Why are the following varieties symplectomorphic? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T07:25:37Z http://mathoverflow.net/feeds/question/43131 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/43131/why-are-the-following-varieties-symplectomorphic Why are the following varieties symplectomorphic? Yuhao Huang 2010-10-22T06:12:46Z 2010-10-23T08:58:28Z <p>I saw a statement somewhere that for the Hirzebruch surfaces $F_n:=\mathbb{P}_{\mathbb{P}^1}(\mathcal{O}\oplus\mathcal{O}(n))$, $F_n$ and $F_m$ are symplectormorphic when $m$ and $n$ have the same parity. </p> <p>My question is: Why is this true?</p> <p>I can see that they are diffeomorphic by Freedman's Theorem: computing the intersection pairing on $\text{Pic}(F_n)=H^2(F_n, \mathbb{Z})$, which is an even form when n is even, and an odd form when n is odd.</p> <p>But this result is deep and abstract, is there any easy way to construct a symplectomorphism? Certainly it is not given by polynomial maps, I'm wondering what it will be.</p> http://mathoverflow.net/questions/43131/why-are-the-following-varieties-symplectomorphic/43136#43136 Answer by Bruno Martelli for Why are the following varieties symplectomorphic? Bruno Martelli 2010-10-22T07:00:18Z 2010-10-22T07:00:18Z <p>Freedman's Theorem tells you that they are homeomorphic, not diffeomorphic. However, you can easily see that they are diffeomorphic by some standard argument as follows. Every Hirzebruch surface is a $\mathbb P^1$-bundle over $\mathbb P^1$. Since $\mathbb P^1$ is diffeomorphic to the 2-sphere $S^2$, these are sphere bundles over the sphere.</p> <p>The sphere bundles over the sphere are classified by $\pi_1(SO(3))$, which is isomorphic to $\mathbb Z_2$. This holds because a bundle is necessarily trivial on the north and south emisphere (because they are contractible), so the only thing that matters is how these two trivialization are glued. The gluing is a continuous map from the equator to $SO(3)$. Homotopic gluings yield isomorphic bundles, so only the homotopy class of this map matters. This is an element in $\pi_1(SO(3))$.</p> <p>It is easy to see that the two elements in $\pi_1(SO(3))$ realize bundles with different intersection form: one is even and the other is odd. Therefore the Hirzebruch surfaces whose intersection forms have the same parity are all diffeomorphic.</p> http://mathoverflow.net/questions/43131/why-are-the-following-varieties-symplectomorphic/43137#43137 Answer by Martin O for Why are the following varieties symplectomorphic? Martin O 2010-10-22T07:14:15Z 2010-10-22T07:14:15Z <p>In Hirzebruch's original paper (F. Hirzebruch, Über eine Klasse von einfachzusammenhängenden komplexen Mannigfaltigkeiten, Math. Ann. 124 (1951), 77–86.) there are polynomial (or rational) formulas for the diffeomorphisms.</p> http://mathoverflow.net/questions/43131/why-are-the-following-varieties-symplectomorphic/43142#43142 Answer by Francesco Polizzi for Why are the following varieties symplectomorphic? Francesco Polizzi 2010-10-22T07:48:03Z 2010-10-23T08:35:55Z <p>In order to obtain an explicit description of the diffeomorphism, one can use the following argument.