Mathematics - 2018-11-21 (page 1 of 3)

Jake Rose

01:56

Anybody here?

Jake Rose

02:11

If S is real symmetri, T is real anti-symmetric show that det(T+iS−1) is non zero

Anybody got an idea as to how to 'deduce' this?

Adam

02:55

anyway I too have an issue with the way in which the supreme fascist says things ought to be, I feel as if I am missing something axiomatic here that will help this sit better with me $$\prod _{j=n+1}^{n-k-1}a_{{j}}=\prod _{j=n-k}^{n}\frac{1}{a_{{j}}}$$

Prashin Jeevaganth

04:31

Does anyone have any knowledge of game theory here?

Have some questions at an elementary level not involving much calculations, just some theoretical concepts to grasp for me

JBis

04:46

@PrashinJeevaganth I would suggest just asking. If someone can (and wants to) answer then they will.

Prashin Jeevaganth

In a zero-sum game, just considering pure strategy, we can find a saddle point due to the existence of a dominant strategy A. Is there any motivation for either party to choose to pursue a mix strategy to try to milk out more gains from the other player?

Secret

a mixed strategy zero sum game always have an equilibrium. It might be useful to check whether the equilibrium is tending towards the pure strategy limit to assess whether extra gain can be obtained by going from pure to mixed

Prashin Jeevaganth

@Secret Is it true by Nash's Theorem, any mixed strategy(zero-sum or variable sum) definitely has an equilibrium, regardless of which in the pure strategy there is a Nash equilibrium

Can't understand why sometimes there's one from each case and sometimes we only have to depend on the mixed strategy one

Nobody recognizeable

06:40

Hey guys i was thinking from definition a harmonic function $\phi$ satisfies $\nabla^2 \phi $=0 so does that indicate sinx is not a harmonic function.

Secret

@PrashinJeevaganth yes, every finite game has a Nash equilibrium, but not necessary a pure strategy Nash equlibrium

1

Q: Why Are Sine and Cosine Called Harmonic Functions

I thought that the definition of a harmonic function f such that $\nabla^2f=0$ In one dimension, doesn't this mean a function who's second derivative is zero? IE $\frac{d^2f}{dy^2}=0$ However, the sine nor cosine function's second derivative does not equal zero, but in many textbooks they are...

because historical reasons

Unknown MathMan

Which introductory book on logic will be best for an Undergraduate? I mean discussed everything clearly

Nobody recognizeable

@Secret thanks.

Secret

07:07

logic questions are best to be asked in the logic room, as 21820 and co are the experts there

Prashin Jeevaganth

07:29

Does anyone understand how x.x gets me x+1?

I dont get how the derivation of x is the multiplicative inverse of x+1 gets me the other result

Secret

basically trial and error. There are only 4 elements in this field and 0 and 1 cannot be the multiplicative inverse of x

if x.x=1, it leads to the contradiction that x=1, thus by elimination, only x+1 remains

Prashin Jeevaganth

@Secret So am I right to conclude that the result has nothing to do with the above derivation that x is the multiplicative inverse of x+1?

Basically when making some weird finite field like this, I have to trial and error out for all cases

Secret

well, for any finite field, there are cases that always holds: x.x cannot be x else inverting gives x=1, and x.x cannot be 0 else inverting gives x=0. Otherwise, unless there are further contradictions, x.x can be anything else

For this finite field, x.x=1 lead to x=1 thus it is also ruled out.

I don't know if there is some general way to test what the possible result when a given field element is multiplied to another

x.(x+1) = 1 as deduced above
Also x.(x+1)=x.x+x by distributivity. This latter equation can be used to find further restrictions

One possible way is to consider the contrapositive. I.e. we knew that if an element has a multiplicative inverse, you cannot be absorbing i.e. ax=a cannot hold.

Thus we can phrase the problem this way in trying to find the inverse of x:

Akiva Weinberger

07:54

Why does the crochet hyperbolic plane not contradict Hilbert's theorem

Secret

Pick any a in the field, you do not want ax=a. Now we an already ignore 0 and 1 since they can never be multiplicative inverses of other elements. That leaves x and x+1. Now suppose x(x+1)=x+1. Expanding we get x.x+x=x+1 and by adding the additive inverse, we get x.x=1, but that is not allowed. We obviously don't want x.x=x either, because then x=1. This means x(x+1)=/=x+1. Now because this is a field, there are no zero divisors, meaning x(x+1)=0 is out. we can also easily ruled out x(x+1)=x+1 and x(x+1)=x as they will then be absorbing elements, which does not exists in fields.

