I'm blanking. $H^7(X; \Bbb Z_2)=\Bbb Z_2$. Then why does the UCT imply that $H^8(X; \Bbb Z)$ is cyclic (and finite)?

@MikeMiller

What is to light as air molecule is to sound?

the aether @leaky

(jk)

@iwriteonbananas It doesn't. Consider $X = \Sigma \Bbb{RP}^7$.

@LeakyNun Light does not require a medium, and can propagate in a vacuum.

@Semiclassical How?

Worse, given a space with $H^7(X;\Bbb Z/2) = \Bbb Z/2$, wedge on a bunch of spheres.

I guess $\Bbb{RP}^8 \vee S^8 \vee S^8$ is another counterexample, eg.

because that's how electromagnetism works, basically

I don't understand it.

If $H_7$ is $\Bbb Z^k \oplus T$, then passing to $\Bbb Z/2$-cohomology your assumption says that either (a) $k = 1$ and $|T|$ is odd; or (b) $k=0$ and $T = \Bbb Z/2^n \oplus T'$, $|T'|$ odd. In both cases $H^8$ is $\Bbb Z^{b_8} \oplus T$, but you can realize any $b_8$ obviously and any $T$ you like subject to the above constraints (either odd cardinality, or $\Bbb Z/2^n$ times something of odd cardinality).

Perhaps this will help, then. With sound, there's a definite medium (air) which carries the oscillation.

That is, there are specific particles which the wave 'waves'

Hi @Semiclassic, Bananas, g'night @MikeM

Morning.

@Semiclassical sure

With light, you can do it in empty space in the absence of any physical particles.

To the extent that there's a medium for light, it's the electromagnetic field itself. But that's not a medium in the sense of air.

How?

If you want to know how a light wave is produced, look up something on antenna.

the point is that once it's going, you've got 1) a time-varying electric field which by Faraday's law induces 2) a time-varying magnetic field, which by Ampere's law (the Maxwell version) induces 1') a time-varying electric field, and so forth

Alright, that's for tomorrow thanks.

It's now 0:30AM here, and there's school toorrow.

that is, the electromagnetic field behaves in such a way that it can carry oscillations

Bye.

night

@semiC i.e it's a self-propagating wave

right?

@Semiclassical related to the thoughts shared these days that Balarka finds tiresome, they lead me to series like $$\sum _{n=1}^{\infty } \frac{\psi (n) \psi \left(n+\frac{1}{2}\right) \psi \left(n+\frac{1}{3}\right)\cdots\psi \left(n+\frac{1}{m}\right)}{n \left(n+\frac{1}{2}\right) \left(n+\frac{1}{3}\right)\cdots \left(n+\frac{1}{m}\right)}, \ m\ge2$$

and at least 5 more users.

Ok.

I am sorry, I can't remove my last statement

It looks rude now but was a reply to your comment that is now (removed)

@iwriteonbananas What was the context?

Hatcher says: $H^7(X;\Bbb Z_2)\cong \Bbb Z_2$ implies by the UCT that $H^8(X;\Bbb Z)$ is cyclic. I couldn't figure it out.

still reading what you wrote above.

awesome example

the $\Sigma \Bbb RP^7$

and all the other ones you gave too

here's another question:

in the Hopf fibration $S^3\to S^7\to S^4$ the inclusion of fiber into total space is nullhomotopic, so we get SES's $$0\to \pi_i(S^7)\to \pi_i(S^4)\to \pi_{i-1}(S^3)\to 0$$

(from the fibration LES)

this thing splits

and i'm trying to prove that

hatcher says to use a nullhomotopy of $S^3\to S^7$ to produce a splitting

@iwriteonbananas can you tell me precisely where he says this?

yes, this is the second paragraph on page 575

(chapter 5)

I meant the cyclic thing

Oh, that's on page 574 :) last paragraph.

So what we need to see is that there's no odd torsion in $H_7$ and that $b_8 = 0$. That puts us in my case (b). Are these perhaps automatic from the spectral sequence setup?

We're ignoring odd torsion without explicit mention

$b_8=0$ hmmm...

