Mathematics - 2016-07-22 (page 1 of 2)

only thing i can think of is that in the next line after that equation, they don't only require $\Vert c'\Vert=0$ which is what i'd expect if they're trying to kill off the second term

but rather as well have $c'(t)=\lambda(t)(\partial/\partial r)$

which makes me wonder if the previous equation is only intended to be valid under that assumption

and that thought seems consistent with the bound on $\Vert c' \Vert$ they give earlier in relation to that

problem is, I don't have a geometric sense of what $c'(t)=\lambda(t)(\partial /\partial r)$ means

Balarka Sen

i woke up. bleh.

0celo7

$\partial/\partial r$ points outwards in the radial direction

It means the tangent is, up to a scale, pointing radially outwards

Balarka Sen

today's riemannian geometry?

Semiclassical

right. wouldn't that require that the curve always be moving radially outwards? (not necessarily at a constant speed)

0celo7

12:09 AM

@Semiclassical Yes.

Semiclassical

i mean, that squiggly curve definitely doesn't satisfy that condition

0celo7

You can adjust the speed issue by redefining the parameter

Semiclassical

I don't like how it's all written out, but I think the key is that tangent-vectors-are-radial condition

because that automatically locks you into moving along one radial ray

you can overshoot and swing back in that case, of course, but you can't wiggle

0celo7

So, we're assuming that $L(c)=r(c(t_0))+\int_{t_0}^1||c'||\,dt$?

@BalarkaSen feel free to contribute

Balarka Sen

<--- dunno anything about riemannian geometry

Semiclassical

12:12 AM

eh, I don't either

Balarka Sen

hence "whereof one cannot speak thereof one must be silent"

Semiclassical

this is really more a purely geometric point, i think

otherwise i wouldn't be able to say shite :P

0celo7

Ah, I know of one more Riem geo book

I'm sitting in the library next to the geometry section

Semiclassical

What I don't get is to what extent they've actually assumed that condition on $c'(t)$

0celo7

ok, let's hope Chavel uses a similar proof

Nope, he uses Lang's proof

Crap

Where is Mike Miller when you need him

Balarka Sen

12:24 AM

@0celo7 i don't get the statement about the two orientation on preimage of subfolds by maps transverse to it. i may ask you tomorrow if i don't figure it out after i wake up tomorrow morning

0celo7

What statement?

Balarka Sen

G-P proposition on 101 page.

0celo7

I found that reading the proof helped me understand it

Balarka Sen

Isn't that applicable to every theorem in the world? :P

Anyhow, I'll try to do it before I read it or ask you.

Semiclassical

knowing the proof doesn't always clarify what's actually being asserted, though

sometimes that's clear regardless of the proof

Balarka Sen

12:27 AM

yeah, the assertion is quite clear

0celo7

It's a technical point that taking $(\partial f)^{-1}(Z)$ and then orienting it is slightly different than taking $f^{-1}(Z)$ and then taking $\partial(f^{-1}(Z))$ and giving it the boundary orientation as the boundary of $f^{-1}(Z)$

Balarka Sen

I can't see immediately why they'd be same upto possible a sign. It might be technical, but it's important - it seems I gotta use that to prove oriented degree is homotopy invariant.

0celo7

Hmm?

Do you not see why they're the same as sets?

The sign is just the orientation

Balarka Sen

@0celo7 Sure, that is obvious.

I am speaking of the orientation...

0celo7

@BalarkaSen Then I don't understand "I can't see immediately why they'd be same upto possible a sign"

If they're the same as sets/manifolds, they can only differ by a sign

Balarka Sen

12:31 AM

I meant the orientation on them.

0celo7

...any two orientations only differ by a sign, ever

See the proposition on page 97

Balarka Sen

That's a triviality. Clearly I meant (-1)^codim(Z) when I said "sign".

It's not clear where that pops in.

0celo7

oh

you have to move a normal vector past $Z$'s tangent space, or something

Semiclassical

evidently not so clearly :P

0celo7

Not a very interesting proof

Refath Bari

12:33 AM

Sup

0celo7

do you know any Riemannian geometry?

