21 votes

Showing that $\bar{\mathbb{B}}^n$ is a manifold with boundary (Lee ITM Probelm 3-4)

I've been worked on this problem for some time, and i think i probably solved it based on the hint given on the book. Maybe this seems a little long, but it is really not. I tried my best to make this ...
Kelvin Lois's user avatar
  • 7,000
12 votes
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Boundary of manifold with boundary has empty boundary: What about corners in the topological category?

Well, manifolds with corners are not one and the same as topological manifolds. Let me come back to that point later, and address the easy parts of your question first. In fact $\partial C=(S^{1}\...
Lee Mosher's user avatar
  • 119k
8 votes
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What is the boundary of two manifolds with boundary?

You're right that "corners" won't be differentiable. In fact, a product of two manifolds $M$ and $N$ with nonempty boundary does not have any natural structure of a differentiable manifold, because ...
Eric Wofsey's user avatar
8 votes

Proving diffeomorphism invariance of boundary

I thought exactly the same and I think I've found a satisfactory solution. Basically your po\text{int } 1. is right, but the argument is indeed quite subtle. I'm gonna divide this proof in 4 parts: ...
Zero's user avatar
  • 3,393
8 votes
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No where vanishing exact $1$-form on compact manifold.

The other two examples provide examples showing that you need to assume your manifold is boundaryless. I want to show where the "usual" proof in the boundaryless case breaks down. So, here ...
Jason DeVito - on hiatus's user avatar
7 votes
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A local diffeomorphism can map a boundary point to an interior point

To say that $f\colon M\to N$ is a local diffeomorphism means that each point of $M$ has an open neighborhood $U$ such that $f(U)$ is open in $N$ and $f|_U$ is a diffeomorphism from $U$ onto $f(U)$. ...
Jack Lee's user avatar
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7 votes
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Is every flat manifold with boundary locally isometric to the Euclidean half-space?

I would say no, because you cannot assure that the boundary will be straight. Consider the disk $$ M = \{ x \in \mathbb R^2 \,:\, |x| \leq 1 \}\,. $$ It is clearly flat, but because the only ...
Martins Bruveris's user avatar
7 votes
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The boundary of a manifold is a closed subset.

This is obvious. Let $x \in M \setminus \partial M$. There exists an open neighborhood $V$ of $x$ in $M$ which is homeomorphic to $\mathbb R^n$. Then any $y \in V$ has the same property, hence $V \...
Paul Frost's user avatar
  • 75.1k
7 votes

Constant Rank Theorem for Manifolds with Boundary

Lee's Introduction to Smooth Manifolds deals with the case of local immersions for manifolds with boundary in Theorem 4.15. So let us suppose that $F:\mathbb{H}^m \rightarrow \mathbb{R}^n$ has $\...
user886204's user avatar
7 votes
Accepted

If the interior of a manifold with boundary is smooth, is the whole manifold smooth?

For instance, you can start with the $E8\oplus E8$ manifold $M$: This is a 4-dimensional closed simply connected manifold with the intersection form isomorphic to $E8\oplus E8$. This manifold has ...
Moishe Kohan's user avatar
  • 95.9k
7 votes
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Confusion over boundary definition in topology

That's right: every neighborhood of any point on the strip intersects both the strip and its complement in $\mathbb{R}^3$, so as a subset of $\mathbb{R}^3$, the strip is its own boundary. However, ...
Арсений Кряжев's user avatar
7 votes
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Is the complement of a quadrant a manifold with corners?

No, the quadrant compliment is not a manifold with corners. Essentially, manifold corners have no notion of angle (i.e. smooth deformations may change the angles of a polygon), but they do have a ...
Kajelad's user avatar
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6 votes
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Can every Riemmanian Manifold be completed?

The metric completion of $M$ might not be a manifold. For an example, take the Alexander horned sphere $A \subset S^3$. There are two complementary components of $A$; let $M$ be one of them. Then the ...
Lee Mosher's user avatar
  • 119k
6 votes
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Are topological manifolds with boundary metrizable?

Michael's answer is the right one in that it works directly for any manifold. But since you say you know the answer for manifolds: Every manifold with boundary is a subspace of its double, which is ...
Pete L. Clark's user avatar
6 votes

Are topological manifolds with boundary metrizable?

