Pick a basis for the Lie algebra of \(SL_2(\mathbb C)\) and show how each basis element can be canonically interpreted as a holomorphic vector field on \[ \widehat{\mathbb C}=\mathbb P(\mathbb C^2). \] Check that the Lie bracket corresponds to the bracket of vector fields.
Solution
Take the standard basis of \(\mathfrak{sl}_2(\mathbb C)\) \[ E=\begin{pmatrix}0&1\\0&0\end{pmatrix},\qquad F=\begin{pmatrix}0&0\\1&0\end{pmatrix},\qquad H=\begin{pmatrix}1&0\\0&-1\end{pmatrix}. \] We use the infinitesimal-action convention \[ X_A(p)=\left.\frac d{dt}\right|_{t=0}\exp(-tA)\cdot p. \] With this sign convention, \(A\mapsto X_A\) is a Lie-algebra homomorphism.
1. If\[ A=\begin{pmatrix}\alpha&\beta\\ \gamma&-\alpha\end{pmatrix}, \]then on the affine chart \(z=[z:1]\) one has\[ X_A =\bigl(\gamma z^2-2\alpha z-\beta\bigr)\frac{\partial}{\partial z}. \]
To first order, \[ e^{-tA} =I-tA+O(t^2) =\begin{pmatrix} 1-t\alpha&-t\beta\\ -t\gamma&1+t\alpha \end{pmatrix}+O(t^2). \] The induced Möbius transformation is therefore \[ z\longmapsto \frac{(1-t\alpha)z-t\beta} {-t\gamma z+1+t\alpha} =z+t(\gamma z^2-2\alpha z-\beta)+O(t^2). \] Differentiating at \(t=0\) gives the formula.
2. The basis elements correspond to\[ X_E=-\frac{\partial}{\partial z},\qquad X_F=z^2\frac{\partial}{\partial z},\qquad X_H=-2z\frac{\partial}{\partial z}. \]
Substitute \((\alpha,\beta,\gamma)=(0,1,0)\), \((0,0,1)\), and \((1,0,0)\) into 1.
3. These vector fields extend holomorphically across \(\infty\).
Use the coordinate \(w=1/z\) near \(\infty\). Since \[ \frac{\partial}{\partial z} =-w^2\frac{\partial}{\partial w}, \] the three fields become \[ X_E=w^2\frac{\partial}{\partial w},\qquad X_F=-\frac{\partial}{\partial w},\qquad X_H=2w\frac{\partial}{\partial w}. \] All three are holomorphic at \(w=0\). Thus they define global holomorphic vector fields on \(\mathbb P^1\).
4. The matrix brackets are\[ [H,E]=2E,\qquad [H,F]=-2F,\qquad [E,F]=H. \]
These follow by direct matrix multiplication.
5. The corresponding vector-field brackets satisfy the same relations.
For vector fields \(P(z)\partial_z\) and \(Q(z)\partial_z\), \[ [P\partial_z,Q\partial_z] =(PQ'-QP')\partial_z. \] Using 2, \[ \begin{aligned} [X_H,X_E] &=[-2z\partial_z,-\partial_z] =-2\partial_z =2X_E,\\ [X_H,X_F] &=[-2z\partial_z,z^2\partial_z] =-2z^2\partial_z =-2X_F,\\ [X_E,X_F] &=[-\partial_z,z^2\partial_z] =-2z\partial_z =X_H. \end{aligned} \] These are exactly the relations in 4.
6. Hence the infinitesimal projective action identifies\(\mathfrak{sl}_2(\mathbb C)\) with the holomorphic vector fields on\(\mathbb P^1\).
The assignment is linear, and 2 shows that its kernel is zero. Its image contains the three linearly independent global fields \[ \partial_z,\qquad z\partial_z,\qquad z^2\partial_z, \] which form the three-dimensional space of global holomorphic vector fields on \(\mathbb P^1\). By 5 the identification respects Lie brackets.