The Unapologetic Mathematician

Mathematics for the interested outsider

Special Linear Lie Algebras

More examples of Lie algebras! Today, an important family of linear Lie algebras.

Take a vector space V with dimension \mathrm{dim}(V)=l+1 and start with \mathfrak{gl}(V). Inside this, we consider the subalgebra of endomorphisms whose trace is zero, which we write \mathfrak{sl}(V) and call the “special linear Lie algebra”. This is a subspace, since the trace is a linear functional on the space of endomorphisms:

\displaystyle\mathrm{Tr}(ax+by)=a\mathrm{Tr}(x)+b\mathrm{Tr}(y)

so if two endomorphisms have trace zero then so do all their linear combinations. It’s a subalgebra by using the “cyclic” property of the trace:

\displaystyle\mathrm{Tr}(xy)=\mathrm{Tr}(yx)

Note that this does not mean that endomorphisms can be arbitrarily rearranged inside the trace, which is a common mistake after seeing this formula. Anyway, this implies that

\displaystyle\begin{aligned}\mathrm{Tr}\left([x,y]\right)&=\mathrm{Tr}(xy-yx)\\&=\mathrm{Tr}(xy)-\mathrm{Tr}(yx)=0\end{aligned}

so actually not only is the bracket of two endomorphisms in \mathfrak{sl}(V) back in the subspace, the bracket of any two endomorphisms of \mathfrak{gl}(V) lands in \mathfrak{sl}(V). In other words: \left[\mathfrak{gl}(V),\mathfrak{gl}(V)\right]=\mathfrak{sl}(V).

Choosing a basis, we will write the algebra as \mathfrak{sl}(l+1,\mathbb{F}). It should be clear that the dimension is (l+1)^2-1, since this is the kernel of a single linear functional on the (l+1)^2-dimensional \mathfrak{gl}(l+1,\mathbb{F}), but let’s exhibit a basis anyway. All the basic matrices e_{ij} with i\neq j are traceless, so they’re all in \mathfrak{sl}(n,\mathbb{F}). Along the diagonal, \mathrm{Tr}(e_{ii})=1, so we need linear combinations that cancel each other out. It’s particularly convenient to define

\displaystyle h_i=e_{ii}-e_{i+1,i+1}

So we’ve got the (l+1)^2 basic matrices, but we take away the l+1 along the diagonal. Then we add back the l new matrices h_i, getting (l+1)^2-1 matrices in our standard basis for \mathfrak{sl}(l+1,\mathbb{F}), verifying the dimension.

We sometimes refer to the isomorphism class of \mathfrak{sl}(l+1,\mathbb{F}) as A_l. Because reasons.

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August 8, 2012 - Posted by | Algebra, Lie Algebras

5 Comments »

  1. […] with the special linear Lie algebras, these form the “classical” Lie algebras. It’s a tedious but straightforward […]

    Pingback by Orthogonal and Symplectic Lie Algebras « The Unapologetic Mathematician | August 9, 2012 | Reply

  2. […] that then the map is an automorphism of . Clearly this happens for all in the cases of and the special linear Lie algebra — the latter because the trace is invariant under a change of […]

    Pingback by Automorphisms of Lie Algebras « The Unapologetic Mathematician | August 18, 2012 | Reply

  3. […] some of the things we’ve been talking about. Specifically, we’ll consider — the special linear Lie algebra on a two-dimensional vector space. This is a nice example not only because it’s nicely […]

    Pingback by An Explicit Example « The Unapologetic Mathematician | August 18, 2012 | Reply

  4. Thanks for mentioning the “common mistake”. I had always thought it was true…

    Comment by Rafael | August 22, 2012 | Reply

  5. Just a comment: When you compute the trace of an arbitrary bracket, what you are actually showing is that the derived algebra of the general Lie algebra is _contained_ in the special one, not that they are equal. To show the equality we may appeal to the dimensionality argument,which you exhibit afterwards.

    Comment by Jose Brox | September 2, 2012 | Reply


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