# Tag Info

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 $\xi$ ...

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

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 ...

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 ...

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, ...

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 ...

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, ...

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 ...

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 ...

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 ...

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 ...

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