Mathematics - 2025-02-04

hbghlyj

00:34

The second page of the paper maths.ed.ac.uk/~v1ranick/papers/kervsmoo.pdf wrote: the trivial group and the binary icosahedral group are the only available examples of finite groups satisfying H_1(\pi) = 0 and having a presentation with an equal number of generators and relators.

Do we know other examples?

I found math.stackexchange.com/questions/3273061/…

Ben Steffan

well open up some of the references in that answer?

I don't understand why you're asking

hbghlyj

The above groups are fundamental group of homology 4-spheres. I have a question: what are the fundamental group of homology 3-spheres?

Ben Steffan

the binary icosahedral group is also the fundamental group of a homology 3-sphere

the rather famous Poincare sphere

as for the general case, who knows

hbghlyj

It is proved in the paper that the properties $H_1(\pi) = 0$ and "having a presentation with an equal number of generators and relators" characterize the fundamental groups of homology 4-spheres. I wonder if this also characterize the fundamental groups of homology 3-spheres.

Thorgott

look up Thurston geometrization

@hbghlyj no, and I'm not sure if the answer to this question is known

a closed $3$-manifold is a homology sphere if and only if its $\pi_1$ is perfect (this is not hard)

the issue is that there is no known general characterization of the groups appearing as $\pi_1$ of closed $3$-manifolds (unlike for $4$-manifolds, where it's all finitely presented groups)

check out this survey to get an understanding of how complicated the situation is: math.ucla.edu/~matthias/pdf/3-manifold-groups.pdf

in particular, (C.3) says that $\pi_1$ of a closed $3$-manifold is automatically balanced (has a presentation with equal number of generators and relations) if it is infinite, so you sometimes get this for free

hbghlyj

01:12

@Thorgott Thanks. My attempt to prove "a closed 3-manifold is a homology sphere if and only if its $π_1$ is perfect": if $\pi_1(M)=0$, by Hurwitz implies $H_1(M)=0$, then by Poincare duality $H_2(M)=0

Ok

Is there a known general characterization of the groups appearing as $π_1$ of 3-dimensional homology spheres?

Now I know that a closed $3$-manifold is a homology sphere if and only if its $\pi_1$ is perfect. But is every perfect group the fundamental group of a 3-dimensional homology sphere?

Thorgott

@hbghlyj correct, the only subtlety is to justify that $M$ is orientable

@hbghlyj as I just said, not to my knowledge

@hbghlyj no, there are a bunch of necessary conditions, the most obvious one being finite presentation, but see the paper I linked for more

hbghlyj

I see. Thanks!

Thorgott

also, it's Hurewicz, not Hurwitz, different guys

Ben Steffan

01:36

that reminds me, in one corner of the MPI there is a framed photograph of the participants of the International Symposium on Algebraic Topology of '56 in front of the Mayan step pyramids in Uxmal with Hurewicz (still) depicted

that photograph must have been shot at most a few hours, if not minutes before his unfortunate end

Thorgott

oh wow

Ben Steffan

the MPI is a strange place

they also have a portrait of Hirzebruch in oil hanging in a rather central position

as you do

Thorgott

well, he was a rather central figure

hbghlyj

01:53

As for my original question "Take a Morse function f on M with a single minimum and a single maximum. Then f possesses an equal number of critical points of index 1 and 2." in the paper $f$ is used to give $\pi_1(M)$ a presentation with an equal
number of generators and relators

Ben Steffan

@Thorgott true, but the painting really doesn't fit in too well with the modern style of the MPI where it hangs

hbghlyj

If $M$ is any 3-dimensional smooth manifold, then $\pi_1(M)$ has a presentation with an equal
number of generators and relators. Is this correct?

Thorgott

write down some examples

hbghlyj

My purported proof: The number of $i$-cells in $X$ is $m_i$, the number of critical points of index $i$.
There is a Morse function $f$ on $M$ with $m_0=m_3=1$.
Since $M$ is an odd-dimensional manifold, $\chi(M)=0$.
By Poicaré-Hopf, $0=\chi(M)=-m_0+m_1-m_2+m_3$, so $m_1=m_2$.
$X$ gives $\pi_1(M)$ a presentation with $m_1$ generators and $m_2$ relators.

Which step of the above proof is wrong?

