Mathematics - 2018-05-23 (page 2 of 3)

geocalc33

4:05 AM

what is that big black blob in the center

DarkRunner

How can we conclude that $lim\quad ln(\frac { 1+\frac { 1 }{ x } }{ 3 } )=ln(\frac { 1 }{ 3 } )$?

GFauxPas

geo that means $|z| \approx 0$ or $|z| = 0$

DarkRunner

I'm very confused

Secret

@geocalc33 what function you are plotting. That might be one of those contours you encountered

geocalc33

contours?

DarkRunner

4:10 AM

If anyone could help with my problem, that would be great

I know I can't use LHopitals, b/c it's not in indeterminate form, so I'm just dumbfounded

Never mind, I got it

geocalc33

so @GFauxPas it's an area where the function is 0?

Secret

@geocalc33 the section on domain colouring beneath mentioned how they make these contours to avoid the colours to desaturate too quickly

Semiclassical

@GFauxPas Poe's law be relevant here

geocalc33

okay so gfauxpas thinks its a zero region and you think its a contour

GFauxPas

i mean it depends on the software youre using so

Secret

4:17 AM

It does look like there's a zero buried somewhere because of the way the colours circle that black patch

@geocalc33 What is your exact function that is plotted?

it looks nothing like 1/log(z)

geocalc33

I just plotted (1+1/2^z+...+1/10^z)^(1/log(z)+1/log(1-z))

Secret

waaaahhh..

checking...

geocalc33

what

Jackozee Hakkiuz

Check mathim.com/2

What.

I didn't write that.

I was going to say I would check that plot as well.

geocalc33

why

Secret

4:26 AM

davidbau.com/conformal/…

It seems there's at least one zero buried there and the region is very close to zero

you can confirm there are zeros there because as you remove the 1/n from high n one by one, the white region splits into two zero regions

geocalc33

yeah like if you just plot it with 1/lnz and take away the ln(1-z)

i see what you mean

@Secret do you think there are more than 1 zero ?

Secret

well I think the crazy graph suggests there are at least two essential singularities there (where the blackness collide with the wavy pattern in the complexgrapher, and where the dark gray (poles) collide with the black/white (near zeros))

it remains to check whether the exceptional value which the essential singularity does not process is indeed zero

geocalc33

so how many zeros could there be in that region? only a couple, or a lot?

Secret

hmm... can I do...:

geocalc33

and you know what kind of function (1+1/2^z+...) is?

Secret

4:38 AM

maybe not the finite version. The infinite is riemian zeta $\zeta (z)$ if I recall

geocalc33

yeah

i just stopped after 10 terms lol

Secret

I cannot think of any good ways to simplify (1+1/2^z+...+1/10^z)^(1/log(z)+1/log(1-z))=0

I cannot take log for example

and is it true the complex function w^z has no zeros?

I knew that is true for reals except the origin (which obviously is not the solution here since you divide by zero), but I am not sure about the complex values

geocalc33

well z^z has no roots

Jackozee Hakkiuz

I'm not sure there is a zero in there.

These are the images I get.

GFauxPas

$e^z$ has no roots because $\cos \theta \ne i \sin \theta$ for any $\theta$

Jackozee Hakkiuz

4:46 AM

Secret

omg what software you are using. It looks beautiful?

Jackozee Hakkiuz

It's just java code.

A friend and me made it like 2 years ago.

Secret

yeah, from the look of those two, it seems we only have two essential singularities, but no roots, thus it does avoided zero among all the possible values

Jackozee Hakkiuz

If you zoom yout you can see the branches.

Secret

hmm, those things that end on the branches, I wonder if there is any significance...

geocalc33

5:10 AM

can you graph that again with the exponent as (1/log(x)+1/log(1-x)) with x real number this time

Secret

davidbau.com/conformal/…;*)%2F2)%2B1%2Flog(1-(z%2Bz*)%2F2))&z=9

it very scary looking

geocalc33

oh

graph (z)^(1/log(z)) :)

davidbau.com/conformal/#z%5E(1%2Flog(z))

skull

@Semiclassical you've had time to hangout in here and write up and prepare for the defence of your phd thesis?

Semiclassical

well, i wasn't here so much the week before last

skull

yeah, I noticed. But still.

