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Balarka Sen
Balarka Sen
12:00 AM
@TedShifrin True!! Let me show you one me and a friend were drawing back at uni
 
Akiva Weinberger
Akiva Weinberger
So, Kirby "calculus"… When do I start taking derivatives
/s
(weird Internet lag issues, didn't mean to post twice)
 
Balarka Sen
Balarka Sen
Same! I think something happened to the chat
 
Akiva Weinberger
Akiva Weinberger
It's similarly disappointing that in the lambda calculus you can't write $d/d\lambda$
@BalarkaSen Weird
 
Ted Shifrin
Ted Shifrin
John Harer and a few other Kirby students tried to teach me the stuff over beers and pretzels.
 
Balarka Sen
Balarka Sen
I swear I drew it in colors...
 
Ted Shifrin
Ted Shifrin
12:05 AM
I only see some horrid pink that looks off-white.
 
Akiva Weinberger
Akiva Weinberger
That sure looks like a tough tangle
 
Balarka Sen
Balarka Sen
Not sure how the camera did that
@TedShifrin Same guy as Harer stability, I assume?
 
Ted Shifrin
Ted Shifrin
Yup.
 
Balarka Sen
Balarka Sen
Cool.
 
Akiva Weinberger
Akiva Weinberger
@TedShifrin But as you bit into the pretzel, you realized to your horror that its inside was attached in a nontrivial way to its outside, and the pretzel represented a topological defect of the universe
Surgery pretzel
 
Balarka Sen
Balarka Sen
12:12 AM
I wanted to make a joke about pretzels and knots, but couldn't find it.
@Akiva I was asked to prove that $\Bbb{RP}^2 \# \Bbb{RP}^2 \# \Bbb{RP}^2$ is homeomorphic to $T^2 \# \Bbb{RP}^2$ in my graduate topology exam, and I gave a 2D Kirby calculus proof
 
leslie townes
leslie townes
i knew a kirby student who tried to explain that to me once. didn't work.
this is why every social circle needs to include at least one 3-manifolds person.
 
Balarka Sen
Balarka Sen
lol
3 is a weird little dimension
its so different from 2 or 4
 
Akiva Weinberger
Akiva Weinberger
@BalarkaSen Cool!
I woulda done some standard-ish cut-and-paste proof
 
Ted Shifrin
Ted Shifrin
Because 2 and 4 are, ironically, less complex? 🤷‍♂️😻
 
leslie townes
leslie townes
boo.
 
Balarka Sen
Balarka Sen
12:17 AM
lol
@AkivaWeinberger I can never remember the polygon one.
There's some tricky cuts to be made that I can't figure out
(incidentally, note that above I have nonorientable 1-handles as well. The analogue in 4D is that there are "twisted" 2-handles, and instead of a $\Bbb Z/2$'s worth of choice of being orientable or not as in 2D, there is now a $\Bbb Z$'s worth of choice for being twisty, i.e., there is a "twisty number" (official name) associated to every 2-handle which is an integer)
 
Akiva Weinberger
Akiva Weinberger
Wild
I guess it corresponds to other slopes of surgeries?
 
Balarka Sen
Balarka Sen
Yes, but only integer slopes.
You're a 3-manifold thinker, it's making me depressed.
Soon you'll be hyperbolizing knot complements
The way I think about it is that there are two real line bundles over the circle, hence $\Bbb Z/2$ is where the twisty number in 2D belongs. Likewise, there are integer many complex line bundles on the sphere $S^2$, hence $\Bbb Z$ is where the twisty number in 4D belongs.
(complex line bundle is the same as an oriented real rank 2 bundle by the way, so you can just think latter)
 
Akiva Weinberger
Akiva Weinberger
12:33 AM
So one of these would be the tangent bundle?
 
Ted Shifrin
Ted Shifrin
I prefer complex line bundle, thank you.
Euler, DogAteMy
 
Akiva Weinberger
Akiva Weinberger
What about him
 
Ted Shifrin
Ted Shifrin
Euler class of bundle generalizes Euler characteristic (=Euler number of tangent bundle)
Did you learn intersection numbers?
@BalarkaSen Is this a frame job?
 
Akiva Weinberger
Akiva Weinberger
12:53 AM
@TedShifrin Don't think so
Fun fact: you can compute a 3x3 determinant using only 8 multiplications instead of the naïve 9
(more additions, but additions are free)
@BalarkaSen Did you ever put more thought into the "two arcs equal three arcs" thing?
 
Ted Shifrin
Ted Shifrin
1:17 AM
Time to read through Guillemin & Pollack, DogAteMy. Or Hirsch.
 
