# Tag Info

## Hot answers tagged geometric-topology

20

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 ...

18

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 ...

14

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 ...

11

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 ...

11

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

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.

9

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

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 $\phi:M\rightarrow BSO(2) = BS^1 = \mathbb{C}P^\infty$. ...

8

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

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

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 ...

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

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 ...

6

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 ...

6

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 ...

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\ ... 5 This is a fantastic lay-person article on Thurston's program by Erica Klarreich https://simonsfoundation.org/features/science-news/getting-into-shapes-from-hyperbolic-geometry-to-cube-complexes-and-back/ 5 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. ... 5 This doesn't directly answer your question, but the computations become simpler I think if you restrict to oriented 2-plane bundles. Now these are classified by maps (up to homotopy) S^1 \to GL_+(2). But by putting a metric on any bundle and making sure our transition functions respect the metric, such bundles are actually classified by maps S^1 \to ... 5 We can make a consistent choice of normal vector at every point of f^{-1}(c), namely the gradient of f at that point. Given a basis v_1,\ldots,v_{n-1} of the tangent space to f^{-1}(c) at some point x, we could define this basis to be positively oriented iff v_1,\ldots,v_{n-1},\nabla f(x) is positively oriented as a basis of the tangent space to ... 4 No. For n \in \mathbb N let K_n = \left\{x\in \mathbb R^2 \mid x_1 \ge 0, \left|x - \left(\frac 1n, 0\right)\right| = \sqrt{1 + \frac 1n^2} \right\} $$and A_i = \bigcup_{n \ge i} K_n. As we have K_j \cap K_k = \{(0, \pm 1)\} for j \ne k, we have \bigcap_i A_i = \bigcap_n K_n = \{(0, \pm 1)\} and so the intersection isn't path connected (it ... 4 One comment about your second question, which is a bit tedious to fit into a comment box: The quotient SL_2(\mathbb R)/SO(2) is isomorphic to H^2, and so SL_2(\mathbb R) is a circle fibration over H^2. In fact, if we forget the group structure, there is a diffeomorphism SL_2(\mathbb R) \cong H^2 \times SO(2). Thus, as a manifold, ... 4 Here is my interpretation of your question: Given two homotopy equivalent (perhaps smooth) n-manifolds M and N, are M and N homeomorphic (or even diffeomorphic)? For n = 1 the answer is of course always yes, and the classification of surfaces shows that the answer is also always yes for n = 2. In higher dimensions, however, the answer to ... 4 The following is essentially a fleshed-out version of the proof of Proposition 4.2 in Ghys's beautiful article Groups acting on the circle. It is convenient to translate the question into a question about continuous functions on \mathbb{R}. Viewing the circle S^1 = \mathbb{R}/\mathbb{Z} as a quotient of \mathbb{R}, a continuous function f\colon S^1 ... 4 You appear to be missing a point-set theoretic tool to do with the quotient topology. Let's build a space by taking two spaces X and Y, take their disjoint union, then identify a subset of X with a subset of Y via a homeomorphism. Let's say A \subset X and B \subset Y and \phi : A \to B is a homeomorphism.$$X \sqcup_\phi Y := (X \sqcup ...

4

For $n<m$ we can isometrically embed $\mathbb R^n\to \mathbb R^m$ by mapping to the first coordinates / filling up with zeroes. This way the limit of all $\mathbb R^n$ is $\mathbb R^\infty$, the set of all sequences $(x_n)_{n\in\mathbb N}$ with almost all terms $=0$, which still gets its topology from the scalar product $\langle ... 4 A handle attachment is the process of gluing a copy of$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$with trivial normal ... 4 I am not sure that your question makes sense. A foliation on a surface decomposes the surface continuously into one-dimensional "leaves". (Think of taking a vector field on your surface; the corresponding flow will give you a foliation.) The simplest example of a foliation is probably to form a torus by identifying opposite sides of a unit square. Take an ... 3 The answer is yes. A variant of this is proven in Milnor's "Morse Theory" text. I don't have it here with me, but what he does is instead of looking at linear functions on the ambient space, he looks at the function that gives the distance from a point in the ambient Euclidean space. He proves this function is Morse on$M$for a generic choice of point ... 3 The general case is very much like the 2-dimensional case, it just takes time to process the picture, to see how you could do the same constructions in the higher-dimensional case. A punctured$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 ...

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