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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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