 
2 hours later…
Bob
Bob
2:56 AM
good evening
 
copper.hat
copper.hat
hello
 
Bob
Bob
How are you tonight copper.hat? You have a PhD in physics right?
 
copper.hat
copper.hat
3:19 AM
@Bob nope, eecs
 
D.C. the III
D.C. the III
I'm back!..............I think.........🤔
 
Ted Shifrin
Ted Shifrin
Were you kidnapped?
 
D.C. the III
D.C. the III
3:34 AM
Nah....I went through a mini-midlife mathematical identity crisis....I was doing a problem in Insel a few weeks ago and I just got frustrated beyond belief. It was a multi-step proof where I would have to use result 1 --> result 2 --> result 3 --> etc. I was trying my hardest to do it without looking at any solution. I got partially there, but not far enough to my satisfaction..... So I went into a spiral of questioning what I'm doing with myself, etc , etc.
 
copper.hat
copper.hat
what is "Insel"?
 
D.C. the III
D.C. the III
I call it "mini" because I was doing everything else in my life with no problems (gym, social activitvies, etc), but my math fire was completely extinguished..
 
Koro
Koro
Friedberg, Insel
and Spence
 
D.C. the III
D.C. the III
Insel, Friedberg, et al - Linear ALgebra
 
copper.hat
copper.hat
@Koro ah thanks! i was thinking a geographical location
 
D.C. the III
D.C. the III
3:36 AM
I suppose I was being hard on myself because I could "recognize" what was needed to do the proof, but I suppose it was just an accumulation of not being where I "envisioned" myself at this point in my life.
 
copper.hat
copper.hat
@D.C.theIII for me it is like a bicycle tyre with a slow leak. if i don't top up regularly then there is (yet another) long refresh cycle somewhere in the future.
 
D.C. the III
D.C. the III
I think I should be good about not needing to refresh much. My mind was just recalling all the results I've worked on up to this point, actually more than I had expected. I just didn't want to get in the "grind"
 
copper.hat
copper.hat
i remember the broad ideas, but need to refresh details.
 
D.C. the III
D.C. the III
It was never an issue of completely dropping things, I love the "art" too much. But I guess I needed to refocus and remember my reasons for doing this.
 
copper.hat
copper.hat
my brain can really only handle one area at a time. so if i am solving a coding problem, my vestigial mathematics skills evaporate.
 
D.C. the III
D.C. the III
3:40 AM
but remembering the broad ideas is what matters no? The details one can always retrieve.
 
Ted Shifrin
Ted Shifrin
Copper’s brain needs an external hard drive
 
D.C. the III
D.C. the III
overflowing with data.
was also doing a bit of cycling. which copper might like alot
 
Ted Shifrin
Ted Shifrin
I would think Shifrin has frustrated you more than Insel. For whatever it’s worth.
 
D.C. the III
D.C. the III
Actually no, cause with Shifrin I can "see" how it is going to be useful. Insel, I needed to take some time to see how it will be applied. I know Linear Algebra is a foundational structure for everything, but sometimes getting caught up in the abstract minutiae I lost sight of it.
 
copper.hat
copper.hat
i like cycling trails. and yes, i wish i could plug in an external drive :-)
 
Koro
Koro
3:54 AM
:(
I recently appeared in an interview for admission to the college I want to go to. I'm awaiting the result and am not sure if I'll get in.
 
D.C. the III
D.C. the III
Well you've made it to the interview stage so congrats on that. Is that a mandatory stage or is it done only for candidates on the cusp?
 
Koro
Koro
about 50% of the candidates are to be selected for admission based on marks in paper 1, paper 2 and interview.
 
D.C. the III
D.C. the III
ah...you'll be alright. At least from what you've displayed here you have a good grasp of things. At least with my limited knowledge...
 
Koro
Koro
They asked me 3 questions from linear algebra, 2 from analysis.
 
D.C. the III
D.C. the III
what did they ask? Did they ask how you woud "solve" these extremely hard questions and you have to provide a thought process? Maybe not have the solution but walk them through what you would do?
 
Koro
Koro
4:05 AM
I answered them but I believe that I was not 'fast' in answering.
the question 1) given n dim. vector space V and its subspace W of dim k, does there exist a linear map f from V to R such that that f(W)=0 but f is not a zero map?
I answered that. Take a basis of W and then extend it to a basis of V. Let $w_1,w_2,...,w_k, v_1,...,v_{n-k}$ Define f to be 0 on every basis vector of V except at $v_1$, where f is to be defined as $f(v_1)=1$.
 
copper.hat
copper.hat
as long as $k<n$.
 
