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4:15 PM
11
A: Not especially famous, long-open problems which anyone can understand

Gabriele FiciIs it true that any word of length $n$ contains less than $n$ squares? (A square is a factor of the form $uu$ for a non-empty word $u$.)

How is this unsolved?
 
how would it be solved?
 
I was confused until I read the comments
 
@MikeMiller Could you enlighten me by telling me how complicated this problem is with an example?
 
and then I was like, "Aaaa."
 
me too
 
4:19 PM
But apparently the deal is that things like abab count, and also that two occurrences of aa don't both count
 
Yep, so you're taking the set of all subwords and counting the ones which are squares.
Not the multiset of them
 
Although, fair point, the guy doesn't provide a source saying it's unsolved
 
Most unsolved problems are not stated somewhere as being unsolved. Experts in the area just know they are.
 
@MikeMiller any length analysis?
 
@DHMO Hwo can I give you an example of why it's hard? For any specific word you could solve it.
 
4:21 PM
@MikeMiller I mean, what example would you give me if I am proposing to solve it by length analysis?
 
What does length analysis mean?
 
Oh, it means analysing the length.
 
How would you analyse the lengths?
That seems to be a word without a meaning to me
 
Something like "each square is at least 2 letters, so $n$ squares take $2n$ letters"
 
Squares can overlap.
 
4:22 PM
That's false. Look at the example in the comment
 
@MikeMiller for example?
@BalarkaSen Yes, and I am requesting more examples like that
which showcase the complicatedness of this problem
 
abaaba, I suppose, also works
Or ababb
 
@AkivaWeinberger nice
 
Neither of those have close to $n$ squares though
 
It's alright
 
4:26 PM
I can't think of any near-counterexamples
 
4:40 PM
I just can't get excited by this flavor of problem
 
Maybe that's why it's still open ^
 
54
A: Widely accepted mathematical results that were later shown wrong?

Micah MillerThe Euler Characteristic V-E+F has an interesting history. It was initially stated that, for all polyhedra, $$V(ertices)-E(dges)+F(aces)=2$$ and its proof was widely accepted, until people found counter-examples. Imre Lakatos' book Proofs and Refutations has an imagined dialogue between t...

The book is really interesting. I'm reading it at the moment.
And this is why I love mathematics so much.
 
@Akiva I hope my taste is not what drives development of mathematics :)
 
5:04 PM
I am being dumb, but what's a Riemannian manifold (ideally a surface in $\Bbb R^3$) which is not complete?
I guess in the noncompact world the unit open disk is a fine one.
 
I don't know that branch of mathematics at all really
but to take a guess
a discontinuous parametric surface
 
Are all infinitely differentiable functions analytic?
 
For example?
 
google it
 
5:13 PM
In mathematics, smooth functions (also called infinitely differentiable functions) and analytic functions are two very important types of functions. One can easily prove that any analytic function of a real argument is smooth. The converse is not true, as demonstrated with the counterexample below. One of the most important applications of smooth functions with compact support is the construction of so-called mollifiers, which are important in theories of generalized functions, like e.g. Laurent Schwartz's theory of distributions. The existence of smooth but non-analytic functions represents one...
Thanks
$\displaystyle f(x) = \begin{cases} e^{-1/x}&x>0 \\ 0&x\le0 \end{cases}$
There is also a function smooth everywhere and analytic nowhere :o
 
@Balarka All compact manifolds are complete.
A submanifold of a complete manifold is complete iff it's closed.
 
Makes sense
 
$\displaystyle f(x) = \sum_{n\mathop=0}^\infty e^{-\sqrt{2^{n}}} \cos(2^nx)$
 
Most smooth functions are not analytic. They are just different classes of things.
 
@BalarkaSen Interestingly, there's always a complete metric on a given smooth manifold.
 
5:27 PM
@0celo7 I believe this.
The examples I can think of all do.
@MikeMiller You mean closed as a subset, yes?
 
@BalarkaSen Easiest way is to Whitney embed and use Mike's comment about closed submanifold.
 
@BalarkaSen Yes.
 
That's what I have in mind.
 
@MikeMiller Did your idea end up working?
 
