Sure, go ahead.

I applied to be at Simons for a month next year. Hopefully they'll fund me.

nice

That would be great.

Let me know how that goes.

Will do.

Consider the intersection form of the torus, $q_{T^4}$.

My internet is weird, so I'm going to reset some stuff. But I'll be right back.

By Freedman's Theorem, there is a simply connected closed four-manifold $M$ such that $q_M = q_{T^4}$. What is it?

@MichaelAlbanese Out of curiousity, what's the statement of Freedman's Theorem?

I can't get my internet to load anything. What's the signature?

@Semiclassical Given any unimodular bilinear form over $\Bbb Z$, there is a simply connected closed oriented topological 4-manifold whose intersection form $H^2(M) \otimes H^2(M) \to H^4(M) \cong \Bbb Z$ is the given form.

For every symmetric unimodular form $q$ over $\mathbb{Z}$, there is a simply connected, oriented, topological four-manifold $M$ such that the intersection form of $M$ is $q$. Furthermore, if $q$ is even, $M$ is unique up to homeomorphism. If $q$ is odd, there are two such $M$ up to homeomorphism, at least one of which does not admit a smooth structure.

Ah. I'm guessing the odd case is what's relevant here?

That's a cool theorem

No, it's even.

The problem is that $T^4$ is not simply connected, so this is not what you get from Freedman's Theorem.

Then wouldn't that say that the only such manifold is $T^4$ itself?

@Semiclassical: Odd means that $I(x,x)$ is odd for some $x$. But on the torus, squaring anything gets you zero.

ohhh

Can anyone calculate the signature of a matrix?

@MikeMiller: Signature should be zero as it is the boundary of a five-manifold, right?

Ooh, I should have thought of that.

Oh, I see what you're doing.

It will be $k E_8\# l H$ and work from there.

So it's the connected sum of three $S^2 \times S^2$s. Right.

...Which is also visible from the matrix itself.

Yeah, that was pretty stupid of me.

Me too. Glad we could be stupid together.

Chat, where being stupid is less public than on the main site.

I wrote a couple mathoverflow answers recently which were wrong. Invariably Igor Belegradek very kindly points out the mistake.

I do my best not to do that, though.

i'm guessing $(A\times B)\# (C\times D)$ is different than $(A\# C) \times (B\# D)$?

Me too, but it happens sometimes. When it does happen, I just try to learn from it (it is usually because I misunderstood something).

Yes, @Semiclassical. You shouldn't expect them to be the same. Think about how the connected sum is defined, in terms of picking discs on both sides and gluing them together.

you're presuming i understand this much more than i actually do :p

@Semiclassical Take all spaces to be $S^1$. One of those will be orientable surface of genus $2$, the other will be the torus

:)

that works.

I wonder if those are ever the same.

But I won't think about it or else I'll be doing nothing productive for a couple hours.

@MichaelAlbanese What are you up to lately?

:P

If i were to ask it on main, I'd probably have to preface it with "This is a very silly question."

That's what I like about this chat...it's a place to be stupid (not too stupid tho)

I was mostly curious because I have no idea how to visualize $(S^2\times S^2)^{\#3}$.

You take a couple $S^2 \times S^2$s and glue em together in a row

right. though that just pushes it back to visualizing $S^2\times S^2$, which I probably shouldn't hope for in the first place :p

Just imagine a big black box

(Hello chat)

other than, y'know, a sphere attached to every point of a sphere

Sounds like a fine pixthre.

yet again, chat proves its worth as a place to be say silly things that don't make sense :p

Reading through the chat history, it becomes apparent that what I said just now was a silly thing that doesn't really make sense

but I don't know how to visualize $S^2\times S^2$ either

I'm algebraic and silly, but I think of it as 'a sphere number of spheres'.

A torus is a circle number of circles.

*thumbs up*

It always depends what you mean by visualize. I have pictures of these things, but they don't necessarily help me do mathematics with them. $S^2 \times S^2$ is two 2-spheres wedged together with a big enveloping cell engulfing them. A sort of curly object.

That is of course completely useless.

@AndrewThompson I suppose that's related to what @Semi said

A sphere attached to every point of the sphere

Hm, yeah.

right. sometimes the simplest answer to "How would I visualize X?" is just "Don't."

I've enjoyed reading topology again because I have to draw these silly pictures for myself.

