# The Unapologetic Mathematician

## Linear Lie Algebras

So now that we’ve remembered what a Lie algebra is, let’s mention the most important ones: linear Lie algebras. These are ones that arise from linear transformations on vector spaces, ’cause mathematicians love them some vector spaces.

Specifically, let $V$ be a finite-dimensional vector space over $\mathbb{F}$, and consider the associative algebra of endomorphisms $\mathrm{End}(V)$ — linear transformations from $V$ back to itself. We can use the usual method of defining a bracket as a commutator:

$\displaystyle [x,y]=xy-yx$

to turn this into a Lie algebra. When considered as a Lie algebra like this, we call it the “general linear Lie algebra”, and write $\mathfrak{gl}(V)$. Many Lie algebras are written in the Fraktur typeface like this.

Any subalgebra of $\mathfrak{gl}(V)$ is called a “linear Lie algebra”, since it’s made up of linear transformations. It turns out that every finite-dimensional Lie algebra is isomorphic to a linear Lie algebra, but we reserve the “linear” term for those algebras which we’re actually thinking of having linear transformations as elements.

Of course, since $V$ is a vector space over $\mathbb{F}$, we can pick a basis. If $V$ has dimension $n$, then there are $n$ elements in any basis, and so our endomorphisms correspond to the $n\times n$ matrices $\mathrm{Mat}_n(\mathbb{F})$. When we think of it in these terms, we often write $\mathfrak{gl}(n,\mathbb{F})$ for the general linear Lie algebra.

We can actually calculate the bracket structure explicitly in this case; bilinearity tells us that it suffices to write it down in terms of a basis. The standard basis of $\mathfrak{gl}(n,\mathbb{F})$ is $\{e_{ij}\}_{i,j=1}^n$ which has a $1$ in the $i$th row and $j$th column and $0$ elsewhere. So we can calculate:

$\displaystyle [e_{ij},e_{kl}]=\delta_{jk}e_{il}-\delta_{li}e_{kj}$

where, as usual, $\delta_{ij}$ is the Kronecker delta: $1$ if the indices are the same and $0$ if they’re different.

We can now identify some important subalgebras of $\mathfrak{gl}(n,\mathbb{F})$. First, the strictly upper-triangular matrices $\mathfrak{n}(n,\mathbb{F})$ involve only the basis elements $e_{ij}$ with $i. If $i so the first term in the above expression for the bracket shows up, then the second term cannot show up, and vice versa. Either way, we conclude that the bracket of two basis elements of $\mathfrak{n}(n,\mathbb{F})$ — and thus any element of this subspace — involves only other basis elements of the subspace, which makes this a subalgebra.

Similarly, we conclude that the (non-strictly) upper-triangular matrices involving only $e_{ij}$ with $i\leq j$ also form a subalgebra $\mathfrak{t}(n,\mathbb{F})$. And, finally, the diagonal matrices involving only $e_{ii}$ also form a subalgebra $\mathfrak{d}(n,\mathbb{F})$. This last one is interesting, in that the bracket on $\mathfrak{d}(n,\mathbb{F})$ is actually trivial, since any two diagonal matrices commute.

As vector spaces, we see that $\mathfrak{t}(n,\mathbb{F})=\mathfrak{d}(n,\mathbb{F})+\mathfrak{n}(n,\mathbb{F})$. It’s easy to check that the bracket of a diagonal matrix and a strictly upper-triangular matrix is again strictly upper-triangular — we write $[\mathfrak{d}(n,\mathbb{F}),\mathfrak{n}(n,\mathbb{F})]=\mathfrak{n}(n,\mathbb{F})$ — and so we also have $[\mathfrak{t}(n,\mathbb{F}),\mathfrak{t}(n,\mathbb{F})]=\mathfrak{n}(n,\mathbb{F})$. This may seem a little like a toy example now, but it turns out to be surprisingly general; many subalgebras will relate to each other this way.

August 7, 2012 Posted by | Algebra, Lie Algebras | 7 Comments