Semiclassical

5:25 AM

and that's because I spent that week with the defense as the main goal

skull

so that's the secret to escaping the blackhole of productivity :-)

Semiclassical

yeah, your adviser hanging over you does help move things along.

skull

lol

Congrats pal.

Semiclassical

thanks

philmcole

6:49 AM

Hi, if I have a monotone function $f: [a,b] \to \Bbb R$ why is $\sqrt{f(x)}^2 = f(x)$ without the absolute value bars $\lvert f(x) \rvert$?

Leaky Nun

@philmcole because it isn't

philmcole

right I was confused too

Silent

7:03 AM

@LeakyNun, let $f:\Bbb R\to \Bbb R$ be a function such that $|f(x)|\le|x|^2$ for all $x\in \Bbb R$, then $f$ is differentiable at zero, right?

Leaky Nun

@Silent yes, its derivative is even 0

Silent

ok, :)

Secret

got sandwiched to zero lol

Sandwich Theorem as its finest

Silent

but f can be discontinuous at other points than 0?

Leaky Nun

@Silent sure

Silent

7:08 AM

thank you!

Tuki

7:37 AM

Hello

Mithlesh Upadhyay

@Tuki hello

Szabolcs

8:51 AM

What software is available for computing with planar graphs, other than Sage?

Is there anything else useful out there that is programmable from some high level language (i.e. not a GUI program and not something that requires C or Java programming)?

Tobias Kildetoft

9:05 AM

@Szabolcs What is missing from Sage since you are looking for something else?

Szabolcs

@TobiasKildetoft I am also looking for inspiration for reasonable design for such functionality. An example of a thing that's missing from sage is dual graphs of non-3-vertex-connected graphs and non-simple graphs.

Tobias Kildetoft

@Szabolcs That sounds like the sort of thing that is missing because it is hard to do efficiently

Szabolcs

Why would it be hard?

Tobias Kildetoft

No idea, but if something is only implemented for certain subclasses of objects, it tends to be because it becomes hard once you move to more general objects

Szabolcs

In this case I don't believe that to be true.

Tobias Kildetoft

9:10 AM

Actually, isn't it also because you need some conditions to make sure the dual is still planar?

Szabolcs

The dual graph is probably restricted because the dual of 3-vertex-connected simple graphs is simple and unique.

Otherwise there may be multiple dual graphs.

Tobias Kildetoft

Well, if it is not unique, what would you want it to return?

Szabolcs

As for dealing with non-simple graphs, one could in principle subdivide those edges (loops and multi-edges) before continuing with the analysis (though I may be missing something)

Anyway, I am trying to design similar functionality, and I am looking for examples of similar software to avoid mistakes in the design.

Also, I'm coming from the "network analysis" side (not pure math), and Sage does not integrate as well with other tools needed for that. Sage is great at pure math, and has probably the most complete graph theory functionality of any system, but it's aimed at mathematicians. Software aimed at network analysis (e.g. igraph) doesn't typically include stuff related to planar graphs.

Tobias Kildetoft

@Szabolcs What other tolls are those? Sage is pretty good at interfacing with other tools, so it might be possible to make it do so with those as well.

Szabolcs

All in due time :-) That's not my priority now, really. But I'll get there eventually (when I'm fluent enough with Sage—which takes time).

I'm most comfortable with Mathematica, which unfortunately doesn't play nice with Sage due to Sage's hard copyleft stance.

Also, I am not simply looking for a tool in order to do certain things. I am looking to see how existing tools implemented this functionality to learn from them when implementing my own package.

Silent

10:12 AM

Let ${x_n}$ and ${y_n}$ be two sequences in $\Bbb R$ such that $\lim\limits_{n\to\infty} x_n=2$ and $\lim\limits_{n\to\infty} y_n=-2$. How to show that: there exists an $m\in \Bbb N$ such that $|x_n|\le y^2_n$ for all $n>m$?

Mary Star

10:31 AM

Hello!! Which is the definition of the region $D$ :

Mary Star

12:00 PM

I posted my question also in the main: math.stackexchange.com/questions/2792731/… Does someone of you have an idea?

Semiclassical

12:49 PM

@MaryStar one concern: the diagram you have only makes sense so long as $x<ct$.

my guess in the case of $x>ct$ is that you draw a diagram like this, but since $x-ct>0$ you never have a reason to draw the reflected line

Mary Star

1:07 PM

I got stuck right now. Coudl you explain further what you mean? @Semiclassical

Semiclassical

take the point (x,t) and move it downwards. if you move it far enough downwards, then the line emanating down and to the left reaches the x-axis before the t-axis. agreed?

