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11

These are two very different and very nice questions. The correct answer is "probably". First I should respond to what might be seen as the naive question - "is Floer homology the homology of the manifold we're plugging in? the space of connections? the moduli space of trajectories?" - the answers to all of these are no. The way you would get a space is by ...


8

This follows from Gauss-Bonnet Theorem: If $f$ is the Gaussian curvature of a compact surface $S$ without boundary, then $$\int_S f=2\pi\chi(S)$$ where $\chi(S)$ is the Euler characteristics of $S$. In particular, if $S$ is $T^2$ the torus, we have $\chi(S)=\chi(T^2)=0$. Therefore, it is impossible for $f>0$ everywhere. BTW, for higher dimensional ...


7

This is $\mathbb{R}P^2\#\mathbb{R}P^2\#\mathbb{R}P^2\simeq\mathbb{R}P^2\#T^2\simeq \mathbb{R}P^2\#K$, where as usual $T^2$ is the torus, $\mathbb{R}P^2$ the real projective plane and $K$ the klein bottle. I will only give an intuitive argument for this. Consider the following sketch: To see this represents your manifold, note that this is a Möbius strip (...


6

By the Gauss–Bonnet theorem, the Euler characteristic of $T^2$ is given by $$\chi(T^2) = \frac{1}{2\pi} \int_{T^2} K dA$$ where $K$ is the curvature and $dA$ is the element area of $T^2$. If $K$ were everywhere positive, then this would be a positive number, for the same reason that the integral of a positive function is positive; but $\chi(T^2) = \chi(...


5

They are absolutely not all $B^4$. Given a smooth 4-manifold with boundary $S^3$, there is a unique way to cap that boundary component off with a copy of $B^4$, thus giving us a smooth closed 4-manifold. (Hidden in this statement is Cerf's quite nontrivial theorem that every diffeomorphism of $S^3$ extends over the 4-ball.) If the original manifold was a ...


5

Once the characteristic curves are given on $\partial H_2$, the construction of the 3-manifold is automatic, by the following procedure: For each characteristic curve $c$, attach a 2-handle $D^2 \times [0,1]$ to $H_2$ by identifying $\partial D^2 \times [0,1]$ to a regular neighborhood of $c$. Doing this for each characteristic curve, let $N$ be the ...


4

As a smooth manifold, it's homeomorphic to $S^1 \times (-1, 1)$, with one homeomorphism being $$ (\theta, u) \mapsto (\cos \theta, \sin \theta, \tan (\frac{pi}{2} u)). $$ It's also homotopy-equivalent to $S^1$, via the deformation-retraction $$ H(\theta, t; s) = (\theta, st) $$ from $S^1 \times (-1, 1)$ onto the subset $S^1 \times \{0\}$. The key points ...


4

Grumpy Parsnip answered the question, but let me summarize the situation and add some other aspects of the story. Denote by $Diff^\partial(D^n)$ the group of diffeomorphisms of the $n$-disc that are the identity on a neighborhood of the boundary, by $Diff^+(S^n)$ the group of orientation preserving diffeomorphisms and by $\Theta_n$ the group of exotic ...


4

Look under "Twisted Spheres" in the Wikipedia article: https://en.wikipedia.org/wiki/Exotic_sphere. For n>6, all diffeomorphisms not isotopic to the identity give an exotic sphere. Edit: As Mike Miller points out and the article as well, all exotic spheres are thus obtainable.


3

No, you can take a point in $R^n$.


1

It's also useful to look at braids here. For example, the maximal self-linking number of an amphichiral link equals the negative of the minimal braid index. (Note: The self-linking number of a braid $\beta$ is $sl(\beta)=w(\beta)-n(\beta)$, where $w$ is the writhe of the braid diagram and $n$ is the braid index. Then we get a knot invariant $\overline{sl}(...



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