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00:22
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Q: Theories of manifolds w/ extra structure and singularities

Will SawinMany different objects in mathematics can be described as manifolds with extra structure. Among the most famous examples of these are smooth manifolds, Riemannian manifolds, complex manifolds, and symplectic manifolds. There have been efforts to unify these examples by defining a general notion o...

nice question by will
Many different objects in mathematics can be described as manifolds with extra structure. Among the most famous examples of these are smooth manifolds, Riemannian manifolds, complex manifolds, and symplectic manifolds. There have been efforts to unify these examples by defining a general notion of what an extra structure to be placed on manifolds is.
Sometimes in mathematics that are "mostly" manifolds with extra structure, but allow some singularities where the space is not a manifold (or, the space may be technically isomorphic to R^n at these points,
but not in a way "compatible" with this extra structure). To be (partially) precise, we have a category of spaces where some subcategory is a category of manifolds with extra structure, and every space contains a dense open subspace isomorphic to an object in that category, but not every space is itself isomorphic to an object in that category
Presumably the ur-example is the category of topological spaces with a finite stratification all whose strata are manifolds. The theory of intersection cohomology of Goresky and Macpherson is applicable in this level of generality, though most of the motivating examples come from complex analytic spaces.
 
6 hours later…
06:54
@Jakobian Hi
 
2 hours later…
08:32
Is there a continuous surjection from the 2-sphere to the 2-torus?
sure, the 2-sphere surjects onto the 2-disk and the 2-disk surjects onto the 2-torus
continuous surjections in general are easy to come by, it's a bit harder to prove but any manifold of positive dimension continuously surjects onto any compact manifold
(add connected)
08:56
I see. Thanks.
 