$\{\vec x\in \Bbb R^6:|x_1|+|x_2|+|x_3|=|x_4|+|x_5|+|x_6|=1\}$?

$S^1 \wedge I$ is one of these red jump-ball things one played with as a kid. The half-spherish thingies that jumped up when you press them down.

$|x_1|$ or $|x_1|^2$?

The first, I think. I'm multiplying together two octahedra.

The second is actual spheres, though.

ah.

Don't know why I did octahedra instead of spheres.

@MikeMiller: Finding almost complex structures on four-manifolds, and related things.

But right now, lunch.

Thanks for the help with the question, sorry for my stupidity.

what is the main difference between the normalizer and centralizer subgroup? I see the centralizer finds elements in $G$ that commute with subsets of $G$. The normalizer subgroup seems to find elements in $G$ that commute with subsets in $G$ too.

That's an algebraic topology question, isn't it?

I haven't read this yet but it seems relevant: en.m.wikipedia.org/wiki/Centralizer_and_normalizer

"Centralizer and normalizer"

Ah. So, the centralizer of the entire group is the center, but the normalizer of the entire group is the entire group?

I think that's if it's abelian @akiva

no @obliv, yes @akiva

It gives this definition:$ $\begin{align}{\rm C}_G(S)&=\{g\in G:gs=sg\text{ for all }s\in S\}\\{\rm N}_G(S)&=\{g\in G:gS=Sg\}\end{align}

So, the normalizer of a normal subgroup will be $G$, but the centralizer won't necessarily be $G$

normalizer is defined as $gAg^{-1} = \{gag^{-1}\mid a \in A\}$, $N_G(A) = \{g \in G \mid gAg^{-1} = A\}$

in my text

which is the same thing I think

$S$ is a subset in which the elements in $G$ commute?

Right, that's the same thing, just multiply $g$ on the right on both sides

no

$S$ above is your $A$; it's just any subset

Hi @tern and DogAteMy

Hi

$S$ is a subset that commutes with an element in $G$

$S$ is just any subset.

Quick question: If you have a multivariable function which is continuously differentiable function then are the partial derivatives continuous?

When they say $gAg^{-1}=A$, they mean that conjugation sends each element of $A$ to some element of $A$ (not necessarily the same one)

Yes @JohnD — that's what cont diff ordinarily means.

(Well, that's just $gAg^{-1}\subseteq A$, technically, but the reverse is easy because of the identity element)

No hi to the two Michaels? My heart is all broken and such.

G'night @MikeM

I'm on my iPad — I only see people if they talk.

There should be a button on the top-right that gives the list of people

these definitions look the same to me :(

though it doesn't show if they're active or not

I don't keep scrolling to the top, DogAteMy

for all $s \in S$ isn't that the same as $S$ itself?

"for all s in S" is a sentence (or part of one), not a set

I promise to be upset whether or not it's justified, @Ted.

Not a sentence, a phrase.

The first essentially says that $gsg^{-1}=s$ for all $s$; the second says that $gsg^{-1}\in S$ for all $s$

I expect no less from you, Mike.

or $a$ and $A$, in your book

Consider the Klein four group $V$ in $S_4$. The centralizer is, if I'm not mistaken, just $V$ (someone double-check that for me?).

I wrote an answer earlier to a question about Floer homology. The OP didn't reapond at all expect to ask if he should post the quesrion on MO as well.

I know I shouldn't be salty, but I am.

I don't know what a klein four group is :\ @akiva I'm going to try going through more definitions and exercises for now and come back to this

Oh

It's the set of permutations on the vertices of a rectangle when you reflect it

it's Z2 x Z2

(1), (12)(34), (13)(24), (14)(23)

$\{e,(12)(34),(13)(24),(14)(23)\}$

No greeting for me, @TedShifrin??? I am hurt and heartbroken.

@MikeMiller three Michaels, technically, though I don't use that as my chat handle for reasons

How about this, @Obliv. True or false: if $n\in N_G(S)$ and $s\in S$ then $ns=sn$.

(note $ns=sn$ and $nsn^{-1}=s$ are equivalent)

I always forget that.

@TedShifrin I can't believe you didn't greet bananaman. Jeez.

heh

@arctic true

Please How to prove that if $t_n\rightarrow t$ then $\int_0^{t_n} f(x) dx\rightarrow \int_0^t f(x) dx$

How rude to our esteemed guest!