Prashin Jeevaganth

@Secret I saw a loophole in this question ... we could define 2 = 1+1 which is actually 0 in the field

Consider x·(x+1) = 1, and from its LHS, x.(x+1)+x = x.(x+(1+1)) = x.(x+0)= x.x

Secret

In fields, the following cannot hold for any elements ab=ba=a, ab=0

We are interested in finding c such that xc=1. Now c cannot be x,0 or 1 because then it will obey any of the above equations (and zero is never a multiplicative inverse in any field, ring etc.). That leaves x+1 as the answer to the inverse of x

Responding to the new paragraph next:

Prashin Jeevaganth

RHS would be 1+x (when we add x to both sides)

hence we have found another multiplicative inverse

Secret

x.(x+1)=1 => x.x+x=1
x+1 = x + x.(x+1) = x.(1+x+1)=x.x
x+1=x.x
x+x+1=x.x+x
1=x.x
x=1
contradiction.
This means 1+1 cannot be 0

if any of the pathways lead to a contradiction, that assignment is forbidden because equality is transitive

wait typo

Prashin Jeevaganth

08:08

@Secret but the question said addition of coefficients are done in the Z/2Z world

Secret

x+1 = x + x.(x+1) = x.(1+x+1)=x.x
x+1=x.x
x+x+1=x+x.x
x(1+1)+1=x(1+x)
0+1=x(1+x)
1=x(1+x)
No contradiction

so no there isn't another inverse of x

Prashin Jeevaganth

@Secret ah ok badly phrased...

oh cool

actually since there are only 3 non-zero elements in this field

and 1 doesn't need to have an inverse(because 1.1 = 1)

we could say informally that x is the multiplicative inverse of x+1 right?

Secret

yes, all two sided inverses are unique. This is why fields are so nice

Prashin Jeevaganth

How big of a field do people usually use?

this is a 4 element one lul

Secret

depends on the domain of study. Polynomials over finite fields are important in algebraic geometry (details I understood nothing of except I am aware of it when mathien and co talk about it in this chat, as well leaky talking about ideals and stuff) and number theory (related to factorisation stuff). Infinite fields such as the reals and complexes made up most of analysis

Prashin Jeevaganth

08:20

Me learning stuff that I don't understand and questioning 1+1=2

-Math

Secret

The field with four elements as an example of finite fields is important in the theory of irreducible polynomials, it is known as GF(4)

Prashin Jeevaganth

The moment I saw "irreducible" XD

Secret

In mathematics, the following is guarentee to give me a BSOD:

1. Having more than 3 inequality signs appearing in the same line of a proof (bonus point if there is one or more absolute value of the form |f(x)=g(a)|

2. Anything that has to do with the word "prime"

3. Irreducible polynomials

I have ZERO intuition on any of these

Prashin Jeevaganth

what's a BSOD?

Secret

knowyourmeme.com/memes/blue-screen-of-death-bsod

basically you computer ceased to respond and hangs

so in this context, my brains hangs when I am exposed to the above 3 things

(Number theory as a whole used to be the 4th, but after I understood it is basically the algebra of factorising numeric expression, I started to get some intuition on it)

My current understanding is that number theory governs the rules of how numbers are partitioned

from fermat last theorem, to sum of two primes problem, to collatz etc., it is the science on figuring how how to split numbers apart in certain ways

I also like general topology more because it will replace almost every $>$ with $\supset$ which strangely, will not cause me BSOD

Prashin Jeevaganth

08:30

Splitting numbers ...

How many ways to express 1 and 0

lol

Secret

and most importantly, the lack of a metric consideration means the absolute sign can rarely appear in topology, which is often give me a lot of headache in mental arithmetic

Akiva Weinberger

5

Q: Galois Field GF(4)

Question: Why is the table of GF(4) look like the one below? I know it has to do with the fact that 4 is composite Let GF(4) = {0,1,B,D} Addition: $$ \begin{array}{c|cccc} + & 0& 1& B & D \\ \hline 0& 0 & 1 & B & D \\ 1 & 1 & 0 & D & B \\ B & B & D & 0 & 1 \\ D & D &...

field-theory finite-fields

Seems related to what you were talking about

Prashin Jeevaganth

@AkivaWeinberger that's the higher level topic XD

Akiva Weinberger

Strange notation^

Prashin Jeevaganth

I hope they dont test Galois fields in my exams XD

haha

Anyone know why the reason why 9,10 are False,True respectively? I just plugged in the formula for random walks to my understanding

got a 10^-5 number for Q9, and 500- 0.04 for Q10

Trying to find my sanity

Secret

08:41

I cannot visualise probabilistic dyanamics very well without reams of paper

and it is at least 8 years since I last did a random walk problem

Prashin Jeevaganth

@Secret omg, it's either your a genius or you are very old

I can't imagine my age -8

Secret

I am a PhD in computational chemistry

I have all the age you need to not give the negative number

lol

Also:

