# Tag Info

## Hot answers tagged low-dimensional-topology

29

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 ...

17

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 ...

14

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.

14

There are precisely two: $S^2 \times S^1$ and a twisted product $S^2 \widetilde \times S^1$. Suppose $M$ is closed, connected, with fundamental group $\Bbb Z$. Then it must be prime (by Poincare and because $\Bbb Z$ is indecomposable under free product), but not necessarily irreducible (meaning that every embedded 2-sphere bounds a 3-ball). Indeed, it ...

11

Lets bump knot theory up a dimension. In general, an $n$-sphere can be non-trivially knotted in $\mathbb{R}^{n+2}$. Obviously, an $n$-sphere can be embedded in $\mathbb{R}^{n+1}$, where it is usually defined, but no knotting can occur. In $\mathbb{R}^{n+3}$, we can use the extra dimension and unknot every $n$-sphere. So, for $n=2$, we have that a ...

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 ...

10

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 ...

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

Due to the theory of quaternions, due to Hamilton, $\bf R^4$ has a structure of a of non commutative field. The only dimension for which $\bf R^n$ is a field are $n=1,2, 4$ . As an application, the special orthogonal group in dimension 4 is not simple : it is the quotient of $U\times U$ by it center $Z/2Z$ where $U$ is the unitary group in complex dimension ...

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 ...

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

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. ...

8

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 ...

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

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 ...

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 ...

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 ...

7

Let $M$ be a closed $n$-manifold, $n>2$. Suppose $S = S^{n-1} \hookrightarrow M$ is a (locally flat) embedded, null-homotopic sphere. Then because it is null-homologous it must separate $M$ into two pieces, so that $M = M_1 \cup_S M_2$. To prove this, if $S$ is smoothly embedded, one can note that if $S$ did not separate $M$, picking points that locally ...

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 $$\pi_1( \mathbb{R}^3 \backslash A) = ... 7 A right-angled hexagon is simply a hexagon in the hyperbolic plane with hyperbolic lines (geodesics) as its edges, and interior angles all right angles: These have lesser total interior angle (and greater total exterior angle) than you'd get in a euclidean hexagon, but that's fine in the hyperbolic plane. There exists a unique one of these given three ... 7 Here is a Kirby Diagram for an exotic \mathbb R^4, taken from Gompf and Stipsicz's book, "4-Manifolds and Kirby Calculus." It's not given in the form of an atlas, but it is a nice explicit description. 6 The key theorem is the following. The braid group B_n is isomorphic to the mapping class group$$ M_n(D^2) \;=\; \mathrm{MCG}\bigl(D^2 - \{p_1,\ldots,p_n\},\partial D^2\bigr) $$of a disk with n punctures rel the boundary. Here "rel the boundary" means that all homeomorphisms and isotopies must fix the boundary of the disk pointwise. The idea ... 6 Only for n=4 does there exist an open set U\subseteq\mathbb{R}^n that is homeomorphic to \mathbb{R}^n but not diffeomorphic to \mathbb{R}^n (a small exotic \mathbb{R}^4). What this means is not too difficult to explain (no need to explain what a manifold is, only what a homeomorphism and a diffeomorphism are between open subsets of ... 6 The real question is how you view the space. You must separate these ideas as you think about the question of what a space is: parametric viewpoint: t \rightarrow (x(t),y(t)) is a curve. For each value of t we obtain a point on the curve C. implicit viewpoint: F(x,y)=k. The curve is the set of all (x,y) which solve the equation F(x,y)=k. Let ... 6 A simple closed curve in a surface bounds a disc in that surface if and only if the induced map on fundamental groups is trivial. So if C is an embedded circle in a surface F, I'm saying$$\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 embedded submanifolds $\emptyset \neq S\subset M$ of codimension $0$ (i.e. $dim M-dimS=0$) of a manifold $M$ are exactly the open subsets of $M$ : Lee, Introduction to smooth manifolds Proposition 5.1, page 99. Hence they are not compact if $M$ is connected and not compact, which applies to $M=\mathbb R^n$.

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 ...

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