Analogue of Infinitesimal Schwarzian for holomorphic $(G,X)$-manifolds

If $\Sigma$ is a closed Riemann surface, a complex projective structure $Z$ on $\Sigma$ is an atlas of charts, refining the holomorphic atlas, with values in $\mathbb{CP}^{1}$ such that the transition functions are locally projective transformations in ${PSL}(2,\mathbb{C})$ acting by Mobius transformations.

Associated to such a structure is a short exact sequence of sheaves \begin{align} 0\rightarrow \mathcal{P}_{Z}\rightarrow \Theta_{\Sigma}\rightarrow K_{\Sigma}^{2}\rightarrow 0 \end{align} where $\mathcal{P}_{Z}$ is the sheaf of projective vector fields (i.e. those vector fields which in a local chart for the projective structure are identified with an infinitesimal Mobius transformation), $\Theta_{\Sigma}$ is the holomorphic tangent sheaf of $\Sigma,$ and $K_{\Sigma}^{2}$ is the tensor square of the canonical sheaf. The first map is given by inclusion. In a projective coordinate, the second map takes a local holomorphic vector field $v(z)\frac{\partial}{\partial z}$ to $v'''(z) dz^{2}.$ Since projective vector fields, expressed in a projective coordinate, are exactly the quadratic vector fields, the above sequence is left exact and it's not too hard to see that the latter map is surjective.

Of course, we know there is some exact sequence of the kind above without ever identifying what the quotient sheaf and quotient map are explicitly: the key here is that we can explicitly identify the quotient sheaf associated to the injection $0\rightarrow \mathcal{P}_{Z}\rightarrow \Theta_{\Sigma}$ with the square of the canonical bundle, a fact which underlies much of the structure of the space of complex projective structures on $\Sigma.$ For those that are somewhat familiar with this story, one can concisely assert that the above described map is the infinitesimal Schwarzian derivative, which explains the title of the question.

I am curious is there is a uniform way to extend this picture to holomorphic $(G,X)$-structures on a given compact complex manifold $M.$ To recall, let $G$ be a complex Lie group acting transitively on a complex manifold $X.$ If $M$ is a complex manifold, a holomorphic $(G,X)$-structure on $M$ is an atlas of charts, refining the holomorphic atlas, with values in $X$ and whose transition functions are locally in $G.$ Now, let $W$ be a holomorphic $(G,X)$-manifold with underlying complex manifold $M.$ If $\mathcal{G}_{W}$ denotes the sheaf of $G$-vector fields, that is, the sheaf whose sections are vector fields which in a local $(G,X)$-chart are identified with the infinitesimal $G$-action on $X,$ then there is a short exact sequence \begin{align} 0\rightarrow \mathcal{G}_{W}\rightarrow \Theta_{M}\rightarrow Q\rightarrow 0 \end{align} where $Q$ is the corresponding quotient sheaf.

$\textbf{My question is the following}:$ As with the case of projective structures on a Riemann surface, can one explicitly identify the quotient sheaf and the quotient map? For instance, is it a locally free sheaf?

The kind of thing I have in mind is the following: I want there to be a differential operator acting on the Pseudo-group of locally defined biholomorphisms of $X$ whose Kernel is precisely $G,$ maybe there is some issue with $G$ not acting effectively but let's ignore that for the moment. The linearization of this operator should be an operator on holomorphic vector fields on $X$ whose kernel consists of the vector fields given by the infinitesimal $G$-action. Since the transition functions of the $(G,X)$-structure are given by elements of $G,$ this data should port over to the $(G,X)$-manifold $W$ and produce a sheaf map from $\Theta_{W}$ to some other sheaf which seems like it should be some kind of jet bundle. If one is familiar with Schwarzian derivative, this is exactly what is happening in the example of complex projective structures. I've found some things in the literature that are in this spirit, but they mostly seem quite ad-hoc, and I am really trying to understand if there is some uniform way to understand this question using the geometry of $X$ as a $G$-homogeneous space.

• I am pretty sure that $Q$ depends on the choice of $(G,X)$-structure for almost every other example. For example, for a flat contact projective structure, I think you can recover the contact structure from $Q$. (That is probably the simplest case to try.) Perhaps the point is that projective spaces are the only complex homogeneous spaces with transitive action of the stabilizer $P \subset G$ on the nonzero elements of $\mathfrak{g}/\mathfrak{p}$. May 25, 2017 at 12:05
• I had an inclination that this is probably the case, I am/was hoping there was at least a rough algorithm to follow that would solve the problem up to Lie theoretic issues concerning the given homogeneous space. I had not thought of your final comment before, thank you for this insight. May 25, 2017 at 12:10
• There is a nice expression for the differential operator if you allow yourself to write it on the total space of the Cartan geometry, but I can't find a good reference right away. May 25, 2017 at 12:13
• That's precisely the kind of thing I had in mind Ben, at the end of the day I should just bite the bullet and do it myself, but I was hoping there might be something hiding in the literature which would save me trouble and perhaps give some insights I have missed. May 25, 2017 at 12:15

Take a (holomorphic) Cartan connection $E \to M$ modelled on a (complex) homogeneous space $(X,G)$, say $X=G/H$. Let $\omega$ be the Cartan connection on $E$. Every (holomorphic) vector field $v$ on $M$ is represented by a unique (holomorphic) $H$-equivariant function $f \colon E \to \mathfrak{g}/\mathfrak{h}$. Compute $df + \omega f = f' \omega$ for a unique $f' \colon E \to \mathfrak{g}/\mathfrak{h} \otimes (\mathfrak{g}/\mathfrak{h})^*$. The differential equation we need to have $v$ be an infinitesimal symmetry is that $f'$ is valued in $\mathfrak{g}$, acting as a linear map on $\mathfrak{g}/\mathfrak{h}$. So $Q^0=E \times^H W$ where $W=(\operatorname{End} \mathfrak{g}/\mathfrak{h})/\mathfrak{g}$.