Koro
Koro
I think they intended that W is a subspace (proper).
as they were happy with the answer.
 
D.C. the III
D.C. the III
How were the analysis questions?
 
Koro
Koro
Then there were two more questions based on this one. One was if set of all such f's is a vector space or not. First I answered No but then I immediately said yes and gave explanation.
The third was finding dim. of that subspace. I answered that.
Then came the analysis question. I was given an n dimensional inner product space V.
They asked how that can be made a metric space. I answered that.
inner product comes with a norm and norm can induce a metric etc.
Then, another question: If W is a subspace of V such that W is open in V, then show that W=V.
 
D.C. the III
D.C. the III
I think you're fine and should get in. When do you find out?
 
copper.hat
copper.hat
4:17 AM
best of luck.
 
Koro
Koro
To that, I said,"0 is in W and since W is open, there is a ball of radius r>0 centered at 0 i.e., B(0,r) that lies completely in W. And if v lies in B(0,r) then every scalar multiple of v lies in W as W is a subspace. I just have to formalize this now."
 
copper.hat
copper.hat
that is almost formalised. If $v \in V$ then ${r \over 2\|v\|} v \in B(0,r)$ so you are finished.
 
Koro
Koro
Then while writing it down- I got stuck! :( So I said," Sir, I think it'll take me some time solve this one." They said -you just solved it when you said (my last comment). But still I was stuck. Then they asked-write out B(0,r).
I did that and then I realized what I was missing and copper I wrote exactly what you have written.
 
copper.hat
copper.hat
excellent.
 
Koro
Koro
Since they said V is finite dimensional, I said if v_i is a basis vector of V then $r/2 v/||v||$ will lie in the ball and hence v lies in W (subspace).
and thus concluded.
they said -good.
Then they said -Ok, your interview is over. But [my name] , tell me why was finite dimensionality of V required?
I immediately realized that finite dimensionality of V was not required at all.
I said -it was given in the question so I thought I should use it.
Ok (or something similar in a positive way, I think), they said.
@D.C.theIII The date is not revealed.
 
Ted Shifrin
Ted Shifrin
4:26 AM
Too much linear algebra for me.
 
Koro
Koro
I have been refreshing the website page every now and then. I have one tab open in laptop and on phone all the time.
@TedShifrin I think it was a terrible interview as I got stuck in the last one even after saying out loud the strategy to solve the question.
 
Ted Shifrin
Ted Shifrin
I’m surprised they didn’t ask if every subspace is closed.
 
Koro
Koro
I myself said the strategy to solve the question but I got stuck while writing it out. :(
 
Ted Shifrin
Ted Shifrin
Everyone has little flubs, Koro.
 
D.C. the III
D.C. the III
Too much linear algebra in the interview Ted?
 
Ted Shifrin
Ted Shifrin
4:29 AM
The analysis was not really much analysis. So very unbalanced according to my view of mathematics.
 
Koro
Koro
I don't know why I got stuck there. I think I knew the answer deep down but I could not write it. I don't know why that happened.
 
Ted Shifrin
Ted Shifrin
So you present proofs at the black/whiteboard?
 
Koro
Koro
yes, at the whiteboard.
it was half an hour interview.
My interview was the last interview.
I think that for some people, it was around 40 mins.
 
Ted Shifrin
Ted Shifrin
This is for undergrad or for graduate?
 
Koro
Koro
for admission to a graduate program
 
Ted Shifrin
Ted Shifrin
4:34 AM
OK
 
Koro
Koro
for admission to undergrad, is open set asked somewhere?
 
porridgemathematics
porridgemathematics
it would be a little unusual to assume knowledge of any point set topology for admissions to an undergrad program in pure math
that being said I wouldnt be surprised if at some fancy universities they may see if you happen to know something like that just for fun
 
Ted Shifrin
Ted Shifrin
I have no experience with interviews for admission at any level. But open sets in a metric space are hardly point set topology.
 
porridgemathematics
porridgemathematics
thats fair enough, I thought 'open set' here meant in the most general set
i.e. assuming some high school student knows the definition of a topological space
that I would say is 'clearly' unusual but I may be wrong
(because it doesnt seem necessary to do well in a undergrad program to know this already)
 
Koro
Koro
Balarka may have known that in their highschool.
 
porridgemathematics
porridgemathematics
4:46 AM
which is fine, but in an interview they want to make sure you wouldnt clearly drop out of the course/fail hard fast
at the grad level of course the standards would be higher, its not enough for them to see you wouldn't fail immediately, lol
 
Koro
Koro
@copper.hat thanks for wishing me luck. :-)
 
copper.hat
copper.hat
:-)
 
Koro
Koro
@D.C.theIII Another thing that comes to my mind is: I understand that other interviewees came straight from college (that is, immediately after completing their undergraduation in Maths). But I completed my UG in 2018 and that too in engineering.
So how about their interview? It seems that it would surely be better than howsoever mine was.
 