Nov 15 at 11:03, by Secret

Prashin Jeevaganth

I didn't understood that either

LOL

Secret

Nov 15 at 11:15, by Secret

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

My learning is very lopsided. I do not follow the usual step by step and my logic make erratic jumps alot from one random math field to another

The only thing which I can say to have a rather solid core on is linear algebra

anything else is basically picked up throughout this chat and some random reading

Prashin Jeevaganth

@Secret would you happen to know why the answer for the following questions is F,T,F?

Secret

08:50

I have not studied anything about geometric partitions, thus I don't know

Prashin Jeevaganth

does this have anything to do with that?

Secret

(and I am surprised this is actually a field of study)

Prashin Jeevaganth

OMG, I don't even understand what I'm getting into

how come all my questions come from a different math field

fields, random walks and now this

Secret

Envy-free cake-cutting

An envy-free cake-cutting is a kind of fair cake-cutting. It is a division of a heterogeneous resource ("cake") that satisfies the envy-free criterion, namely, that every partner feels that their allocated share is at least as good as any other share, according to their own subjective valuation. When there are only 2 partners, the problem is easy and has been solved in Biblical times by the divide and choose protocol. When there are 3 or more partners, the problem becomes much more challenging. Two major variants of the problem have been studied: Connected pieces, e.g. if the cake is a 1-...

wtf

what is in your exam syllabus?

Prashin Jeevaganth

Some pretty insane shit, taught on a very skimpish level, but the questions escalate quite quickly

Secret

09:06

pretty applied stuff. Is this commerce mathematics?

Elsa

I'm stuck at the Second Isomorphism Theorem , precisely I'm confused with the meaning of HN/N , where H is a subgroup of G and N is normal subgroup of G. Shouldn't HN/N be equal to H/N ?

Tobias Kildetoft

@Elsa No, why should it?

Elsa

Well , HN/N is a subgroup of H/N for sure , right ?

Tobias Kildetoft

no, why?

when you multiply subgroups, you get something bigger

Elsa

Yeah but , if you took (hn)/N element of HN/N , you get (hn)/N = hN which is an element of H/N ?

Where am I going wrong then

(hn)N **

Tobias Kildetoft

09:13

Ohh, the problem here is that $H/N$ does not make sense

Prashin Jeevaganth

@Secret the name of the module is called Living with Mathematics, definitely it's about showing how abstract stuff can be used in real life ...

but the exam is still abstract

Elsa

@TobiasKildetoft could you please elaborate why ?

Prashin Jeevaganth

Logic and Sets, Functions, Voting System, Game Theory, Counting, Graph Theory, Number Theory,Counting

Tobias Kildetoft

@Elsa Because $N$ is not a subgroup of $H$

Elsa

OF COURSE yea, thanks a lot :)

Secret

09:15

that subejct really seems like it is geared towards politicians and businesses

Prashin Jeevaganth

@Secret At least through this module I can really see where my theoretical math is going

how banks can charge people in excess

or how governments adopt systems to ensure they are always in power

lol

Well the questions are mainly True or false ... it's hard to research on them because they are all over the place

I literally placed a toe in every field

Michael.P

09:37

Is the first statement true in all cases? Because it seems to me that if S = B then sup B not contained in B won't be present in S either

Balarka Sen

@TobiasKildetoft The correct statement along those lines here is that $HN/N$ is the image of $H$ by the quotient map $G \to G/N$, I think.

So $HN/N$ is the correct meaning of "$H/N$" which is per se meaningless

Tobias Kildetoft

@BalarkaSen Sure

Danu

10:26

@BalarkaSen o/ long time no see

Balarka Sen

hi

Danu

how's uni?

Balarka Sen

good

Danu

What kinda stuff are you learning about

Balarka Sen

various things

Danu

10:29

hmkay

here's a little something I've been puzzling about for the past few hours

I'm trying to understand this theorem that any simply connected surface with $b_2=1$ is biholomorphic to $\Bbb CP^2$

It suffices to prove that it is homeomorphic to it

Since simply connected 4-manifolds are classified up to homeo by the intersection form, and we know $b_2=1$, only have to exclude intersection form $(-1)$ or equivalently $\sigma=-1$

The way to do this must be to use the signature formula $\sigma=1/3p_1(X)$ where $p_1$ is the first Pontryagin number

for complex surfaces this takes the form $p_1=c_1^2-2c_2=c_1^2-2\chi=c_1^2-6$

So $\sigma=1/3c_1^2-2=-1\implies c_1^2=3$ should lead to a contradiction somehow

but I don't know how to get one

Any ideas? :D

Balarka Sen

idk this stuff

Danu

If I could prove the ring structure on cohomology is the same as $\Bbb CP^2$ we're done

ah, but doesn't that follow from poincare duality?