You can also conformally change any metric to make it complete. This should not be particularly surprising; the idea being that you pick a compact exhaustion and make lengths longer as you go out to infinity.
@AndrewThompson Yeah. I had a nice meeting with someone not-my-advisor yesterday to talk about Tate cohomology and had a lot of great ideas as a result. Then I looked at my notes and I have no idea what they're talking about.
 
5:31 PM
@MikeMiller That wasn't the original idea.
 
Haha, I don't think that's too unusual. Glad you're doing well.
 
@0celo7 I never said that's how it was originally published, and I don't much care, either. It's the first idea that ever came to mind to me about it, and it worked.
 
The idea I have in mind goes along 0celo7's comment, by Whitney embedding properly.
 
@MikeMiller Could you describe it in more detail? I've heard that idea before but couldn't make anything out of it.
 
@BalarkaSen That doesn't help you for making a conformal change to the metric.
 
5:35 PM
Oh, I wasn't referring to your approach, just wrote out the proof modulo the fact you said. I'd like to hear more about that too.
 
@0celo7 Pick a compact exhaustion of submanifolds, a small open neighborhood of each submanifold, and a smooth function that is a sufficiently large constant $c(n)$ on $W_n \setminus U_{n-1}$. Then replace $g$ by $e^fg$. The point is that you want to show that any geodesic that goes through $W_n$ and out $W_{n+1}$ takes at least one unit of time.
 
I'm not particularly interested in writing all the details out.
 
Hopf-Rinow is a nice theorem.
 
It is indeed.
 
5:52 PM
So I can give an example of where Hopf-Rinow fails in the noncomplete world, eg $\Bbb R^2 - 0$ : given any path joining a pair of antipodal points, I can find another one which is of smaller arclength (the point is the straightline does not exist on the surface). But is there an example where a locally arclength minimizing path exists between $p, q$, but it's not the geodesic?
 
Hello.
 
So in particular those two has to have the same arclengths.
I also wonder if it's possible that a locally arclength minimizing path exists between $p, q$ but a geodesic doesn't.
 
"Locally arclength minimizing" is synonymous with geodesic.
 
Guys what are your recommendations for set theory ?
 
The relevant theorem I am aware of is geodesic is locally arclength minimizing, whenever it exists. Do any of those questions contradict that?
 
 
1 hour later…
7:10 PM
Hi chat
 
hi @Semi
 
7:27 PM
@Semiclassical hello
 
7:38 PM
Hello everyone in the order that they appear in the list.
 
What does the (10/4) mean in these answers? math.stackexchange.com/q/1988872/330822
 
@JPhi1618 Binomial coefficients $\binom n k$?
You may be more familiar with them as "$n$ choose $k$" or "how many combinations of $k$ elements can I make from $n$ total items."
 
Does anybody of you know why 1/3 * 3 = 1 but 0.3 (which is 1/3) * 3 = 0.999?
 
@Axoren, Thanks. The more I look at questions on this site, the more I realize I don't know very much about math at all.
 
because $0.\bar{9}=1$
 
7:41 PM
It's very humbling, lol.
 
@JPhi1618 The biggest hurdle is matching notation to concepts you actually do know.
Because there's plenty of those cases where you know something, but it doesn't look like you do.
 
I also have no idea where the 210 comes from in that first answer. I guess the answers here mostly assume that you have some clue what's going on.
 
@MikeMiller Got a really boring and probably obvious question for you
(accidentally pinged Mahmoud with that initially, oops)
I'm looking at Wikipedia's definition of Lefschetz maps here
...actually, I think I've convinced myself of what the resolution is. nm
 
basically, if someone writes $H_{DR}(M)$ where $M$ is a manifold, should I take that to be the de Rham complex? Otherwise I don't get what's meant by $[\alpha]\mapsto [\alpha\wedge \omega]$ is a mapping from $H_{DR}(M)$ to itself.
yeah, it makes more sense now
 
7:57 PM
sorry, take it to be the direct sum over all the cohomology groups
 
so they're saying a k-form maps to the appropriate k+l-form, up to exact forms
 
Right.
And more generally, each form is mapped somewhere in the direct sum
I'm trying to do something dangerous, which is read notes which are above my level and piece together what's being said
I'm trying to break down what the "Main Result" on page 4 means
(since I have no idea what the hodge filtration is, I'm probably doomed. :/ )
 
me neither
 
admittedly, I only want the case of $n=2$
so that's maybe a bit easier
 
8:20 PM
Does the inverse limit functor and fundamental group functor commute (in a topological context)?
 
