Some examples of Lie algebras

October 26, 2012

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 $i<j=k<l.$ On the other hand, if $\delta_{il} \neq 0,$ it is because $i=l$ and then $k<l=i<j$ 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.