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Lie algebras of the additive, infinitesimal and general linear groups

Example

Assume the Axiom of Choice for the finite-type assertions inherited from the matrix-group supplier. Let $k$ be a field, let $n \geq 1$ , and let $Lie$ be as in The Lie algebra of a group scheme. (a) $Lie (G_{a}) ≅ k$ and $Lie (G_{m}) ≅ k$ , generated by the functionals dual to the cotangent classes $[t]$ and $[t - 1]$ , respectively; the bracket is zero because these groups are abelian, so the Lie algebras are the one-dimensional abelian Lie algebras. (b) If $char k = p > 0$ , then $Lie (α_{p}) ≅ k$ and $Lie (μ_{p}) ≅ k$ with generators dual to the corresponding cotangent classes, so there are isomorphisms of $k$ -Lie algebras $Lie (α_{p}) ≅ Lie (G_{a})$ and $Lie (μ_{p}) ≅ Lie (G_{m})$ . (c) $Lie ({GL}_{n}) = {g l}_{n} = M_{n} (k)$ , the isomorphism sending $X$ to $I + ε X$ , and $[X, Y] = X Y - Y X$ (The Lie algebra of the general linear group, The Lie bracket from infinitesimals and the adjoint action).

Facts & Assumptions

Given: The Axiom of Choice and a field $k$ , an integer $n \geq 1$ and, in part (b), an integer $p = char k > 0$ .

[F1]

Additive and infinitesimal group schemes: $G_{a} = Spec k [t]$ , $G_{m} = Spec k [t, t^{- 1}]$ , and in characteristic $p$ the group schemes $α_{p} = Spec k [t] / (t^{p})$ and $μ_{p} = Spec k [t, t^{- 1}] / (t^{p} - 1)$ , whose coordinate rings are both isomorphic to $k [s] / (s^{p})$ , with $s = t$ respectively $s = t - 1$ ; these groups are commutative.

[F2]

Cotangent space at a rational point and The Lie algebra of a group scheme: for a $k$ -rational point $e$ , $Lie (G) = {Hom}_{k} (m_{e} / m_{e}^{2}, k)$ and $m_{e} / m_{e}^{2} ≅ Ω_{G / k} \otimes κ (e)$ , a nonzero class $[u]$ with $m_{e} = (u)$ is a cotangent basis, whose dual functional generates the Lie algebra.

[F3]

Lie algebras over a field: a one-dimensional $k$ -Lie algebra has zero bracket, since $[a X, b X] = a b [X, X] = 0$ by bilinearity and alternation.

[F4]

The Lie algebra of the general linear group and The Lie bracket from infinitesimals and the adjoint action: $Lie ({GL}_{n}) = {g l}_{n}$ via $X \mapsto I_{n} + ε X$ , and the bracket is the matrix commutator $[X, Y] = X Y - Y X$ .

Verification

technique · direct

1.1F1F2algebra

Cotangent spaces and dual generators. At the identity of $G_{a}$ and $G_{m}$ , respectively, the local rings are $k [t]_{(t)}$ and $k [t, t^{- 1}]_{(t - 1)}$ , with maximal ideals generated by $t$ and $t - 1$ . Their quotients by the squares of these ideals are $k [u] / (u^{2})$ , with $u = t$ or $u = t - 1$ : any denominator outside the maximal ideal has a nonzero constant term and is invertible in this square-zero quotient. Thus $[t]$ and $[t - 1]$ are cotangent bases. In characteristic $p \geq 2$ , both infinitesimal coordinate rings are $k [s] / (s^{p})$ by [F1]; every element outside $(s)$ is a unit by a finite geometric sum, so this ring is already local. Its cotangent quotient $(s) / (s)^{2}$ has basis $[s]$ , regardless of whether higher powers survive in the ring. By [F2] the Lie algebras are the duals of these one-dimensional cotangent spaces, generated by the functionals $u^{*}$ with $u^{*} ([u]) = 1$ .

1.2F4given

The general linear group. By [F4] the identification $Lie ({GL}_{n}) = {g l}_{n} = M_{n} (k)$ sends $X$ to $I + ε X$ and the bracket is $[X, Y] = X Y - Y X$ .

2.1F1F2F3step 1.1step 1.2∎

Zero brackets and the isomorphisms. By the bracket construction in parts (a)-(d) of The Lie bracket from infinitesimals and the adjoint action, with Choice inherited through its matrix-group and adjoint-representation suppliers, these tangent spaces carry Lie brackets. They are one-dimensional by step 1.1, so [F3] makes each bracket zero. Sending the dual generator $s^{*}$ of $Lie (α_{p})$ to $t^{*}$ of $Lie (G_{a})$ is therefore a Lie-algebra isomorphism; likewise sending the dual generator for $s = t - 1$ to $(t - 1)^{*}$ gives $Lie (μ_{p}) ≅ Lie (G_{m})$ . Together with step 1.2 this proves (a), (b) and (c).

Depends on

Used by

The Lie algebra does not detect nonsmooth group schemes Counterexample

Dependency tree · two levels

62 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. S. Milne, Algebraic Groups (corrected 2022 printing, Cambridge University Press) (standard reference, not scraped)
The Stacks Project, Groupoid Schemes chapter (standard reference, not scraped)