D.C. the III
D.C. the III
I see, but you've done work to bring yourself up to par. You can only worry about what you have control over. You did your part in the interview, now just have to wait.
 
porridgemathematics
porridgemathematics
for what its worth, doesnt sound like your interview went badly , anyway no point worrying about it now
 
Koro
Koro
4:57 AM
I want to know from your experience-what would be a bad interview then?
1) Not knowing the strategy to solve a problem and getting stuck from the start of the problem.
2) Knowing the strategy to solve a problem but still getting stuck while writing it out.
In case of 1), they will give some hints, which I believe interviewee will build upon to get to the complete solution.
Same is true for 2). So 1) and 2) are the same. So 'taking hints in an interview' is bad?
 
copper.hat
copper.hat
nothing you can do now except wait. learn from the past but do not dwell.
 
Koro
Koro
5:17 AM
thanks @copper.hat @D.C.theIII @porridgemathematics :-)
 
leslie townes
leslie townes
i think i've said this before. some of the interviews that i thought went the 'worst', in my career, ended up leading to good places. and sometimes a good feeling after an interview is just a vibe and not meaningful.
 
copper.hat
copper.hat
i generally know in an interview how it went.
 
porridgemathematics
porridgemathematics
i tend to feel the same way, unless the interviewer is extremely opaque
 
copper.hat
copper.hat
doing psqs to maintain my user deleted & downvotes rep, i'll never get the jump suit at this rate
 
porridgemathematics
porridgemathematics
which happens from time to time
 
copper.hat
copper.hat
5:26 AM
i had one interview that went great until near the end.
i asked a little about the interviewer and it turned out that i knew his advisor very well, the interview basically stopped there.
 
porridgemathematics
porridgemathematics
wrong handshake?
ah.
 
copper.hat
copper.hat
i don't know what the issue was really. maybe it disturbed the 'i'm your boss' perspective.
experience has its downsides
 
porridgemathematics
porridgemathematics
most likely did, maybe he thought it was too incestuous to hire you
 
copper.hat
copper.hat
a lesson in sales is when you hit the crucial point just shut up.
which can be pretty difficult for me :-)
 
porridgemathematics
porridgemathematics
lol, I feel like that could be a lesson in life too
I have some friends who behave like salesmen 24/7
like, with opinions..
 
leslie townes
leslie townes
5:50 AM
don't you just hate that? it's good we have lesliecoin, a stable cryptocurrency to rely upon. instead of our no-good friends.
 
porridgemathematics
porridgemathematics
6:08 AM
at least you stopped at the crucial point
 
 
1 hour later…
Wave
Wave
7:23 AM
Can maybe someone help me here math.stackexchange.com/questions/4481249/…
 
Mad Spaces
Mad Spaces
8:16 AM
Good morning.
@Wave, you need to remove the 0 from your radius, since by theorem it is not included
In addition, i do not understand how you are plotting a function of two arguments on two d space.
Also try writing $e^{i\phi \alpha}$ using eulers theorem in Sine and cosine and check for the sign
 
 
2 hours later…
Mad Spaces
Mad Spaces
10:32 AM
A monoid is usually defined with an inner operation, is there a name for an object that is similiar to a monoid, but with an outer operation?
IE fullfils associative properities, the existence of 1 element.
We can say a vector space is a set and two operations, one of them build an abelian group, i was thinking if we can say something elegantly similiar to the other set of operations. such as the set adn the multiplication are a commutative outer monoid with distrubutive properities
 
 
1 hour later…
monoidaltransform
monoidaltransform
11:44 AM
 
Wojciech Kulma
Wojciech Kulma
12:42 PM
this definition of stochastic boundess is incorrect isn't it?
https://imgur.com/a/klycTv5

the M should be kept constant and n should go to inf, right?
 
