Last time, we defined Lie algberas and basic notions related to them. This time, I’ll show you some actual Lie algebras.

First, remember that given an associative algebra $A,$ we can get a Lie algebra by defining the bracket, using multiplication in $A,$ as $[x,y] = xy - yx.$ This gives us a ton of examples already, the most important being that of the general linear Lie algebra:

Given a $\mathbb{K}-$vector space $V,$ the space of all $\mathbb{K}-$linear maps from $V$ to $V,$ noted $\text{End}_{\mathbb{K}}(V),$ is an algebra over $\mathbb{K}.$ Then define the general linear algebra $\mathfrak{gl}(V)$ as the Lie algebra you get when you redefine multiplication as the bracket $[X,Y] = XY - YX.$

Similarly, $\mathfrak{gl}(n,\mathbb{K})$ is the corresponding Lie algebra of $n \times n$ matrices with entries in $\mathbb{K}.$ All subalgebras of $\mathfrak{gl}(V)$ or $\mathfrak{gl}(n, \mathbb{K})$  are called linear Lie algebras. For example, we have the space of skew-symmetric $n \times n$ matrices with coefficients in $\mathbb{K}$ :

$\mathfrak{so}(n, \mathbb{K}) = \{X \in \mathfrak{gl}(n,\mathbb{K}) : X + X^t = 0 \}$

This is indeed a subalgebra because if $X,Y$ are skew-symmetric,

$[X,Y]^t = (XY - YX)^t$

$= (XY)^t - (YX)^t$

$= Y^tX^t - X^tY^t$

$= YX - XY$

$= -[Y,X]$

so $[X,Y]$ is also skew-symmetric. It is as easy to show that these are also linear Lie algebras :

$\mathfrak{u}(n) = \{ X \in \mathfrak{gl}(n, \mathbb{C}) : X+ X^* = 0 \},$

$\mathfrak{sl}(n,\mathbb{K}) = \{ X \in \mathfrak{gl}(n,\mathbb{K}) : \text{Tr}(X) = 0 \},$

$\{X \in \mathfrak{gl}(n,\mathbb{K}) : JX + X^tJ = 0 \}$

where $J \in M_n(\mathbb{K}).$

There are also the linear Lie algebras of upper-triangular matrices, stricly upper-triangular matrices and diagonal matrices, respectively noted $\mathfrak{t}(n,\mathbb{K}),$ $\mathfrak{n}(n,\mathbb{K})$ and $\mathfrak{d}(n,\mathbb{K}).$ It is easy to see that these are linear subspaces of $\mathfrak{gl}(n,\mathbb{K}),$ to see that the diagonal matrices are closed under the bracket operation, note that for any two commuting elements $X,Y,$ we have $[X,Y] = 0$ so actually

$[\mathfrak{d}(n,\mathbb{K}), \mathfrak{d}(n,\mathbb{K})] = 0.$

Such a Lie algebra is called Abelian. For the two others, introduce the standard basis elements $E_{ij},$ the $n \times n$ matrices consisting of a $1$ in position $(i,j)$ and of $0$ elsewhere. This is a basis of $\mathfrak{gl}(n,\mathbb{K})$ and we calculate

$E_{ij}E_{kl} = \delta_{jk}E_{il}$

meaning that $E_{ij}E_{kl}$ is the zero matrix if $j \neq k$ and is $E_{il}$ if $j = k.$ So for this basis, the bracket multiplication reads

$[E_{ij},E_{kl}] = \delta_{jk}E_{il} - \delta_{il}E_{kj}.$

Then for $E_{ij}, E_{kl}$ with $i < j$ and $k < l$ (those are the basis elements appearing in the decomposition of a matrix in $\mathfrak{n}(n,\mathbb{K})$), only one of the two terms in the bracket expansion can appear. Indeed, if $\delta_{jk} \neq 0$, it is because $j=k$ but then we can’t have $i=l$ because $i On the other hand, if $\delta_{il} \neq 0,$ it is because $i=l$ and then $k forces $j \neq k.$ We showed that for two strictly upper triangular matrices $X,Y,$ the expression of $[X,Y]$ involves only elements of the basis of $\mathfrak{n}(n,\mathbb{K})$ and so this subspace is closed under the bracket operation. A similar argument shows that $\mathfrak{t}(n,\mathbb{K})$ is a linear Lie algebra.

In the next post I will hint at the link between Lie algebras and the differential geometry of Lie groups.

References:

1. Humphreys – Introduction to Lie Algebras and Representation Theory
2. Knapp – Lie Groups Beyond an Introduction