# Tag Info

24

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 ...

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 ...

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 ...

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 ...

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

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

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

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

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 ...

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

Edit: The answer is still negative. What you have to use is the Lusternik-Shnirelmann category $cat(X)$: Definition. $cat(X)$ for a topological space $X$ is the least number of contractible open sets needed to cover $X$. It is known that $cat(T^n)=n+1$, see here. Thus, you cannot cover 2-torus with two simply-connected open sets (since such sets are ...

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 $S^1 \to ... 5 For the compact case, I believe the answer is, as you said in the comments above, any closed surface with a single puncture, i.e. a disk removed. I claim that this completely classifies compact surfaces with boundary$S^1$. This is because you can glue a disk to the surface along its boundary to obtain a closed surface, and there is a unique way to do this ... 5 A simple closed curve in a surface bounds a disc in that surface if and only if the induced map on fundamental groups is trivial. So if$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 ... 4 In general it's impossible to keep$H_t(f(S^1))=f(S^1)$. This is because the image of$S^1$under a$C^1$embedding is only a$C^1$manifold in general, while the image under a$C^\infty$embedding is a$C^\infty$manifold. For a concrete example, the curve$y=x\sqrt{|x|}$in the$xy$-plane is the image of$\mathbb R$under the$C^1$-embedding$x\mapsto ...

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

Maybe the best-looking example of this is the Koch snowflake: The iteration does indeed go on forever, but there is no limit to the length of the curve! If you look carefully, the snowflake's perimeter increases by a factor of $\frac{4}{3}$ each iteration, so it tends to infinity. Don't think of the size of the measuring stick. Think instead of errors in ...

4

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 ...

4

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 ...

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

A relevant theorem is the Hales-Jewett theorem. Basically it states that the higher-dimensional, multi-player, $l$-in-a-row generalization of game of tic-tac-toe cannot end in a draw, no matter how large $l$ is, no matter how many people $k$ are playing, and no matter which player plays each turn, provided only that it is played on a board of sufficiently ...

4

The answer is yes. Decomposing $3$-manifolds using compression bodies instead of handlebodies results in "generalized Heegaard splittings." If you follow the link to the compression body Wikipedia page, you'll see two dual definitions of "compression body." For the definition of generalized Heegaard splitting, I'm going to use this: a compression body ...

3

The concept of a characteristic element is purely algebraic. Let $G$ be a finitely generated free abelian group and let $$Q: G \times G \longrightarrow \Bbb Z$$ be a symmetric bilinear form on $A$. Then $x \in G$ is a characteristic element of $G$ for $Q$ if $$Q(x, \alpha) \equiv Q(\alpha, \alpha) \pmod 2 \text{ for all } \alpha \in G.$$ So to define ...

3

Part I Non-parametric discussion of thinness. each point in the interior is very close to a point on the boundary. This says precisely that the inradius is small. The inradius (for inscribed radius) is the maximal radius of an open disk enclosed by the curve. For example, the rectangle of dimensions $a,b$ has inradius $\frac12 \min(a,b)$. It remains ...

3

The direct analogue of Kirby diagrams exist of course but they tend to be things people have difficulty visualizing. For example, attaching a 3-handle to a 5-ball amounts to embedding an $S^2$ in $S^4$. But embeddings of $S^2$ in $S^4$ are things that don't have easy diagramatic representations. There is Kamada's "Braids and knot theory in dimension 4", ...

3

If the conjecture is what I think it is (i.e. knots are determined by their compliments), then not, at least when knots are considered up to isotopy. A solid torus is an exterior of the unknot, so take a Whitehead link and twist around one of the components. The exterior of this link stays the same, and it is the exterior of the second component in the solid ...

3

The condition that $i_*$ is trivial is neither necessary nor sufficient to to guarantee that $i(\mathbb{T}^2)$ bounds a solid torus. To show that it is not necessary, consider the manifold $M = S^2 \times S^2$. If $E$ is the equator of $S^2$, then $E\times S^1$ is a solid torus in $M$. However, $\pi_1(M) = \mathbb{Z}$, and the homomorphism $i_*\colon ... 2 I think I can prove: For any pair of closed orientable surfaces$N_1$and$N_2$, there is a smooth function$f:S^3\rightarrow S^1$with$f^{-1}(-1) \cong N_1$and$f^{-1}(1) \cong N_2$Here, I'm thinking of$S^1\subseteq\mathbb{C}\$ as the unit complex numbers. The proof will occur via a couple of lemmas. Lemma 1: For any two closed orientable ...

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