inverse limits of good spaces for algebraic topology don't necessarily give good spaces for algebraic topology
do you mean direct limit?
 
inverse limit of circles by say an nfold covering map is a horrible space.
the fundamental group is more complicated than the n-adics.
usually one takes the homotopy colimit or something, which is the right notion of colimit in the homotopy category.
I think that does commute with the fundamental group
 
8:36 PM
Does the fundamental group even exist?
 
did really well in my functional analysis exam today
so happy.
 
Of that inverse limit thing I said? Yes, it's path-connected.
 
Like, is that thing even at all path-connected?
I'm skeptical.
 
@MikeMiller Do you know why the 4-manifold invariant in HF is nonzero for symplectic manifolds?
 
Also, inverse limits are limits. I have no idea whether or not fundamental group commutes with homotopy limits. It commutes with directed colimits of CW complexes already, as long as everything is an inclusion.
 
8:38 PM
@MikeMiller You can probably write down a path between two points in the inverse limit, because the structure maps are covering maps (by path lifting). It's called the n-adic solenoid.
 
@BalarkaSen I still don't believe you.
@PVAL-inactive Because it's equal to Taubes' GW invariant and there are holomorphic curves. But there's a purely HF proof that I never learned.
 
@MikeMiller That argument is factoring through ECH?
 
Ya.
 
I don't see why you don't. Pick two points $p, q$ in the first $S^1$, join by a path and lift them up to each $S^1$ in the limit. These gives map from $I$ to the inverse system which commutes with the structure maps. That gives a map from $I$ to the inverse limit.
 
@Fred +1 for your anwer for Proving that $l^2...$, nice, additionally your city has the same flag than Spain!!!
 
Hey everyone, has anyone here read Linear Algebra Done Right by Axler and then gone on to Abstract Algebra by Dummit and Foote, Artin or Lang?
 
the problem is that you need to be able to lift your paths so that the endpoints always agree with the projections of the points in the inverse limit
 
@MikeMiller At a glance that looks like one of those AI spewing random gobbledegook (I'm sure it isnt).
 
ops
 
it's obvious that if you pick $p, q$ at some finite stage, there are points in the solenoid that project to them that have a path connecting them
@PVAL-inactive also known as an REU
 
8:44 PM
yeah, good point
 
That kinda looks like Hilbert-Smith stuff. That stuff is on my list of stuff to read I never will.
 
I feel that way too.
 
Once upon a time I knew that this is not path connected. Evidently I forgot. (I don't have any regrets)
 
I met with my advisor today and was explaining a proof to him until I realized I missed something important. I told him I was scared about it and he told me "It's near halloween, it's OK to be scared."
 
Hah.
 
9:03 PM
Well I was thinking about the inverse limit of an n-fold torus
I thought maybe understanding something about its fundamental group would give me insight
 
$n$-fold torus, as in, $S^1 \times \cdots \times S^1$ (n many copies of $S^1$)?
 
No
n-times connected sum of $\Bbb T$
 
ok, the surface of genus $n$.
 
So $T_1 \# T_2 \# \cdots \#T_n$
 
Inverse limit of these by what maps?
 
9:05 PM
Yeah there aren't really any homomorphisms i can think of
Somebody asked me this question over lunch
My thinking was that the fundamental group would be a free product of n circles
And then that would have an inverse limit maybe?
 
I don't really understand your space but ok
 
Which bit do you not understand?
 
what are the structure maps in the inverse limit?
 
Hello @amWhy
Are you familiar with harmonic functions?
 
@BalarkaSen yeah I am not really sure, when I was listening to the question it was kind of brushed over
 
9:10 PM
The direct sum doesn't make sense to me anymore, I don't think.
 
@Axoren why?
 
At some point, the components of an element must be zero onward, right?
Assuming infinite components.
Where does that restriction come from and when is it useful?
I generally work in spaces where it's equivalent to the direct product.
 
What are you summing, modules?
 
Vector spaces, let's say.
 
