# Tag Info

2

I know this is an old post but the disk is a once-perforated sphere, and the Euler char formula becomes 2-2g-n, where g is the number of handles and n is the number of perforations.

3

I'll prove the statement I have in my comment. That any CW-complex with the homology of $S^n$ suspends to a space homotopy equivalent to $S^{n+1}$ for $n\geq 1$. If $X$ is such a CW-complex, by Mayer-Vietoris, $SX$ has the homology of $S^{n+1}$. As $X$ is $(0)$-connected, $\pi_1(SX)=0$ by the Freudenthal suspension theorem. By applying Hurewicz repeatedly, ...

0

If $\mathbb{C} \setminus U$ has only finitely many components then you can proceed roughly as follows: Choose $\varepsilon > 0$ less than the distance of $K$ to any other component of $\mathbb{C} \setminus U$. Then consider the set of all squares $$Q_{k, l} = \{ x+iy \mid (k-1)\varepsilon \le x \le k\varepsilon, (l-1)\varepsilon \le y \le l\varepsilon ... 3 Any manifold book worth its salt should prove this. Put a Riemannian metric on X; then for every point in the boundary, there is a unique tangent vector in T_p X that is orthogonal to the boundary, of norm 1, and points inwards. This provides a trivialization of the normal bundle of \partial X in X. Now, how would one prove the tubular neighborhood ... 2 A necessary condition for a closed n-manifold M to (smoothly) embed or immerse into \mathbb{R}^{n+1} is that its tangent bundle T becomes trivial after adding a single line bundle L (namely the normal bundle of the embedding). This condition is sufficient for an immersion by Hirsch-Smale theory, but the question of embedding is more delicate. The ... 3 Yes, this is correct. You could prove it as a corollary of the annulus theorem (sketch of a sketch: at each stage the closure \overline U_n is an open n-ball, and you're attaching an annulus, and the limit of this procedure is \Bbb R^n), but this wasn't known in all dimensions until the 80s. An early proof of this theorem was given in 1961 by Morton ... 1 The same answer works, essentially. Suppose X is infinite and Hausdorff. Then X has a discrete (in itself) infinite subspace S. Pick p \in S. Then \overline{S\setminus\{p\}} is proper (as p is not in the closure of S\setminus \{p\}), closed, and has all points of S \setminus \{p\} as components so has infinitely many components (they're still ... 0 The Klein bottle can be defined as the quotient space$$ K=I^2 /{\sim}, \quad (x,0)\sim(x,1), \; (0,y)\sim(1,1-y), \; \forall x,y\in I $$Let p:I^2\to K be the canonical quotient map. The Möbius band is the quotient space$$ M=I^2 /{\sim}, \quad (0,y)\sim(1,1-y) $$with a quotient map q: I^2\to M. We have a product map$$ q' = q\times 1_I : I^2\times I ...

3

It depends on what you mean by a "simplicial subdivision". I can think of two different definitions, leading to two different outcomes for your question. In one definition, a "simplicial subdivision" is obtained from the original decomposition into simplices by repeating some kind of elementary subdivision. The iterated barycentric subdivision is like this, ...

0

See the accepted answer here: http://mathoverflow.net/questions/63439/how-can-we-detect-the-existence-of-almost-complex-structures In particular this paragraph: A closed oriented 6-dimensional manifold $X$ without 2-torsion in $H^3(X,\Bbb Z)$ admits an almost complex structure. There is a 1-1 correspondence between almost complex structures on X and ...

0

This turned out to be false. By Example 3.1 in the paper of James Davis "Manifold aspects of the Novikov conjecture", there is a homotopy equivalence $h:S(E')\to S^4\times S^4$ such that $\sigma(h^{-1}(pt\times S^4))=16$, but $\sigma(h^{-1}(S^4\times pt))=\sigma(S^4)=0$ since $h$ preserves the fibers.

2

Yes, this is true. We may as well suppose $f$ is a homeomorphism that already preserves one of the fibers. Now cut the homeomorphism open along that fiber. Then both manifolds are now homeomorphic to $\Sigma \times [0,1]$, and we may identify your homeomorphism with a homeomorphism $\Sigma \times [0,1]$ to itself; assume it preserves $\Sigma \times \{0\}$ or ...

6

For Question 1 note first that it is important that the base space of the vector bundle be finite-dimensional, as one can see by looking at the canonical line bundle $\gamma$ over ${\mathbb R}P^\infty$ since $w_k(\gamma^{\oplus k})$ is nonzero for all $k\geq 1$, so $\gamma^{\oplus k}$ is not even stably trivial. On the other hand, for a vector bundle $\xi$ ...

4

By taking the orientable cover if necessary, we assume that $M$ is orientable (Note that the Betti number is not changed after taking the orientable cover). The Bochner formula for one form is $$\tag{1} \Delta_d \alpha = -\nabla^*\nabla \alpha + \text{Ric}(\alpha),$$ for all $\alpha$. Let $\alpha$ be a harmonic one form, then integrating the above formula ...

1

No, this would imply that the boundary of a simply connected manifold is simply connected. But this isn't true for a number of reasons; there's the disc $D^2$, say. If you demand higher-dimensional things, simply connected 3-manifolds have simply connected boundary automatically, but as soon as you're in dimension 4 there are contractible manifolds whose ...

5

Any $M$-bundle over $S^1$ is a manifold known as a mapping torus. You obtain it by picking a diffeomorphism, $\varphi: M \to M$, and letting $M_\varphi$ be the quotient manifold $M \times \Bbb R/(x,t) \sim (\varphi(x),t+1)$. The bundles $M_\varphi \to S^1$ are classified by the isotopy class of $\varphi$. For the case of $M = S^2$, there are two isotopy ...

4

As mentioned in the comments, such a thing needs to be a rational homology sphere, and in particular a $\Bbb Z[1/d]$-homology sphere, where $d$ is the degree. It also needs to have fundamental group of order coprime to $d$. As in the comments, we may as well assume the manifold is simply connected by passing to the universal cover. First, there are no ...

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