Mary Star

Ok

Semiclassical

in that case, your line never has a need to reflect

to put the point more succinctly: How would you draw that diagram if the point (x,t) weren't where it is but were instead at, say, the spot where D is?

you have to use the same slopes as in the original picture, since said slopes are set by c

Mary Star

You mean both sides of the triangle have slope c?

Semiclassical

yes.

(well, +c on one side and -c on the other)

Mary Star

1:25 PM

Ok! But how can we use this information? I am confused @Semiclassical

Semiclassical

1:40 PM

I already told you. (Try to) Redraw the diagram but with the point (x,t) shifted down to where the label D is.

If you do that, you'll find that the construction of the region no longer makes sense. That's why I'm asking.

You may have been told to assume $x>ct$. If so, the issue I’m getting at is excluded. But that’s a relevant part of the problem statement

So either you haven’t stated all the relevant assumptions or you’ve only indicated how the region is defined in a specific case. In either instance, the problem statement is incomplete

Bah, I said it backwards: You may have been directed to assume x<ct.

Mary Star

1:59 PM

It doesn't say anything at the exercise statement, but I will check it again

Evinda

Hello.. I have the following question

0

Q: energy of initial and boundary value problem

I want to show with the energy method the uniqueness of the solution of the following initial and boundary value problem. $\left\{\begin{matrix} u_{tt}=u_{xx}+f(x,t), & 0<x<L,t >0\\ u(x,0)=\phi(x), & 0 \leq x \leq L\\ u_t(x,0)=\psi(x), & 0 \leq x \leq L \\ u_x(0,t)-2u_t(0,t)=g(t) & , t \geq 0...

pde

does anyone have an idea?

Liad

2:16 PM

im trying to show there is injection from $\Bbb N \ ^ {\Bbb N}$ to $2 \ ^ { \Bbb N}$, someone can help?

mercio

inject it into $(2^{\Bbb N})^{\Bbb N}$ first

Liad

i thought about showing there is an injection to $\Bbb R$ , taking $f: \Bbb N \to \Bbb N$ to $0.f(0)f(1)\dots$

but it wont be one to one

@mercio i think i can do that

mercio

there are also some bijections from $\Bbb N^{\Bbb N}$ to $2^{\Bbb N}$ that are really easy to describe

Liad

i only need injection because the other way around is easy. can you describe one ? my attempts weren't so good.

mercio

can you make an almost-bijection into {"add one" ; "go to the next line and write a 0"}^N ?

though I guess making it a bijection makes it difficult and painful to describe

Liad

2:34 PM

again ,i dont need a bijection ^^

Sha Vuklia

2:46 PM

hey @Semi, do you have time for a question?

kevinz

4:13 PM

Can someone help me with this question cs.stackexchange.com/q/92218/88822

Balarka Sen

4:29 PM

Hey @CooperCape, @Daminark

CooperCape

Alrighty :p

how's you?

Balarka Sen

Surviving

CooperCape

I guess... that's good.

Balarka Sen

From the Darwinian perspectives, yes

CooperCape

What's poppin?

Balarka Sen

4:32 PM

Planning on doing some math

Alternatively I could watch a movie

CooperCape

Nice one...

I still gotta do organic chemistry...

Balarka Sen

Ugh. Well, I can recommend an album if you want to get through that stuff while listening to something (that's what I used to do)

CooperCape

I dunno. music keeps me grounded in reality but not to what I should be doing.

Balarka Sen

Heh, fair.

CooperCape

Nothing keeps me grounded to what I should be doing.

I swear I'm gonna get AAC

But life goes on ygm

It'll do.

Balarka Sen

4:37 PM

Same. Time will wither us but procrastination shall still be an incurable disease

CooperCape

At least next year it'll be something I do to procrastinate now... Hopefully.

I need to get in a good frame of mind. I don't think the "There's literally less than two weeks to go" panic has hit yet.

Then again you probs don't wanna hear another person mope about exams. so I shall shut.

