User hbghlyj | chat.stackexchange.com

hbghlyj

yst 17:44

What is the weak sequential closure of unit sphere in $\ell^\infty$?

hbghlyj

Tue 18:51

Is the number n_G the order of G?

hbghlyj

Tue 18:48

Related to this question: math.stackexchange.com/questions/2816786/… I have a question: Let Σ_g be a genus g compact connected oriented surface, for a finite group G, is the set {g | Σ_g has an action by G that is trivial on H_2 but no element of G acts trivially} the same as the natural numbers that are $1\mod n_G$?

hbghlyj

Tue 13:11

Is the unoriented S^2 bundle over S^1 diffeomorphic to unoriented S^1 bundle over S^2?

hbghlyj

Feb 15 16:14

The numbers don't match. I must be wrong.

hbghlyj

Feb 15 16:13

But the first homology group of ∂H_2 is Z^{2g}

hbghlyj

Feb 15 16:12

given $g$ disjoint curves which represents linearly independent homology classes in ∂H_2, as characteristic curves

hbghlyj

Feb 15 16:12

Yes

hbghlyj

Feb 15 16:11

I try to prove the resulting 3-manifold has boundary $S^2$, by showing its first homology group is $0$.

hbghlyj

Feb 15 16:07

Attaching $g$ 2-handles to $H_2$, the resulting 3-manifold should have first homology group $0$.

hbghlyj

Feb 15 16:05

I thought about it and the number doesn't match

hbghlyj

Feb 15 16:04

Its first homology group is Z^{2g}

hbghlyj

Feb 15 16:03

∂H_2 is homeomorphic to the genus $g$ orientable surface

hbghlyj

Feb 15 16:02

I try to prove component ∂N only have 1 component and is homeomorphic to S^2, given $g$ disjoint curves which represents linearly independent homology classes in ∂H_2, as characteristic curves

hbghlyj

Feb 15 15:56

But in the "Added" part of the answer "then attach B
(which is a 3-handle)." seems to mean there is only one component.

hbghlyj

Feb 15 15:52

Does it only have 1 component?

hbghlyj

Feb 15 15:52

In the answer math.stackexchange.com/questions/1806042/… in step 2, "Each component of ∂N is homeomorphic to a 2-sphere." I wonder how many component ∂N would have.

hbghlyj

Feb 12 14:59

Yes, the usual PD with $\mathbb{Z}$ coefficients assumes orientability

hbghlyj

Feb 12 14:36

for a compact connected manifold of dimension 3.

hbghlyj

Feb 12 14:30

https://topospaces.subwiki.org/wiki/Fundamental_group_determines_homology_groups_for_compact_connected_orientable_3-manifold : For a compact connected orientable manifold of dimension 3, the fundamental group is sufficient to determine the isomorphism classes of all the homology groups.

I wonder if the fundamental group is sufficient to determine orientability

hbghlyj

Feb 12 08:44

Does there exist a closed 3-manifold with fundamental group $\mathbb Z^2$?

hbghlyj

Feb 12 01:46

The deficiency of a finite presentation $\langle S | R \rangle$ is $|S| - |R|$.
The deficiency of a finitely presented group $G$ is the maximum of the deficiency over all presentations of $G$.
https://math.stackexchange.com/questions/478841/finitely-presented-group-with-fewer-relations-than-generators proved that a finitely presented group with positive deficiency has to be infinite. I wonder if there exists an infinite finitely presented group with negative deficiency.

hbghlyj

Feb 10 11:38

@Thorgott If I understand correctly, an $n$-manifold is orientable iff every loop has a neighborhood homeomorphic to $S^1 \times \mathbb{R}^{n-1}$. A non-orientable $n$ manifold must contain a subset homeomorphic to the product of a Möbius band and $\mathbb{R}^{n-2}$. Is it true?

hbghlyj

Feb 9 22:44

Thanks

hbghlyj

Feb 9 22:37

Is there a "n-dimensional Möbius strip" that is the source of all non-orientability in n-dimensional manifolds?

hbghlyj

Feb 9 22:37

A two-dimensional manifold is orientable iff the surface contains no subset that is homeomorphic to the Möbius strip. Thus, for surfaces, the Möbius strip may be considered the source of all non-orientability. I wonder if there is a similar result for higher-dimensional manifolds.

hbghlyj

Feb 8 01:42

I noticed that subgroup lattice of $\mathbb Z$ is distributive. Is there a name for the groups whose subgroup lattice is distributive?

hbghlyj

Feb 8 01:37

Ok

hbghlyj

Feb 8 01:37

Let $H,K,L$ be three distinct lines through the origin in $\mathbb{R}^{2}$. Then $H \cap L=K \cap L=\{0\}$, so that $(H \cap L)(K \cap L)=\{0\}$, while $(H K) \cap L=L$.

hbghlyj

Feb 8 01:25

hbghlyj

Feb 8 01:25

A Course in the Theory of Groups p.15 remarked that the distributive law doesn't hold for subgroup lattices in general. Can you give an counterexample?

hbghlyj

Feb 7 13:20

I don't understand why (1) the integral is independent of the bounding disk and (2) by the area inequality, it gives a lower bound on the area of such a bounding disk.

hbghlyj

Feb 7 13:20

I have a question about the proof of this example (the cubic nature the Heisenberg group)

hbghlyj

Feb 7 13:20

hbghlyj

Feb 6 16:18

@Thorgott I wonder if it is a manifold

hbghlyj

Feb 6 11:02

I think it has real dimension =4. Can we prove a neighborhood B of (0,0,0) is not homeomorphic to R^4 by removing one single point and connectedness?

hbghlyj

Feb 6 11:01

I mean, real manifold

hbghlyj

Feb 6 10:59

need to analyze a neighborhood of the x=y=z=0 point

hbghlyj

Feb 6 10:58

How do you show that the "complex cone" $\{(x, y, z) ∈\mathbb C^3 : x^2 + y^2 - z^2 = 0\}$ is not a manifold?

hbghlyj

Feb 4 02:45

Ok, I see.

hbghlyj

Feb 4 02:37

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

hbghlyj

Feb 4 02:37

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, ...

hbghlyj

Feb 4 02:33

Ok

hbghlyj

Feb 4 02:28

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?

hbghlyj

Feb 4 02:25

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?

hbghlyj

Feb 4 02:17

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?

hbghlyj

Feb 4 02:14

I see. Thanks!

hbghlyj

Feb 4 02:10

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

hbghlyj

Feb 4 01:57

Which step of the above proof is wrong?

hbghlyj

Feb 4 01:57

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.

hbghlyj

Mathematics

$Mathematics$ Mathematics