Balarka Sen

@Semiclassical It's unlikely that I'd ask about a triviality (aka, manifolds admit two orientations).

Or anyone would, really.

0celo7

@BalarkaSen no, it was not clear

Refath Bari

@0celo7 Oh god no.

0celo7

@BalarkaSen that's not trivial

Balarka Sen

12:34 AM

I disagree.

But whatever.

Semiclassical

trivial is a fighting word :P

Refath Bari

@0celo7 Closest I ever got was Number Theory

0celo7

@RefathBari That's the opposite side of math

Refath Bari

@0celo7 No wonder. I'm more around the comptetition side of math

I.e. MATHCOUNTS

0celo7

@Semiclassical so, what do you think is going on in that proof?

what's the conclusion

Semiclassical

12:36 AM

the one above?

0celo7

I've lost most of my evening to this

@Semiclassical the Riem geo one

Semiclassical

i'm not sure. i do think it comes down to their use of that one condition

Refath Bari

Sorry for interrupting, but do you guys know any room specifically for math competitions?

Semiclassical

but it's not clear to what extent they actually employ it. so i find the argument confusing.

0celo7

I think it's nontrivial that if $c'$ satisfies that condition then the two $t_1$s are equivalent.

You'd have...let's see

Semiclassical

12:38 AM

maybe. geometrically it seems apparent

Balarka Sen

@0celo7 Arguing about trivial statements being nontrivial and clear-from-context statements being unclear is not helpful, though more annoying.

Semiclassical

since if you can't wobble back and forth, then the counterexample you proposed is forbidden

0celo7

$\int_0^{t_1}||c'||\,dt=\int_0^{t_1}\lambda\,dt$

assuming you have that condition

I don't see why that's $||v||$

$\lambda$ has nothing to do with $v$

Semiclassical

that only gets to the $\Vert c' \Vert=1$ part, though

you've also got it always in the direction $\partial/\partial r$ which seems geometrically the more important point

0celo7

uh $c$ does not have arc length parametrization

the_new_guy

12:41 AM

hello, would you guys mind helping me with a proof? I think I have it but just wanted to get a sanity check. Trying to prove $\lim \limits_{x \to 0}\sqrt[3]{4x}\left(\sin\frac{5}{x^2}\right)=0$

0celo7

it's really $||c'||=\lambda$

because $||\partial_r||=1$

Semiclassical

yeah, you're right

it's the second one i meant

careless on my part

0celo7

hmm?

Semiclassical

i meant the $\partial_r$ version

0celo7

1990s GTM books are so yellow

Semiclassical

12:43 AM

speaking quite loosely, i think the point is that if $c'(t)=\lambda(t)\partial_r$, then $\lambda(t)$ is roughly equivalent to $dr/dt$

and in that case you would have $\int \lambda\,dt=\int dr =\Vert v \Vert$

0celo7

Ah, that might be the chain rule?

the_new_guy

So with my proof I came up with $|n|\lt \frac{\epsilon^3}{4}$

Semiclassical

@0celo7 that's the heuristic, at any rate

SteamyRoot

@the_new_guy what is $n$?

Semiclassical

what i wrote may not really be sensible taken literally

the_new_guy

12:45 AM

@SteamyRoot Sorry, it should be $|x|$

0celo7

One more book: Helgason

the_new_guy

ugh $x$ sorry getting my variables mixed up!

SteamyRoot

And you're trying to prove it using epsilon-delta or so?

the_new_guy

correct

0celo7

literally no clue what I just read

that notation

the_new_guy

12:48 AM

@0celo7 you're talking about what I typed up?

0celo7

lol it's on the exponential map for Lie groups

not Riemannian stuff

@the_new_guy no

the_new_guy

@0celo7 ahhh ok lol

0celo7

@Semiclassical nope, Helgason has the standard proof :(

Semiclassical

i'd find someone who knows this stuff and ask them :)

0celo7

I've checked do Carmo, Lee, Lee, Petersen, Walschap, Lang, Milnor, Jost, Helgason, Kobayashi

Semiclassical

12:49 AM

rather than finding a physicist who can only handwave

SteamyRoot

@the_new_guy Yeah, it does sound logical then that you get something along the lines of $|x| < \delta = \epsilon^3/4$ gives you limit $< \epsilon$

0celo7

I'm running out of diff geo books to check

@Semiclassical no one around at 9PM

Semiclassical

is this for an assignment?