Balarka Sen

Think of two weeks as a stretch of time to put whatever you have learnt into order. That's what I did pre-exams, arranging the pieces of the puzzle in order so that I have my catalog of informations in arrangement

A well-arranged catalog (regardless of if it's complete or not) helps pulling the relevant bits out when needed (in exams)

@CooperCape That's literally what I did here before my exams lol. Feel free to ask for advice or just general convo

CooperCape

@BalarkaSen In which case everyone else doesn't want to hear another person mope about exams :p

@BalarkaSen And yeah I need to timetable myself or something

Balarka Sen

Start writing super short summaries of your textbook material - I forget a lot so that's what helped me bookmark the informations in my head

But yeah fix times to spend on stuff

CooperCape

It really is just organic chemistry for me. Whatever happens I just can't get it into my head. I'll see what I can do, though. Got time :p

Balarka Sen

4:49 PM

I feel ya. Don't forget about the coggle map, that actually helped me a lot

I ended up doing quite well in my chemistry exam

surprisingly

CooperCape

In which case, congrats.

(forgot about the coggle map oops)

Balarka Sen

tnx

GFauxPas

I'm not seeing the difference between axiom of replacement and axiom of subsets, can someone pls ELI5

in ZFC

the latter is an axiom schema so it's stronger but i dont see how

B. Mehta

subsets lets you filter, replacement lets you map

Secret

If you have {1,2,3,4} replacement allows you to make {w1,w2,w3,w4} which cannot be done by subset

GFauxPas

4:58 PM

ah, but if you make a map that sends a set to a subset of itself, then replacement implies subsets?

Secret

and the map used in replacement does not have to be constructed beforehand (impredicative), thus axiom schema of replacement is a much stronger thing

yeah, to my understanding, subset is just one of the definable maps

B. Mehta

yeah, replacement implies subsets

Alessandro Codenotti

@GFauxPas they're both a schema

B. Mehta

and the map is allowed to be partial

GFauxPas

got it, thanks for explainoing

Sirmimer

5:17 PM

Doing some probability 101 exam prep: Let $X~Exp(3)$ and Let $Y=e^{2X}$. Find $EY$. I know the answer should be $EY=3$

I've seen a similar question where the answer claims that $EY=e^{2EX}$. I know of the property that $X~Exp(\lambda)$ then $EX=\frac{1}{\lambda}$. But that won't give me EY=3. Any hints?

GFauxPas

\sim for $\sim$

philmcole

5:31 PM

Hi, maybe this question is stupid but if we know that $\lvert a - b\rvert \lt \varepsilon$ for all $\varepsilon \gt 0$ then we consider $a=b$. Why? Technically we never really know if $\lvert a-b \rvert = 0$. Has it something to do with axioms of the real numbers?

Eulb

@BalarkaSen hey!

B. Mehta

@philmcole Say for contradiction that $|a-b| > 0$, then there is some $\epsilon$ with $0 < \epsilon < |a-b|$ contradicting that $|a-b| < \epsilon$

philmcole

I see thx

Balarka Sen

5:47 PM

@Eulb Hey

What's up

Any cool topology up?

Mary Star

6:02 PM

I want to check the convergence of $\displaystyle{a_n=\frac{2-n}{-4n^2}}$ and give n as a function of $\epsilon$.

For a fixed $\epsilon>0$ we have to find a number $n_{\epsilon}$ from whiich the condition $|a_n- a| < \epsilon$ holds: \begin{equation*}|a_n-a|<\epsilon \iff \left |\frac{2-n}{-4n^2}-0\right |<\epsilon \iff \left |\frac{2-n}{-4n^2}\right |<\epsilon \iff \frac{|2-n|}{4n^2}<\epsilon\iff |2-n|<4n^2\epsilon\iff 4\epsilon n^2-|2-n|>0\end{equation*}

For $2-n>0\Rightarrow n<2$ the $n$ can take only the value $1$.

Eulb

@BalarkaSen working on it

Silent

6:14 PM

The number of group homomorphisms from $\Bbb Z/12\Bbb Z$ to $\Bbb Z/13\Bbb Z$ is $1$. Am i right?

Tobias Kildetoft

@Silent Yes

Silent

Thank you!

Balarka Sen

@Eulb Anything in particular?

Eulb

no not yet.

Balarka Sen

kk keep me notified

Eulb

6:23 PM

If you enter a building before the end of business hours and stay inside after business hours then for the interval of time between entering and getting forcefully removed the building could be considered clopen.

topology has such beautiful real world applications.

Balarka Sen

Lmao

Note that there's also neither-open-nor-closed: kinda like a half-open door

Eulb

Professor uses this one weird topological trick to fool students into thinking he's free for office hours. Students hate him!