Thorgott

you can write down a $3$-manifold with fundamental group $\mathbb{Z}$

Ben Steffan

02:04

you should probably add "closed", "connected," and perhaps "orientable"

but then it should be ok

there is some question of whether the 2-handles/cells you attach actually represent new relations

Thorgott

they don't

I just wanted to shift focus onto the fact that the statement is clearly wrong before worrying about why the proof doesn't work

but if the abelianization of $\pi_1$ is torsion (in particular, if $\pi_1$ is perfect), you can show the relations all have to be non-trivial

hbghlyj

@Thorgott Yes, $S^1\times S^2$ has fundamental group $\mathbb Z$

Thorgott

(general fact: the deficiency of any presentation of a group $G$ is bounded above by the dimension of $G_{ab}\otimes\mathbb{Q}$)

@hbghlyj yeah, so clearly this can't be right

and the issue with the argument is, as Ben pointed out, that there may be redundancy among the relations imposed by the $2$-cells

but this does show that any $\pi_1$ of a closed orientable $3$-manifold has deficiency at least $0$

hbghlyj

I see. Thanks!

This show that any $π_1$ of a closed orientable 3-manifold has deficiency at least 0. For the original question, how do you show that $π_1$ of a closed orientable 3-manifold with the same homology as $S^3$ has deficiency equal to 0?

Thorgott

02:21

I explained this just above

hbghlyj

Let $G=\pi_1$ we have $H_1=G_{ab}$, under the assumption $H_*(M)=H_*(S^3)$ we have $H_1=0$, so $G_{ab}=0$. Is this right?

Thorgott

yes

hbghlyj

I wonder which step in the above proof of "$\pi_1$ of a closed orientable 3-manifold has deficiency at least 0" used the fact that $M$ is orientable?

Thorgott

after a second of thought, none

I was just being cautious, but Euler characteristic does not depend on choice of coefficients, so it works out

hbghlyj

Ok

9

Q: Not all finitely-presented groups are fundamental groups of closed 3-manifolds

It is a well-known result that, for any finitely-presented group $G$ and any integer $n \geq 4$, there exists a closed $n$-manifold whose fundamental group is isomorphic to $G$ (a sketch of proof can be found here). It is also well-known that such a result becomes false when $n \leq 3$. However, ...

From this answer, the fundamental group of every
closed, connected 3-dimensional manifold has a deficiency zero presentation.

Thorgott

02:43

I suppose Qiaochu wants to allow redundant relations

hbghlyj

Ok, I see.

Zac U.

09:42

Is there any meaningful way to ask how many functions can I input a specific integer and get out a different specific integer?

Like any applicable definitions that might make this actually seem relevant

By definition I mean a constraint on the sort of functions that might be allowed

Or maybe would it be simpler and more apt to say how many functions don't include a specific integer ... Losing train of thought

Or maybe how many functions have a removed point at a specific integer where that removed point could have remained a function if the output were a specific different integer

Ok the last I don't think is relevant but the previous is take a vertical line at x and horizontal line for y and find out how many functions are possible?

Soumik Mukherjee

10:31

Can you be more clear in what you're asking?

Zac U.

10:42

I'm trying to find out if there's potentially any worth or way in exploring a scenario where you have two distinct integers that we'll say are coincidentally related somehow & if functions could lead to some kind of interesting find if the values served as discontinuities

Say we didn't allow any x or y to be positive or negative 5 and 8 or something

Soham Saha

15:07

Hi everyone. I had a question: Let $A=\{1,2,3\}$. The number of relations on $A$ containing $(1,2)$ and $(2,3)$ which are reflexive and transitive but not symmetric are _______?

Should be only $1$ relation right?

$\{(1,1),(1,2),(1,3),(2,2),(2,3),(3,3)\}$?

Thorgott

that's the smallest one, but are you sure there aren't larger ones?