0celo7

No, the assignment is a different book that proves the same thing a different way

And I understand that

but I was reminded of this one

Semiclassical

i mean, it sucks if you can't clarify it right now. but if there's not a deadline then you can just be patient :P

0celo7

12:51 AM

Patience is not my strong suit

Semiclassical

No, I suppose not :)

probably best to find something else to procrastinate with

0celo7

I'm not procrastinating, I'm obsessed

Semiclassical

eh, two variations on the word 'compulsion'

the_new_guy

@SteamyRoot OK great! I basically used the fact that $\left|\sin^3\frac{5}{x^2}\right|\le|x|\lt\frac{\epsilon^3}{4}$

SteamyRoot

Whoa what?

0celo7

12:58 AM

@SteamyRoot the first inequality is true

SteamyRoot

True, sure, but necessary?

the_new_guy

@SteamyRoot I don't understand, what's wrong with it?

SteamyRoot

Well, by that inequality, did you try to put the sine under the cubic root or so?

I would've just used that $|\sin x| \leq 1$ for all $x$, hence you only need to worry about the cubic root of $4x$.

the_new_guy

I see

SteamyRoot

Well, either way, the proof is pretty much the same

So you're right

the_new_guy

1:12 AM

Given that $\epsilon$ is not dependent upon $x$, does that mean this is uniformly convergent on $(0,2)$?

0celo7

0

Q: Geodesics in geodesic balls

It is well-known that in a geodesic ball containing points $p$ and $q$, the radial geodesic between them is the unique minimizing curve. I'm trying to follow the proof of this given in Cheeger & Ebin (AMS Chelsea edition, page 8). They seem to take a slightly different approach than usual. Let $...

Forever Mozart

1:50 AM

tell me something good

0celo7

2:03 AM

@ForeverMozart the nullity of the energy Hessian is the dimension of the space of Jacobi fields with conjugate points

Mazura

4:40 AM

What do you call equations that the answer of is just a smaller equation? E.g., where the 'answer' is X+2Y ?

Balarka Sen

11:21 AM

@TedShifrin I was in fact given the problem about dimension of space of homogeneous polynomials today, parsed in a different way (choose fixed no of balls out of a collection of balls allowing multiplicity). I worked out an elementary argument by the end of school, suddenly realizing what you mean by box and carets

If I pick k balls out of n (caring about order) allowing multiplicity, it's same as partition a bunch of collection of balls by a few walls, say. You need precisely k - 1 walls, you so you have n + k - 1 objects, and you're choosing k of them out of there the plain vanilla way.

That's C(n + k - 1, k) :)

Great technique. Took me a long while to figure it out, probably couldn't have if I didn't hear "box and carets".

0celo7

12:20 PM

@Semiclassical 1000% sure there's a typo.

The $=$ should be a $\ge$.

You get equality by assuming $c'=\lambda\partial_r$, then you cut off the curve at $t_1$.

The rest of the proof works then.

Semiclassical

1:41 PM

@0celo7 okay, that makes more sense.

Linear Christmas

Hey, everyone! A quick question on notation. Say $$\mathbf{A}\subseteq\Bbb{R}\ \land \mathbf{B}\subseteq\Bbb{R}$$ and $f:\mathbf{A}\rightarrow\mathbf{B}$. What does then the notation $$f(\mathbf{A})=\mathbf{B}$$
mean? Is it equivalent to the function $f$ being surjective?

Samuel Yusim

$f(\mathbf{A}) := \{f(a) : a \in \mathbf{A}\}$. The equality is an actual equality of sets, so yes it is equivalent

(two sets are equal when they have exactly the same elements)

Linear Christmas

Ah, that makes a ton of sense. Thank you, sir.

Semiclassical

"For every element $b$ of $B$ (the codomain of $f$) there is an element $a$ of $A$ (the domain of $f$) such that $f(a)=b$."

which is, yeah, literally the definition of surjective.

