Some question about algebraic group and Lie algebra action.

Question

I have some basic question about algebraic group and Lie algebra actions which confuse me. Suppose that algebraic group $G$ acts on a variety $X$. Then $G$ acts on $\mathbb{C}[X]$ by $(g\cdot f)(x) = f(g^{-1} \cdot x)$, $x \in X$. Does the Lie algebra $\mathfrak{g}$ of $G$ act on $\mathbb{C}[X]$? If yes, how does $\mathfrak{g}$ act on $\mathbb{C}[X]$? Thank you very much.

Answer 1

It is basically the same as in differential geometry, except done algebraically.

The action of $G$ on the (infinite-dimensional) $\mathbb{C}$-vector space $\mathbb{C}[X]$ is an algebraic group homomorphism $G\to GL(\mathbb{C}[X])$. So taking derivative gives $T_eG\to T_e(GL(\mathbb{C}[X]))$. Identifying $T_eG$ with $\mathfrak{g}$ and $T_e(GL(\mathbb{C}[X]))$ with $\mathfrak{gl}(\mathbb{C}[X])$ then gives a Lie algebra homomorphism $\mathfrak{g}\to\mathfrak{gl}(\mathbb{C}[X])$.

If you are looking for a formula, it is the same as the differential geometry case: for $A\in\mathfrak{g}$, $f\in\mathbb{C}[X]$, $x\in X$, $v\in T_xX$, we have $$ (A\cdot f)(x)=(v\mapsto -(T_x f)(Av)) $$

Answer 2

$\DeclareMathOperator{\g}{\mathfrak{g}} \DeclareMathOperator{\complex}{\mathbb{C}}$ I can't parse the formula proposed by user10354138, so I am adding the formula I came to. For $A \in \mathfrak{g}$, $f \in \complex[X]$, $x \in X$, we have $$ (A \cdot f)(x)=-d_x f(A x), $$ where $d_x f (a)= \sum_{i} \frac{\partial f}{\partial t_i}(x) a_i$. Note that $(A \cdot f)\in \complex[X]$, as expected.

Note that this agrees with the familiar formula from differential geometry

$$ (A \cdot f)(x)=\frac{d}{d t}\left[f(e^{-t A} x)\right](0)=-d_x f(A x), $$

where the last equality follows from the chain rule.

How do we see this formula? First note that the image of $G$ in $GL(\complex[X])$ is an algebraic group of $\complex$-algebra automorphisms of $\complex[X]$, so the image of $\mathfrak{g}$ in $\mathfrak{gl}(\complex[X])$ is a Lie algebra of derivations of $\complex[X]$. It follows that the action is uniquely defined by the action on the coordinate functions $(A \cdot t_i)$.

The above formula is clearly a derivation, so it suffices to compute $(A \cdot t_i)$ and show it agrees with the formula. WLOG assume $G\subseteq GL(n, \complex)$. Let $\phi: G \rightarrow GL(\complex[X])$ be the OP's homomorphism $\phi(g)(f)(x)=f(g^{-1}x)$. Then

$$ \phi(g)(t_i)=\sum_{j=1}^n g^{-1}(i,j) t_i. $$

Since $\phi(g)$ preserves the subspace of linear polynomials for all $g \in G$, it suffices to compute the differential of $\phi$ as a map $G \rightarrow GL(\text{span}\{t_1,\dots, t_n\})$. With respect to the basis $\{t_1,\dots, t_n\}$, we have $\phi(g)=g^{-1}$, so $d_e \phi (A)=-A$, i.e.

$$ (A \cdot t_i)=d_e \phi(A)(t_i)=-\sum_{j=1}^n A(i,j) t_j, $$

where the first equality is by definition. Now we just need to check that this agrees with the proposed formula: $-d_{e_j} t_i (A e_j)=-t_i(A e_j)=-A(i,j)$. QED

All we need is an algebraically closed field here.