23
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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 ...
- 7,280
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}\...
- 129k
9
votes
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Does such a manifold exist??
No such manifold $W$ exists.
To see this, note to start that $H_1(W) = H_2(W) = 0$ implies that $H^1(W) = H^2(W) = 0$ as well, by the universal coefficient theorem. Since you want $W$ to be orientable,...
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8
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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 ...
- 338k
8
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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:
...
- 3,543
8
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 \...
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8
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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 ...
- 51.1k
8
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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 ...
- 105k
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 ...
- 1,640
7
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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 $f(U)$.
...
- 49.1k
7
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) \...
- 105k
7
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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 $\...
- 71
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, ...
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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 ...
- 15.2k
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 ...
- 55.5k
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, $H_i(\mathrm{int}(M))\rightarrow H_i(M)$ ...
- 14.7k
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}^...
- 3,253
5
votes
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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 ...
- 338k
5
votes
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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.
- 88.6k
5
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Why do we sometimes call the boundary of the body by $\partial$?
Consider the ball $B_r$ of radius $r$. This is a manifold with boundary whose boundary is the sphere of radius $r$. Now consider specifically the difference between this ball and a slightly larger ...
- 447k
5
votes
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What are the possible boundaries of connected compact manifolds?
Well...its not clear exactly what constitutes an answer for your question, but I can say one or two things.
"$M$ is null-cobordant" is the same as "$M$ is the boundary of some manifold $N$" (...
- 97k
5
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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 ...
- 160k
5
votes
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Is the closed hemisphere diffeomorphic to the closed disk?
Whatever meaning you assign to diffeomorphism for manifolds with boundary, you are going to have to admit that the hemisphere and the disk are diffeomorphic: just use the stereographic projection ...
- 45.8k
5
votes
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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 ...
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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 ...
- 9,538
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 ...
- 3,303
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 ...
- 129k
5
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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 ...
- 51
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.
- 58.8k
5
votes
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Manifolds with boundaries: Why map boundary points onto boundary points?
Get rid of that image of a manifold given by an embedding into some $\Bbb R^n$. An atlas describes the manifold without reference of any surrounding space (if such a space exists at all). Think of a ...
- 379k
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