Editing Lie algebra (section)

=== Subalgebras, ideals and homomorphisms ===
The Lie bracket is not required to be [[associative]], meaning that <math>[[x,y],z]</math> need not be equal to <math>[x,[y,z]]</math>. Nonetheless, much of the terminology for associative [[ring (mathematics)|rings]] and algebras (and also for groups) has analogs for Lie algebras. A '''Lie subalgebra'''  is a linear subspace <math>\mathfrak{h} \subseteq \mathfrak{g}</math> which is closed under the Lie bracket.  An '''ideal''' <math>\mathfrak i\subseteq\mathfrak{g}</math> is a linear subspace that satisfies the stronger condition:<ref>By the anticommutativity of the commutator, the notions of a left and right ideal in a Lie algebra coincide.</ref>
:<math>[\mathfrak{g},\mathfrak i]\subseteq \mathfrak i.</math>

In the correspondence between Lie groups and Lie algebras, subgroups correspond to Lie subalgebras, and [[normal subgroup]]s correspond to ideals.

A Lie algebra '''homomorphism''' is a linear map compatible with the respective Lie brackets:
:<math> \phi\colon \mathfrak{g}\to\mathfrak{h}, \quad \phi([x,y])=[\phi(x),\phi(y)]\ \text{for all}\ x,y \in \mathfrak g. </math>
An '''isomorphism''' of Lie algebras is a [[bijective]] homomorphism.

As with normal subgroups in groups, ideals in Lie algebras are precisely the [[kernel (algebra)|kernels]] of homomorphisms. Given a Lie algebra <math>\mathfrak{g}</math> and an ideal <math>\mathfrak i</math> in it, the ''quotient Lie algebra'' <math>\mathfrak{g}/\mathfrak{i}</math> is defined, with a surjective homomorphism <math>\mathfrak{g}\to\mathfrak{g}/\mathfrak{i}</math> of Lie algebras. The [[first isomorphism theorem]] holds for Lie algebras: for any homomorphism <math>\phi\colon\mathfrak{g}\to\mathfrak{h}</math> of Lie algebras, the image of <math>\phi</math> is a Lie subalgebra of <math>\mathfrak{h}</math> that is isomorphic to <math>\mathfrak{g}/\text{ker}(\phi)</math>.

For the Lie algebra of a Lie group, the Lie bracket is a kind of infinitesimal commutator. As a result, for any Lie algebra, two elements <math>x,y\in\mathfrak g</math> are said to ''commute'' if their bracket vanishes: <math>[x,y]=0</math>.

The [[centralizer]] subalgebra of a subset <math>S\subset \mathfrak{g}</math> is the set of elements commuting with ''<math>S</math>'': that is, <math>\mathfrak{z}_{\mathfrak g}(S) = \{x\in\mathfrak g : [x, s] = 0 \ \text{ for all } s\in S\}</math>. The centralizer of <math>\mathfrak{g}</math> itself is the ''center''  <math>\mathfrak{z}(\mathfrak{g})</math>. Similarly, for a subspace ''S'', the [[normalizer]] subalgebra of ''<math>S</math>'' is <math>\mathfrak{n}_{\mathfrak g}(S) = \{x\in\mathfrak g : [x,s]\in S \ \text{ for all}\ s\in S\}</math>.<ref>{{harvnb|Jacobson|1979|p=28.}}</ref> If <math>S</math> is a Lie subalgebra, <math>\mathfrak{n}_{\mathfrak g}(S)</math> is the largest subalgebra such that <math>S</math> is an ideal of <math>\mathfrak{n}_{\mathfrak g}(S)</math>.

==== Example ====
The subspace <math>\mathfrak{t}_n</math> of diagonal matrices in <math>\mathfrak{gl}(n,F)</math> is an abelian Lie subalgebra. (It is a [[Cartan subalgebra]] of <math>\mathfrak{gl}(n)</math>, analogous to a [[maximal torus]] in the theory of [[compact Lie group]]s.) Here <math>\mathfrak{t}_n</math> is not an ideal in <math>\mathfrak{gl}(n)</math> for <math>n\geq 2</math>. For example, when <math>n=2</math>, this follows from the calculation:
<blockquote><math>\begin{align}
\left[
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix},
\begin{bmatrix}
x & 0 \\
0 & y
\end{bmatrix}
\right] &= \begin{bmatrix}
ax & by\\
cx & dy \\
\end{bmatrix} - \begin{bmatrix}
ax & bx\\
cy & dy \\
\end{bmatrix} \\
&= \begin{bmatrix}
0 & b(y-x) \\
c(x-y) & 0
\end{bmatrix}
\end{align}</math></blockquote>
(which is not always in <math>\mathfrak{t}_2</math>).

Every one-dimensional linear subspace of a Lie algebra <math>\mathfrak{g}</math> is an abelian Lie subalgebra, but it need not be an ideal.