Minimal genus, adjunction inequality - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T19:23:47Z http://mathoverflow.net/feeds/question/83278 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/83278/minimal-genus-adjunction-inequality Minimal genus, adjunction inequality Nikita Kalinin 2011-12-12T20:25:30Z 2011-12-15T20:53:18Z <p>Let's consider closed simply-connected 4-manifold $M$ and some $a\in H^2(M)$. It is very natural question to estimate minimal $g$ that $a$ can be presented as embedded surface of genus $g$.</p> <p>As I know there is the adjunction inequality for estimation of minimal genus via Seiberg-Witten theory.</p> <p>Question 1: Does there exist other methods to estimate minimal genus ?</p> <p>I heard that there are homeomorphic but not diffeomorphic 4-manifolds $M,N$ such that for some $a\in H^2$, $a$ has different minimal genus in $M,N$.</p> <p>Question 2: Could you give me such examples? As I understand it should be some manipulations with Seiberg-Witten invariants...</p> http://mathoverflow.net/questions/83278/minimal-genus-adjunction-inequality/83292#83292 Answer by Andrey Gogolev for Minimal genus, adjunction inequality Andrey Gogolev 2011-12-12T22:06:37Z 2011-12-13T03:21:10Z <p>This is regarding question 1. There is a much earlier approach through Atiyah-Singer G-signature theorem that works for certain divisible classes. </p> <p>So if $g$ is the genus of an embedded surface $S$ representing $a$ then (under certain assumptions) one gets the following inequality $$\beta_2+2g\ge\left|\frac{1}{2}a\cdot a-\sigma(M)\right|$$ where $\sigma(M)$ is the signature of the intersection form on the second homology and $\beta_2$ is the second Betti number. (I am not sure about the coefficient 1/2 on the right, it is more complicated actually, but I think morally it's ok.)</p> <p>This inequality applies to even classes $2nx$ in $\mathbb{CP}^2$ and $2nx+2my$ in $S^2\times S^2$ resulting in $$g\ge n^2-1,\;\;\; and\;\; g\ge 2nm-1$$ respectively. (Also it applies to other divisible classes.)</p> <p>Assume $a=kb$, where $k$ is an integer. One has to look at $k$-fold cover $\tilde M$ that branches over $S$ and apply the G-signature theorem to this cover. This gives a formula for $\sigma(\tilde M)$ $$\sigma(\tilde M)=k\sigma(M)-\frac{(k^2-1)a\cdot a}{3k}.$$ Then the final estimate comes from comparing Euler characteristics, signatures and second Betti numbers. </p> <p>I think the proof can be found in Rohlin's "Two dimensional submanifolds of 4 dimensional manifolds" or in Hsiang, Szczarba, "On embedding surfaces in four-manifolds".</p> <p>It is interesting that Rohlin remarks that he had the above display formula from "a corresponding version of cobordism theory" without using Atiyah-Singer. </p> http://mathoverflow.net/questions/83278/minimal-genus-adjunction-inequality/83312#83312 Answer by Mike Usher for Minimal genus, adjunction inequality Mike Usher 2011-12-13T04:26:38Z 2011-12-13T04:26:38Z <p>Regarding Question 2, Corollary 2 of <a href="http://www.ams.org/journals/proc/1999-127-02/S0002-9939-99-04457-3/home.html" rel="nofollow">this paper by Li</a> shows that any symplectic four-manifold which contains a smoothly embedded homologically essential sphere with nonnegative intersection is obtained by blowing up either $\mathbb{C}P^2$ or an $S^2$-bundle over a surface some nonnegative number of times. So you could let $M$ be $\mathbb{C}P^2$ # $k \overline{\mathbb{C}P^2}$ for suitable $k$ (by now $k\geq 2$ will do) and $N$ be any of <a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.jdg/1214441784" rel="nofollow">the</a> <a href="http://www.springerlink.com/content/t836j2265775775u/" rel="nofollow">wide</a> <a href="http://front.math.ucdavis.edu/0311.5395" rel="nofollow">variety</a> <a href="http://front.math.ucdavis.edu/0412.5126" rel="nofollow">of</a> <a href="http://front.math.ucdavis.edu/0412.5216" rel="nofollow">examples</a> <a href="http://front.math.ucdavis.edu/0701.5664" rel="nofollow">of</a> <a href="http://front.math.ucdavis.edu/0703.5480" rel="nofollow">symplectic</a> <a href="http://front.math.ucdavis.edu/0702.5211" rel="nofollow">exotic</a> <a href="http://front.math.ucdavis.edu/0703.5065" rel="nofollow">rational</a> <a href="http://front.math.ucdavis.edu/0701.5829" rel="nofollow">surfaces</a>. Then if $a\in H^2(M)\cong H^2(N)$ is Poincare dual to the pullback of the hyperplane class in $\mathbb{C}P^2$, $a$ will have minimal genus zero in $M$ but positive minimal genus in $N$.</p> <p>There are also examples with $b^+>1$, in which case as you say one has the Seiberg-Witten adjunction formula $2g(\Sigma)-2\geq |K\cdot\Sigma|+\Sigma\cdot\Sigma$ for any surface $\Sigma$ of positive genus and nonnegative self-intersection. If you take for $M$ the $K3$ surface, viewed as an elliptic fibration with a section of square $-2$, and let $\Sigma$ be obtained from a fiber and the section by smoothing the intersection between them, then $\Sigma$ has square zero and genus 1 (and $\Sigma$ can't be represented by a sphere); this is consistent with the adjunction formula since the only basic class for the $K3$ surface is the zero class. But there are many exotic smooth structures on the $K3$ surface (for instance the ones <a href="http://front.math.ucdavis.edu/9612.5114" rel="nofollow">here</a>) for which some positive multiple of the fiber class is a basic class, and the minimal genus for the homology class of $\Sigma$ in one of these exotic $K3$ surfaces would be larger than one since there would be a basic class having positive intersection with $\Sigma$.</p> http://mathoverflow.net/questions/83278/minimal-genus-adjunction-inequality/83322#83322 Answer by Jim Bryan for Minimal genus, adjunction inequality Jim Bryan 2011-12-13T07:13:48Z 2011-12-13T07:13:48Z <p>To add a bit to Andrey's answer, you can improve the method of Hsiang and Szczarba (also Kotschick and Matic deserve credit here) a bit using Furuta's approach to Seiberg-Witten theory. The method can give genus bounds on four manifolds where the adjunction formula doesn't apply (for example on <code>$\mathbb{CP}^2 \# \mathbb{CP}^2$</code>). I wrote a short paper on this: Math. Res. Lett. 5 (1998), no. 1-2, 165–183. The arXiv version is here: <a href="http://arxiv.org/pdf/dg-ga/9704010v1" rel="nofollow">http://arxiv.org/pdf/dg-ga/9704010v1</a>. Theorem 1.6 gives a typical genus bound from this method.</p> <p>The basic idea is the following. If your class has certain divisibility properties and it is represented by an embedded surface, then the four manifold obtained by taking a cyclic cover branched along the surface will be a new four manifold with certain properties which can be used to constrain the genus of the embedded surface. Under appropriate hypotheses, the cover will be a spin four manifold with a cyclic group action. Refining Furuta's proof of the 10/8ths theorem for spin four manifolds to include the presence of a cyclic action, one obtains a bound on the size of the signature in terms of the second Betti number which then translates into a genus bound on the original manifold.</p>