Samuel Yusim

that notation is pretty common, so it's a good thing to be comfortable with

Linear Christmas

1:52 PM

Agreed. Sometimes notation is the only confusing aspect about a proof :p

Samuel Yusim

in my experience it's most often the notation, for sure

0celo7

That's the best way to write surjective in symbols

Semiclassical

it does nicely capture that surjectivity is as much about $A$ and $B$ as it is $f$

@0celo7 sounds like you're satisfied re: that equation having a typo

0celo7

@Semiclassical Yes. I will bring it up with my prof in our next meeting. He loves that book, but he probably didn't bother ever reading the "review" part ;)

Semiclassical

lol

0celo7

2:01 PM

@Semiclassical Yeah, it shows that if $B$ is any larger, $f$ will definitely no longer be surjective, and if $A$ is any smaller, $f$ might not be surjective.

Semiclassical

right.

0celo7

Reading Milnor is a struggle. I need like 5 reference texts

I wish I had Hermione's magic bag, but for books

Semiclassical

bookbag of holding?

0celo7

Is that what her bag was called?

Semiclassical

nah

0celo7

2:04 PM

Did you read/watch them?

Semiclassical

I'm making a pun on the D&D item "bag of holding"

0celo7

Showing your age, bro

Semiclassical

pfft

i haven't actually played D&D all that much. not a lot of opportunities for it

0celo7

If you don't know what my avatar is, you're also showing your age

Semiclassical

it looks like a bunch of diamonds

0celo7

2:06 PM

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Semiclassical

i see.

not the sort've thing i pay attention to whatsoever

0celo7

Heine-Borel is such a black box theorem

Akiva Weinberger

2:33 PM

@0celo7 What about $A\xrightarrow fB\to0$? (Not actually)

0celo7

@AkivaWeinberger What is that

What is $B\to 0$

Akiva Weinberger

Notation for surjective. It's an "exact sequence", not sure if you've heard of them

though I think it only really makes sense for groups

0celo7

I know what an exact sequence is

What is $0$

Empty set?

Tobias Kildetoft

@AkivaWeinberger it makes sense in abelian categories (which groups are not actually)

Akiva Weinberger

The trivial group

Oh, OK. So they need to be abelian.

0celo7

2:35 PM

Most objects are not groups

Akiva Weinberger

$(\{0\},+)$

Tobias Kildetoft

it takes less for it to make sense, but it need no longer imply surjective (as far as I recall)

hmm, or rather the other way around, not all surjective (i.e. epi) morphisms can be put into such an exact sequence

Akiva Weinberger

You sure? Exactness means that the image, which is $f(A)$ in this case, is equal to the kernel, which is all of $B$.

Tobias Kildetoft

@AkivaWeinberger well, or something that looks like that when taken in more generality (I can never remember precisely how it goes)

Akiva Weinberger

Hm.

Tobias Kildetoft

2:38 PM

I think you are right that it implies surjective, because it means that the map is a cokernel (and these are epi). But not all epi morphisms need be cokernels in general

(this is the dual version of what goes wrong for groups where not all monomorphisms are kernels)

0celo7

@AkivaWeinberger Ah, right, that works for vector spaces

Akiva Weinberger

Hm. But "epi" is slightly different from "surjective" in some categories

0celo7

$0\to A\to B\to C\to 0$ means $A\to B$ is injective and $B\to C$ is surjective

Tobias Kildetoft

@AkivaWeinberger right, because "surjective" is not defined in categorical terms

Akiva Weinberger

Yep. Which means, for vector spaces, I believe, that $B\simeq A\oplus C$.

0celo7

2:40 PM

But $f(A)=B$ is certainly better

Akiva Weinberger

It is.