Alessandro Codenotti
Alessandro Codenotti
12:52 PM
@Koro Here’s a nice generalization: suppose $E$ is a Banach space and $F$ a subspace of $E$ which has the Baire property and is not meager. Then $E=F$.
(The point is that this already implies that $F$ is open)
 
 
1 hour later…
Jakobian
Jakobian
2:05 PM
@MadSpaces what's an outer operation? And how would you make a theory of them
 
Mad Spaces
Mad Spaces
2:41 PM
I am roughly translating, inner operations are operations acting on the set and go to the set itself, such as the ones defined on groups and monoids, an outer operation is for example scalar multiplication, where you are taking an element that is not in the set (fore xample from the field not the vector space) and going into the vector space
 
Alessandro Codenotti
Alessandro Codenotti
3:06 PM
So you want a monoid action on some set?
 
Mad Spaces
Mad Spaces
I am just asking for a nice naming
since i would like to say "a vector space is a set with an operation that creates an abelian group and another operation that makes ".........naming here "
 
Alessandro Codenotti
Alessandro Codenotti
A vector space is an Abelian group with a ring action of a field by endomorphisms
 
Akiva Weinberger
Akiva Weinberger
3:23 PM
@JoeShmo Did you ever do this?
 
geocalc33
geocalc33
for a braid on n strands if you further identify the two sets of distinguished points on the parallel planes P_0 and P_1 to become just two distinguished points say a and b, then you lose the permutation structure?
 
chaaru
chaaru
3:49 PM
Hello
 
geocalc33
geocalc33
Hello
 
chaaru
chaaru
I am trying to obtain outage probability for the given expression but not getting it clearly.

$P_o = \text{Pr}\biggl(XY \leq \frac{\gamma_t(\mu_3Z+\mu_4)}{\zeta-\frac{\gamma_t \mu_1}{Y}-\frac{\gamma_t \mu_2}{X}}\biggr)$, where $X,Y,Z$ are independent exponential random variables having CDF $F_X(x), F_Y(y), F_Z(z)$ and PDF $f_X(x), f_Y(y), f_Z(z)$ and all other things are constant.
Any logic to solve this problem would be highly appreciated.
 
Violet Flame
Violet Flame
4:23 PM
On pascal triangle does it hold that sum followed by differenece of a row is equal to 0. For example in the row "1 4 6 4 1" it holds that 1-4+6-4+1=0, is this true of all rows?
 
Joe Shmo
Joe Shmo
@AkivaWeinberger yes, I sent him an email. Still waiting for him to get back to me.
 
Koro
Koro
4:42 PM
Hi
 
copper.hat
copper.hat
Hi!
 
leslie townes
leslie townes
violet: yes consider the binomial expansion of (1 - 1)^n
 
Koro
Koro
Today the results came out. And I am happy to inform that I am in the merit list :-)
 
leslie townes
leslie townes
and ignore the first row, where there's no alternating anything to sum up
 
Koro
Koro
Finally I'll join my dream college for masters :-)
 
leslie townes
leslie townes
4:44 PM
koro: does that mean accepted?
 
copper.hat
copper.hat
I got one -10 because serial upvote was removed. Which is curious, because if it was a serial upvote, surely in the vernacular, serial implies more than one?
 
Joe Shmo
Joe Shmo
good job @Koro !
 
leslie townes
leslie townes
oh, cool. congrats
 
Koro
Koro
@leslietownes yes :-)
 
Joe Shmo
Joe Shmo
what college is that
 
copper.hat
copper.hat
4:44 PM
Congrats @Koro well done
remember me when you are head of tata
 
Joe Shmo
Joe Shmo
hah
 
Koro
Koro
Thanks a lot @leslietownes @copper.hat @robjohn @TedShifrin and the others for guiding me :-)
I have also submitted my resignation today to the company I'm employed at.
:-)
 
Ted Shifrin
Ted Shifrin
Congrats, Koro. On both counts! Now the serious work begins :D
I got -60 or -70 for correction of serial voting, @copper.
 
Koro
Koro
Thanks a lot dear professor @Ted.
@TedShifrin me -20 today.
 
copper.hat
copper.hat
@TedShifrin that makes sense, at least serial implies a sequence of prior upvotes. But -10 implies a single vote, hardly a pattern.
It just amuses me, other than interfering with my jumpsuit progress.
 
leslie townes
leslie townes
4:54 PM
copper: same happened to me. what about my jumpsuit progress?
i'm beginning to think i'll never hit 100k
this is why i'm grateful for lesliecoin, where i have well over 100k, and the indelible blockchain ensures that all transactions are irreversible
 
copper.hat
copper.hat
:-)
 
Ted Shifrin
Ted Shifrin
I still have the empty jumpsuit.
And no lesliecoin.
 