OK, so we just need to know why $b_8 = 0$

(That means that homology group isn't necessarily actually cyclic, just cyclic modulo odd torsion. I guess the point is we're only trying to calculate the 2-part of the homotopy groups?)

yeah, exactly.

@iwriteonbananas If $f_t: D^4 \to S^7$ is a null-homotopy, composing that with the Hopf fibration gives you a map $S^4 \to S^4$ since the $S^3$ gets killed. So send $\eta: S^{i-1} \to S^3$ to $\eta_t: D^i \to S^7$, and compose that with the fibration to get $\eta': S^i \to S^4$

That's probably a section of your last map

Ahh, yes. that's good. couldn't figure that out.

actually sorry, i'm a bit confused. what exactly is $\eta_t:D^i\to S^7$?

hmm

Define $\tilde \eta_t: S^i \times I \to S^3 \times I$ to be $\eta$ for all $t$, and then compose with $f_t: S^3 \times I \to S^7$. Because $f_1$ is constant, $f_1 \tilde \eta_1: S^i \times \{1\}$ is constant, so I can kill it off to get a map $\eta_t: D^i \to S^7$.

right, makes sense

guys I need a little help in homotopy theory

I asked this yesterday too but didn't get a very understandable explanation

yes

that one

You define the notion of fundamental group, concatenate the two paths, and show it's not zero in the fundamental group (by developing tools for calculating fundamental groups like covering spaces and van Kampen's theorem).

I was flagged and banned for telling my honest, civilized and balanced opinion. Now, a simple question to all of you: can you tell me who of you consider himself/herself a real mathematician? Just to know for my future questions, I have some (right now). Only if possible, of course.

ohhhhhhhhhhhh

thanks a lot @MikeMiller

thanks for the help, @MikeMiller. i don't see why $b_8=0$ right now. probably i'm missing something that should be obvious. i'll try again tomorrow.

I was like at the 2nd chapter of the book of algebraic topology where they just begin to explain basic definitions like homotopies and I kept wondering WHY THE HELL CAN'T I PROVE THIS SIMPLE THING

I mean I thought that should have been obvious

I see now it was highly non trivial

@MikeMiller I rarely talk to you, but I'm curious if you consider yourself a real mathematician. If I don't ask for too much (it might also be a private piece of information), sure.

Let me prepare some series in the meantime.

Calculate in closed form

$$\sum_{n=1}^{\infty} \frac{H_n H_{2n} H_{3n}}{n^3}$$

It's sort of strange that you need machinery to do anything with the fundamental group, but it's true. I think somehow it's "obvious" that $\pi_1(S^1) = \Bbb Z$, but it still takes like a while in the first chapter of Hatcher to prove that. If you're content with working with smooth paths and smooth homotopies, you can prove this with Green's theorem (which more or less says that the integral of an irrotational vector field only depends on the homotopy class of the loop you're integrating).

@Semiclassical you might like the one above (in case you're not so terribly busy with your stuff).

Then you would write down a loop in $\Bbb R^2 \setminus \{0\}$, a nice irrotational (but not conservative!) vector field on $\Bbb R^2 \setminus \{0\}$, and integrate to see that the integral is not zero. Thus your loop was not null-homotopic.

You might also like to try this version

$$\sum_{n=1}^{\infty}(-1)^{n-1} \frac{H_n H_{2n} H_{3n}}{n^3}$$

@MikeMiller if a student comes to you and asks you for a tiny hint for the one above, what would you like to answer back?

@MikeMiller I seriously do not know most of the terminology you're using

like I just passed my 4th year undergraduate majoring in math and I still feel like a real noob

Anyway, I think I have some stuff to finish.

BBL

@DanielFischer: sorry for disturbing you, but how could I find limit $\lim_{z \to -1} \frac{z+1}{(z^2+a^2)(\log z - \pi i)}$, where $a>0$ and $\log z$ is branch of logarithm with $0 \le arg z < 2\pi$ (principal one). Basicly, I am interested in finding $Res(\frac{1}{(z^2+a^2)(\log z - \pi i)}, -1)$.

