# The Unapologetic Mathematician

## 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.

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

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

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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 […]

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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 […]

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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