Jakobian
Jakobian
If $\mathcal{M}$ is a family of $w(F, E)$-bounded subsets of $F$, and we create local base for $E$ by taking finite intersections of multiples of the sets $M^\circ$ with $M\in \mathcal{M}$, where $M^\circ$ denotes the (real) polar of $M$, then why is the multiplication continuous?
(E, F) here is a dual pair (in the sense that elements of F separate elements of E, and conversely)
I get stuck in trying to bound $\text{Re} \langle tsz, m\rangle$ where $z\in M^\circ$, $t > 0$ and $s\in B(r, a)$ where $a> 0$
If M were to be absorbing, then I think I'd know how to finish this, but here I'm clueless
The only thing that I can change here is t and a, and the bound is supposed to be independend of m (so I'm trying to bound the supremum over all of m)
 
Mad Spaces
Mad Spaces
5:25 PM
Stupid question incoming
So as i undertsand, the definiteness of a bilinear form is dependendent on the choice of basis. Why does this attribute then matter, if it is not globaly inherited?
 
leslie townes
leslie townes
the notion can be phrased in ways that look basis dependent, but the notion is not basis dependent
which hopefully answers the second question. this is not a stupid question, btw
 
Mad Spaces
Mad Spaces
What do you mean the notation can be phrased .. can you clarify.
 
leslie townes
leslie townes
the notion. typo.
 
copper.hat
copper.hat
@MadSpaces it i like determinant of a linear map
 
leslie townes
leslie townes
a bilinear form either is or isn't definite. independent of basis.
some of the ways of characterizing definiteness do make use of bases, but the end result does not depend on them
 
Mad Spaces
Mad Spaces
5:29 PM
The determinant is not dependendent on choice of basis. well maybe it changes orientation
 
copper.hat
copper.hat
neither is definiteness
 
leslie townes
leslie townes
people often do care about particular bases, also, because e.g. diagonalizing the form can make calculations easier. i wouldn't say that bases 'don't matter' for purposes of thinking about bilinear forms
 
Mad Spaces
Mad Spaces
I do not seem to understand, like i have a question infront of my eyes, where it states, find basis such that the bilinear form is once = 0 once positive definit once negative, so i am not sure i understand what you mean, it is not dependent on basis choice
 
leslie townes
leslie townes
maybe include the question verbatim
 
Mad Spaces
Mad Spaces
But it is not about the question per se.
It is more about, if i can find basis, where i can change the definitness. why do we bother?
Am i being stupid
 
leslie townes
leslie townes
5:31 PM
well, you can't, which is why it might help to include this question verbatim
so people can figure out what it's really asking, or if it's wrong
it very much is about this question per se if it's telling you that definiteness is basis-dependent
 
Mad Spaces
Mad Spaces
So are you saying, that a bilinar form, has an intrinsic definiteness. but it is not examanable given a certain base, since another base could give you a different one?
 
leslie townes
leslie townes
no, not at all
you can use a basis to compute stuff relating to a bilinear form
one of those things being whether it is definite or not
this doesn't mean that the result depends on the basis. it doesn't.
 
copper.hat
copper.hat
why don't you post the question otherwise this is just going to be another loop
 
leslie townes
leslie townes
you keep saying that it does, and i can't tell if that's because you have some independent reason for thinking this, or if it's just some (as yet unstated) question that's raising this confusion
 
Mad Spaces
Mad Spaces
Well.. Honestly because i have to go catch a bus in 5 minutes
 
leslie townes
leslie townes
5:34 PM
include it when you get back :)
 
Mad Spaces
Mad Spaces
lol
Yea, i am going to straight into bed. so i guess tomorrow.
 
Ted Shifrin
Ted Shifrin
Give the definition of definiteness, too.
 
Mad Spaces
Mad Spaces
But thanks for the try, i will mark you when i write it
 
leslie townes
leslie townes
ted's suggestion is, as always, a good one. as i mentioned above, you can certainly define definiteness in a way that makes it appear as though it might depend on a choice of basis. but this doesn't mean that definiteness does depend on the choice of basis.
like determinants in coppers example
 
Jakobian
Jakobian
I don't think anyone will help me. I guess I'll just be stuck like always
Polar topology
In functional analysis and related areas of mathematics a polar topology, topology of G {\displaystyle {\mathcal {G}}} -convergence or topology of uniform convergence on the sets of G {\displaystyle {\mathcal {G}}} is a method to define locally convex topologies on the vector spaces of a pairing. == Preliminaries == A pairing is a triple ( X , Y , b )...
I need something like this but instead of absolute polars, I am working with real polars
 
Ted Shifrin
Ted Shifrin
5:50 PM
I'm not being rude, @Jakobian. I understand very little of what you typed.
 
