# Symplectic blow-up

Blow-ups of points can also be performed in the symplectic category; for a given point $p\in (X,\omega)$ we choose a Darboux chart around $p$ and then use the symplectic cut corresponding to the standard hamiltonian S^1-action (Diagonal action) to define the blow-up.

It should be possible to define blow-ups of general symplectic submanifolds (which is J-holomrphic with respect to some fixed compatible almost complex structure) in a similar way. Given $X\subset Y$ (a symplectic and J-holomorphic submanifold) we need to put a symplectic structure on a neighborhood of $X$ in $N_X^Y$ (normal bundle) which by symplectic neighborhood theorem is symplectomorphic to a neighborhood of $X \subset Y$ and such that the action of $S^1$ (multiplication by $e^{i\theta}$) is Hamiltonian.

I can't find a reference in the literature where the details are written? Is it any where out there?

## 3 Answers

I don't know a place where this is written up explicitly, but it's not hard as soon as one has an appropriate standard model for a neighborhood of a symplectic manifold for use in the Weinstein Neighborhood theorem.

A model that I've found useful in other contexts (see also p. 12 of this paper) is as follows. Let $X\subset Y$ be a symplectic submanifold of the symplectic manifold $(Y,\omega)$, and let $\pi: E\to X$ denote the symplectic normal bundle to $X$ in $Y$. $E$ is then naturally a symplectic vector bundle, and choosing a compatible fiberwise complex structure $I$ on $E$ induces a Hermitian metric $\langle\cdot,\cdot\rangle$ on $E$. Choose a unitary connection on $E$, inducing a horizontal-vertical splitting $TE=T^h E\oplus T^v E$, with respect to which we may write an arbitrary tangent vector $u\in T_eE$ as $u^h+u^v$ with $u^v$ naturally identified as an element of the fiber $E_{\pi(e)}$. Now define $L:E\to \mathbb{R}$ by $L(e)=\frac{\langle e,e\rangle}{4}$, and define a $1$-form $\theta\in \Omega^1(E)$ by $\theta(u)=dL(Iu^v)$ and then a $2$-from $\Omega$ on $E$ by $$\Omega = \pi^*(\omega|_X) - d\theta$$

There is an obvious identification of $TE|_X$ with $TY|_X$, and under this identification the restrictions of $\Omega$ and $\omega$ are equal. Hence the Weinstein neighborhood theorem we get a symplectomorphism from a neighborhood of $X$ in $(Y,\omega)$ to a disk-bundle in $(E,\Omega)$. Moreover the Hamiltonian $L$ on $E$ generates a circle action which just rotates the $\mathbb{C}^k$ fibers of $E$ in the standard way. Performing the symplectic cut using this circle action replaces an appropriate-radius disk-normal bundle to $X$ with the quotient of the sphere normal bundle by the Hopf map (i.e. with the complex projectivization of the normal bundle), thus seeming to give what you want. Evidently this reduces to the standard construction of the blowup when $X$ is a point.

• Tanks, your definition of \theta looks completely natural, but: 1- I think your \theta is missing a factor of L behind (i.e. it should be LdL(Iv). 2- Forgetting that, how do you show that along X they coincide? Did you actually check that or you guess they should. – Mohammad Farajzadeh-Tehrani Jun 14 '13 at 14:17
• Lets X be a point. Then your formula says $\Omega = - d\theta$ (where $\theta= dL \circ J$ or $\theta= L dL \circ J$ if my point is correct) and you want this to be $\omega$ at that point. But a simple calculation shows your $\theta$ is $d\theta_1+\cdots+d\theta_n$ where $r_1,\theta_1,\cdots,r_n,\theta_n$ are polar coordinates on $C^n$. – Mohammad Farajzadeh-Tehrani Jun 14 '13 at 14:50
• I guess some sort of trace formula should be involved in the definition of $\theta$. – Mohammad Farajzadeh-Tehrani Jun 14 '13 at 14:52
• If X is a point, so E is the standard $R^{2n}$, then $L=\sum (x_{i}^{2}+y_{i}^{2})/4$, $dL = \sum(x_i dx_i+y_i dy_i)/2$, and so $dL\circ J = \sum(y_i dx_i-x_idy_i)/2$, which is the negative of a primitive for the standard symplectic form. So by my calculation $\theta$ is $-\frac{1}{2}\sum r_{i}^{2}d\theta_i$. – Mike Usher Jun 14 '13 at 16:47
• More generally, if you work in a unitary trivialization you can check that $-d\theta$ restricts to each fiber as the standard symplectic form, and that, on $TE|_X = TX\oplus E$, all elements of the $TM$ summand include trivially into $d\theta$ (it helps somewhat that the connection is unitary). So $\Omega$ indeed does coincide with $\omega$ along $X$. – Mike Usher Jun 14 '13 at 16:49

I believe that you will find a careful discussion of symplectic blow-up of symplectic submanifolds in D. McDuff, Examples of simply-connected symplectic non-Kählerian manifolds, J. Differential Geom. 20 (1984), 267–277.

• Thanks, I think you had mentioned this paper in another post of mine but I could not find it through my old posts. – Mohammad Farajzadeh-Tehrani Jun 13 '13 at 17:56
• Now, I remember. Introduction of blow-up as a symplectic cut was after this paper of McDuff; so she uses some sort of cut and paste argument. Do you know of any reference where it uses the new language of symplectic cut. – Mohammad Farajzadeh-Tehrani Jun 13 '13 at 18:57
• @Mohammad: You might try looking at Eugene Lerman, Symplectic cuts, Mathematical Research Letters 2 (1995), 247–258. It has been a while since I looked at that, but I think that maybe it is there. Certainly, the Wikipedia page "Symplectic cut" seems to suggest that this is so. I don't have time to go look it up myself right now. – Robert Bryant Jun 13 '13 at 21:12
• that does say anything about the general situation. – Mohammad Farajzadeh-Tehrani Jun 14 '13 at 0:08

I think "Birational equivalence in the symplectic category" - Invent.Math (by V.Guillemin and S.Sternberg) covers all details that you want. (Section 8).