man ! I enjoyed your videos <3

Sharath Zotis

Following your step $a ( 1 + 1 + 1 + 1 + \cdots + 1 ) = 0 _ { R }$

since rings are closed the pile of ones is some element lets say $x \in D$

Secret

If $m \not\in D$ then by definition it does not contribute to $D$'s characteristic

Sharath Zotis

06:23

so we get $a(x) = 0_R$

however as it is an integral domain then this can't be possible without a or x being $0_R$

Is this right?

Secret

yup because $D$ has no zero divisors

so $ax=0_R$ implies $a=0_R$ or $x=0_R$

Sharath Zotis

However, how are we using the fact that char(D) = 0 to show that nonzero elements in D have infinite order

It seems to me we are just using the fact that D is an integral domain to show that nonzero elements in D have infinite order?

Secret

But isn't zero characteristics is defined to be adding infinitely many ones to get zero and hence one can never add any element finitely many times to get zero and hence all elements have infinite order?

I don't see anything to prove in "char(D) = 0 to show that nonzero elements in D have infinite order"

it's basically how ring characteristics are defined

Expressing in contrapositive. If there exists at least one element that has finite order, then char() > 0

Sharath Zotis

Im super confused

so basically this implication "$\operatorname { char } ( D ) = 0 , \Longrightarrow \text { all nonzero elements of } D \text { have infinite additive order. }$"

we prove by just using the fact that $D$ is an integral domain

we don't use that $char(D) = 0$?

Secret

Ok maybe I should phrase it this way:

What we have proved is:

Secret

06:42

ok fuck I am not sure anymore. Turns out I need to understand how prime ideals work in order to correctly understood characteristic, and I have been struggling with anything that has the word "prime" in it for ages. Beh, I am out of here

0

Q: Showing that the characteristic of a commutative ring R without zero divisors is 0 or prime

Question: Suppose that $R$ is a commutative ring without zero-divisors. Show that the characteristic of $R$ is either $0$ or prime. I have established that every element in a commutative ring $R$ without zero divisors have the same additive order $n$. Now, if no such additive order n exi...

abstract-algebra ring-theory integral-domain

Sharath Zotis

No worries thank you so much for your help.

Alessandro Codenotti

07:20

@SharathZotis did you see my ping yesterday?

@CaptainAmerica16 Have you seen the proof that no finite field can be ordered?

Will Hunting

09:51

@Jacksoja Actually now there are 6 altogether, but the last 2 are not songs but vocal exercises. =D

Tobias Kildetoft

I have now also started smuggling a bit of algebraic geometry into the exercises of the algebra course.

Leaky Nun

@TobiasKildetoft such evilness :P

Tobias Kildetoft

@LeakyNun Yeah, the students will just have to put up with the evils of learning something useful :)

Leaky Nun

@TobiasKildetoft ok so now a group representation is a continuous homomorphism from a locally compact hausdorff group to the group of unitary operations of a hilbert space

Secret

algebraic geometry in a nutshell = too long to describe

Tobias Kildetoft

10:04

I do need to not do too much of it, as not all of them have taken topology (and those that have are taking it right now)

Leaky Nun

> I do need to not do

that's a mouthful

Leaky Nun

10:22

needless needles

2

Daminark

Nice

Will Hunting

@TobiasKildetoft I guess it's some commutative algebra?

Leaky Nun

@WillHunting ^^^

Tobias Kildetoft

@WillHunting Yeah, basically just the definition of Spec and its topology

as well as mSpec

Will Hunting

Hiii @LeakyNun.

Tobias Kildetoft

10:25

Then showing that mSpec of a polynomial ring over the complex numbers can be identified with $n$-tuples of complex numbers (assuming Zariski's Lemma, which I did not want to go through)

Leaky Nun

@WillHunting "^^^" means "look 3 messages above"

@TobiasKildetoft or just Nullstellensatz lol

Tobias Kildetoft

@LeakyNun Yeah, I really don't want to get into that at this stage.

Will Hunting

@LeakyNun I see. I never thought of needless needles. Interesting.

Leaky Nun

@TobiasKildetoft that's a pity

Tobias Kildetoft

But showing that maximal ideals have the required form mainly needs Zariski's Lemma as the non-trivial part. The rest is a fine exercise with a few guiding steps

Will Hunting

10:27

@LeakyNun Do you want to use Hangouts with me?

Leaky Nun

@WillHunting sure

Will Hunting

@LeakyNun But I don't have your Google address. I think we need that instead of the Yahoo address.

Tobias Kildetoft

We are done with ring theory now anyway, and moving into rep theory

Daminark

:0

In rep theory now actually we're starting rep theory of S_n

Fun stuff

Will Hunting

@LeakyNun Maybe you can send me a mail using your Google address, then I will send you a Hangouts invite.

parvin

10:30

hi

I've got a question,
Why does 2s complement give a negative form of the number?

Leaky Nun

@Daminark it sure is

@WillHunting I sent a message to your address in Hangouts

@parvin the intuition is that if x=-4 then x+4=0

but (2^n-4)+4 also = 0 for large enough n that causes overflow

Tobias Kildetoft

@Daminark Yeah, rep theory of symmetric groups is quite amazing

Will Hunting

@LeakyNun I don't see it anywhere...

Leaky Nun

@WillHunting somehow it is still sending..

Will Hunting

@LeakyNun I set my Hangouts settings to people with my mail can send me an invitation, not send me a message directly. Maybe that is why you can't send me anything.

Leaky Nun

10:35

how to send an invitation?

oh phone?

Will Hunting

@LeakyNun I tell you what. Since you don't know how Hangouts works very well, just send me a mail from your Google address, is that OK? Then I will send you the invite instead.