Koro
Koro
what's up with 'serial voting was reversed'?
There is a topology in which Q is not open but the set of irrationals is open 😮😮
 
Ted Shifrin
Ted Shifrin
Think of $\Bbb Q$ as $\Bbb Z$.
It is far from the order topology. I have never thought about this before.
 
Koro
Koro
6:05 PM
Or I consider R and $U\subset R$ to be open if $U=\emptyset$ or $R\setminus U$ is countable. Then the set of all such U's gives me the topology on R.
 
Xander Henderson
Xander Henderson
@Koro It is likely related to the following:
94
Q: Upcoming cleanup of duplicated votes

Yaakov EllisIt was reported on MSO that it is possible for a user to upvote or downvote more than once on a single post (resulting in reputation changes for the author for each vote). After some data examination, we were able to confirm that this vulnerability exists (it is happening due to race conditions s...

discussion featured voting announcements duplication
(though it needn't be---it could also be that you had a fan / enemy whose votes have been overturned).
 
Ted Shifrin
Ted Shifrin
“Race conditions” sounds particularly bad these days.
@Koro So any countable set of irrationals is closed.
 
Koro
Koro
I have not yet studied closed sets in general topology :(
but I believe that would be complement of open sets
as in metric space topology
 
Akiva Weinberger
Akiva Weinberger
Yes
 
Ted Shifrin
Ted Shifrin
Yes, that’s a proposition, not a definition.
 
Koro
Koro
6:13 PM
Oh
The definition would involve limit points?
 
Akiva Weinberger
Akiva Weinberger
Hrm? How do you define closed sets in general topology?
 
Ted Shifrin
Ted Shifrin
The way Munkres does.
Yes, Koro.
 
Koro
Koro
I came across Furstenberg's proof of infinitude of primes. In the proof, closed sets were used.
 
Akiva Weinberger
Akiva Weinberger
@TedShifrin I dare you to guess my next question
 
Ted Shifrin
Ted Shifrin
Right. That’s sorta famous, Koro.
 
Akiva Weinberger
Akiva Weinberger
6:16 PM
@Koro If you're free to choose any topology you want on the reals, all that matters is cardinality, so it's just about getting an uncountable space with a nonopen countable subset with open complement
@TedShifrin How does Munkres define them
 
Ted Shifrin
Ted Shifrin
Actually, I was wrong. Haven’t taught topology in 20 years. My book used what I said. Munkres used what DogAteMy wants.
 
Akiva Weinberger
Akiva Weinberger
Oh, with limit points. I missed that
 
Ted Shifrin
Ted Shifrin
My book uses a sequential statement, because that’s useful for the analysis in $\Bbb R^n$.
 
Akiva Weinberger
Akiva Weinberger
Is that even valid in general topology?
 
Ted Shifrin
Ted Shifrin
No, of course not. Containing limit points is, though.
 
Jakobian
Jakobian
6:20 PM
$A'\subseteq A$? Yes, of course
 
Akiva Weinberger
Akiva Weinberger
@Jakobian I assumed the "sequential statement" was "a set is closed iff it contains all limits of sequences in the set"
which in fact fails for the cocountable topology
 
Jakobian
Jakobian
That one is not true in general, but it's true if we replace the word "sequence" with "net"
 
Akiva Weinberger
Akiva Weinberger
Sure
 
Koro
Koro
@AlessandroCodenotti Thanks for the comment. I don't yet know Banach spaces :(
 
Akiva Weinberger
Akiva Weinberger
A Banach space is a (usually infinite-dimensional) vector space with a norm, which is complete as a metric space with respect to that norm
 
Ted Shifrin
Ted Shifrin
6:26 PM
I had my pedagogical reasons for multivariable analysis, of course.
 
Koro
Koro
@AkivaWeinberger doesn't every vector space have a norm?
 
Akiva Weinberger
Akiva Weinberger
An example is the set of all infinite sequences of real numbers $(a_n)_n$ such that $\sum|a_n|$ converges. With the norm $\|a\|=\sum|a_n|$, this is called $\ell^1$
@Koro The norm is part of the data of the Banach space
 
Koro
Koro
hmm, I'm not yet sure if all infinite dimensional vector spaces have norms.
 
Akiva Weinberger
Akiva Weinberger
You choose the norm
@Koro I think you can show this using Hamel bases (and the axiom of choice)
 
Koro
Koro
For finite dim., there are basis vectors which can create orthogonal basis using Gram Schmidt.
 