Yes, I think it does, because the matrix representing the map $c\in H^2(X,\Bbb Z)\mapsto (c\cup c)[X]$ is unimodular

so a generator $c$ must square to $\pm $ a generator in $H^4$

Balarka Sen

when you write $b_1$ do you mean $b_2$

Danu

ohhhh derp

yes I do haha

sorry about that contradiction :D

Balarka Sen

yes, that was confusing

Danu

10:40

llmao you must'\ve been sitting there like

WHAT IS THIS GUY SMOKING

Balarka Sen

heh. so yeah cup square is $+1$

its a complex manifold

Danu

ah wait why does it have to be +1 and not -1?

this is equivalent to the signature haha

Balarka Sen

Intersection numbers are always positive in complex manifolds, right?

Danu

I thought only if your class is represented by a complex submanifold

Balarka Sen

Ah fair.

Danu

10:44

anyways, I thought we could argue like this

Balarka Sen

Yeah, sure you can look at the zero section of $O(-1)$

Danu

since we know a generator, say $\alpha$, squares to $\pm$ a generator

Balarka Sen

Which compactifies fiberwise to $\Bbb{CP}^2 \# \overline{\Bbb{CP}^2}$.

Danu

then we know that if $c_1=d\cdot \alpha$ (plus possibly torsion)

then $c_1^2=d^2\alpha^2=\pm d^2$ times the positive generator of $H^4$

so it can never evaluate to $3$ since it must be a square

that works, no?

Balarka Sen

I don't know too much about characteristic classes but if I trust you on $c_1^2 = 3$ that sounds good

My immediate approach is that the submanifold that represents $\alpha$ must have an $O(-1)$ tubular neighborhood. So I'd blowdown it.

Danu

10:48

:)

Balarka Sen

I suppose it's understood how Betti numbers change under blowup. I don't know too much about it.

Danu

yeah

for 4-manifolds connected sum just is additive on the level of intersection form

eurocoder

11:12

Has anyone here read Adam's book on Sobolev spaces? I have a headache at the moment due to a section where he says a discontinuous function is in a space of continuous functions! I asked about it here - math.stackexchange.com/questions/3007565/… - if you are familiar with this book it would be great if you could clarify what he is saying here as its really annoying me!

Akiva Weinberger

13:16

Do paths in a topological space (up to homotopy) make a category?

Where the objects are points of the space and the morphisms are paths

Balarka Sen

Yes.

Well, morphisms are paths upto homotopy, if you want to do that.

It's called the fundamental groupoid of the space

Akiva Weinberger

OK

I knew it was a thing, I didn't realize that the thing was a category

since I'm used to morphisms in categories being functions

Alessandro Codenotti

And you can build a whole tower of higher categories in a similar way

Akiva Weinberger

but I guess it's really just an algebraic structure like any other

Balarka Sen

its a nice formalism but idk what one does with it

its just... there

Alessandro Codenotti

13:21

Thing about posets as categories for examples where the morphisms are not functions

I hear there's a version of SvK where the intersection is not connected that works with groupoids @Balarka, do you know anything about that?

Balarka Sen

The categories where objects "look like sets" and morphisms "look like functions" are called concrete categories.

Akiva Weinberger

Like symmetry groups

where the objects "look like permutations"

Balarka Sen

There's a theorem out there which says that hTOP is not concretizable (remember that morphisms in hTOP are not maps, but maps upto homotopy)

@AlessandroCodenotti I have heard of it but I don't know anything about it

Akiva Weinberger

Every group is isomorphic to a symmetry group

Balarka Sen

@Mathein does

Akiva Weinberger

13:23

so I guess that's one difference

Balarka Sen

Right

Alessandro Codenotti

I'm trying to remember the name of the guy, he has a whole book on this kind of stuff

Balarka Sen

Ronnie Brown

Akiva Weinberger

The reason I bring it up is I was reading this thing http://www.cs.ox.ac.uk/samson.abramsky/tambook.pdf

Alessandro Codenotti

@BalarkaSen Exactly! Thanks

Akiva Weinberger

13:25

and they introduced the "Temperley-Lieb category"

and I was like, "That's not a category

"…oh wait it is"