@Obliv no, false.

blushes

If I have the vertices labeled $\begin{array} 11&3\\2&4\end{array}$, the first is the identity, the second is reflecting vertically, the third is reflecting horizontally, and the fourth is reflecting across the origin (rotating 180 degrees).

the true statement is: if $n\in N_G(S)$ and $s\in S$ then there exists a $s'\in S$ (not necessarily $s$) such that $nsn^{-1}=s'$

@arctic so the normalizer just permutes the elements in $S$? So it doesn't conclude anything about commutativity

exactly

unless $s' = s$

Yeah. It permutes the elements of $S$ when it conjugates it.

@Obliv Well, the identity is a permutation…

@TedShifrin can you tell me: How to prove that if $t_n\rightarrow t$ then $\int_0^{t_n} f(x) dx\rightarrow \int_0^t f(x) dx$

So the centralizer is a subset of the normalizer.

the centralizer is a set of elements that must commute. The normalizer of a subset can commute but it doesn't have to.

yeah

And, if $A$ is a normal subgroup, then ${\rm N}_G(A)$ is…?

$A$

er

$G$.

what is a normal subgroup?

...

What makes $A$ normal is that $gAg^{-1}$ is always $A$.

so the normalizer of any subset of $G$ should be the entire group $G$?

@Semiclassical Did you never dream to reach the performance of the great people in science or even get higher results? (having in mind now Ramanjan, of course) Or is it wrong to aim very high?

The normalizer of any normal subgroup should be the the entire group!

@Obliv But the normalizer of an arbitrary subset doesn't have to be the entire group

see you folks in a bit

see you @mike

Not sure. If I did, I don't remember it.

Take care Mike

My main ambition has simply been to learn interesting things and do interesting problems.

@Obliv "Normal" being a mathematical term here, not just meaning "ordinary"

Take, say, $\{e,(12)\}$ as a subset of $S_3$

(or just $\{(12)\}$ since it doesn't need to be a subgroup)

hi guys. is there anyone willing to indulge a computer sciencey, big O type question?

@Semiclassical Yeah, that was my abition too, but as you become better you benefit of more interesting things and do more interesting problems. That's the point with aiming very high, not for the sake of being on top but for the simple fact you can enjoy EPIC STUFF.

I forgot that $gA = Ag$ has to be the same permutation of $A$. makes sense

$gA=Ag$ is an equality not a permutation

the map $a\mapsto gag^{-1}$ is a permutation of $A$ whenever $g\in N_G(A)$

@pingOfDoom just ask

@pingOfDoom Nice username

$gA \to \star$ and $Ag \to \star$ has to be the same mapping is what I meant to say

my ambitions are decidedly not epic at the moment

lets say you have a computer running at a billion instructions per second. If you wrote an algorithm that ran in O(log(n)) time, how long would it take to execute if the input size was 10,000

@AkivaWeinberger thanks man!

@Obliv what does $gA\to \star$ mean?

@Semiclassical It's also true that I learned pretty late that aiming very high can be very wrongly perceived, wrongly catalogued, sometimes it might be preferred not even to share such thoughts, but at the same time I'm sure that everyone here would like to the solve most badass problems ever, even if some were so scared of top performance that they would never publically admit it.

It doesn't mean much mathematically. In my mind $\star$ is the product of $gA$ and $Ag$ must be the same product. It doesn't make sense to say mapping to the product though, I think @arctic

@Semiclassical Keep your spirit and goals very high! :D

I also had bad days when I crawled in front of some beast math problems, but then I remained there and decided to fight back and put down all the problems I wanted to. :-)

Can someone help me please

There is much beauty even when a problem apparently puts you down. Those lessons you learned at those hard times are amazing, then you're never the same after you go through them.

Eh. I think we're talking past each other to some extent.

:D

What comes to mind is the following bit of popular psychology, namely Maslow's hierarchy of needs

belonging is below self esteem??

@vrouvrou top right: "Just ask; don't ask to ask."

When I say my goals are not so grand at the moment, what I probably really mean is that my goals are not at the top of that pyramid but a bit farther down

@Obliv i ask my question but no one answer me... : How to prove that if $t_n\rightarrow t$ then $\int_0^{t_n} f(x) dx\rightarrow \int_0^t f(x) dx$

@Semiclassical We learned about it since high school, one of the cool things taught in the psychology hours (you usually never forget). :-)

"Esteem" includes self-esteem, but also includes respect from others @pingOfDoom

So in that respect, at least, the order makes sense. Feeling like you belong to some community usually comes prior to having a certain level of reputation within it.

i guess thats true

@Obliv are you there ?

