Generators of the Diffeomorphism Group

Question

So what are the generators of a Diffeomorphism Group?

For simplicity, let's consider $ Diff(R^2) $ (diffeomorphisms of the euclidean plane.)

Diffeomorphisms are differentiable, invertible transformations (right?) So $Diff(R^2)$ would be a group made of all the differentiable, invertible functions on $R^2$, correct?

What would the generators be, then? All I can think of would be $x_1^i, x_2^j$ and $x_1^k*x_2^l$ so that you could build up your functions in a sort of power series fashion. ($x_1$ and $x_2$ being your $R^2$ coordinates.)

Yes, I know that Wikipedia says that the generators are

$L_h=h^\mu(x) \frac{\partial}{\partial x_\mu}$,

but what the heck is $h$ then? Any arbitrary function?

Answer 1

Think of an infinitesimal Diff as of a translation where the the shift is space dependent, $x^\mu\rightarrow x^\mu+\epsilon^\mu(x)$. Now, you get that the generators are $L_\epsilon=\epsilon^\nu(x)\partial_\nu$ since $L_\epsilon x^\mu=\epsilon^\mu(x)$. They form an infinite space since $\epsilon^\mu(x)$ is a function that can be expanded in infinitely many constant parameters given by the derivative of $\epsilon$ in zero, $\epsilon^\mu(x)=\sum_n x^{\nu_1}\cdot x^{\nu_n} \partial^n_{\nu_1\ldots \nu_n}\epsilon^\mu(x=0)/n!$. So a basis of generators would be $$ L_\nu= \partial_\nu \,,\qquad L^{\mu_1}_\nu=x^{\mu_1} \partial_\nu \,,\qquad L^{\mu_1\mu_2}_\nu=x^{\mu_1}x^{\mu_2} \partial_\nu\,,\qquad \ldots $$ where one recognizes some generators of various finite dimensional subgroups ($L_\mu$ generates translations, $L^\mu_\mu$ generates scale transformations, $L^{\mu}_\nu-L^{\nu}_\mu$ generates rotations,...)

Answer 2

Formally speaking, given a (differentiable, finite dimensional) manifold $M$, then the (infinite dimensional) Lie group of (globally defined) diffeomorphisms (with composition $\circ$ as group structure) has the set $\Gamma(TM)$ of (globally defined, differentiable) vector fields as corresponding Lie algebra.

This (infinite dimensional) Lie algebra $\Gamma(TM)$ is endowed with the usual Lie bracket of vector fields $[\cdot,\cdot]$. (Basis) elements for a Lie algebra are often called generators.

Answer 3

To add to Qmechanic's Answer and TwoBs's Answer and answer "....what the heck is h then? Any arbitrary function?": $h$ is pretty much arbitrary. It is wontedly taken to be at least differentiability class $C^1$ (all first derivatives continuous) so that the Lie bracket of vector fields is defined as in Qmechanic's answer. You need to assume it is of class $C^\infty$ ("smooth", i.e derivatives of all orders exist) to kit $\Gamma(TM)$ (notation as in Qmechanic's answer) with the basis that TwoBs's answer defined for you. The exact conditions depend on the application, but $C^2$ will get you the ability to define the curvature tensor properly, and therefore to make the Einstein field equations meaningful in GTR, for example.