Tobias Kildetoft

@AkivaWeinberger Sure, it is just a categorification of the Temperley-Lieb algebra

Akiva Weinberger

'cause it was also not concrete

Is "categorification" an actual technical term

Tobias Kildetoft

in some contexts, yes. Mostly it is used in a vague sense

Akiva Weinberger

or is it just "this is a thing that's like another thing but has categories in it now"

Aha

Tobias Kildetoft

As in, there are papers going into great detail defining what a categorification of a certain objects needs to be

Balarka Sen

13:27

:S

2abstract7me

Tobias Kildetoft

But mostly, people just mean something like "it is a (insert adjectives) category which decategorifies (via chosen method) to the original object

Akiva Weinberger

@TobiasKildetoft "If it looks like this other thing in this particular way, but also has categories in, it needs to look like this"

@TobiasKildetoft Wait, sorry, "decategorifies"?

Tobias Kildetoft

Where most commonly the decategorification is via Grothendieck group

Akiva Weinberger

Ah

Tobias Kildetoft

yeah, that at least is a well-defined term (or terms, as there are a couple of ways to do it)

Akiva Weinberger

13:28

@TobiasKildetoft Don't know what that is but I trust you

Tobias Kildetoft

I only know of one other though, and I have never really used it (trace decategorification)

Akiva Weinberger

Something like matrix traces?

Tobias Kildetoft

yeah, something like that, though obviously much more abstract

so in either case, you need more than just a category (you need some additive structure)

Akiva Weinberger

(Why are matrix traces even called that, I don't see any connection between them and the English word "trace")

Tobias Kildetoft

Though there is also the stupid example people always start with, where the category of vector spaces with linear maps is a categorification of the natural numbers

Akiva Weinberger

13:31

confused

Like, the dimension?

Balarka Sen

ye

Tobias Kildetoft

hmm, actually there the decategorification is also just the Grothendieck group

Akiva Weinberger

Oh

Secret

traces of its action as the amount of volume changed?

Tobias Kildetoft

(well, Grothendieck monoid)

Balarka Sen

13:31

@Secret That's not what the trace measures

Tobias Kildetoft

@BalarkaSen isn't it? Trace is the derivative of determinant

Akiva Weinberger

at the identity in the direction of the matrix

Balarka Sen

To me determinant measures the amount of volume distortion

Akiva Weinberger

$\frac\partial{\partial A}\det(x)|_{x=I}$

${}=\operatorname{tr}(A)$

Tobias Kildetoft

@BalarkaSen ahh, right, so trace is how fast the distortion changes

Balarka Sen

13:34

Yeah, I'd never thought of it like that. Guess that's true.

Tobias Kildetoft

anyway, at least by now there are enough examples of "practical" uses of categorification that it is not that hard to justify studying it (compared to just 5 years ago)

Akiva Weinberger

Is that the proper notation of directional derivative? Or do I write $D_A$

$A\cdot\nabla$

Balarka Sen

Write that

$D_A$

$D_A \text{det}(I) = \text{tr}(A)$

Akiva Weinberger

$D_A\det(x)|_{x=I}$

Yeah

@TobiasKildetoft I have no idea what sort of math has happened between 2013 and 2018

All my math is too old

All basis-invariant functions of matrices can be written in terms of the coefficients of the characteristic polynomial, right?

(aka symmetric functions of the eigenvalues)

FuzzyPixelz

0

Q: Finding the area of a polygon with complex numbers.

If we consider a regular polygon defined by the $n$-tuple $Z=(z_0, z_1, \dots, z_{n-1})$, and set $$A(Z)=\frac{1}{2} \Im \left ( \sum_{k=0}^{n-1} \bar{z_k} z_{k+1} \right )$$ Given that the number $\Im(\bar z_0 z_1)$ represents the signed area of the triangle $T=(0, z_0, z_1)$ (that's not nece...

complex-numbers area polygons

If you're interested enough :P

Akiva Weinberger

13:54

Re: my last question, probably true in characteristic 0 at least

Adam

14:21

if y divides the numerator of the Euler Product of x and x divides the denominator of the Euler Product of y, then x=y=1.

Jasmine

14:31

Hi!

Is it true that if any vector is collinear to 2 distinct vectors then that vector is $\vec {0} $?

Can anyone please help!

Secret

14:51

what do you mean by distinct, two parallel vectors of different magnitude is also distinct

Jasmine

@Secret distinct by direction basically two non collinear vectors

But now I understand the statement

It is true as $0* \vec {a}= \vec {0}$

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$Mathematics$ Mathematics