Hot answers tagged

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 ...


19

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 ...


16

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.


16

If by a "circle" you mean the set of all points inside a circle (e.g., points whose distance from some center $C$ is less than or equal to 1), then the answer is "yes" and one solution is called "stereographic projection;" another is "vertical projection". If you have a point $(x, y)$ in the unit disk (the "filled in circle"), the corresponding point, ...


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 can'...


12

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 ...


12

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 $2$-...


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

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

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

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 ...


10

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 ...


9

If $M$ is a connected closed (i.e. compact without boundary) 3-dimensional manifold, then $\pi_1(M)$ cannot be isomorphic to $\pi_1(S)$, where $S$ is an orientable surface of genus $\ge 2$. You can see this by first noting that $\pi_2(M)=0$ (otherwise, by the sphere theorem, $M$ is a nontrivial connected sum which will imply that $\pi_1(M)$ is a nontrivial ...


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

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 $\mathbb{Z}^...


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

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

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 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 $S^...


8

I think you mean to say disk instead of circle. If you mean disk then you can do it. Just imagine of punching the disk from center, it will give you a hemisphere. And if you want precisely the map then you can get it from Stereographic projection. EDIT It will look something like this when you punch a disk.


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.


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 $n$-...


7

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 ...


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

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

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) = \bigcup\...


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

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

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 $\mathbb{R}^n$)....


7

Yes. Deleting a point from a manifold of dimension $n \geq 3$ doesn't change the fundamental group, so the result is still simply connected. (Either use van Kampen or transversality.) Mayer-Vietoris shows that the homology of the resulting space is the same as $S^2$, whence $\pi_2(U \setminus p)$ is isomorphic to $H_2(U \setminus p) = \Bbb Z$, generated by a ...



Only top voted, non community-wiki answers of a minimum length are eligible