3
$\begingroup$

In studying the transformation groups generated by holomorphic vector fields V(z) d/dz on ℂ, I've noticed the (surely well-known) fact that the complex quadratic vector fields:

            {(a z2 + b z + c) d/dz  |  (a,b,c) ∊ ℂ3}

form precisely the Lie algebra whose nonzero elements generate the linear fractional transformations, i.e., PSL(2,ℂ).

  • Is there some underlying reason for this? (Beyond direct calculation, which provides an easy proof.)

Other than the further Lie algebras {0}, {a d/dz}, and {(a z + b) d/dz} over ℂ of trivial, constant, and linear (affine) vector fields, there seem to be no other polynomial finite-dimensional Lie algebras of vector fields on ℂ.

  • Is there some easy-to-explain reason for this?

The next "simplest" such polynomial Lie algebras of vector fields on C seem to be those defined by all polynomial and all rational functions:

(*)        VP := {P(z) d/dz  |  P(z) ∊ C[z]}   and   VR := {R(z) d/dz  |  R(z) ∊ C(z)}.

  • Contrariwise, do there exist finite-dimensional Lie algebras of vector fields on ℂ defined by rational functions -- other than the polynomial ones mentioned above?

  • In case VP and/or VR generate well-studied (infinite-dimensional) Lie "groups" of transformations from open sets of ℂ into open sets, then what are these groups? Properly, these are pseudogroups, but perhaps they behave like Lie groups.

[Note: It's not hard to compute formulas for such transformations -- the flows -- directly from an expression for the vector field in terms of its zeroes (and poles, if any).]

  • In any case, are there standard names for the Lie algebras VP and VR ?

  • References to the above matters would also be appreciated.

$\endgroup$
4
  • 2
    $\begingroup$ The Lie algebra you've denoted $V_P$ is the complexification of the Lie algebra of the poly vector fields on the real line, $W_1,$ which is one the 4 families of Cartan's transitive pseudogroups. While AFAIK $V_R$ doesn't have another name, its subalgebra consisting of the Laurent polynomials in $z$ times $d/dz$ is the complexification of the Lie algebra of polynomial vector fields on the circle. In a certain sense, it corresponds to the diffeomorphism group of the circle, whereas $W_1$ corresponds to the diff group of the line or the group of holomorphic automorphisms of the closed unit disk. $\endgroup$ Commented Aug 26, 2010 at 1:25
  • $\begingroup$ Drinfeld-Krichever construction in the theory of integrable systems involves a generalization of the polynomial vector fields: for a complete algebraic curve $X,$ a point $x_\infty$ and a local coordinate $z$ at this point, consider the Lie algebra of vector fields on $X\setminus x_{\infty}$ expressed in the coordinate $z.$ Its analytic version is the Lie algebra of complex vector fields on a small circle $\partial D_\infty$ around $x_{\infty}$ that extend to holomorphic vector fields on its exterior $X\setminus D_\infty.$ $\endgroup$ Commented Aug 26, 2010 at 1:38
  • $\begingroup$ The answer to your first question is yes. If you differentiate the action of the group $PSL_2$ on the Riemann sphere at the identity to give a Lie homomorphism from the Lie algebra $\mathfrak{sl}_2$ to vector fields on the Riemann sphere. The image of this map on a coordinate chart obtained by removing a point at infinity is your Lie algebra of complex quadratic vector fields. $\endgroup$ Commented Aug 26, 2010 at 16:13
  • $\begingroup$ What you mean exactly under "Lie algebras of vector fields"? If any Lie algebra admitting realization as vector fields with polynomial/rational/etc. coefficients (i.e. subalgebras of appropriate $W_n$) - there are many. This topic goes back to Sophus Lie with a tremendous amount of more recent literature. But your examples probably suggest that you are interested in "full", in some sense, algebras -- of the form $f(z) d/dz$ ? what about many variables? -- but the exact meaning is not clear to me. $\endgroup$ Commented Jan 5, 2011 at 22:31

2 Answers 2

3
$\begingroup$

There is a local statement: Suppose you have a finite-dimensional vector space of germs of tangent vector fields at a point in a one-dimensional complex manifold, and suppose that some element of it is nonzero at that point. There is a coordinate function $z$ such that this field is $\frac{\partial}{\partial z}$. So, expressing all the elements as $f(z)\frac{\partial}{\partial z}$ for function germs $f$, you have a finite-dimensional vector space of germs of functions of $z$, having $1$ as a member and closed under the operation $f,g\mapsto fg'-gf'$. In particular the space is closed under differentiation. This forces it to consist of exponential polynomials, in fact to be the direct sum, over some set of complex numbers $\omega$ including $0$, of the space spanned by $z^je^{\omega z}$ for $0\le j\le m(\omega)$ for some integers $m(\omega)\ge 0$. For this to be closed under that bracket operation it must be either $3$-dimensional with basis $1,z,z^2$ or $1,e^{\omega z},e^{-\omega z}$ for some $\omega\ne 0$, or $2$-dimensional with basis $1,z$ or $1,e^{\omega z}$ for some $\omega\ne 0$, or $1$-dimensional with basis $1$. Under a further change of coordinates this becomes the example you mentioned or a subalgebra thereof.

It's pretty much the same over the real numbers.

$\endgroup$
0
$\begingroup$

The germs of Lie algebra actions on curves and surfaces were classified by Lie. In particular, the action of $\mathfrak{sl}(2,\mathbb{C})$ on $\mathbb{C}$ comes from the action of $\mathbb{P}SL(2,\mathbb{C})$ on $\mathbb{CP}^1$, differentiated and written out in an affine chart. Lie proved that this action does not embed into any holomorphic effective action of any larger complex Lie group. The proof is not difficult: if you add any other holomorphic vector field $f(z)\partial_z$, you need the Lie brackets in there too, so you get $z^2f'(z)\partial_z$ in there. So if $f$ vanishes at the origin to some order larger than 2, then $z^2f'(z)\partial_z$ vanishes to higher order still, so you get an infinite dimensional Lie algebra. If $f$ vanishes somewhere to order larger than $2$, use the transitivity of the Lie group action. Moreover, the same has to hold out at $z=\infty$, so you can easily prove that $f$ is a polynomial of degree 2. For more about what little I know about the generalization to complex surfaces, see my paper: Complex homogeneous surfaces

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .