This is a follow-up to some earlier questions about flatness of schematic fixed points of certain deformations. Since I could never come up with good enough hypotheses in those examples, let me try a much more restrictive set of examples.


  • $X$ is a quasi-projective symplectic variety which is projective over its affinization (algebraically convex) with $H^i(X;\mathcal{O}_X)=0$ for $i>0$ over a field $k$ of characteristic 0,
  • $L$ is a line bundle on $X$,
  • there is a Hamiltonian action of a torus $T$ on $X$, and a commuting action of $\mathbb{G}_m$ which is conformal of weight $n$ (the symplectic form is a weight vector of weight $n$ for it) and has positive weight on all global functions.

A theorem of Kaledin tells us that $X$ possesses an essentially unique twistor deformation, that is a flat deformation $\mathcal X\to S$ where $S=\mathrm{Spec} ( k[[t]])$ with an extension of $L$ to a line bundle $\mathcal{L}$ such that

  • $\mathcal X$ is Poisson with the coordinate $t$ being Poisson central.
  • $\mathcal L$ is a Poisson module with $\{t,-\}$ inducing the identity map.
  • The action of $T\times \mathbb{G}_m$ extends to an action of the Lie algebra of this group on $\mathcal X$ lifting the action on $S$ where $T$ acts trivially and $\mathbb{G}_m$ acts with weight 1 on $t$ (here you see why I had to use the Lie algebras instead of the algebraic groups).
  • The global functions $A=\Gamma(\mathcal X)$ are also flat over $k[[t]]$, $t$-adically complete and normal.

If you're having trouble imagining what this looks like, the point is to "invert Hamiltonian reduction." The total space of $\mathcal{L}$ minus its 0-section is symplectic, and its induced $\mathbb{G}_m$-action (by scalar multiplication; this has nothing to do with the one we introduced before) is Hamiltonian with moment map $t$. The variety $X$ is the Hamiltonian reduction for this action.

We can take schematic fixed points $B$ of the Lie algebra $\mathfrak t$ acting on $A$; this is the quotient of $A$ by the ideal generated by $y\cdot a-a$ for $y\in \mathfrak t, a\in A$. This is an algebra over $k[[t]]$.

Is $B$ flat over $k[[t]]$?

  • $\begingroup$ This thing with the bounty doesn't seem to work that well, does it? $\endgroup$ – André Henriques Feb 4 '11 at 17:06
  • $\begingroup$ Nope, apparently I ask questions that are too hard. It was worth a shot, right? $\endgroup$ – Ben Webster Feb 4 '11 at 19:00

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