## Special Linear Lie Algebras

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

Take a vector space with dimension and start with . Inside this, we consider the subalgebra of endomorphisms whose trace is zero, which we write and call the “special linear Lie algebra”. This is a subspace, since the trace is a linear functional on the space of endomorphisms:

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:

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

so actually not only is the bracket of two endomorphisms in back in the subspace, the bracket of any two endomorphisms of lands in . In other words: .

Choosing a basis, we will write the algebra as . It should be clear that the dimension is , since this is the kernel of a single linear functional on the -dimensional , but let’s exhibit a basis anyway. All the basic matrices with are traceless, so they’re all in . Along the diagonal, , so we need linear combinations that cancel each other out. It’s particularly convenient to define

So we’ve got the basic matrices, but we take away the along the diagonal. Then we add back the new matrices , getting matrices in our standard basis for , verifying the dimension.

We sometimes refer to the isomorphism class of as . Because reasons.

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

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[...] 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 |

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

Comment by Rafael | August 22, 2012 |

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 |