6 hours later…
14:31
hi
 
1 hour later…
15:52
In the definition of connected sum of manifolds, they identify the punctured balls with cylinders and glue the cylinders reversing the vertical coordinate:$$B\setminus \{m\} \simeq S^{n-1}\times (0,1)\owns(s,t) \mapsto (s,1-t) \in S^{n-1}\times (0,1) \simeq B'\setminus \{n\}.$$
I wonder if changing the map $(s,t)\mapsto(s,1-t)$ to $(s,t)\mapsto(s,t)$ would give a different manifold. I think it would not, but I am not sure.
topological manifold
Oh, I think if changing the map $(s,t)\mapsto(s,1-t)$ to $(s,t)\mapsto(s,t)$ the resulting space would not be manifolds because the resulting space would have a corner at $t=1$.
well, $t=1$ is not part of where you glue anymore
so rather than creating a corner, it creates a non-hausdorff space
a corner would be fine anyhow
16:16
I would not agree with that
on a topological manifold?
@BenSteffan You forgot "clear", but maybe that is what you removed?
"That result is obvious. It is so clear that a toddler could see it. It is completely trivial. You are stupid."
@XanderHenderson The joke was that there were three messages before that one: One that said obvious, one that said trivial, and then one by Mats Granvik that was removed.
@BenSteffan taking a connected sum should not introduce new boundary
maybe that message read "Clear," who knows
16:20
hi @SineoftheTime
Things on the starboard are contextless.
@XanderHenderson yes, so the onus of navigating to the message and figuring out the context is on anybody that wants to reply :)
@Thorgott Do people use "manifold with corners" to mean "with boundary?"
I was not thinking that formally
@SineoftheTime I took the algebra and geometry exam days ago
Corners are more than boundaries.
how was it?
@XanderHenderson How so?
It is the difference between a square and a disk.
Corners are not smooth.
16:23
We're talking about the topological category.
Then "corners" makes no sense.
@SineoftheTime pretty standard, a linear system, an exercise on ker, image, associated matrix etc..., an exercise on Grassmann's formula, a geometry exercise and an exercise on diagonalization
Does it have also an oral part?
yes
16:26
It should be around February
Oral exams > written exams.
Much easier to either BS, or grope your way to a correct response.
And much harder to get completely hung up on some detail that you just forgot.
"corner" has a formal meaning
almost all exams I did were written+oral
but topologically a manifold with corners or with boundary are the same thing
no corners, then
16:29
@Pizza are you planning to do methods?
@Thorgott I see. It is similar to a branching line
@SineoftheTime yes if I pass these exams
did you do physics?
yes but I'm waiting for the results, then there's the oral exam too
16:33
instead of algebra and geometry I have to take the oral exam because I passed
I think i should also bring up a topic of my choice
what inspires you?
mm I don't know I saw that some had chosen linear systems
what would you suggest?
that's up to you :)
16:36
something that has a proof?
I guess
and some applications
you have a very... interesting exam system over there
@XanderHenderson remind me: How do we handle questions that were asked both on MSE and MO and now have an accepted answer on MO? These should go, right? Do we vote to close or flag for moderator attention or something else?
sometimes you can start from a topic you choose
@BenSteffan Generally speaking, the rules against cross-posting are enforced a lot more laxly on Math. However, if the exact same question is asked in both places, it is worth flagging for moderator attention.
16:42
@SineoftheTime Yes, what I meant was that you can present a topic as you like, then the teacher asks you questions about the entire program
yeah I know, I was telling Ben
sounds involved
written exam + oral exam + you have to come up with something you want to talk about
though oral examiners here also sometimes ask you to talk about something of your choosing I guess
but they won't tell you before the exam whether they're going to or not so :)
Hi
@BenSteffan The oral exam for potential PhD candidates at my PhD institution is essentially "talk about what you want to talk about" (because you are outlining your intended research path).
16:45
can someone go on chatgpt for a moment?
phd interviews are a different beast
@Binky Hell, no.
what on earth would you want somebody else to go on chatgpt for??
somebody I know is going through phd interviews now and they're painting a very... interesting picture
I finished chatgpt 4o. daily, and I will get them back tomorrow morning :(
16:49
@Pizza Cayley-Hamilton's theorem? It's a theorem with a fun proof I think, a theorem with applications in e.g. ODEs, but I haven't followed what the topic should be about. Maybe it is off-topic to just talk about a single theorem?
I'm sure you're using it only for good things, like asking it how many 'r's there are in the word "strawberry"
@Binky I don't know what this means, but you are likely in the wrong place to discuss it. I think that most of the people here are pretty hostile to GPT.
because chatgpt gave me a solution to an exercise so if one of you can also check what solution it gives you so to compare the two
asking chatgpt for solutions to exercises is a great way to fail a course
16:53
:(
@Binky No. Do your own work, and recognize that GPT is not a math engine, and can't be trusted to do math.
@hbghlyj exactly
@BenSteffan that's what I'm saying yeah
If you want to check use wolfram
Or GeoGebra. Or Maple. Or Mathematica. Or R. Or Sage. Or a gosh darn calculator.
Or ask Xander
17:00
Don't ask me. I probably don't know.
Or ask Sine :^)
Ok
how can i simplify arsinh(2x*sqrt(x^2-1)) using t = arsinh(x) ?
what do you mean precisely?
simplify the function
17:07
@psie mm I don't think I did it
that is, or at least I can't find anything on my material with that name
but I looked online to find out what was it talking about
anyway im going bye :-)
@Binky why do you think it can be simplified?
@Binky I don't think this simplifies. Are you sure it's not $t = \operatorname{arcosh} x$ or $x^2 + 1$ instead of $x^2 - 1$?
@Pizza bye!
@Pizza ok, bye :) maybe one day you'll encounter it. I studied it in connection with ODEs and the proof was on an oral exam, but it was very much a math course.
Pizza studies engineering
17:13
everyone with a background in algebra is hurting a little
ah, ok
"Caley-Hamilton has applications in ODEs" well it's also a sort-of fundamental theorem in commutative algebra
@VladimirLysikov I was thinking of using t=arcsinh(t), but okay I can try that!
18:06
@XanderHenderson I can proudly say that till date I have not searched a single thing on chatgpt.πŸ—ΏπŸ—ΏπŸ—Ώ
@SoumikMukherjee we need more people like you :D
@BenSteffan yeah, the determinant trick is almost worth calling the fundamental theorem of commutative algebra
it implies Nakayama!
18:56
@SoumikMukherjee You don't "search" things on GPT. You provide an input, and it generates a random output, where the underlying distribution of what it outputs is based on the statistics of some huge dataset upon which GPT has been trained.
The experience mimics that of asking a question and getting an answer.
(Yay for stochastic parrots...)
And, as an instructor and SE moderator, I have used GPT, but largely only to understand the kinds of things that it is likely to output (and to compare student work and SE posts to GPT output, to get some sense of how likely it is that purportedly student (or user) written was actually written by an LLM).
@XanderHenderson right, I used the word 'search' to mean that I never used chatgpt.
19:13
I have a function $f:\mathbb R^n\to S^{n-1}$ defined by $x\mapsto x/|x|$ and an isometry $g:\mathbb R^n\to\mathbb R^n$. Is it true that $$f\circ g=g\circ f?$$I'm tempted to say yes, but I'm not sure what it is one needs to prove here.
You don't have a function $f$...
@psie what is the value of f at origin?
also your $g$ need not preserve $S^{n - 1}$. Do you mean to say linear isometry?
yes, I forgot, $0$ excluded. If I'm not mistaken, every isometry on $\mathbb R^n$ is linear?
you are mistaken
consider translations
19:18
affine linear*
anyways, with every issue fixed you should try and prove or disprove this yourself
it's not hard.
 
4 hours later…
23:31
For $x\in\mathbb R^d$, define $r=|x|$ and $x'=x/|x|$. Let $\Phi:\mathbb R^d\setminus \{0\}\to (0,\infty)\times S^{n-1}$ be given by $x\mapsto (r,x')$. This is a homeomorphism. I wonder, since it is a homeomorphism, are the Borel $\sigma$-algebras on the respective spaces equal? This seems absurd, but this is what I gather from reading this answer, though I'm not really sure what they mean by the notation $\sigma(X)$, so I might be all wrong.
The question makes no sense. They are different spaces, how can their Borel $\sigma$-algebras be equal?
They can be isomorphic of course, and they are.
Nothing absurd about it.
Well, perhaphs one space could be identified with the other, much like $A\times (B\times C)=(A\times B)\times C$.
the key is the OP's "If we identify X and Y via the bijection" (which still isn't at the level of making literal sense but is pointing the way)
Sure, and then you would identify their Borel $\sigma$-algebras in the process, of course :)
Ok. πŸ‘
23:51
I always appreciate it when Dr Lee replies to posts.
Dr Lee?
Not Prof. Lee?
Do you not use Prof. for emeriti?
americans generally care/know less about any distinction between prof and dr, and a lot of them do not make it
which doesn't stop some profs from being very particular about being called profs :)
hm, I see
its even fairly common to call people without doctoral degrees or academic titles or even semi-permanent academic jobs "professor" if they are doing the kind of teaching job that someone who did have those things might also do
so some people might even think of "dr" as more honorary because people generally do reserve that for phds/degree holders
one of us should comment and find out which way Lee comes out on this

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