@Cortizol If $g$ has a simple zero at $z_0$, and $f$ is holomorphic there, then the residue of $f/g$ at $z_0$ is $\frac{f(z_0)}{g'(z_0)}$. Let $f(z) = \frac{1}{z^2+a^2}$ and $g(z) = \log z - \pi i$. Thus your residue is $\frac{-1}{1+a^2}$.

@DanielFischer Of, I forgot for that rule. Thank you! just to ask, is there some more computation way?

@Cortizol Well, Taylor-expand the logarithm. $\log z = \log \bigl((-1) + (z+1)\bigr) = \log \bigl((-1)\cdot\bigl(1-(z+1)\bigr)\bigr) = \pi i + \log \bigl(1 - (z+1)\bigr) = \dotsc$.

HI everyone

@DanielFischer Thank you. I had never Taylor expand complex logarithm. For some reason, I am afraid when I think about it. I will look for some reference first

someone can explain me, what does mean when say that the isometries of the sine curve is a group?

I was thinking in $\begin{matrix} && 1 && 2\pi \\ v && v && 0 \\ -v && -v && 0\end{matrix}$

with the binary operation $\sin(v*2\pi) = 0$

is that correct?

@SoumyoB We've all been there. You should probably know what Green's theorem is (that's back from multivariable calculus - but I forgot multivariable not long after I learned it, too ;) ), but you'll (re)learn a lot of this in time, if you want to.

@SoumyoB It is not "highly" nontrivial, but trivial only after getting introduced to the right machinery. You're essentially wanting to prove, e.g., two arcs in S^1, starting and ending at the north and south poles, are not homotopic which is equivalent to trying to prove S^1 has nontrivial fundamental group.

hi @BalarkaSen

hi

Hi guys

@Obliv Exactly so.

Small technical question: In class, my prof. presented a proof that for CW complexes $X\vee Y$ has the direct sum of the cohomology of $X$ and $Y$ as cohomology. Today, I proved the same, under the assumption that the point where the wedge sum is formed is a defo retraction of a neighborhood of it, in each space (no CW complex assumption). In my prof's proof, he started by stating that $X\vee Y$ is homotopy equivalent to the space where the "shared point" is "thickened" to a line segment.

This somehow must exploit the CW assumption, because after that the whole proof is quite simple, but of course it doesn't work for just any pair of top. spaces. I'm wondering how this simple step uses the CW structure. My prof. couldn't tell me when I asked him (though he acknowledged that he must be using it, somehow).

I am curious to see examples where the shared point cannot be thickened to a line.

Well, I think you need at least the assumption I made (Hatcher makes that assumption, too), for stuff to work out in the end.

But I, too, have a hard time seeing where you really use anything to say that this is possible.

@Danu What do you mean with "thickened to a line segment"? $(X \vee [0,1]) \vee Y$, where $0$ is identified with the point in $X$ and $1$ is identified with the point in $Y$?

So... what can we say? First of all, the "thickened" space can be retracted to the wedge sum. So where does the converse process go wrong?

@DanielFischer Pictorially: OO $\mapsto$ O-O

(where O represents X and/or Y)

Yeah, that's what I meant too.

@DanielFischer Oh, yes, that's obvious, actually. Sorry.

It feels like it'll always be homotopy eq to X v Y.

@MikeMiller So the prof. did the $S^2\vee S^4\not\simeq \Bbb CP^2$ example in class, today. Hah.

@BalarkaSen But it's not true :\

Ex?

@BalarkaSen Any wedge sum of spaces that's not summative in homology.

I don't remember one.

Maybe take something with a shitty topology on it

You're the one who cares, though.

I just wanted to see an explicit example. But it's ok if coming up with one is too tedious or annoying: I am not thinking too hard about it either :)

I'm just wondering if any of you sees where the assumption should come in

It suffices to see if $(X \vee [0,1] \vee Y, [0,1])$ has the homotopy extension property by Hatcher ch0. Presumably having one of the basepoints be an NDR is sufficient.

Of course $[0,1]$ is a CW subcomplex of that, which makes it automatic.

NDR?

neighborhood deformation retract

probably the wrong word but w/e

You mean it has a neighborhood which defo retracts onto it?

