An Explicit Example
Let’s pause and catch our breath with an actual example of 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 representative of some general phenomena, but also because the algebra itself is three-dimensional, which helps keep clear the distinction between as a Lie algebra and the adjoint action of on itself, particularly since these are both thought of in terms of matrix multiplications.
Now, we know a basis for this algebra:
which we will take in this order. We want to check each of the brackets of these basis elements:
Writing out each bracket of basis elements as a (unique) linear combination of basis elements specifies the bracket completely, by linearity. We call the coefficients the “structure constants” of , and they determine the algebra up to isomorphism.
Okay, now we want to use this basis of the vector space and write down matrices for the action of on :
Now, both and are nilpotent. In the case of we can see that sends the line spanned by to the line spanned by , the line spanned by to the line spanned by , and the line spanned by to zero. So we can calculate the powers:
and the exponential:
Similarly we can calculate the exponential of :
So now it’s a simple matter to write down the following element of :
In other words, , , and .
We can also see that and themselves are also nilpotent, as endomorphisms of the vector space . We can calculate their exponentials:
and the product:
It’s easy to check from here that conjugation by has the exact same effect as the action of : .
This is a very general phenomenon: if is any linear Lie algebra and is nilpotent, then conjugation by the exponential of is the same as applying the exponential of the adoint of .
Indeed, considering , we can write it as
where and are left- and right-multiplication by in . Since these two commute with each other and both are nilpotent we can write
That is, the action of is the same as left-multiplication by followed by right-multiplication by . All we need now is to verify that this is the inverse of , but the expanded Leibniz identity from last time tells us that , thus proving our assertion.
We can also tell at this point that the nilpotency of and and that of and are not unrelated. Indeed, if is nilpotent then is, too. Indeed, since and are commuting nilpotents, their difference — — is again nilpotent.
We must be careful to note that the converse is not true. Indeed, is ad-nilpotent, but itself is certainly not nilpotent.