Soham Saha

@Thorgott well (2,1) and (3,2) can't be there

oh ok wait

i was thinking antisymmetric

it's just not symmetric

Thorgott

right

Soham Saha

We can add (2,1) to the set I wrote, and it would be OK

right?

same for just adding (3,2)

Thorgott

15:26

yes, but that's still not all

Soham Saha

so we can independently add (21) (32) and (31)

so that's 8 cases

but we can't add all of them, so 7 cases

we've already checked 3

4 left to check

can't add only (31)

can't add (21) (31)

I think we can add (21) (32)

no we can't

nor can we add (32) (31)

@Thorgott I think those 3 are the only cases we are gonna get

others invalidate transitivity somehow

Thorgott

yeah I agree, you can add (21), (32) or (31)(21)

so together with the minimal one, that should yield 4

Soham Saha

if you have (31) (21), then you can combine (31) with (12) to get (32)

so can't have that one

Astyx

15:42

1<2<3, 1=2<3, 1<2=3

Soham Saha

@Astyx could you elaborate how to convert from my question to your form? I knew equivalence relations could be counted by just counting the partitions, but this one is new to me

Astyx

This is only in this specific case

You can draw a directed graph with nodes 1,2,3 and arrows 1->2 and 2->3

Now the question you are asking is what other arrows can I add until I can go from any node to any other

Soham Saha

ya, while remaining "not symmetric"

got it now

Astyx

This amounts to grouping nodes that are "equivalent"

Soham Saha

yes

@Thorgott and @Astyx , any ideas how this can be extended to arbitrary size of $A$?

Maybe we can come up with some generic formula of $\operatorname{card}(A)$

Astyx

15:48

depends with which relations, but if you keep 1<2<..<|A| the same approach should work

then it becomes a combinatorics problem

Soham Saha

Wait how are you converting from the graph to the inequality like thingy?

Astyx

if a<a+1<...<b and you can go from b to a, then a, a+1, a+2, ..., b are all equivalent

Thorgott

you mean the number of relations on $\{1,\dotsc,n\}$ that are reflexive, transitive, not symmetric and have $1<2<\dotsc<n$?

Soham Saha

@Thorgott just the number of relations on $\{1,…,n\}$ that are reflexive, transitive, not symmetric

Astyx

Oh

Soham Saha

15:51

@Astyx I think I would need some time to digest that...

Thorgott

hmm

Astyx

drawing the graph should make it clear

Thorgott

this shouldn't be too bad

following Astyx' logic, this is just the number of disconnected digraphs on $n$ vertices

Astyx

to be clear: what I said relied on the assumption that Thor also made, i.e. 1<2<...<n

Soham Saha

@Thorgott too much graph theory for someone who just came to know that graphs aren't just things on Desmos

Astyx

15:53

graph means something different here

Soham Saha

I will try drawing the graphs and see, anyhow

@Astyx yep

Astyx

Graph theory

In mathematics and computer science, graph theory is the study of graphs, which are mathematical structures used to model pairwise relations between objects. A graph in this context is made up of vertices (also called nodes or points) which are connected by edges (also called arcs, links or lines). A distinction is made between undirected graphs, where edges link two vertices symmetrically, and directed graphs, where edges link two vertices asymmetrically. Graphs are one of the principal objects of study in discrete mathematics. == Definitions == Definitions in graph theory vary. The following...

Soham Saha

@Astyx I meant that I am relatively new to graph theory

Astyx

Yes I understood, which is why I emphasized that graphs in Desmos are a different mathematical notion than what I meant

Thorgott

hmm no, what I said is wrong, distinct digraphs can present the relation

Astyx

15:55

just to avoid confusion

Soham Saha

I was trying out with adjacency matrices previously

@Astyx oh thanks

If we have reflexivity and transitivity, we must have all the principal diagonal elements and the upper right triangle right?

Thorgott

ok, I think this problem reduces to the variant where we assume $1<\dotsc<n$

Soham Saha

@Thorgott Ah got it now, so that's what having the upper right triangle means

I think I have it wrong again. Reflexivity=>Need all principal diagonal elements. Not symmetric=>Can't stop there, need at least something more

We can add just one element and be done with it, like for card=4, $\{11,22,33,44,13\}$ works

@Thorgott So I don't think that this reduction works?

Thorgott

the number of such relations is, I believe, $\sum_{n=k_1+\dotsc+k_m}\sum_{i=1}^m\sum_{k_i=l_1+\dotsc+l_{r_i},r_i\ge2}\#\{\text{DAGs on $r_i$ vertices}\}$

Soham Saha

DAG?