So the question is whether you can pull $X \vee Y$ apart. Take neighbourhoods, $U,V$ of the gluing point in $X$ resp. $Y$. If you can pull the two apart, then you must be able to choose $U,V$ so that $U\vee V$ is mapped into $X \vee [0,\varepsilon)$ resp. $(1-\varepsilon] \vee Y$. That means on at least one end, the neighbourhood is sucked into the interval, which means the point is a deformation retract.

I guess Hatcher calls that $(X,x_0)$ being a good pair

If it has, then HEP kicks in.

Which shows quotienting [0, 1] doesn't matter for the homotopy type.

Hatcher assumes all base poitns to be good pairs.

what is HEP?

(High Energy Physics?!)

Homotopy extension property.

Homotopy extension property. A good property for (X, A) such that the qt map X --> X/A becomes a htyp eq.

@DanielFischer Ok.

Ah, is that (X,A) being a "good pair"?

yes.

It's discussed at length in chapter 0 somewhere in Hatcher.

Thanks @DanielFischer for being so explicit

In any case, this is all true for any space you'll think of unless you're explicitly trying to think of spaces where things aren't true.

standard answer of Mike whenever he sees horrendous spaces coming up :D (I don't object to it!)

If it's true for Poincare then it's true enough for me :)

I don't know what that means.

if it's not a manifold then I don't care

manifolds are nice, but IMO there are nice spaces which aren't manifolds.

I tried reading Poincare's famous paper one time. It was not even that incomprehensible! :D

That doesn't sound like Poincare's philosophy or mine. But that's fine.

@BalarkaSen Sure... In physics maybe we can stretch it to orbifolds, but that's as far as I'll go ;)

@MikeMiller (you killed the humor)

What humor?

ba-dum-tsss

@Danu It's a good paper. I wrote a blog post about those ideas somewhere.

It's wrong, but it's still fantastic.

Hello!

@MikeMiller Which was wrong? The paper or your blog post?

The paper, Analysis Situs.

@TobiasKildetoft Lol, "my blog post is fantastic" :D

Maybe the blog post is wrong too. I don't have the gall to - yeah

@MikeMiller Maybe in time you will develop the delusions of grandeur necessary for such statements :)

hehehe

@HirotoTakahashi not sure what you mean by that. An isometry of the sine curve is precisely that: An isometry that maps the sine curve to itself.

and the set of isometries of any space is a group under composition, essentially by definition.

I guess the question is why the latter is true

good small exercise! :)

@TobiasKildetoft I have this problem: . Describe the group of isometries of the sine curve (the graph of $y = \sin x $): list its elements and construct a (compact) multiplication table.

@TobiasKildetoft it's because I was thinking in the above group.

The table you wrote doesn't make much sense to me, unfortunately. I expect the problem is asking you to find a "name" for this isometry group.

It's something you know.

@Danu So, I have a photon.

@MikeMiller I think is related to periods of sine. Like a cyclic group?

@BalarkaSen Do you now? ;)

That's massless. Momentum is 0. Any uncertainty in momentum is 0 since I just declared what it is. That means by Heisenberg, I can't ever measure it's position with nonzero uncertainty or can't ever tell where it is, or what?

so many questions on main, but so few that i have any interest in answering.

@HirotoTakahashi Do you at least see one non-trivial isometry?

That seems a bit unbelievable to me.

@BalarkaSen Momentum is not zero.

The formula $p=mv$ that you probably know is not the most general one.

If mass = 0, mass * velocity = 0?

See the above.

@Danu Oh.

Want to elaborate on that?

$p=mv$ is the nonrelativistic momentum of a massive particle

Can someone explain maximal element by an example?

@BalarkaSen For photons one needs special relativity, where one should use $E^2=p^2c^2+m^2c^4$

Hence $p=E/c$ for a photon.

(I asked this here but uncertain about it.)

@TobiasKildetoft No :( can you explain me, please?

If you've ever seen Snell's law, that actually provides a hidden 'proof' that the momentum of a photon isn't $mv$.

@HirotoTakahashi what about mirroring around the $y$-axis?

@Semiclassical Doesn't anything that follows straight from Maxwell's equations? :P

Yeah, I have seen Snell.