Ted Shifrin

I've used that one in class numerous times, a @Balarka.

Balarka Sen

@TedShifrin Yup, I caught that from you :)

Ted Shifrin

Oh.

GFauxPas

6:28 PM

When is a door not a door? When it's $D^3$

(A jar)

Eric Silva

is a half open door not just open

Ted Shifrin

Howdy Eric.

Mike Miller

I dunno that I would call $D^3$ a jar

Eric Silva

yoyo

GFauxPas

I was thinking without the lid

Balarka Sen

6:30 PM

If your ball looks like a jar you need to see a doctor

ASAP rocky

Ted Shifrin

@MaryStar: You should get in the habit of doing simplifying estimates. For example, you know that $|n-2|\le n$ for all $n\in\Bbb N$, so just estimate $n/(4n^2) = 1/(4n)$.

I was just at the doctor, @Balarka :P

Balarka Sen

@TedShifrin Was it the terrifying dentists again

Mary Star

@TedShifrin Ah ok! Thank you!

Ted Shifrin

LOL, no, my regular doctor for annual physical. Mostly OK.

GFauxPas

You wouldn't say D3 and a jar (sans lid) are homeomorphic?

Balarka Sen

6:32 PM

Ah gotcha. Glad to know!

Ted Shifrin

So, Balarka, are you back to topology/geometry world now?

Balarka Sen

Slowly returning back to my habitat, yeah. Mike gave me some homological exercises, have been pacing through obstruction theory a little

Trying to recover my lost algebro-topological knowledge

Ted Shifrin

Ah, cool.

Mike Miller

I am trying to remember how geometry works after doing so much algebra

Ted Shifrin

Well, maybe you can recover mine as well.

Mike Miller

6:34 PM

There was a gauge theory question yesterday that ate a couple of hours

Ted Shifrin

Did it get well digested?

Mike Miller

it got an answer eventually, yeah

Ted Shifrin

Cool.

Mary Star

And if the absolute value is in the denominator? For example $\frac{3}{|1-n|}<\epsilon $ ? Do we take here cases or can we find here also an other bound? @TedShifrin

Ted Shifrin

@Balarka: Eventually you should understand why the $j$th Chern class of a rank $k$ bundle gives (by duality) the cycle where $k-j+1$ generic sections become linearly dependent. (I think I got that right.)

Eulb

6:37 PM

@BalarkaSen oh, and if thy spherule ocularly resembles a jar, thy shall make haste to a medico – Aesop Rock

Balarka Sen

LOL

Ted Shifrin

@MaryStar: No, don't do cases. Remember you only care about large $n$. So if you need to assume $n\ge 10$ to get an estimate, then take $n\ge\max(10,\text{whatever})$ at the end.

Mary Star

Ah ok!

Ted Shifrin

For example, if you're in the denominator, you want a lower bound on the denominator, so certainly $(n-1)\ge n/2$ for large $n$ (indeed, for $n\ge 2$?).

@Balarka: I like understanding that in terms of Schubert cycles and the universal bundle on the Grassmannian, but of course you can do it naturally with obstruction theory, too.

Secret

I misread that as "If you're a denominator"

and then I go "lolwut?"

Ted Shifrin

6:40 PM

Keep misreading, @Secret.

You can be a denominatrix next.

Balarka Sen

@TedShifrin Wow, I did not know this (Sorry for the delay, I have been juggling a lot of conversations in-chat and off-chat)

Ted Shifrin

I think I've mentioned this to you before, @Balarka.

I might even have sent you my proof a long while ago.

Semiclassical

Dumb Fourier transform question which I should remember but don't

Mike Miller

@TedShifrin I really need to understand this as well

Someone told me this once but I didn't quite understand it

Leaky Nun

hi @ted @balarka

Ted Shifrin

6:44 PM

Well, I won't offer to send you pages from my lecture notes, because you'll say they're illegible, @MikeM :)

Semiclassical

Suppose I have a discontinuous function on the real line e.g. the usual unit box function and I want to smooth it out. An obvious way to do that is to convolve it with a Gaussian.

Mary Star

Ah ok!