Thorgott

16:07

ah wait, minor error in the formula still

Soham Saha

@Thorgott How many are you getting for $n=3$ (just with reflexivity, transitivity and not-symm criteria)?

Thorgott

second attempt $\sum_{n=k_1+\dotsc+k_m}\left(\sum_{r\ge1}\#\{\text{DAGs on $r$ vertices}\}\cdot\left(\sum_{i=1}^mp(k_i,r)\right)-1\right)$

Astyx

is the idea of the formula to group in connected components, and for each component do the 1<..<n case?

Soham Saha

Wait, what we need is just $\operatorname{partitions}(n)-\# (\text{only reflexive and transitive relations})$?

right?

because partitions will count all equivalence relations

Thorgott

@Astyx yeah, any relations generates an equivalence relation, so the first sum is over those (equivalently, their components). then, any reflexive transitive relation contains a largest equivalence relation (given by $x\le y$ and $y\le x$), so that's where the second partitions come from. then, the components over this second equivalence relation yield a connected directed acyclic graph.

I forgot to specify connected above

Astyx

16:20

@SohamSaha but you want more than equivalence relations

Thorgott

the idea is sound, but I think I'm still not quite counting it correctly

Astyx

there are symmetries playing tricks I think

Soham Saha

@Astyx #(Ref+Symm+Transitive)-#(Ref+Transitive)=#(Ref+Transitive+not symm) right?

Astyx

with a minus sign but yeah, sure

Soham Saha

Wait minus where?

Astyx

16:23

#(Ref+Symm+Transitive) is less than #(Ref+Transitive)

Soham Saha

oh sorry yes

And then ref+symm+transitive=equivalence?

And we can count that with Bell's number?

So we only need to think about counting Ref+transitive?

Thorgott

@Astyx yeah, if we only count the reflexive transitive relations for simplicities' sake, the formula is $\sum_{n=k_1+\dotsc+k_m}\sum_{r\ge1}\sum_{i=1}^mp(k_i,r)\cdot\#\{\text{DAGs on $r$ vertices}\}$

but the catch is that we don't count DAGs up to isomorphism

we have to count them up to equality on the fixed vertex set $\{1,\dotsc,r\}$

and, again, I should specify them to be connected

Soham Saha

@Astyx and @Thorgott Can't counting ref+transitive be simplified to just counting the transitives by some symmetry?

Astyx

reflexivity isn't hard to take into account

Thorgott

@Thorgott I don't think the above formula simplifies further, but it should be efficiently computable by some recursive algorithm

Soham Saha

16:34

@Astyx And counting transitivity seems non-trivial...

Astyx

Huh, I wasn't expecting it to be that hard

Thorgott

the number of deleted answers to that question is truly a sight to be hold

Soham Saha

Not enough rep

Thorgott

I gave the formula above, but as I said you shouldn't expect a simplification

Astyx

how many deleted answers?

Soham Saha

16:38

I think we can find a representation with A000110 and A006905

Thorgott

@Astyx 4

Astyx

that's a nice ratio

on a post with 2 upvotes

Thorgott

the only thing unclear to me is whether there's a nice method of counting the number of DAGs (including symmetries, as above)

but this is no doubt something CS people have already figured out, so I don't feel like thinking about it

Soham Saha

Should I open this up as a post on MSE?

Astyx

Do you expect anything more than the MSE post you already linked?

or do you mean about relations to the OEIS sequences?

Soham Saha

16:42

@Astyx I will see if I can get anything interesting from the sequences. And that post is only about transitive

Thorgott

personally, I wouldn't expect something nicer than the formula above, but I'm repeating myself

Soham Saha

Hm...

Astyx

@Thorgott Sure, but there might be a nice relation with another sequence that also doesn't have a nice closed form

Thorgott

I suppose

then I would start by writing down a recursive algorithm computing the formula to put in the OEIS

Soham Saha

Right

I think I should try that first

Soham Saha

17:29

@Thorgott and @Astyx I think what we have is A000798$(n)$-A000110$(n)$

Doesn't seem to be in OEIS, though I will check superseeker once

Binky

21:19

But a subset that has a certain property P is minimal only if it has an element, right?

Jakobian

23:50

Does there exist such thing as "differentiably connected" space?

where we can guarantee that paths are differentiable

Thorgott

what does differentiable mean

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