Mathematics

12:27 AM

maybe this is a silly question, but is a quotient of a manifold is always hausdorff?

rarely

for instance, take the quotient of $\mathbb R$ that identifies all points with $x>0$ and all points with $x \leq 0$

true

this gives $\{\leq 0, +\}$ the topology $\{\varnothing, \{+\}, \{\leq 0, +\}\}$

even under fairly strong conditions on the quotient map you can get non-hausdorff things

so then how do you characterize when the quotient will be hausdorff? or is there even a characterization?

i don't know one that's not trivially equivalent to "the quotient is hausdorff"

12:33 AM

well then

quotients are bad

12:56 AM

I figure it's more that topological spaces don't have enough structure for anything to be good

@SamuelYusim Any upper semicontinuous decomposition space of a Hausdorff space is Hausdorff (somewhat trivially I guess). The study of these spaces leads to a lot of surprising/important results ("exotic" involutions of S^3, The generalized Schoenflies theorem, the double suspension theorem, the topological 4-d Poincare conjecture). A good introduction is available here : maths.ed.ac.uk/~aar/papers/daverman.pdf

that's a lot of words I don't know

There defined in the first couple sections of the link.

Or at least "upper semicontinous decomposition space" is.

@PVAL: out of curiosity, have you read the 82 Freedman paper?

@Chris'ssistheartist @SamuelYusim We have $t^n - 1 = (t-1)(t^{n-1} + t^{n-2} + \dots + t + 1)$, so $(t^n - 1)/(t-1) = t^{n-1} + t^{n-2} + \dots + t + 1$.
$(t^n - 1)/(t-1) \mod (t-1) \equiv (t^{n-1} + t^{n-2} + \dots + t + 1) \mod (t-1)$

I haven't really understood how we get $n \mod (t-1)$. Could you explain it to me?

1:08 AM

@MikeMiller No. I haven't tried. I think Ancel and Edwards were probably the only people who claimed to understand that paper. Everyone says its pretty much unreadable. The Freedman and Quinn book and Ancel's paper on the other hand are fine and I have read those.

Do they prove the result? I didn't know it was done anywhere else.

Freedman and Quinn prove that every infinite capped tower of gropes is homeomorphic to an open 2-handle. This is a slightly different result than every casson tower is homeomorphic to an open 2-handle, but I dont think there are any topological results you can get from the latter that you cannot get from the former (i.e. they prove the disk theorem and the topological h-cobordism theorem).

The decomposition theory arguments in that book are VERY hard to read, but Ancel reproduces what I think is the key one in his article (and there it is very easy to understand).

what's the article? (and what's the disk theorem?)

I don't know anything about the paper other than the main theorem and that it involved things called gropes, to be honest

oh, is that that oriented embeddings $D^4 \hookrightarrow M$ are all isotopic? is this done by proving the annulus theorem?

"Approximating Cell-Like maps of S4 by homeomorphisms" is the paper of Ancel. The disk theorem states that if $D^2 \to M^4$ is a framed-immersion (extends to an immersion $D^2 \times D^2$) with isolated intersections (with S^1 in the boundary of M) and there exist a framed-immersed sphere which intersects the image of the disk in one point, then there is a topological locally flat embedding of a disk with the same boundary.

It is more carefully written as theorem 1 here : mathunion.org/ICM/ICM1983.1/Main/icm1983.1.0647.0665.ocr.pdf or in the first few pages of the F-Q book.

It's a statement about extending immersions of 2-disks into embeddings.

1:24 AM

is this the fundamental unit of the h-cobordism proof?

Yes.

thanks, appreciate it

not something I plan to tackle tomorrow but it'd be nice to do at some point

The D^4 statement you said sounds like a consequence of Cerf's theorem.

Cerf's theorem that $\pi_0 \text{Diff}^+(S^3) = 1$? how is that related?