@Danu Hmm.

eh, fair enough. though that requires careful use of boundary conditions

@BalarkaSen I know it's a bit annoying. High school physics is full of lies.

which do follow from Maxwell's equations, but are a bit of an arse regardless

I'm amused by the fact that the first twoish years of a US physics education is just wrong, but a good primer.

@HirotoTakahashi what? How does a parabola show up? What are you answering?

@balarka a.k.a the energy is dependent on frequency so the frequency gives it momentum in a way

So why did I exactly need "special relativity" and not $p = mv$ for computing mass of a photon?

@BalarkaSen You mean why?

because that formula is not actually true

Because $p=mv$ is just a first order approximation to the real equation in the limit $v\to 0$.

@Obliv I know $\lambda = h/p$ but I thought that only works for electrons inside atoms, and not photons.

@Danu Oh boy.

and an approximation which is never sensible if $m=0$.

...because then $v=c$ always

@TobiasKildetoft Well, It's symmetric with respect to $y$-axis

photoelectric effect is a good example of the correspondence of energy with frequency and not mass (intensity)

point.

@HirotoTakahashi I don't think you understand what this question is asking about

$\lambda=h/p$ is universal in its meaning: Every particle has a de Broglie wavelength.

@TobiasKildetoft maybe you right

now, whether that's a useful length scale will depend on the context

Hm.

@HirotoTakahashi the question asks about a specific curve. Why would you bring up a different one then?

@TobiasKildetoft Well, the sine curve is symmetric with respect to the $x$-axis

Meh, physics.

No, with respect to the $y$-axis

it's beucase I was thinking period of $2\pi$

Woo, physics

meh, physics education

^

the curve of a function can never be symmetric around the $x$-axis

High school physics made me 100% convinced I didn't wanna study that shit.

can't argue with that

if we identified math with math education...

@TobiasKildetoft yes, you right. sorry.

i liked my physics teacher in high school, which helped

@mikeM I think i once asked a physics teacher in HS if a photon had 0 momentum. She said yes l0l

Then I took a course "by accident" in uni and liked it much better :)

but i wouldn't say I really learned general physics until I had it in college. and even there i'd say i wasn't a master of it until I had to TA for it a while

I asked around a bunch of people. Everyone said photon had momentum 0. Blah.

If photons had zero momentum, solar sails wouldn't be a thing :)

high school teachers should probably not cover quantum mechanics

how can you have momentum without mass?

oboy

hehe

@BalarkaSen Yikes...

@MikeMiller This is not a QM issue, rather a relativity issue.

@samuelYusim if you define momentum only as $p = mv$ you can't. Thankfully, momentum isn't defined that way in modern physics

...though the notion of photon is a QM construct

(tricky)

whence my use of the latter term

well then

in any case I should teach neither

how is it defined if not by that?

for a massive particle, $p=\gamma m v$

e = mc^2.

e = mc * c = p * c.

where $\gamma=1/\sqrt{1-v^2/c^2}$ is relativistic gamma

p = e/c. you define it as e/c.

derive it from $E^2 = p^2c^2 + m^2c^4$ @samuelYusim

@BalarkaSen Yikes, no :P

You gotta be very careful with stuff like that

I mean, those were motivations.

For m = 0 of course that doesnt make sense.

dimensionally, it's fine. physically, it's nonsense.

In physics, you can somehow get away with morally very wrong manipulations all of the time :D

@SamuelYusim See the above discussion. When $v$ is extremely small, aka anything we observed until post-1800, the formula $p = mv$ holds by any measurement we can make. But for massless particles or things where that go vroom vroom, the equation stops being so close to true.

I exploited that.

the universe is so weird

Anyway, the definition is $E/c$ thus, am I right?

F***ing photons, how do they work

@BalarkaSen No, the definition is $E^2=p^2c^2+m^2c^4$

@SamuelYusim I don't bother too much with it

For massless particles, I mean.

