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24

Disclaimer: I'm by no means knowledgeable in this field and I haven't read the papers or books I mention below. I found these by digging in the literature and hope these pointers are useful. The answer to your question is no. The first example was given by R.H. Bing, The Cartesian Product of a Certain Nonmanifold and a Line is $E^4$, Ann. of ...


22

There is exactly one way in which one can convince oneself that a statement is not obvious: try to prove it and look at your attempts very, very critically. If you think you can come up with a proof of the curve theorem, edit it into the answer and we can help you dissect it :) Later. Asaf observes that it may be the case that you are refering to ...


16

I think perhaps the biggest thing that blinds one's intuition is that when one imagines embedding a circle in the plane, it's very easy to "lose the plot" and instead imagine embedding a disc in the plane, with the circle on the boundary. So not only do you see immediately the inside and outside, but you also see the Schoenflies theorem -- that a circle in ...


16

The Jordan Curve theorem is actually pretty easy to prove if you assume the curve is smooth or piecewise linear. The difficulty arises when you try to handle the general case. This includes nowhere-differentiable curves like the boundary of the Koch snowflake, and even wilder curves which can't even be drawn by hand, like Mariano says. It's kind of a magic ...


12

The $E_8$ manifold is pretty easy to construct. We may describe it by the following diagram: The meaning of the diagram is as follows. For each dot, we take the disk bundle over $S^2$ with Euler number $2$. This gives us eight $4$-manifolds with boundary. Now we plumb together each disk bundle as indicated by each edge. The result of the plumbing is a ...


12

Stillwell's book Classical Topology and Combinatorial Group Theory is a good first place to start to get a feel for the techniques of geometric topology. If you want to get your feet wet in the world of $4$-manifolds, there's a great book called The Wild World of $4$-manifolds by Scorpan which could serve as a source of further papers for you to look at. For ...


11

In general, the tangent bundle of a smooth $n$-manifold $M$ is classified by a (homotopy class of) map $\phi:M\rightarrow BO(n)$. The manifold $M$ is orientable iff there is a lift of this map to $BSO(n)$. For $n=2$, and $M$ orientable, we see see that the tangent bundle to $M$ is classified by a map $\phi:M\rightarrow BSO(2) = BS^1 = \mathbb{C}P^\infty$. ...


11

No for $n=2$. Consider a disk with two circular holes (red in the picture below) versus a disk inside an annulus (blue) versus three disks (green). In all cases the boundary consists of three disjoint circles.


11

You can't do it. Suppose $G$ were an $n$-dimensional manifold which is a topological group. Recall that an orientation of a topological manifold $M$ is a consistent choice of generator for $H_n(M,M\setminus\{x\})\cong \mathbb Z$ for each $x\in M$. But the left-multiplication homeomorphisms $\ell_g\colon G\to G$, $x\mapsto gx$ give canonical isomorphisms from ...


10

There's the path fibration $\Omega B \to PB \to B$, where for basepoint $* \in B$, $PB = \{\gamma:[0,1]\to B \ |\ \gamma(0) = *\}$ is called the path space of $B$, and $\Omega B = \{\gamma:[0,1]\to B \ |\ \gamma(0) = \gamma(1) = *\}$ is the loop space of $B$. The map $p:PB \to B$ is the endpoint map $\gamma \mapsto \gamma(1)$. I'm not quite sure what ...


10

As I said in the comments above, this has a positive solution due to a recent preprint of Lars Louder, which can be found here. The paper proves that surface groups have a "single Nielsen equivalence class of generating $2g$-tuples". I'll explain what this means, and then I will explain why this solves the problem. (In the comments below, Lars Louder has ...


10

The connected sum of two disks is an annulus. If you think of an annulus as being a hole, then I suppose a disk is half a hole.


10

Trying to use $\chi = 2 - 2g$ to describe things that aren't closed orientable surfaces is missing the point, I think. In my opinion one should think of the Euler characteristic of a compact space as a homotopy-invariant refinement of the cardinality of a finite set; see this blog post. A closed disk is contractible, so has Euler characteristic $1$, and ...


10

A vector bundle on $S^2$ can be constructed by gluing two trivial vector bundles over $S^2_+$ and $S^2_-$, the closed hemispheres. This is called the clutching construction; see, for example, Husemoller's book. The «gluing instructions» are a map from the equator, a cicle $S^1$, to $\mathrm{GL}_n(\mathbb R)$, and the result depends only on the homotopy ...


9

This answer is meant to complement Jim's. Guillemin and Pollack's Differential Topology text is a great start that's not too specialized any any particular direction. Once you've got some basic algebraic topology background, you can start to link up a lot of basic notions via Guillemin and Pollack (Poincare duality, intersection theory). A lot of ...


8

The intuition is basically that $4$ dimensions is large enough that $4$-manifolds exhibit great variety (one example: there are $4$ manifolds with arbitrary finitely-generated fundamental groups), but small enough that the high-dimensional-manifold surgery theory apparatus doesn't apply to smooth manifolds, only topological manifolds. The key idea for the ...