Next, I want to show that the sequence $\frac{2-n^2}{-4n}$ diverges using the definition. So, we assume that it converges to a number a. Then we have the following:
$$|a_n-a|<\epsilon \iff \left |\frac{2-n^2}{-4n}-a\right |<\epsilon \\ \iff \left |\frac{2-n^2}{-4n}-a\cdot \frac{-4n}{-4n}\right |<\epsilon \\ \iff \left |\frac{2-n^2+4an}{-4n}\right |<\epsilon\\ \iff \frac{|2-n^2+4an|}{4n}<\epsilon$$ How can we find here a contradiction?
@TedShifrin

Semiclassical

How does such a convolution affect the L2 norm?

Ted Shifrin

Do you believe in Chern classes of the universal bundle being P.D. to certain Schubert cycles, @MikeM?

Semiclassical

I can't remember if there's a nice answer to that. It'd be convenient if there was, but I honestly don't remember

loch

6:44 PM

I think this is also done in 3264 and all that - at least in the algebro-geometric context.

Balarka Sen

@TedShifrin Ah, so this is in your notes that you sent me

Ted Shifrin

@MaryStar: Can you show that given any $M$, there is $N$ so that $n>N \implies a_n>M$?

Secret

@Semiclassical Are bump functions always $L^2$?

Ted Shifrin

@Balarka: I've sent you so many I've lost track, but I think I recall our discussing this before.

Mike Miller

@TedShifrin I do not know the story but it seems plausible enough to me. Presumably you can prove this algebraically if you wanted, identifying the Schubert cells in the Grassmannians and running some sort of spectral sequence.

Ted Shifrin

6:45 PM

@Secret: Is your definition of a bump function that it's compactly supported?

Secret

yeah that's how I usually read it

Mike Miller

(I'm thinking inductively)

Semiclassical

if it's compactly supported, it's definitely L2 isn't it?

Ted Shifrin

@MikeM: I don't know how Hatcher would do this. I do it the way Chern did, by integrating the invariant forms representing the cohomology classes over Schubert cycles. That's actually quite easy.

Secret

I think so, I cannot think of any compactly support function to blow up when squared

Ted Shifrin

6:47 PM

Does Milnor/Stasheff discuss Schubert cycles? I no longer own the book.

Mike Miller

I don't remember. I never loved that book, which I feel is scandalous to most.

loch

At least the baby case would be that for the top chern class is the euler class which is PD to the zero locus of a generic section..

Ted Shifrin

Nor I.

Mike Miller

@loch I know exactly two of the cases

Ted Shifrin

Correct, @loch. That's "well known" in the case of (complex) line bundles.

Secret

6:47 PM

convolving a discontinuous but compactly support function with a gaussian should give smoe kind of bump function, but for the noncompact case I am not sure

Mike Miller

One is that, and the other is $c_1(E) = c_1(\det E)$

Balarka Sen

I stopped reading Milnor-Stasheff after it introduced Steenrod squares without motivation

Everything feels like magic there

It's also why I forgot most of characteristic classess

Mike Miller

Steenrod squares are not magic but they are complicated

We can talk about it sometime

Ted Shifrin

I don't know the book well at all, but there's a several-volume book by several Canadian geometers that does bundles, characteristic classes, etc. Anyone remember what I'm talking about?

Balarka Sen

That'd be quite nice

Ted Shifrin

6:49 PM

@Balarka: I'll discuss char classes with you using curvature forms :P

Balarka Sen

That'd also be pretty cool! I'll recover some forgotten geometry

Mike Miller

@TedShifrin Personally I would define Chern classes axiomatically and prove their existence by studying the cohomology of $BU(n)$, and I would do that by running some spectral sequenes.

Ted Shifrin

And then you have essentially no feeling for what they mean ... :)

But the obstruction theoretic interpretations should be clear.

Mike Miller

Not define to kids, define to myself.

I really like the interpretation you give. Once I 'get' it, it will be my preferred definition.

Ted Shifrin

Oh, Loring Tu has a new book on curvature, bundles, and characteristic classes. I bet it's pretty good.

Mike Miller

6:50 PM

I think that understanding these fibrations is probably a) not too difficult b) concretely related to Schubert cycles as you say.

Semiclassical

@Secret convolving a compactly-supported function with a gaussian doesn't give a function with compact support

Ted Shifrin

Yeah, if you're algebraic enough with homogeneous spaces, @MikeM, I'm sure that's so.

The group representation guys do all sorts of Schubert variety stuff.

