In Linear Algebraic Groups 2, we defined the Lie algebra of an algebraic group $ G$ to be the Lie algebra of all left-invariant derivations $ D:k[G]\to k[G]$ where $ k[G]$ is the representing algebra of $ G$. However, we were left trying to figure out exactly how a morphism $ \varphi:G\to H$ determines a morphism $ d\varphi:\mathcal{L}(G)\to\mathcal{L}(H)$. It turns out that the answer is slightly tricky, and thus the Lie algebra in terms of left-invariant derivations $ k[G]\to k[G]$ really can’t be the “real” or “most natural” definition!

Now, I could just give the formula for morphisms right away, but I think it would be a bit unmotivated. So before looking at the formula, let us first look at another way of defining the Lie algebra of an algebraic group. In fact, the next defintion is much more natural in that the functoriality of the Lie algebra construction will be clear, whereas in this case it was not.

The Second Definition

To keep things brief, I will assume the reader has seen Hopf algebras and their relation to affine group schemes. All the basic material can be found in Waterhouse’s book, “An Introduction to Affine Group Schemes”.

As usual, let $ G$ be the group scheme and $ k[G]$ be the representing algebra. We denote by $ \delta:k[G]\to k[G]\otimes k[G]$ the comultiplication map. This is the map that defines the multiplication on $ G$. We denote by $ \epsilon:k[G]\to k$ the counit, which defines the unit element in $ G$.

Now, we consider differentiations of $ k[G]$. These are $ k$-linear maps $ f:k[G]\to k$ that satisfy $ f(ab) = \epsilon(a)f(b) + f(a)\epsilon(b)$ for all $ a,b\in k[G]$. Remember, $ \epsilon:k[G]\to k$ is the counit morphism. We define the Lie algebra of the algebraic group $ G$ to be the algebra of differentiations, which we will denote by $ \mathcal{L}_{\mathrm{dif}}$ (I’ll explain the multiplication later), together with the Lie bracket $ [f,g] = fg – gf$.

The reader should think about the following before continuing: why is this definition more natural than the previous one?

Why indeed? A morphism $ \varphi:G\to H$ of algebraic groups induces the comorphism $ \varphi^*:k[H]\to k[G]$. Now, if $ f:k[G]\to k$ is a differentiation then $ f\circ \varphi^*$ is a differentiation! Thus functoriality is clear in this case.

What About That Multiplication?

So I left a little something out of the above description. In fact, I didn’t say what the multiplication is in the set of differentiations. We would like a multiplication so that the set of differentiations becomes a ring and thus using the Lie functor we can make it into a genuine Lie algebra.

The differentiations are in fact a subset of $ k[G]^* = \mathrm{Hom}_{k}(k[G],k)$. So this dual space, $ k[G]^*$ has an algebra structure, but what is it? (Note, that we don’t need this now to define how a morphism of algebraic groups is taken to a morphism of Lie algebras in the left-invariant derivation setting, but this will be helpful when we want to work with these differentiations directly).

Let $ f,g:k[G]\to k$ be two differentiations. We get a third $ k$-linear map $ f*g:k[G]\to k$ via $ (f*g)(x) = (f\otimes g)(\delta(x))$. Here $ f\otimes g$ is a homomorphism $ k[G]\otimes_k k[G]\to k\otimes_k k$ and $ k\otimes k$ is identified with $ k$. It is left an an exercise to verify that this is also a differentiation.

Note that this is just a special case of giving the set $ \mathrm{Hom}_k(C,A)$ an algebra structure whenever $ C$ is a coalgebra and $ A$ is an algebra.

Morphisms Again!

So, recall that we have a morphism $ \varphi:G\to H$ of algebraic groups, and we would like to determine how this gives rise to the morphism of the corresponding Lie algebras, which we will denote by $ d\varphi:\mathcal{L}(G)\to\mathcal{L}(H)$.

We know how this works in the differentiations case (i.e. with the second definition). Thus, all we need to do is demonstrate that the two definitions are equivalent (i.e. show that the two Lie algebras for $ G$ are isomorphic) and use this equivalence to see what $ d\varphi$ actually is. Of course, we could just guess what $ d\varphi$ is directly, but then naturality would not be clear.

Let $ \mathcal{L}$ be the Lie algebra functor defined as derivations, and $ \mathcal{L}_{\mathrm{dif}}$ be the one defined as differentiations. For an algebraic group $ G$, define a map $ \Phi:\mathcal{L}(G)\to\mathcal{L}_{\mathrm{dif}}(G)$ by $ \Phi(f) = \epsilon\circ f$. This we claim is well-defined, which we leave as a short exercise. The inverse map is $ \Psi:\mathcal{L}_{\mathrm{dif}}(G)\to\mathcal{L}(G)$ defined by $ \Psi(f) = (1_{k[G]}\otimes f)\circ\delta$. This lands in $ k[G]\otimes_k k$, which we identify with $ k[G]$!

I recommend that the reader check that these are well-defined, homomorphisms of Lie algebras, and inverse to each other.

Now, we can say finally what happens to a morphism $ \varphi:G\to H$ under the Lie algebra functor $ G\mapsto \mathcal{L}(G) = \mathcal{L}(G)$. This functor should take $ \varphi$ to $ d\varphi:\mathcal{L}(G)\to\mathcal{L}(H)$. Given a derivation $ f:\mathcal{L}(G)$, the corresponding element in $ \mathcal{L}_{\mathrm{dif}}(G)$ is $ \epsilon\circ f$.

Under $ d\varphi$ here, this is just sent to $ \epsilon\circ f\circ \varphi^*$ where $ \varphi^*$ is the comoprhism $ \varphi^*:k[H]\to k[G]$ associated to $ \varphi$. Finally, sending this triple composition back to $ \mathcal{L}(G)$ gives $ (1_{k[G]}\otimes \epsilon\circ f\circ \varphi^*)\circ\delta$. So we can say explicitly: if $ d\varphi$ denotes the image of $ \varphi$ under the Lie algebra functor as defined by derivations, then

$ d\varphi(f) = (1_{k[G]}\otimes\epsilon\circ f\circ\varphi^*)\circ\delta$.

(Wouldn’t this be nice to have on a t-shirt?) We call $ d\varphi$ the differential of $ \varphi$ because formally it is similar to the derivative.

What’s Next?

After having spent some time in the realm of algebra, and seeing two definitions of the Lie algebra of an algebraic group, it’s time to connect this up to geometry to answer a few basic questions about this Lie algebra functor, as well as do a few computations to see what kind of beast it is. In the remaining parts of this series on this Lie algebra business, we will look at tangent spaces, examples of differentials of a morphism, and a bit about how these Lie algebras are related to representations of linear algebraic groups.

So in the next post we will spend some time on tangent spaces and other related things.

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