8

For your first question, it is true that fundamental groups of closed hyperbolic manifolds cannot contain copies of $\mathbb{Z}^2$. However, a knot complement is not a closed manifold! The hyperbolic structure on the knot complement will be a complete hyperbolic manifold with finite volume, but with a cusp. The fundamental group of the cusp is ...


8

What you get when you do that quotient is a space homeomorphic to $\mathbf P^3(\mathbb R)$. In particular, it does not have a boundary. A way to see this is to remember that $\mathbf P^3(\mathbb R)$ is more usually built as the quotient space of a $3$-sphere $S^3\subseteq\mathbb R^4$ by identifying antipodal points, and noticing that when you do this, the ...


8

Here are explicit equations for nonsmoothable manifolds (all of which admit triangulations). I do not know if these are the "easiest" but they are surely much more explicit than a description of the E8-manifolds, which is constructed as a result of some infinite, and very implicit, process (Freedman's work). Consider the homogeneous equation $$ z_1^5 + ...


7

Proposition 1 answers the "revised" question and Proposition 2 the original one. For completeness we give a self-contained proof of Proposition 2. Proposition 1 Let $n\in\mathbb{N}$ and let $p(x)=\alpha x^2+\beta x +\gamma$ be a polynomial with real coefficients where $\alpha >0$ such that $p'(n)> 0$ and $p'(n+1)<1$. Then for $k\in \mathbb{Z}\,$ ...


7

Yes. Your space is homeomorphic to the standard unit sphere $S^2\subseteq\mathbb R^3$ with three open disks removed. These disks can be chosen so that the rotation of $\mathbb R^3$ by $\frac{2\pi}3$ around some axis cyclically permutes them. This rotation, however, has two fixed points, where the axis of rotation intersects $S^2$. Compose it with a ...


7

The fibrations $O(n-1)\to O(n) \to S^{n-1}$, $U(n-1)\to U(n) \to S^{2n-1}$, and $Sp(n-1)\to Sp(n) \to S^{4n-1}$ from Bott periodicity are fairly important. Also mapping tori are fiber bundles. For example, the complement of a fibered knot in $S^3$.


7

Consider the topologist sine curve $$y = \sin \bigg(\frac{1}{x}\bigg),\ 1\geq x>0$$ together with the interval $\{(0, t): |t|\leq 1\}$ and a curve joining this interval with the graph. This is simply connected but is not contractible. You may find the proof of noncontractibility in http://math.ucr.edu/~res/math205B-2012/polishcircle.pdf


7

The general case is very much like the 2-dimensional case, it just takes time to process the picture, to see how you could do the same constructions in the higher-dimensional case. A punctured $S^1 \times S^1$ looks like a wedge of two circles, but fattened up a little bit. Precisely, around each circle you have an annulus neighbourhood. To immerse the ...


7

Here's an alternate proof which doesn't use invariance of domain. It also gives a slightly stronger result. Theorem: Let $M^n$ be compact without boundary. Then there is no immersion $f:M\rightarrow \mathbb{R}^n$. Proof: (sketch). Assume for a contradiction there is such an $f$. Since $M$ is compact, $f$ is a closed map, that is, it maps closed sets to ...


7

A handle attachment is the process of gluing a copy of $D^k\times D^{n-k}$ to $\partial X$. A (normal) framing gives a recipe for performing such a gluing, by specifying (up to ambient isotopy) a collar of $\partial D^{k}\times \{0\}$ in $X$. Gompf-Stipsicz express this data as: An embedding $\varphi_0\colon\, S^{k-1}\to\partial X$ with trivial normal ...


7

The answer depends on what you mean by a solid genus-2 handlebody, and the trouble is that this is ambiguous and requires some interpretation. Had you asked about "the complement of the solid genus-2 handlebody then I would assume you meant this and the answer would be yes, as in the answer of @QuangHoang. But since you asked about "the complement of a ...


6

For the equivalence of these definitions, I would look here: Local triviality of principal bundles. The existence of a $G$-equivariant cover is equivalent to the existence of $G$-valued transition functions: Suppose $(U_\alpha,\Phi_\alpha)$, $\Phi_\alpha : P\vert_{U_\alpha} \to U_\alpha\times F$, is a trivializing cover. This defines a collection of maps ...


6

Poincaré's conjecture follows from Perelman's proof on the Thurston Elliptization Conjecture. To put it simply, Thurston's Geometrization Conjecture claims that if you have a closed prime orientable $3$-manifold than you can cut it along a suitable collection of embedded tori so that each of the pieces you are left with can be endowed with a "nice" geometry. ...


6

Serge Lang, Algebraic Number Theory, page 85, defines $$G(a,b)=\sum_{x{\rm\ mod\ }b}e^{2\pi iax^2/b}$$ for $a$, $b$ non-zero integers, $b\gt0$, $\gcd(a,b)=1$, and states on page 87 $$\eqalign{G(1,b)&=(1+i)\sqrt b{\rm\ if\ }b\equiv0\pmod4,\cr &=\sqrt b{\rm\ if\ }b\equiv1\pmod4,\cr &=0{\rm\ if\ }b\equiv2\pmod4,\cr &=i\sqrt b{\rm\ if\ ...



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