Well non-isotopic S^3 in a 4-manifold would lead to a non well-defined connect sum. If Cerf's theorem were false we would also have a non well-defined connect sum. I guess they are not more related than that. According to wikipedia, that result is of Palais from 1960, but it isn't related to Freedman's work afaik.

1:35 AM

Sure, I should have said topological. Smooth is not so hard.

Isotope them to be in the same coordinate patch with $0$ mapping to a chosen point with chosen derivative. So we may as well show $f: D^4 \hookrightarrow \mathbb R^4$ with $f(0) = 0$, $f'(0) = I$, are all isotopic to the standard embedding. But take $f_t(x) = \frac 1t f(tx)$ with $f_0$ defined as the limit (the standard embedding). I guess I'm not actually sure this is smooth as a function of $t$, but I think one can fix that.

I guess you're right. I think I mean to say locally flat $D^4$s are locally flat isotopic. I just want well-definedness of connected sums.

OK, right, that's definitely what I mean.

1:54 AM

I do not know if the topological statement is used anywhere. It might follow from the annulus conjecture. I think I recall seeing the smooth result in Milnor's notes now that you mention the theorem. I think locally flat embeddings $D^4$ are always isotopic to smooth one in a neighborhood of the image (probably due to Brown), and then the topological case follows the smooth case.

Well anyway, I need to leave now.

Thanks for your time.

@MaryStar recall that you can evaluate that expression by taking each term you're summing mod n, and then putting them together. how many terms are there?

2:17 AM

@SamuelYusim There are $n$ terms. We have that $1 \equiv 1 \mod (t-1)$, $t=(t-1)+1 \equiv 1 \mod (t-1)$. But why does it stand that $t^{n-1} \mod (t-1) \equiv t^{n-2} \mod (t-1) \dots \equiv t^2 \mod (t-1) \equiv 1 \mod (t-1)$ ? I got stuck right now...

pjs36

Remember, if you're trying to show that $t^k \equiv 1 \pmod {t - 1}$, this is equivalent to showing that $t - 1$ divides the difference $t^k - 1$. Does that make things any better?

Also, I get fooled by \cong vs \equiv every time. Am I alone in this?

$t^k-t^{k-1}=t^{k-1}(t-1)$

2:41 AM

$(t^n-1)/(t-1) - n = (t^{n-1}-1) + (t^{n-2}-1) + \dots + (t-1) + (1-1) = (t-1)(t^{n-2} +\dots + 1) \equiv 0 \mod (t-1)$

So, $(t^n-1)/(t-1) \equiv n \mod (t-1)$.

Is this correct? @pjs36 @anon

user147690

@Deathslice I already suggested that you learn Havel-Hakimi, it would have solved this problem for you :P

pjs36

It took me a while to process where the subtracted $n$ went, but yes, that seems correct to me, @MaryStar.

Ok... Thank you!! :-) @pjs36

columbus8myhw

3:23 AM

^

mia.avery

4:37 AM

I get yelled at when I type up a problem and don't submit the screenshot, then I get yelled at when submit the screenshot (and am called 'lazy' for not typesetting the problem).

Yay moderators.

columbus8myhw

5:17 AM

Why would you get yelled at for the first one??

I have not looked into mia's situation, but there are many OPs that prove they simply cannot be trusted to communicate a problem intelligibly, so we just get them to cite a source. If it's something they can't cite (like a handout), then a screenshot would be in order.

Ted Shifrin

How are your annuluses, focuses, calluses, @anon? :)

There's no point in typing up a problem when you have a screenshot of course, unless someone thinks they can write the problem even better. One might be criticized from PSQ and not showing any thoughts, efforts, etc.

@Ted I'm fine.

Ted Shifrin

Good :)

trying to get tan without burning these days

5:24 AM

@Ted: I forgot your time zone shifted...

maybe someone (@Ted?) can give a better answer to this question than my comment did

visual intuition is rough enough when you can think inside small Euclidean spaces