Jakobian
Jakobian
6:30 PM
@Koro Every vector space admits some norm, but it's not "part of the data" of a vector space
 
Akiva Weinberger
Akiva Weinberger
Another example is, in general, you can have all infinite sequences where $\sum|a_n|^p$ converges. This is a metric space as long as $p>1$. This is called $\ell^p$ space
Wait
The norm is $(\sum|a_n|^p)^{1/p}$
Yeah
 
Jakobian
Jakobian
@Koro Sure, if $(x_i)$ is a Hamel basis, then define $\|\sum a_ix_i\| = \sum |a_i|$
 
Akiva Weinberger
Akiva Weinberger
There's also $\ell^\infty$ where the norm is $\sup|a_n|$. This is the limit of the $\ell^p$ norm as $p\to\infty$ (note that if $p<q$ then $\ell^p\supset\ell^q$)
$\ell^2$ is special 'cause you can also give it an inner product
 
Koro
Koro
@AkivaWeinberger ah yes. I know that space :). We have Minkowski inequality also in this space.
 
Akiva Weinberger
Akiva Weinberger
@Koro The Minkowski inequality is precisely the statement that $\ell^p$ is a metric space, yeah
 
Jakobian
Jakobian
6:33 PM
Iirc $l^p$ is equivalent to a Hilbert space iff $p = 2$
 
Koro
Koro
I think I said the following thoughtlessly
 
Akiva Weinberger
Akiva Weinberger
Hilbert spaces are Banach spaces with inner products (that agree with the norm)
 
Koro
Koro
5 mins ago, by Koro
For finite dim., there are basis vectors which can create orthogonal basis using Gram Schmidt.
 
Jakobian
Jakobian
Well, that's kind of obvious looking at the duals of those spaces
 
Koro
Koro
I should know for my statement to be valid: "Every finite vector space has an inner product."
 
Akiva Weinberger
Akiva Weinberger
6:35 PM
Yes
Over $\Bbb C$ or $\Bbb R$
 
Jakobian
Jakobian
What does that mean though
 
Koro
Koro
yes over C or R.
 
Akiva Weinberger
Akiva Weinberger
@Koro However, there are normed finite vector spaces where the norm does not come from an inner product
 
Koro
Koro
@Jakobian I mean that suppose we have an n dimensional vector space V over F (F is either R or C), can we say that V is an inner product space?
 
Akiva Weinberger
Akiva Weinberger
Eg $\Bbb R^2$ with the norm $\|(a,b)\|=|a|+|b|$
@Koro Yes, it can be equipped with an inner product
 
Jakobian
Jakobian
6:37 PM
@Koro Uh... kind of not really? Inner product is not part of the "data" for a vector space
 
Koro
Koro
yes, I think $l^p$ norms with p<1 also doesn't come from inner product.
 
Akiva Weinberger
Akiva Weinberger
@Koro If p<1 it's not a norm
 
Jakobian
Jakobian
They are quasi-norms
 
Akiva Weinberger
Akiva Weinberger
@Jakobian Yeah it doesn't have one chosen but you have options
 
Koro
Koro
@Jakobian yes. I mean -can we define one inner product on V?
 
Jakobian
Jakobian
6:39 PM
@AkivaWeinberger It's always linearly isomorphic to one though
 
Akiva Weinberger
Akiva Weinberger
@Koro Not uniquely but yes
 
Koro
Koro
Oh. Is there an obvious one that we can define?
 
Jakobian
Jakobian
the dot product
just choose a basis
 
Koro
Koro
what if I define <,>: V cross V --> R as <v,u>=0 for every v not equal to u?
@Jakobian and then I declare them to be orthogonal?
 
Alessandro Codenotti
Alessandro Codenotti
@AkivaWeinberger it's equivalent to closed for first countable spaces
In general the spaces for which closure=sequential closure are called Frechet-Urysohn
 
Ted Shifrin
Ted Shifrin
6:47 PM
@Koro This is a different question than starting with a norm.
 
Koro
Koro
I wanted to say -inner product norm following my comment about using Gram Schmidt.
(I may have missed "inner product" and written only 'norm' above.)
 
Adil Mohammed
Adil Mohammed
hello folks
can i say the imaginary number "i" cant be measured and thats the reason why i cant compare two complex numbers and say which one is greater
 
Ted Shifrin
Ted Shifrin
There is length and distance. What does “can be measured” mean?
 
Adil Mohammed
Adil Mohammed
thats a good question
i mean something that isnt just the absolute value from 0 on the argand plane
 
Ted Shifrin
Ted Shifrin
That’s not answering.
 
Adil Mohammed
Adil Mohammed
6:58 PM
ik, but i was under the notion it cant be measured
 
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