Balarka Sen

@loch I find the Euler class fact quite intuitive by thinking about $E = TM$. A perturbation of the zero section is a vector field on $M$, whose zeroes are counted by $\chi$ using the Poincare-Hopf battery

Mike Miller

By algebraic, you mean by writing out the matrices? I think so.

Eric Silva

@Semiclassical do you want like actual quantitative info, it's obviously gonna be bounded as long as your guy is bounded

Semiclassical

6:52 PM

Quantitative if possible

Secret

ah right, step function convolve gaussian will give something that spans an open interval, and hence no longer compact

Ted Shifrin

Oh, maybe it's Greub's multi-volume book I'm thinking of?

Semiclassical

it'll have support on the real line, I should think

Eric Silva

you can bound it by some p-norm of the gaussian and the bound of your original guy

Semiclassical

@EricSilva Sounds right.

Ted Shifrin

6:53 PM

No, I don't think so ...

Ah, Greub-Halperin-Vanstone.

That's it.

@Secret: That's no longer a "bump function."

Semiclassical

But that has the sound of an inequality not an equality. Doesn't mean it's wrong, but it's a bit frustrating

loch

@BalarkaSen yeah - actually off the top of my head i guess a proof of poincare hopf is using the fact that capping the euler class of $TM$ and the fundamental class gives you the euler characteristic. i don't quite rmb how one is proves poincare hopf without this

Balarka Sen

I like the Lefschetz fixed point theorem way, by flowing along the vector field. Zeroes of the vector fields are the fixed points of the flow.

Mike Miller

@BalarkaSen Actually Mosher-Tangora is probably where I got my intuition for Steenrod squares. Steenrod's idea is to start with $\text{Sq}^n(x) = x \cup x$ in degree $n$. Then note that $C^*(X;\Bbb Z/2)$ is homotopy-commutative but not commutative. The homotopy itself carries interesting information, and one can derive $\text{Sq}^{n-1}(x)$ from it.

Ted Shifrin

By showing that that the sum of the indices of a v.f. is the self-intersection of the zero-section of $TM$ @loch.

Semiclassical

6:54 PM

What I really want is a smooth family of functions $\phi(x;\epsilon)$ with unit L2-norm such that $\phi(x;\sigma)$ converges pointwise to a discontinuous function as $\sigma\to 0^+$

Mike Miller

In fact it is homotopy-commutative in all degrees, and these homotopies all carry actual information about the space, in that you can define all of the $\text{Sq}^{n-k}(X)$ from them.

Semiclassical

I had hoped I could just say "convolve with a gaussian with width $\sigma$"

Balarka Sen

Mmm I see. That it's cup square in degree $n$ is basically all I know

Semiclassical

And I guess I can do that if I just agree to then normalize each time

Balarka Sen

I'll look in Mosher-Tangora

Semiclassical

6:56 PM

But that's a bit less convenient.

Ted Shifrin

@Semiclassic: The classic construction converging to the $\delta$-function doesn't converge to a function at all :P

Semiclassical

heh, true

loch

@TedShifrin hmm i see. now that i think of it maybe ive never learnt how to prove it (before knowing things about char classes). i should probably revisit them some time

Semiclassical

Fun fact, the phrase 'chern #' is written on one of the whiteboards nearby

Ted Shifrin

@loch: I think the proof I suggested is in Hirsch, but I've presented it myself about 6 or 7 times in lecture and haven't looked in books.

Only one of 'em, @Semiclassic?

Mike Miller

6:58 PM

@BalarkaSen There are a number of approaches, some cup-down, some bottom-up. I think Mosher-Tangora prefer bottom-up

Mary Star

@TedShifrin We have that $\frac{2-n^2}{-4n}=\frac{n^2-2}{4n}>n^2-2>N^2-2$. Is this correct?

Semiclassical

"Chern # $\neq 1$ is not equivalent to having an edge mode"

is basically what's written there

Balarka Sen

Today seems like a productive day for math. I am going to sit down with a notepad and a pen and starting writing down the informations that's flowing in this chat

Ted Shifrin

No, @MaryStar. Pay attention to your algebra. And use intelligent estimates.

Balarka Sen

Lest I forget

loch

6:58 PM

@TedShifrin thanks! i might look into it later on.

Ted Shifrin

@Semiclassic: The question is, which Chern number? In general, there are lots (for a given bundle and given manifold).

Semiclassical

Ah

Ted Shifrin