# Tag Info

28

While I'm sure an expert could give a much more informative answer, let me give a naive one. In mathematics we are always interested in classification results. If we want to understand an object that occurs in some problem and we have a classification result for that type of object, we can use it to gain traction on the problem. For example we use the ...

14

If you want to visualize the action of $\mathbb{Z}_p$ on $\mathbb{Z}^3$, you may be more successful using stereographic projection to think of the action as on $\mathbb{R}^3$. I've found it helpful to think of the $\mathbb{Z}_p$ action as a discrete analogue of the Hopf fibration. This isn't exactly true, but the actions are similar enough that if you can ...

11

The easiest way I know to do this is to think of $S^3$ as the boundary of $D^4$. This is the same as the boundary of $D^2\times D^2$, which is $S^1\times D^2 \cup D^2\times S^1$. By a similar trick you can cook up decomposition of spheres of other dimensions too.

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 ...

10

Exotic $\mathbb R^4$ There are infinitely many non-diffeomorphic smooth structures on the topological space $\mathbb R^n$ if and only if $n=4$. (Otherwise there is only one diffeomorphism class.)

10

This is a well-known phenomenon among topologists, and although I'm not an expert, I'll give one standard answer: 3- and 4-dimensional topology are very different from topology in 5 or more dimensions because surgery theory works in 5 or more dimensions. In 3 and 4 dimensions one does not have enough "wiggle room" for surgery theory to be effective and this ...

9

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 ...

8

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 ...

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

There is a very specific reason why one needs 3 dimensions or more for the Banach Tarski paradox. In dimension 3 or higher one can make rotations in independent directions, and so the group $SO(3)$ of rotations of space contains a copy of $F_2$, the free group on two generators. This fact is what underlies the Banach Tarski paradox. (The group $F_2$ is ...

8

Smale observed a version of what's now known as The Smale-Hirsch Theorem. see: http://en.wikipedia.org/wiki/Immersion_%28mathematics%29 This states that if a manifold $M$ has dimension strictly less than the dimension of the manifold $N$, then the space of all immersions of $M$ in $N$ has the same homotopy-type of the space of fibrewise-linear bundle ...

7

A right-angled hexagon is simply a hexagon in the hyperbolic plane with hyperbolic lines (geodesics) as its edges, and interior angles all right angles: These have lesser total interior angle (and greater total exterior angle) than you'd get in a euclidean hexagon, but that's fine in the hyperbolic plane. There exists a unique one of these given three ...

7

Consider the orientations on the circles $S^1 \times 0$ and $S^1 \times 1$ coming from a choice of orientation on $S^1$. If the gluing map $f : S^1 \times 1 \to S^1 \times 0$ preserves orientation, the quotient is homeomorphic to a torus. If not, it is homeomorphic to a Klein bottle. The proof requires one to know that two self-homeomorphisms of $S^1$ are ...

7

In his book Knots and Links, Rolfsen shows that, writting Antoine's necklace $A$ as an intersection of chains of tori $\bigcap\limits_{n \geq 0} C_n$ as usual, the inclusions $\mathbb{R}^3 \backslash C_n \hookrightarrow \mathbb{R}^3 \backslash C_{n+1}$ are $\pi_1$-injective, so that the fundamental group $$\pi_1( \mathbb{R}^3 \backslash A) = ... 7 The fundamental idea behind algebraic topology is to translate topological problems into algebraic problems. By so doing, one is able to strip unnecessary structure away from the study of manifolds, reducing it to a subject which can be tackled by group theoretical methods. A typical problem in topology would be, given a group G, to classify the ... 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 compact case, I believe the answer is, as you said in the comments above, any closed surface with a single puncture, i.e. a disk removed. I claim that this completely classifies compact surfaces with boundary S^1. This is because you can glue a disk to the surface along its boundary to obtain a closed surface, and there is a unique way to do this ... 6 I don't know if that's what you're looking for, but here are some equivalent ways to visualize spheres. I added some sketches, I hope they help somewhat. One standard way to think of an n-dimensional sphere is by taking an n-dimensional ball and identifying all the points of its boundray. For example, for S^2, we have a 2-dimensional disk, which has ... 6 I'm not going to give you an actually formal proof, but a way to think about this, which anyway I find quite satisfactory. What you are asking for is known as a Heegaard splitting of a 3-manifold (in your case it's S^3), i.e. a decomposition of the manifold in two copies of the same "easier" 3-manifold with boundary, such that gluing these two copies along ... 6 EDIT: this was worked out for me by the same team of gerbils that does Jordan Normal Form of matrices when I need that. Amazingly versatile. Take the coordinates as named (w,x,y,z) to reduce subscripts. The sphere is$$ w^2 + x^2 + y^2 + z^2 = 1. $$The common boundary is the Clifford Torus,$$ w^2 + x^2 = y^2 + z^2 = \frac{1}{2}. $$One solid torus is ... 6 Here is one way of possibly deciding which is a mixture of local and global, but is all internal. It "only" requires knowledge of every complete Riemannian metric on your manifold (so is completely unpractical, but is still completely internal). It's local in the sense that you can look at your manifold pointwise and get the answer (which is nice because, ... 6 Milnor's book on characteristic classes is almost surely exactly what you want. 6 Here is a general theorem that does the job: Theorem. If \Omega is an open connected subset of R^3 then \pi_1(\Omega) is torsion-free. Proof. Suppose not. It is a theorem of D.B.A. Epstein (see theorem 9.8 in the book "3-manifolds" by J.Hempel) that if M is a connected oriented 3-manifold whose fundamental group has nontrivial elements of finite ... 6 Poincaré's conjecture follows from Perelman's proof on the Thurston Elliptization Conjecture. To put it simply, Thurston's Geometrization Conjecture claims that if you have a closed prime orientable 3-manifold than you can cut it along a suitable collection of embedded tori so that each of the pieces you are left with can be endowed with a "nice" geometry. ... 6 A simple closed curve in a surface bounds a disc in that surface if and only if the induced map on fundamental groups is trivial. So if C is an embedded circle in a surface F, I'm saying$$\pi_1 C \to \pi_1 F$$is the zero map if and only if C bounds an embedded disc in F. You can break the proof up into steps. Step 1: the curve bounds a ... 6 The real question is how you view the space. You must separate these ideas as you think about the question of what a space is: parametric viewpoint: t \rightarrow (x(t),y(t)) is a curve. For each value of t we obtain a point on the curve C. implicit viewpoint: F(x,y)=k. The curve is the set of all (x,y) which solve the equation F(x,y)=k. Let ... 6 Here is a Kirby Diagram for an exotic \mathbb R^4, taken from Gompf and Stipsicz's book, "4-Manifolds and Kirby Calculus." It's not given in the form of an atlas, but it is a nice explicit description. 6 The key theorem is the following. The braid group B_n is isomorphic to the mapping class group$$ M_n(D^2) \;=\; \mathrm{MCG}\bigl(D^2 - \{p_1,\ldots,p_n\},\partial D^2\bigr)  of a disk with $n$ punctures rel the boundary. Here "rel the boundary" means that all homeomorphisms and isotopies must fix the boundary of the disk pointwise. The idea ...

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 $S^1 \to ... 5 The signature can be thought of as the signature of the intersection form$H_2(M)\otimes H_2(M)\to\mathbb Z.$This may be the way it is defined for non-closed 4-manifolds. In that case, it's easy to see that$H_2(M)$is generated by$S^2×\{0\}\$, and this has trivial self-intersection since there is a parallel push-off which is disjoint, implying that the ...

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