</p> <p>Take $B=\mathbb{C}^{n-1}$, with coordinates $t_1, \ldots, t_{n-1}$, and consider the complex space $\mathcal{X}$ obtained glueing $\mathbb{P}^1 \times \mathbb{C} \times B$ with $\mathbb{P}^1 \times \mathbb{C} \times B$ by the identification of </p> <p>$(y_0, y_1, z, t_1, \ldots, t_{n-1})$ with $(y_0', y_1', z',t_1, \ldots, t_{n-1})$</p> <p>if </p> <p>$z'=z^{-1}, \quad y_1'=y_1z^{-n}, \quad y_0'=y_0+y_1 \sum_{i=1}^{n-1}t_iz^{-i}$.</p> <p>Let us denote by $\pi \colon \mathcal{X} \to B$ the family obtained in this way.</p> <p>Let now $T_k \subset B$ be the determinantal locus given by rank $M \leq k$, where $M$ is the matrix</p> <p>\begin{bmatrix} t_1 &amp; \ldots &amp; t_{k+1} \cr t_2 &amp; \ldots &amp; t_{k+2} \cr \cdot &amp; \cdot &amp; \cdot \cr \cdot &amp; \cdot &amp; \cdot \cr \cdot &amp; \cdot &amp; \cdot \cr t_{n-k-1} &amp; \ldots &amp; t_{n-1} \end{bmatrix} </p> <p>Then, if $t \in T_k - T_{k-1}$ we have $X_t :=\pi^{-1}(t) \cong \mathbb{F}_{n-2k}$.</p> <p>By using Ehresmann theorem, one concludes that </p> <p>$F_n$ is diffeomorphic to $F_{n-2k}$.</p> <p>Geometrically speaking, we are considering all the rank $2$ vector bundles $V$ which fit into the short exact sequence</p> <p>$0 \to \mathcal{O}_{\mathbb{P}^1} \to V_n \to \mathcal{O}_{\mathbb{P}^1}(n) \to 0$.</p> <p>They are classified by $H^1(\mathbb{P}^1, \mathcal{O}(-n)) \cong \mathbb{C}^{n-1}$, and we consider the family of ruled surfaces $\mathbb{P}(V_n)$, thus obtained, as a deformation of $\mathbb{F}_n = \mathbb{P}(\mathcal{O} \oplus \mathcal{O}(n))$.</p> <p>This argument shows that even Hirzebruch surfaces are diffeomorphic to $S^2 \times S^2$, whereas odd Hirzebruch surfaces are diffeomorphic to $\mathbb{CP}^2 \sharp \overline{\mathbb{CP}^2}$.</p> <p>The uniqueness of the symplectic structrure in each case is a more difficult story, and was proven by Lalonde and McDuff [J-curves and the classification of rational and ruled symplectic 4-manifolds, Contact and Symplectic Geometry (Cambridge, 1994)].</p> http://mathoverflow.net/questions/43131/why-are-the-following-varieties-symplectomorphic/43154#43154 Answer by David Speyer for Why are the following varieties symplectomorphic? David Speyer 2010-10-22T09:36:49Z 2010-10-23T08:58:28Z <p>Let $n>0$. It is enough to show that the vector bundles $\mathcal{O} \oplus \mathcal{O}(n)$ and $\mathcal{O}(1) \oplus \mathcal{O}(n-1)$ are isomorphic as complex vector bundles with smooth transition maps. (They are NOT isomorphic as complex vector bundles with holomorphic transition maps.) Once we show that, we get that $\mathbb{P}(\mathcal{O} \oplus \mathcal{O}(n))$ is diffeomorphic to $\mathbb{P}(\mathcal{O}(1) \oplus \mathcal{O}(n-1))$, and we have $\mathcal{O}(1) \oplus \mathcal{O}(n-1) \cong \mathcal{O}(1) \otimes ( \mathcal{O} \oplus \mathcal{O}(n-2))$, so $\mathbb{P}(\mathcal{O}(1) \oplus \mathcal{O}(n-1)) \cong \mathbb{P}(\mathcal{O} \oplus \mathcal{O}(n-2))$.</p> <p>We have a short exact sequence <code>$$0 \to \mathcal{O} \to \mathcal{O}(1) \oplus \mathcal{O}(n-1) \to \mathcal{O}(n) \to 0$$</code> where the maps are given by <code>$\left( \begin{smallmatrix} x \\ y^{n-1} \end{smallmatrix} \right)$</code> and <code>$\left( \begin{smallmatrix} -y^{n-1} &amp; x \end{smallmatrix} \right)$</code>.</p> <p>Now, in the category of smooth vector bundles, every short exact sequence $0 \to A \to B \to C \to 0$ splits. The proof is as follows: use a partition of unity argument to put a positive definite Hermitian structure on $A$, $B$ and $C$. Then the adjoint of the map $B \to C$ provides a splitting. </p> <p>So $\mathcal{O}(1) \oplus \mathcal{O}(n-1) \cong \mathcal{O} \oplus \mathcal{O}(n)$ and we are done.