2
$\begingroup$

Given a possibly singular, connected, symplectic algebraic variety with a torus action, every fiber of the moment map admits a torus action. Is each fiber of this moment map connected? Any examples or counter-examples? Thanks!

$\endgroup$
  • 3
    $\begingroup$ Example: Let $X\subset G/P$ be Schubert variety, then $\Phi:X\to Lie(T)^*$, Let $K^\mathbb C=G$, and $T$ denote a maximal torus in $K$, then fibers of this momentum map are connected subspaces of Schubert variety $X$ see arxiv.org/abs/math/0606474 $\endgroup$ – user21574 Oct 25 '17 at 21:38
  • 3
    $\begingroup$ There are examples such that fibers of moment map may not be connected , for example non-compact symplectic toric manifold. See Atiyah M.F., Convexity and commuting Hamiltonians, Bull. London Math. Soc. 14 (1982), 1–15., and Guillemin V., Sternberg S., Convexity properties of the moment mapping, Invent. Math. 67 (1982), 491–513. $\endgroup$ – user21574 Oct 25 '17 at 21:52
  • 3
    $\begingroup$ Theorem of Kirwan :Let $M$ be a Hamiltonian $G$-manifold, if $M$ is connected and compact then the level sets of moment map are connected. Kirwan, F.: Convexity properties of the moment mapping III. Invent. math. 77 (1984), 547-552 $\endgroup$ – user21574 Oct 25 '17 at 22:01
  • 3
    $\begingroup$ Konp extended Kirwan theorem for Hamiltonian G-varieties, see Knop, Friedrich: A connectedness property of algebraic moment maps. J. Algebra 258 (2002), no. 1, 122–136. $\endgroup$ – user21574 Oct 25 '17 at 22:23
  • 3
    $\begingroup$ Example: For singular symplectic varieties in the sense of Beauville see Theorem 5.3 of numdam.org/article/AMBP_2006__13_2_209_0.pdf and use theorem 2.6, and Theorem 6.3 of arxiv.org/pdf/math/0112144.pdf and also p.8 of theorem of Mostow 1955 mat.ug.edu.pl/kwwk/2010/presentations/imykytyuk.pdf $\endgroup$ – user21574 Oct 25 '17 at 23:12
2
$\begingroup$

In the category of symplectic algebraic varieties moment maps have in general disconnected fibers. Easy example go as follows: Let $T={\bf G}_m$ act on the affine plane ${\bf A}^2$ by $t\cdot(x,y)=(tx,t^{-1}y)$. Then symplectic form $\omega_0=dx\wedge dy$ is $T$-invariant. The corresponding moment map is $m_0(x,y)=xy$.

  1. Let $X_1:={\bf A}^2\setminus\{(0,0)\}$. Then the zero-fiber of $m_0$ becomes disconnected.

  2. A less trivial example is as follows: Let $f(z)$ be an arbitrary non-constant polynomial with derivative $f'(z)$. Now rescale the symplectic form to $\omega=f'(xy)\omega_0$. Then $\omega$ is nondegenerate on the open subset $X_2=\{f'(xy)\ne0\}\subseteq{\bf A}^2$. The corresponding moment map is $m(x,y)=f(xy)$. So, unless $f$ is linear, the generic fibers of $m$ are disconnected.

  3. A similar example can be obtained as follows: Start with any moment map $m:X\to\mathfrak t^*$. It is dominant if the action is effective. Let $f:Y\to\mathfrak t^*$ be any étale morphism. Then $\tilde X:=X\times_{\mathfrak t}Y$ is Hamiltonian where the symplectic form is the pull-back from $X$. The moment map is the composition $\tilde X\to Y\to \mathfrak t^*$. It will have disconnected generic fibers unless $f$ is birational.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Interesting answer! . In 3. why the pullback of symplectic form is a symplectic form, since as far as I know pullback of symplectic form isnot symplectic(for Kahler form, in the sense of current the pullback is highly non-trivial see arxiv.org/abs/math/0606248). Also you mean form or current? since your example is for non-smooth! $\endgroup$ – user21574 Oct 29 '17 at 15:49
  • $\begingroup$ I assumed that $X$ is smooth. Then the pull-back of a symplectic form is clearly symplectic for an étale morphism. $\endgroup$ – Friedrich Knop Oct 29 '17 at 20:56
  • $\begingroup$ Closedness of $(1,1)-$ current in singular setting is more complicated see Theorem 1.26, of thichthichiu.files.wordpress.com/2011/07/… . This is a reason that for symplectic variety , we define symplectic form in regular part!. May you add some additional examples in singular setting $\endgroup$ – user21574 Oct 30 '17 at 15:06
  • $\begingroup$ I have nothing to say in the singular setting. Also, I assumed that the setting is the algebraic category: algebraic varieties and algebraic forms not necessarily over $\mathbb C$. Also if something doesn't work (like connectedness of fibers) for regular varieties it won't work for singular ones, either. $\endgroup$ – Friedrich Knop Oct 30 '17 at 16:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.