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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 ...


9

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


9

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 ...


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 ...


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

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

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

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. ...


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

Milnor's book on characteristic classes is almost surely exactly what you want.


6

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.


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

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 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 ...


5

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, ...


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 ...


5

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 ...


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

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 ...


5

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 ...


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

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

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

I think Alexander Duality is what you are looking for. I gather that you are a non-expert, so I will attempt to describe in fairly informal terms how Alexander duality deals with the questions that you are interested in. Consequently, I'll suppress the inevitable technicalities, since they don't enter into the very geometric situations that you are ...


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

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

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 ...



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