Hi, bananas. I came back to say hi just so you wouldn't sulk. @iwriteonbananas

in any case, though, i doubt anyone would argue that it's a strictly 'true' model of human needs.

if it's a useful tool, then that's enough.

self actualization is more important than esteem

give me a sec lol

I don't think it's a matter of what needs are more important, actually. More that some needs are necessarily prior to others

@vrouvrou I think you can use squeeze theorem $\lim_{t_n\to{t^-}} \int_0^{t_n} f(x) dx <\int_0^t f(x)dx < \lim_{t_n\to{t^+}} \int_0^{t_n} f(x)dx$

not really sure though

well this only holds true if $f(x)$ is positive for the bound I think

it'd have to be modified if $f(x)$ was negative. the inequalities would be flipped.

i dont understand your proof

the integral function is continuous

differentiable in fact

@vrouvrou well it's just a simple demonstration that as $t_n$ approaches $t$, the integral becomes that of $t$ instead of $t_n$

@iwriteonbananas You might find this amusing: Apparently, there's a paper somewhere that deals with the category of "Banach analytic manifolds", called $\sf BanAnaMan$.

so you just need to prove the general theorem: If $g:X\to Y$ is continuous and $x_n\to x$ then $g(x)\to g(x_n)$

@vrouvrou replace the bounds with what the limit is approaching to, $\int_0^{t^-} f(x) dx <\int_0^t f(x)dx < \int_0^{t^+} f(x)dx$ I think is sufficient to show that the integral bound becomes $t$

no i don't see

Do you know how limits work?

yes but how from t^- and t^+ we obtain t_n ?

@JasperLoy Hi Jasper! Well, not really secretive, my number is related to the golden ratio, it is not put at random. I might change it at some point. How are you doing these days? :-)

@JasperLoy Good to hear that. All that depends on me is done right in time, and the work to it is great so far. :-)

@JasperLoy Sure. They are on the desk along with many other plans that depend on many variables. :-)

@JasperLoy :-))))))))), sure

@JasperLoy How about your studies plans? I remember that you bought many books at some point ...

(not decided when to start though)

@JasperLoy I learned a golden rule some time ago, maybe it may help you. Do math every day, even a minute, but educated yourself not to skip any day. That tiny minute you put in mind may create great changes in time. :-)

@Akiva There's not. That was in an unpublished version; after refereeing it was BanMan.

@JasperLoy sometimes we need to be more confortable with ourselves, and then let's do it, but keep that rule. :-)

@JasperLoy Hope your condition is improved now.

@JasperLoy Not really, we are both very busy persons (and we like different areas in math). :-))))

@JasperLoy I thought that if you give a person a mask then he will tell the truth, therefore I removed my photo.

"when something is hard, that means it's interesting."

How does one create anti-degenerate mathematics?

@JasperLoy I like that too.

@MikeMiller Awww :(

hard work and obsession

yeah but if its too easy, it can't be very interesting

not for long

@vrouvrou sorry I was busy. Well you're asking, if $t_n$ approaches $t$, show that $\int_0^{t^n}f(x)dx \to \int_0^{t}f(x)dx$, no?

so I'm doing this by taking the left and right limit as $t_n \to t$ to show by squeeze theorem

@JasperLoy degenerate is when an ellips becomes a circle, a circle becomes a point

or so it says in mathworld

Generalize, think outside the box, that is anti-degenerate mathematics. But to generalize is to be an idiot.

very deep, much platitude

my bibtex is not working :(

So, a topological space is a set of points $X$ and a family of sets of points $\tau$ with certain conditions (intersection of two things in $\tau$ is in $\tau$, etc.), and a function from $(X,\tau_X)$ to $(Y,\tau_Y)$ is continuous if the inverse image of stuff in $\tau_Y$ is in $\tau_X$. This definition still makes sense if we throw away the conditions, though. What do we get if we allow $\tau$ to be any family of sets of points of $X$?

We'd still get a category, right?

Similarly, what happens what we go up a level, and consider things like $(X,\tau)$ where $\tau$ is a family of families of sets of points?

you have asked one question (to which the answer is yes) and not the more important one: is this interesting?