0celo7

@AkivaWeinberger Yes, although I've never been really convinced of that

I should prove it again

Akiva Weinberger

I think you can make an argument using bases

Tobias Kildetoft

@AkivaWeinberger yes, all short exact sequences of vector spaces split (fields are semisimple)

0celo7

You just show that $A\oplus C$ has the same dimension as $B$

It might work for infinite-dimensional too

But I'm unconcerned with those for the time being

Akiva Weinberger

2:41 PM

And now I'm wondering about spaces without bases (in a choice-less world)

Tobias Kildetoft

@0celo7 in fact, the statement can be strengthened (the sequence splitting is stronger than just having an isomorphism)

Akiva Weinberger

(since you need the axiom of choice to prove all vector spaces have a basis)

0celo7

@AkivaWeinberger of course

I dislike choice but I like vector spaces having bases

Tobias Kildetoft

@AkivaWeinberger right, and I think it also lurks somewhere when trying the more theoretical approach using semisimple algebras, since we need it to show that all rings have maximal ideals

Akiva Weinberger

I like my infinite sets to be able to be split into two infinite subsets.

The fact that that can fail without choice is just so bizzare.

0celo7

2:43 PM

And well-ordering isn't?

Akiva Weinberger

And I like ultrafilters. I think, if I had to give up choice, I would still insist on the ultrafilter lemma.

@0celo7 Forget well-ordering, even; orders at all aren't guaranteed without choice. $\cal P(\Bbb R)$ might not have a total order.

0celo7

Typos are the worst

@AkivaWeinberger I don't know what that means, set theory is my enemy

Akiva Weinberger

(Incidentally, you can prove that everything has a total order using the ultrafilter lemma.)

0celo7

I do know that some results in geometry and general relativity use e.g. Zorn

Akiva Weinberger

@0celo7 Just any order. A $\le$ relation that acts like $\le$ relations should.

$\cal P(\Bbb R)$ is the set of subsets of $\Bbb R$.

Tobias Kildetoft

2:46 PM

@AkivaWeinberger well, I would not usually require that a $\leq$ relation be total

0celo7

@AkivaWeinberger With choice, what is the ordering?

Tobias Kildetoft

anyway, it is possible to construct an uncountable well-ordered set without choice. It is just not possible to show that it has the same cardinality as the reals without choice

0celo7

@TobiasKildetoft Interesting.

Akiva Weinberger

@0celo7 Well-order it. Well orders are types of orders (aka total orders aka linear orders).

Tobias Kildetoft

I recall this being an exercise in Munkres' topology book

0celo7

2:47 PM

@AkivaWeinberger I was asking if one can actually find the order.

Because it does not seem to me that it should have an ordering.

Akiva Weinberger

Well, you need some amount of choice for it, so I can't give you one explicitly.

But I still feel like every set should have an ordering (like $\Bbb R$ does).

0celo7

Sometimes I wonder if I should go down the constructivist rabbit hole.

But I want to live life without ever seeing the ZFC axioms

Akiva Weinberger

Do you know about the set $\omega+\omega$, by any chance?

Apparently, you can't prove it exists without the "F" part of ZFC.

Tobias Kildetoft

@AkivaWeinberger me?

Akiva Weinberger

(Which I believe is for the axiom of replacement.)

0celo7

2:51 PM

@AkivaWeinberger Nope, what does that set do?

I'm completely disinterested in set theory for the sake of set theory.

Akiva Weinberger

Essentially, you start building this hierarchy of sets:

$0$ is the empty set.

Tobias Kildetoft

Never thought much about that distinction. I mean, a set like $\omega + \omega$ is so easy to describe using the usual techniques.

Akiva Weinberger

$1$ is $\{0\}=\{\emptyset\}$.

Tobias Kildetoft

Basically, you need axiom of infinity and being able to take products

Akiva Weinberger

$2$ is $\{0,1\}=\{\emptyset,\{\emptyset\}\}$.

0celo7

2:52 PM

@AkivaWeinberger I know how the natural numbers are constructed.

Akiva Weinberger

Etc. So $n$ is $\{0,\dots,n-1\}$.

Then you define $\omega=\{0,1,2,\dots\}$, the set of all natural numbers.

So, {{},{{}},{{},{{}}},…}

And then $\omega+1$ is the set of everything from $0$ to $\omega$; $omega+2$ is the set of everything from $0$ to $\omega+1$, etc.

And, finally, $\omega+\omega$ is the set $\{0,1,2,\dots,\omega,\omega+1,\omega+2,\dots\}$.