They are metrisable. One way to see this is to use the following theorem. Urysohn's Metrisation Theorem: Every Hausdorff, second countable, regular space is metrisable. Let $M$ be a connected ...
Michael Albanese's user avatar
6 votes
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Orientable manifold $M$ ,then $\partial M$ is orientable

First, it suffices to consider the case when $M$ and $\partial M$ are both connected. (If $M$ is connected and $\partial M$ contains several components $D_i$, consider manifolds $M_i=(M - \partial M) \...
Moishe Kohan's user avatar
  • 95.9k
6 votes
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Euler characteristic of a manifold is odd

$\newcommand{\Q}{\mathbb{Q}}$Poincaré duality tells you that there are non-degenerate pairings $H^i(M) \otimes H^{n-i}(M, \partial M) \to \Q$ for all $0 \le i \le n$. Using the long exact sequence of ...
Najib Idrissi's user avatar
6 votes
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On the homology of manifolds with boundary

The inclusion $\mathrm{int}(M)\rightarrow M$ is a homotopy equivalence. This follows from the existence of a collar neighborhood of $\partial M$ in $M$. Thus, $H_i(\mathrm{int}(M))\rightarrow H_i(M)$ ...
Thorgott's user avatar
  • 11k
6 votes
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Submanifold of ball is entire ball

First note, by compactness, there is a $ \varepsilon > 0 $ such that the $ \varepsilon $-neighborhood of $ \partial \mathbb{B}^n $ (the set of points within $ \varepsilon $ of $ \partial \mathbb{B}^...
Jake Mirra's user avatar
  • 3,188
5 votes
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Smoothness of the boundary is the only obstruction for being a submanifold with boundary?

I'm assuming that your first sentence was meant to say "Let $N$ be a smooth manifold, $\dots$ ." Yes, it is always the case that $\overline S$ is a smooth submanifold with boundary in $N$. However, ...
Jack Lee's user avatar
  • 46.4k
5 votes

Compact 3-manifolds with boundary an orientable surface

Definitely not. For instance, you could take your example $N\subset\mathbb{R}^3$ and form a connected sum $N\mathbin{\#} P$ where $P$ is any closed oriented 3-manifold. By some simple long exact ...
Eric Wofsey's user avatar
5 votes
Accepted

Why is the boundary of an oriented manifold with its (opposite oriented) copy the empty set?

You're a bit confused about what this definition means. First of all, the $n$-manifolds we're considering here all do not have boundary. So $M$ should be an $n$-manifold without boundary, and you ...
Eric Wofsey's user avatar
5 votes
Accepted

If the boundary $\partial M$of a manifold with boundary is connected,is $M$ connected?

Take the disjoint union of any manifold without boundary and a manifold with a connected boundary like a torus to which you remove a disc.
Tsemo Aristide's user avatar
5 votes

Show that $\mathbb{S}^d$ is homeomorphic to the the boundary of the cube $\partial I^{d+1}$.

$S^d=\{x\in\Bbb R^{d+1}:\|x\|_2=1\}$ and $\partial I^{d+1}=\{x\in\Bbb R^{d+1}:\|x\|_\infty=1\}$. There are continuous maps $x\mapsto \|x\|_\infty^{-1}x$ and $x\mapsto \|x\|_2^{-1}x$ in both ...
Angina Seng's user avatar
5 votes
Accepted

When both $U$ and $W$ are open in $\mathbb{H}^k$ and $\mathbb{H}^l$, respectively, then why $U\times W$ cannot be open in $\mathbb{H}^{k+l}$

For example, $[0, 1)$ is open in $\mathbb{H}$, but $[0, 1) \times [0, 1) \subseteq \mathbb{H}^2$ (the interior of a square together with two of the sides) isn't open. The reason this doesn't fit with ...
SCappella's user avatar
  • 2,516
5 votes
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Spin structure and bordism

A good reference for a lot of this is "Characteristic Classes" by Milnor and Stasheff. I can definitely answer 2, 3, 4 and partially answer 1. 1) As Peter points out in your comments, every closed ...
William's user avatar
  • 9,260
5 votes

Classification of surfaces

If I read correctly you are trying to determine $c$ and $g$, given a compact connected surface $M$ for which you know $b$ the number of boundary components, $\chi(M)$ the Euler characteristic the and ...
Adam Chalumeau's user avatar
5 votes

Are there metrics of nonnegative Gaussian curvature on these surfaces?

There do indeed exist such metrics. The idea is that any compact surface with nonempty boundary can be smoothly immersed in the plane, and hence in $S^2$, and so you can get a metric of constant ...
Lee Mosher's user avatar
  • 119k
5 votes

Does a manifold without boundary necessarily have to be the boundary of some other higher dimensional manifold?

Tsemo's comment seems to be irrelevant. A manifold which is the boundary of another manifold is called null-bordant. There are many obstructions to being null-bordant, but every oriented manifold up ...
epsteinbrain's user avatar
5 votes

Is $\mathbb{D} = [-1,1]^3$ a compact manifold?

Indeed, a cube is homeomorphic to a closed ball in the usual metric, so anything topological that applies to the closed ball also applies to a cube.
Matt Samuel's user avatar

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