@MikeMiller Eh, I have to object to that phrasing a little. People could observe the wave nature of light by that point---Snell's law is pretty old---and in that sense the formula $p=mv$ already didn't apply. It's just that they weren't in a position to realize it didn't make sense.

like, I think it's entirely bizarre that there are nice rules that govern anything we interact with, but then we discovered that if you scale up these rules are way nastier

@balarka for massless particles you get $E^2 = p^2c^2$ which you can rearrange to get $p = E/c$

I'll back out from talking about something I know nothing about. Sorry.

That's what I said.

depends on what you mean by 'nastier.' in certain ways, QM is simpler than than classical physics. it's just those simplifications aren't conceptual :/

just to clarify what i mentioned earlier, here's what i meant re: Snell's law

Wow, this really has turned into Physics.SE. :P

And PSE has turned into... :\

Nothing in particular?

0celo7's room

@TedShifrin Yeah. Not that I understand much less of this than of the usual algebraic topology going on. But with that at least I know where I would have to look to understand it.

(and hi, btw)

the statement there is that $\frac{v_1}{\sin\theta_1} = \frac{v_2}{\sin \theta_2}$ where $v_1,v_2$ are the incident/transmitted speeds of light and $\theta_1,\theta_2$ are the angles of incidence/transmission

LOL, hi, @Tobias. I know a little bit of physics and a little bit of algebraic topology.

how to find the distance between two vertices in a graph: make each edge have length 1, hold one of the two vertices up, and measure how far down gravity pulled the other one

cute

now, assume that a photon has mass and therefore momentum $p=mv$. In that case, I can multiply both sides by $m$ and therefore replace $v\to p$

in that case, I get something which almost looks like momentum conservation.

how to draw a graph with minimal vertex overlap: give each vertex a positive electric charge. Increase charges until vertices repel each other enough to be far apart

this is actually done in sage to draw graphs, if I recall correctly

specifically, if it was $v_1\sin \theta_1=v_2\sin \theta_2$ i'd be able to interpret that as such: The photon comes in with some momentum, and in passing through the interface the transversal component $p_1 \sin \theta_1$ isn't changed.

that's not unreasonable, since that's precisely true for reflected light.

it changed direction, but it keeps the same angle and therefore the same transversal momentum upon reflection.

@TedShifrin Can you think of a way to get a tubular neighborhood that depends continuously on the choice of submanifold $M$? I'm thinking of putting a metric on the total space, pulling it back to the submanifold, and exponentiating the normal bundle, but I'd need some sort of "normal injectivity radius" (or at least a lower bound thereof) that depends continuously on the submanifold $M$, and I don't know how I'd get that.

but, as I noted earlier, that conflicts with Snell's law if we assume $p\propto v$

the only way to make it work is if instead $p\propto 1/v$.

That's the only way I can think of doing it, @MikeM. So you don't have compactness in play?

My submanifolds are properly embedded, but let's pretend they're compact and I can work out the noncompact case myself.

and, once you account for the differing speed of light in material versus vacuum, that's exactly what $p=E/c$ amounts to.

With non-compact, you probably need an $\epsilon$ that depends on position.

Right, but I think if I can figure out how to do the compact case I can figure out how to do the noncompact one.

@danu I've done the above argument before, but eh. not sure i said it in the best way.

(Instead of getting a lower bound on an injectivity radius, I'd get a lower bound on the injectivity radius function, or something, that depended continuously on the manifold.)

I don't think you need anything so fancy as injectivity radius. I think it's just the souped-up inverse function theorem (which is an exercise in G&P).

Hi all

@Semiclassical Not very clear :P

No?

@TedShifrin I don't understand what you mean.

yeah :/

I probably should've started from momentum conservation and gone from there to show that $p=mv$ would contradict Snell's law

@MikeMiller D'oh, misread. Sorry.

Something like exercises 10 (p. 19) and 14 (p. 56), Mike.

To clarify, I of course know how to produce tubular neighborhoods, but I want a way to define a tubular neighborhood as a function $\text{SubMan}(\Bbb R^n) \to \text{SubMan w/ TubNeighb}(\Bbb R^n)$. I want the tubular neighborhood to depend continuously on my choice of submanifold.

You're just exponentiating a smoothly varying subbundle.

I don't have time right now to think about it too carefully.

