I'm currently reading about the Witt algebra, and I'm trying to understand in what sense the Witt algebra basis $L_n = -z^{n+1}\partial _z$ generates conformal maps in dimension $2$.

From what I've read, a generator is a vector field $X = (X_1,X_2)$ whose flow $\Phi_t(z)$ gives a conformal map around $z=0$ (Generator of the Special Conformal Transformation). So I'm guessing $L_n$ represents a vector field whose flow $\Phi_t(z)$ gives a conformal map for every $t$. Is that correct?

If so, what is the vector field $L_n = -z^{n+1}\partial_z$ in coordinates $(X_1,X_2)$? And what are the flows of $L_n$?

  • $\begingroup$ In your convention which basis is associated with your notation $X = (X_1,X_2)$? $\endgroup$
    – Qmechanic
    Oct 11, 2018 at 11:56

1 Answer 1


Take a scalar field $g(z,\bar{z})$, which means that $$\Delta g=\delta g+dg=0$$ i.e. its total variation $\Delta$ is zero and then $\delta g=-dg$.

Now consider an infinitesimal conformal transformation: $$z\to z'=z+\epsilon(z)$$ $$\bar{z}\to \bar{z}'=\bar{z}+\bar{\epsilon}(\bar{z})$$ and evaluate: $$\delta g(z,\bar{z})=-dg(z,\bar{z})=-g(z',\bar{z}')+g(z,\bar{z})=-g(z+\epsilon(z),\bar{z}+\bar{\epsilon}(\bar{z}))+g(z,\bar{z})\simeq -g(z,\bar{z})-\epsilon \partial g-\bar{\epsilon}\bar{\partial}g+g(z,\bar{z})=(-\epsilon\partial-\bar{\epsilon}\bar{\partial})g(z,\bar{z})$$ where I have indicated $\epsilon=\epsilon(z),\bar{\epsilon}=\bar{\epsilon}(\bar{z}),\partial=\partial_z,\bar{\partial}=\partial_{\bar{z}}$ for shortness.

Now, since the conformal transformations are olomorphic transformations (the components of the transformations satisfy Cauchy-Riemann equations), we can write $\epsilon$ and $\bar{\epsilon}$ in a Laurant expansion of the type: $$\epsilon=\sum_{n=-\infty}^{\infty}\epsilon_nz^{n+1}\;\;\;\;\;\;\;\;\;\;\bar{\epsilon}=\sum_{n=-\infty}^{\infty}\bar{\epsilon}_n\bar{z}^{n+1}$$ In this way we get: $$\delta g(z,\bar{z})=\sum_{n=-\infty}^{\infty}(-\epsilon_nz^{n+1}\partial-\bar{\epsilon}_n\bar{z}^{n+1}\bar{\partial})g(z,\bar{z})=\sum_{n=-\infty}^{\infty}(\epsilon_n\ell_n+\bar{\epsilon}_n\bar{\ell}_n)g(z,\bar{z})$$ Where $$\ell_n=-z^{n+1}\partial\;\;\;\;\;\;\;\;\;\bar{\ell}_n=-\bar{z}^{n+1}\bar{\partial}$$ And so these $\ell_n$ and $\bar{\ell}_n$ are the generators of the conformal group acting on scalars.

If you want the vector fields that generates the conformal transformation in the function space, you have to consider: $$v(z)=-\sum_{n=-\infty}^{\infty}\epsilon_n\ell_n\;\;\;\;\;\;\;\;\;\;\bar{v}(\bar{z})=-\sum_{n=-\infty}^{\infty}\bar{\epsilon}_n\bar{\ell}_n$$

And using this field, requiring its regularity around 0 it is found that $n\geq-1$, moreover, since we are considering our theory on the Riemann sphere, it must be regular even for $z\to -\frac{1}{z}$ and this fix the $n\leq 1$.

In turn, those conditions state that for a conformal transformation to be defined over all the Riemann sphere, only the terms $\ell_0,\ell_{-1},\ell_{1}$ are admitted.


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