Leaky Nun

ok

done

parvin

@LeakyNun a very silly question ! if x+y=0 what is the x relative to y? and if (in base 10) x+y = 10 again what is x relative to y?

Leaky Nun

I don't understand your question

let's say you take x+13=0

then you can solve it as x=...99987

parvin

how is that possible!?

Leaky Nun

10:43

well you only store a finite number of digits

for the case of the 2's complement

parvin

is that number complement of 13?

Leaky Nun

yes, in base ten

it's basically asking yourself, how to calculate 10000000 - 13

you write 13 as 0000013, and then flip the digits to get 9999986, and then add one to get 9999987

parvin

I see!

thank you

Secret

Am I reading some kind of floating point magic?

(checks transcript to see what this topic is about)

Leaky Nun

@Secret 2's complement

Secret

10:52

never heard of it before...
Hmm... google said it is something related to radix

might investigate it later

I wonder if it has relation with the more well known "add to 10" arithmetic

Method of complements

In mathematics and computing, the method of complements is a technique used to subtract one number from another using only addition of positive numbers. This method was commonly used in mechanical calculators and is still used in modern computers. The nines' complement of a number is formed by replacing each digit with nine minus that digit. To subtract a decimal number y (the subtrahend) from another number x (the minuend) two methods may be used: In the first method the nines' complement of x is added to y. Then the nines' complement of the result obtained is formed to produce the desired result...

confirmed

detail

Basically, a biased monte carlo sampling iterated until full understanding of a given maths domain is reached

another detail is that I first learnt about axioms, definitions, elementary derivations, general cases, and then examples and counterexamples, before converging to the theorems themselves

Liad

11:22

@TobiasKildetoft hi

Tobias Kildetoft

@Liad Hi

Liad

im trying to understand Georges's answer here :math.stackexchange.com/questions/106464/…

does his answer fine to you? @TobiasKildetoft

Tobias Kildetoft

Sure

Liad

ok so what's not clear to me is: $a$ is a homogeneous ideal in $S = k[x_0,\dots,x_n]$

if $P \in Z(a)$ then for all $f\in a $$f(\lambda a_0, \dots,\lambda a_n) = 0$ for all $\lambda \ne 0$ and $[a_0: ,\dots,:a_n] =P $

so $f$ vanishes on the line of $(a_0,\dots,a_n) \in \Bbb A ^{n+1}$. right? @TobiasKildetoft

Tobias Kildetoft

right

Liad

11:29

why this implies that $P \in V(a) \subset \Bbb A^{n+1}$?

Tobias Kildetoft

By definition?

Liad

$P$ don't live in $\Bbb A^{n+1}$

$a$ isn't either

this what's not clear to me

Tobias Kildetoft

@Liad $P$ corresponds to a set of points in $\mathbb{A}^{n+1}$

Liad

right. a line.

Tobias Kildetoft

So I guess using $\in$ is not quite correct

And I see that George also does not write that

Liad

11:37

ok so maybe if i say this: $(a_0,\dots,a_n) \in V(a)$ iff $f(a_0.. a_n) =0$ for al l$f\in A$ iff $f(\lambda a_0...\lambda a_n) $ for all $\lambda \ne 0$ iff $f([a_0,..,a_n] ) =0$ iff $[a_0,..,a_n] \in Z(a)$ , what do you think?

the second iff is true only because $f$ is homogenuous

Tobias Kildetoft

You have quite a few typos there which makes it slightly hard to tell what precisely you wanted to write

Liad

ah. cant edit

Jacksoja

Is L( V,W) same as HOM (V,W) ?

Liad

forgot some "," this is what you mean ?@TobiasKildetoft

Tobias Kildetoft

@Jacksoja Usually, yes

Liad

11:40

and $f\in a$

Jacksoja

okay thanks

Tobias Kildetoft

@Liad You also have both a capital A and an a and a missing =0

Liad

right

Jacksoja

and how does one prove that L(V,W) is infinite dim ?

can we find base for that vector space?

Liad

i shouldn't rush to write it.. sorry.

without the typos, does it look right to you? @TobiasKildetoft

Tobias Kildetoft

11:41

@Jacksoja Well, assuming either of the spaces is infinite dimensional, you reduce to the case where the other is 1-dimensional

@Liad Looks right

Liad

in my book the hint is "interpret the problem in terms of the affine n+1 space whose affine coordinate ring is S" @TobiasKildetoft

(actually its the same question as in the link i sent)

Jacksoja

@TobiasKildetoft I have this situation, V is finite dim, Dim V > 0 and W is infite dim, need to show that L(V,W) is infinite dim

Tobias Kildetoft

@Liad Sure, and that is precisely what this does

Jacksoja

but not sure what is a base for such v.s

the elements of L(V,W) are linear maps

Liad

@TobiasKildetoft where did we interpret S as $K(Y)$ ?

Tobias Kildetoft

11:43

@Jacksoja You don't need a basis, just an infinite collection of linearly independent vectors

Jacksoja

@TobiasKildetoft okay thanks let me try that

Tobias Kildetoft

@Liad In passing to lines in $n+1$-dimensional affine space

Liad

that's looking at $S$ as $K[x_0,..,x_n] / I(Y)$ for some $Y$ ?

Jacksoja

@TobiasKildetoft Hi again, to show the existence of linearly indep set of functions that are linear from V-->W

I am not sure what is the span of (S)

S being a linear function

Tobias Kildetoft

same as it would be in any other vector space

Jacksoja

11:51

am lost with this exercice, clealy missing something

so i take linear combinations of linear functions

but if W is infinite dim

we can never have finite functions from V--W

Liad

12:51

is it possible that $V(a) = \emptyset $ but $a \ne k[x_0,..,x_n]$ ?

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$Mathematics$ Mathematics