Tobias Kildetoft

@AkivaWeinberger which can be identified with $\mathbb{N}\times\{0,1\}$.

0celo7

@AkivaWeinberger I've never understood what "everything from $0$ to $\omega$" means

I should probably head to work

Akiva Weinberger

$\omega+\omega$ is, then, {{},{{}},{{},{{}}},…{{},{{}},{{},{{}}},…},{{},{{}},{{},{{}}},…,{{},{{}},{{},{{}}‌},…}},{{},{{}},{{},{{}}},…,{{},{{}},{{},{{}}},…},{{},{{}},{{},{{}}},…,{{},{{}},{{‌},{{}}},…}}},…}

if I didn't mess up

@0celo7 $\omega\cup\{\omega\}$, or $\{0,1,2,\dots,\omega\}$

0celo7

Most of my mind rejects that notion :P

Akiva Weinberger

2:57 PM

It's a weird infinite fractal thing made of {s and }s

Also, you can go much higher with this.

Tobias Kildetoft

@AkivaWeinberger but of course $\omega + 1\neq 1+\omega = \omega$.

0celo7

@TobiasKildetoft Oh god

Akiva Weinberger

These are called "ordinals." And, the set of countable ordinals is another ordinal (an uncountable one, otherwise it would contain itself).

0celo7

I know what ordinals are

Akiva Weinberger

In fact, that's the uncountable well-ordered set Tobias mentioned earlier, I believe.

Ordered by $\in$.

0celo7

2:59 PM

I don't want to know why it's uncountable

Akiva Weinberger

Well, it's the set of countable ordinals. If it were countable, it would contain itself, being a countable ordinal.

It's like how the set of finite numbers is infinite.

Samuel Yusim

what's the cardinality of the set of all ordinals containing only ordinals of strictly smaller cardinality than themselves

Akiva Weinberger

@0celo7 Also, using ordinal addition, over the ordinals we have that $1+x$ is continuous but $x+1$ isn't

(Order topology)

Tobias Kildetoft

@AkivaWeinberger what do you mean by continuous?

Akiva Weinberger

@SamuelYusim Proper class, I believe. Not a set. You want things of the form $\omega_\alpha$.

Tobias Kildetoft

3:05 PM

you mean as a function of $x$?

Akiva Weinberger

@TobiasKildetoft I just said. Order topology.

Yeah.

Tobias Kildetoft

but defined on what?

set of countable ordinals?

Akiva Weinberger

$\lim_{x\to\omega}1+x=1+\omega$ but $\lim_{x\to\omega}x+1\ne\omega+1$

Samuel Yusim

why is the thing I mentioned not a set, though? It certainly doesn't contain itself

Akiva Weinberger

@SamuelYusim For one thing, the union of its elements does contains itself.

For another, it's equinumerous to the entirety of the ordinals, which is a proper class.

@TobiasKildetoft Fine, maybe you can't have a function whose domain isn't a set. But it still works if it's defined over any ordinal larger than $\omega$.

Samuel Yusim

3:09 PM

sure, I believe the first one.

Akiva Weinberger

And one of the axioms of ZFC says you can take the union of the elements of a set, I believe.

Samuel Yusim

and after some thinking the second one too

yep that's true

Balarka Sen

are we discussing about weird set theory again

Akiva Weinberger

Maybe

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

3:26 PM

How's your health these days? @BalarkaSen

0celo7

@BalarkaSen I have a stupid point-set question that might just be the tube lemma

Never mind

Balarka Sen

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ It's ok.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Good, good.

How's mike doing? @BalarkaSen

Balarka Sen

Last talked to him 3 days ago. So no idea.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

I see.

0celo7

3:32 PM

It's a shame he left

He was the only geometer who is up when I'm working on math

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Yup.

Balarka Sen

I respect his decision: this chat is not a constructive or productive way to spend time on the whole.

in my honest opinion, that is to say

0celo7

Maybe not for him

I once did not come to the chat for a month

I didn't feel like I was more productive or anything

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

By choice?