Sure, I agree. My only concern is exponentiating in such a way that the image is an embedded submanifold. That's why I was concerned about injectivity. Let me take a look at what you suggested.

I must prove $N_G(H) = C_G(H)$ where $H$ is a subgroup of $G$ with order $2$ but I'm not convinced this is true. $N_G(H) = \{g \in G \mid gHg^{-1} = H\}$ and $C_G(H) = \{ g \in G \mid ghg^{-1} = h\}$ i.e $C_G(H)$ is the set of elements in $G$ that commute with $h$ whereas $N_G(H)$ is the set of elements in $G$ that satisfy $gHg^{-1} = H$. If $H$ was order 1, I would see easily that they are the same.

I feel like there could be twice as many elements that satisfy the normalizer condition

@SamuelYusim That sounds like it would only give some sort of local minimum

@Obliv $H=\{e,h\}$ right? So $gHg^{-1}=\{e,ghg^{-1}\}$...

@Obliv A good thing to know is that $C_G(H)$ is normal in $N_G(H)$ and the quotient is isomorphic to a subgroup of $Aut(H)$.

(thought this particular one can be done more directly)

wait I think I forgot a crucial part of the definition of $C_G(H)$ ... It's actually $C_G(H) = \{g \in G \mid ghg^{-1} = h, \forall h \in H\}$ right?

@MikeM: You'd better think about what you mean by "nearby" submanifolds. $C^0$ is clearly no good. I assume you need at least $C^1$ or probably better close.

I have an explicit topology on the space, coming from the $C^\infty$ topology.

ok ... so nearby submanifolds will be diffeomorphic. That helps.

@Obliv yes

@arctictern oh right it has to have the identity in it.. duh

Heya @tern

@TedShifrin Indeed let's just fix a manifold and talk about submanifolds diffeomorphic to it; that would be sufficient for me. Then my space of submanifolds is the same as $\text{Emb}(M,\Bbb R^n)/\text{Diff}^+(M)$; so if it helps (not clear that it would) we could define a tubular neighborhood function upstairs that's Diff-invariant.

Well, but then my comment about exponentiating smoothly varying subbundles (although there's an extra base-point moving function in there) makes perfect sense.

Yes, that was my plan. But the problem is - and the original question about injectivity - is I need to restrict to an appropriate subset of the bundle so the exponential is an embedding.

Presumably picking out what subset of the bundle I want also needs to be done continuously. Whence the question about a continuous bound on "normal injectivity radius" (which is obviously the wrong word... but oh well.)

Right. But I'm claiming that's just compactness plus inverse function theorem stuff.

I guess I'm just not seeing it right now. Maybe I'm too sleepy. Thanks.

hello

hi

heya @Ted

oh god i'm tired. i just wrote thatisfy instead of that satisfy

thasnice

lol

@arctictern Sorry about my error the first time. independently: I looked a little at Qiaochu's note but got too spooked.

np

he has blog posts explaining groupoid cardinality that helped

@AkivaWeinberger you're right, but in what sense is there a "best" way to draw a graph? good enough is good enough

You can only have a locally metrizable space that isn't globally metrizable if your space isn't Hausdorff, right?

I'm trying to figure out if those dreams where you run towards something but end up "further away" are possible in any space.

add a dimension of time

It's obviously a breach of the triangle inequality, but if you're just constantly entering differently metrized neighborhoods as you travel, you're never breaking the local metric.

@arctictern In forward time, no temporal shenanigans.

they say the universe is expanding after all. so it's possible to go towards something sufficiently distant and yet the gap is ever increasing

I'm talking like running towards a door in a hallway and it gets further away.

yeah, just make space expand

That's cheating.

Because then I'd have to be exploding.

If I were exploding, there are much larger issues than the door getting further away.

@Axoren math.stackexchange.com/questions/909040/…

@MikeMiller Thanks.

@MikeMiller I once asked you what you studied to which you replied "Analytic Topology", although explaining that you made up that term

After all this time I still don't know what you study, really

I'm going to make that term real. My colleagues like it.

In that case, you should be able to explain what it is and with what field of mathematics it concerns itself, no?

You know what the cup product is in cohomology?

Yes