But then we need higher order conditions. Define $\mathfrak{g}^{(-1)}=\mathfrak{g}/\mathfrak{h}$. Define $\mathfrak{g}^{(0)}=\mathfrak{g}$. Define the prolongations $\mathfrak{g}^{(k)}$ to be the constant coefficient tensors $\phi \in \operatorname{Sym}^{k+1} (\mathfrak{g}/\mathfrak{h})^* \otimes \mathfrak{g}/\mathfrak{h}$ for which $\phi(v,*,\dots,*) \in \mathfrak{g}^{(k-1)}$. Define $f^{(-1)}=f$. Define $f^{(k)} \in \operatorname{End}(\mathfrak{g}/\mathfrak{h})^{(k)}$ by $df^{(k)}+\omega f^{(k)} = f^{(k+1)} \omega$. Define $Q^{(k)}=\operatorname{End}(\mathfrak{g}/\mathfrak{h})^{(k)}/\mathfrak{g}^{(k)}$. Then I think we get the differential operator $f \mapsto f^{(k)} + \mathfrak{g}^{(k)} \in Q^{(k)}$. For a flat $(X,G)$-geometry, these operators all vanish just when a vector field lies in the Lie algebra $\mathfrak{g}$, i.e. a local symmetry vector field. But what a mess. My choice of $k$ is maybe not what it should be, because it is one less than the order of the operator, but that is probably not unusual in the literature.

To add to what Ben said: Typically the resolution of the sheaf mapping ${\frak{g}}\to\Gamma(TX)$ is the Spencer resolution, which is a canonical $G$-equivariant exact sequence of differential operators $D_i:\Gamma(V_i)\to \Gamma(V_{i+1})$ for $0\le i< n = \dim X$, where $V_0 = TX$ and $\mathrm{ker}(D_0) = {\frak{g}}$. When you have a flat $(G,X)$-geometry on a manifold $M$, this sequence pulls back (under the local charts) to define the corresponding sequence on $M$.

For example, when $G$ is the group of isometries of a (simply-connected) $3$-dimensional Riemannian manifold $X$ of constant curvature, one finds that the ranks $r_i$ of the bundles $V_i$ over $X$ are $$(r_0,r_1,r_2,r_3) = (3,6,6,3)$$ and that the order of the differential operators $D_0$ and $D_2$ is $1$ while the order of the differential operator $D_1$ is 2. When $G$ is the group of conformal transformations of compactified flat $3$-space~$X = S^3 = \mathbb{R}^3\cup\{\infty\}$, then the ranks are $$(r_0,r_1,r_2,r_3) = (3,5,5,3)$$ and the order of the differential operators $D_0$ and $D_2$ is $1$ while the order of the differential operator $D_1$ is 3.

As another explicit example, if you take $G$ acting on $X = G$ by left-translation, then the Lie algebra of symmetries is the right-invariant vector fields on $G$, and there is a unique (flat) connection $\nabla$ on $TG$ whose parallel sections are the right-invariant vector fields. In this case, $V_i = TG\otimes \Lambda^i(T^*G)$ and $D_i$ is just the $\nabla$-twisted exterior derivative $\mathrm{d}^\nabla:TG\otimes \Lambda^i(T^*G)\to TG\otimes \Lambda^{i+1}(T^*G)$, so $r_i = n{n\choose i}$ and the order of $D_i$ is $1$ for all $i$.

Look in any book that treats the Spencer resolution or the topic of Lie equations for references. (I'm traveling, so I don't have access to the literature.) One general purpose reference that treats this topic from a much more general point of view (and contains the important references to the earlier literature from the 50s, 60s, and 70s) is Chapter X of the book Exterior Differential Systems by Bryant, Chern, Gardner, Goldschmidt, and Griffiths (Mathematical Sciences Research Institute Publications, volume 18, 1991). It has long been out of print, but you can download a copy of the whole book from http://library.msri.org/books/Book18/MSRI-v18-Bryant-Chern-et-al.pdf. [Chapters IX and X can be read independently from the first eight chapters.] (Caution: There are many misprints, but there was never a second edition published, so the reader has to be somewhat careful.)

• Robert, thank you for this elucidating information; I had a suspicion that the "right" resolution would not only consist of three terms. If only to prompt your memory, my literature search has not turned up much useful information yet regarding the Spencer resolution or Lie equations, so if you do have a reference in mind, I would appreciate if you would share it when you have time. It's a little surprising to me that this topic is buried so deeply, since the theory of projective structures on surfaces is so classical and central, though this wouldn't be the first time I was surprised. Jun 7, 2017 at 20:22
• @AndySanders: I've added a reference, which contains further references to the literature. Jun 8, 2017 at 7:46