I felt that was part of the "What do we get" thing.

Has this been studied before?

I'm gonna write something out to see if I can get my thoughts in order. $S = P(G)$ is the collection of subsets of $G$ Let $G$ act on $S$ by conjugation (i.e $g: B \to gBg^{-1}$ for all $b\in B$ and $B \subset G$) It is clear to see the normalizer $N_G(A)$ is precisely the stabilizer of $A$, where $A \in S$. So this means.. $A$ gets acted on so that $A \to gAg^{-1}$. The def. of the stabilizer of $A$ is $\{g \in G \mid g = gA\}$ hm.. so since $gA = Ag$, by the group action, $g = gA = Ag$ but

the normalizer is defined as $\{g \in G \mid gAg^{-1} = A\}$ .. which means it's the reverse mapping of the group action..

@JasperLoy and math (at any level). :-)

@JasperLoy wait a bit ...

@JasperLoy ^^^

:-)

@JasperLoy There is a misconception that people loving math so much (or not necessarily referring to math) they live in a lab away for any kind of social life. I actually think these people have enough imagination to get more fun than all their critics.

@JasperLoy I don't, but I come on chat once in a while to get some fresh air and stay a bit away from hard work (which is not a thing that always happens - to say the truth). It's good to take little breaks once in a while.

@JasperLoy :-)))))

@JasperLoy Maybe you didn't give the opportunity to the others to talk to you. Is that really a big issue for you? :-)

@JasperLoy No idea. He might be busy these days.

@JasperLoy Edited (briefly). :-)

Trying to finish an article. I'm back a bit later.

This afternoon I will see the video in YouTube: Cédric Villani - Of triangles, gases, prices and men , provide us by the official channel of the Institut des Hautes Études Scientifiques (IHÉS) I say it, if some user is interested in see it.

hi @Huy

what are you doing with your life

o/

h

@MikeMiller I wonder why you type "h" instead of an actual word

It's short for hi

Alright :P

Question: Is there a way to determine the cohomology (ring) of the $n$-torus that does not involve annoying work on spaces with a circle factor first?

de rham cohomology

you can explicitly write down an isomorphism $\Lambda^k(\Bbb R^n) \to H^k_{dR}(T^n)$ by sending $\wedge_I x_I \mapsto \wedge_I dx_I$

where the latter makes sense because you're just pushing down a form that lived upstairs on $\Bbb R^n$

OK, so you need de Rham's theorem

or you could just be happy with de Rham cohomology :)

you could always just try to be explicit... send $\wedge_I x_I$ to the poincare dual to the sub-torus in the plane spanned by the $x_I$

But then I don't know if it's the cellular/singular cohomology

You'll live.

The PD thing I just said can be done without actually using Poincare DUality. Look at where Hatcher calculates the cohomology ring of a genus g surface; this is what he's implicitly doing there.

...after almost dying from typing up this discussion of spaces with a circle factor

Mimic that but with what I said in mind

You mean that intersection number stuff?

After I saw him cheating me out of a proper discussion of how to compute the cohomology of the unorientable surfaces (I'm sure you remember how painful it was for me to find Ext=0), I'm not talking to Hatcher for a while.

You can compute it explicitly simplicially.

Though really you should do it simplicially for projective space and then prove a theorem about the ring structure on connected sums.

Well, the cellular version was not so terrible either... after one understands what a module is :P

So I talked a bit to my gauge theory lecture and he said most Floer homology stuff is about 3-manifolds. Not really so much about 4-manifolds. He also said it is very difficult. Is this your general impression, too?

You cannot compute the ring structure in general in cellular cohomology.

@MikeMiller ??

You cannot do it straight from the cells. There is no process to do so.

You're still talking only about connected sums of RP^2?

CW complexes in general.

Oh.

I'm sure that's true

You won't see such a thing written down in any book you look at. The cup product will be defined singularly, which is isomorphic to cellular homology, but I don't think there's a cup product on the cellular chain complex itself. (I bet it's a theorem that there's not one.)

Floer homology is an invariant of 3-manifolds but functorial under (4-dimensional) cobordisms. This is why it so often is useful for saying something either about the way 3-manifolds play with 4-manifolds, or 4-manifolds themselves.

I don't think I do anything particularly nontrivial, but it took some setup time to get here.

Hmkay.