Why must components be zero onward?
@BalarkaSen maybe they meant direct limit? I have no idea. I have no idea about this stuff.
 
9:14 PM
"In more technical language, if the summands are ${\displaystyle (A_{i})_{i\in I}}$, the direct sum ${\displaystyle \bigoplus _{i\in I}A_{i}}$ is defined to be the set of tuples ${\displaystyle (a_{i})_{i\in I}}$ with ${\displaystyle a_{i}\in A_{i}} $ such that ${\displaystyle a_{i}=0}$ for all but finitely many $i$."
The direct sum is an operation from abstract algebra, a branch of mathematics. For example, the direct sum R ⊕ R {\displaystyle \mathbf {R} \oplus \mathbf {R} } , where R {\displaystyle \mathbf {R} } is real coordinate space. Then R ⊕ R {\displaystyle \mathbf {R} \oplus \mathbf {R} } is the Cartesian plane, ...
Wikipedia doesn't copy-paste nicely with LaTeX
Doubles every expression
 
18
Q: Is $A \times B$ the same as $A \oplus B$?

LJRWhen $A, B$ are $K$-modules, then $A \times B$ is the same as $A \oplus B$. Let $A, B$ be two $K$-algebras, where $K$ is a field. Is $A \times B$ the same as $A \oplus B$? Thank you very much. Edit: $A \times B$ is direct product and $A \oplus B$ is direct sum. Edit: I am asking this qu...

 
@AliCaglayan I agree they're not the same.
 
If you read the answer it gives you the exact reason why
 
I'm going to keep reading, but I'm already not sure if this post answers my question
Still reading.
 
That is the reason for the distinction
The tuples kind of hide how it is constructed
 
9:20 PM
@AliCaglayan I'm pretty sure you don't want inverse limit. I think what you're trying to think about is the "infinite genus torus", where you connect sum forever to the right.
 
@MikeMiller Maybe you are right. Maybe I misheard. What can be said about this torus then?
 
which is homotopically a lot easier object.
 
Let's not bother until you know what it is.
 
Is its fundamental group an infinite free product of S1?
 
the fundamental group is the free group on countable many elements.
I think.
 
9:22 PM
That's correct.
@AliCaglayan Infinite free product of Z.
 
sorry yes
 
'cause it deformation retracts to it's 1-skeleton which is wedge of infinite many S^1's.
 
Is that the thing they say is not the hawaiian earring
 
It's not the Hawaiian ring, no.
Because you've got the CW-topology on it.
 
It's like lining up a bunch of circles of the same radius side-by-side.
 
9:27 PM
Right
 
10:10 PM
Hey all :-)
Does anyone know why the First Fundamental Form only depends on the parameterisation of the surface and the point at which you're calculating it?
 
@Kari You should check that. Pick a different parameterization: these two are related as $f = \phi \circ g \circ \phi^{-1}$. Does that change the first fundamental form? Checking $E, F, G$ are invariant suffices, note.
You should also be able to tell me intuitively why it doesn't depend on any parameterization and is intrinsic to the surface, without checking definition.
 
@BalarkaSen Why the $\phi^{-1}$?
I thought we only needed the composition $f \circ \phi$ for a diffeomorphism $\phi$.
 
You're right, typo.
 
Ah, ok! I'll check it out now :-)
Maybe you were thinking of a transition map or something similar!
I think it has a similar form with the inverse as well.
 
Yah, I didn't think much when I wrote that down, also it's crack of a dawn. It doesn't take too much to check it.
 
10:27 PM
Crack of a dawn?
You mean the crack of dawn :P
 
I should really sleep now. Before I go: @Kari, as a hint, it's just because the tangent space doesn't change under parameterization.
 
Thanks, @Balarka!
Good night :-)
 
10:57 PM
Hey guys, need some help here
I have to solve a system of equations using a matrix
I've been struggling but I cant find a way to represent the system as a matrix

$2a - b = 0$
Sorry, the whole system is
2a - b = 0
-ax - cx = x
-ax^2 + bx^2 = 0
cx^3 = 0
$2a - b = 0$
$-ax - cx = x$
$-ax^2 +bx^2 = 0$
$cx^3 = 0$
 
11:17 PM
loool
how about you look up the definition of matrix multiplcation?
 
:(
I did, it's too hard
 

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