0celo7

No

Balarka Sen

3:36 PM

I did mention it was my opinion... though I do believe many of us think the same

0celo7

Did you figure out the orientation thing yet?

Balarka Sen

Didn't get the time to look at it.

0celo7

I need to read section 6.4 in Hirsch

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Did JD just call Jim a crack pot?

0celo7

yes.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

3:39 PM

sigh

There, he humbly apologized :-)

0celo7

I should make some notes on polarization of quadratic forms

@BalarkaSen Riemannian geometry is really quite beautiful

Balarka Sen

I believe that.

0celo7

I'm about 10 pages away from another CW monstrosity in Milnor, just with an infinite-dimensional manifold this time

Balarka Sen

CW complexes are nice objects.

0celo7

I recall you not completely understanding Milnor's argument last time I encountered them in this book

Or at least saying it was highly nontrivial

Balarka Sen

3:54 PM

Not sure what are you referring to, but sure, I didn't study Morse theory so that's quite possible.

Samuel Yusim

nontrivial does not imply not nice

Balarka Sen

I'd rather parse it as "I cannot understand it does not mean it's a monstrosity". But then again, I was just referring to CW complexes being a nice class of topological spaces than anything particular out of Milnor which involves CW complexes.

0celo7

Oh, I didn't mean CW complexes are monstrosities

I mean that the proof will be a monstrosity

and it will involve CW complexes and higher homotopy groups and bleh

I'll probably take it on faith

Balarka Sen

@0celo7 Oh, now that I look at it, one doesn't actually need that theorem to conclude oriented degree (or in general intersection no) is htpy invariant.

0celo7

Do GP use it?

Balarka Sen

4:05 PM

I was being silly.

@0celo7 No idea.

0celo7

What do you mean, no idea?

Didn't you read the proof?

Balarka Sen

I didn't look in GP.

I just came up with a proof.

0celo7

Did you prove the extension lemma?

Balarka Sen

If map extends to a manifold which bounds the original domain manifold, then it's got 0 degree? Yes.

0celo7

Yeah, did you use that in your proof of homotopy invariance?

Balarka Sen

4:09 PM

Of course. It's a special case of the extension lemma with the manifold $X \times I$.

0celo7

Yup

Balarka Sen

But that's not new.

Same idea as in mod 2.

0celo7

Do they use the same proof in mod 2?

Ok

Balarka Sen

Well, mod 2 is easier.

0celo7

So I'm not sure where that boundary result actually pops up

Mod 2 depends crucially on the classification of 1-manifolds IIRC

Balarka Sen

4:10 PM

Yes, so does this. But one needs to take care of orientation.

Orientaton no. at boundary of an interval cancel, is the gist. I was confuzzled how this means so does the preimage orientation, but I see now.

0celo7

Right, I think GP has an "observation" somewhere on that

Gotta go

Balarka Sen

I don't really pay attention to G&P on every line. I am an impatient reader.

Lembik

4:31 PM

if $k<N$, what is ($1-k/N)^N$ approximately?

assuming $k,N > 1$

I feel it is $e^{\text{something}}$

We know that $(1-1/N)^N \approx e^{-1}$

Lembik

4:49 PM

hmm. dead chat :)

the answer is $e^{-k}$

user61230

I posted an answer that I realize might be wrong, but the reason why is very, very unclear to me.

user61230

Could I impose on someone to look it over and see if the reason becomes clearer?

Owatch

Hm, can anyone help me here:

$
H=
\begin{bmatrix}
5 & 6 & -10.2530 & 4.2738 & -14.9394 & -19.2742 \\
-6 & 5 & -0.1954 & 1.2426 & 7.2023 & -8.6299 \\
0 & 0 & 2.2585 & -5.4807 & -10.0623 & 4.4380 \\
0 & 0 & 1.0188 & -0.2584 & -5.9762 & -9.6872 \\
0 & 0 & 0 & 0 & 4 & 4.9224 \\
0 & 0 & 0 & 0 & 0 & 3
\end{bmatrix}
$

Apparently, the evals here are: $(1 \pm 2i, 3, 4, 5 \pm 6i)$

I can see that 3 and 4 are on the diagonal.

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