</p> http://mathoverflow.net/questions/43131/why-are-the-following-varieties-symplectomorphic/43264#43264 Answer by Allen Knutson for Why are the following varieties symplectomorphic? Allen Knutson 2010-10-23T04:55:10Z 2010-10-23T04:55:10Z <p>Let $F\to E\to B$ be a smooth proper fibration over a connected manifold $B$, where the total space is symplectic and the fibers are smooth and symplectic. Then one can use Moser's theorem to generate symplectomorphisms of the fibers.</p> <p>More specifically, let $E_+\to B_+$ be a complex, projective map with $E_+$ K\"ahler. Then we can restrict to the regular values $B$ in $B_+$, and get the fibers to be smooth; since they are complex they will be symplectic. Even if you don't figure out what $B$ is, you can still say that any two smooth fibers are symplectomorphic, because $B$ will be connected.</p> <p>It remains to write down a family which has both $F_m$ and $F_n$ as fibers. I will not do this in detail. One can think of $F_m$ as the toric variety associated to a quadrilateral in ${\mathbb Z}^2$ with two vertices on $y=0$ and two on $y=1$, such that the difference in the lengths of the rows is $m$. By moving some dots from one row to the other, one can get the quadrilateral for $F_n$, iff $m\equiv n \bmod 2$. Then one can write down quadratic relations with a parameter $t$, such that at $t=0$ they define the toric variety $F_m$, and at $t=1$ they define $F_n$. </p> <p>(Actually, since one can set it up so the dots move only vertically, one can get a circle to act on each fiber of the family, showing that $F_m$ and $F_n$ are $S^1$-equivariantly symplectomorphic.)</p> <p>Example: $$F_0 = Spec\ k[{\atop d} {a\atop e} {b\atop f} {c\atop }],\qquad F_2 = Spec\ k[{\atop d} {a\atop e} {b\atop f} {\atop c}]$$ and consider the equations $$e^2 = df,\quad af=be,\quad ae=bd,$$ $$cd = (t)eb + (1-t)ef,\quad ca = (t)b^2 + (1-t)bf,\quad ce=(t)bf + (1-t)f^2$$</p> <p>At $t=1$ we get $F_0$, at $t=0$ we get $F_2$, as hoped.</p> http://mathoverflow.net/questions/43131/why-are-the-following-varieties-symplectomorphic/43273#43273 Answer by Diego Matessi for Why are the following varieties symplectomorphic? Diego Matessi 2010-10-23T08:54:39Z 2010-10-23T08:54:39Z <p>I think that $F(n)$ with a Kahler form such that the base $\mathbb{P}^1$ has area 2 and the fibre has area 1, is symplectomorphic to $F(n+2)$ with the Kahler form such that base and fibre have area 1. </p> <p>There is a "visual proof" of this which uses an idea of Leung and Symington (in an article called "Almost toric manifolds", on the arxiv). The operation is called "nodal trade", and is pictured below. </p> <p>Figure A) is the image of the moment map of the toric variety $F(n)$. You can perturb the moment map to another Lagrangian fibration, so to replace a "corner" with a singular Lagrangian fibre (a pinched torus) over the red cross in figure B). Then you can slide this fibre along the dashed straight line until it hits the opposite edge and becomes another "corner". This gives figure C) which is the polytope of $F(n+2)$. </p> <p>Leung and Symington explain these operations carefully in the above paper.</p> <p><img src="http://www.freeimagehosting.net/uploads/a5a7b4ad8e.jpg" alt="alt text"></p>