Last time in this series, we saw the definition of the Lie algebra of a linear algebraic group $ G$ over a arbitrary field $ k$ as the set of differentiations $ f:k[G]\to k$; these are the $ k$-linear maps satisfying $ f(ab) = \epsilon(a)f(b) + f(a)\epsilon(b)$ where $ \epsilon:k[G]\to k$ is the counit morphism corresponding to the identity in $ G$.

In this post we will look at the geometric definition of the tangent space, which is natural when we consider the $ k$-points of $ G$ as a subset of affine space. Furthermore, we shall see an example of the adjoint representation, and how morphisms of algebraic groups correspond to morphisms of Lie algebras in the explicit case of $ G$ embedded into $ \rm{GL}_n$. This will allow us to write down an explicit formula for the adjoint representation.

This is important because over an arbitrary commutative ring $ k$, if a group scheme is represented by a finitely generated algebra then it must be the subgroup of the $ \rm{GL}_n$ group scheme over that ring for some $ n$. Thus all algebraic groups are matrix groups. Furthermore, it is easy to define subgroups of $ \rm{GL}_n$ so many examples we will see will be given in the form of a matrix group.

### The Geometric Tangent Space

Consider the classical situation of an affine algebraic variety $ X$ defined over an algebraically closed field $ k$. We will consider this variety embedded in $ k^n$ for some $ n$ so that it is given by the zero set of a finite set of polynomials $ f_1,\dots,f_n\in k[T_1,\dots,T_n]$. As usual, we write $ k[X]$ for the coordinate ring $ k[T_i]/(f_i)$ of $ X$ and by abuse of notation $ f\in k[X]$ denotes an equivalence class of polynomials represented by a polynomial $ f$.

For any $ f\in k[T_i]$ and $ x = (x_1,\dots,x_n) \in k^n$, we define $ d_x(f) = \sum_{i=1}^n \partial f/\partial T_i|_x(T_i – x_i)$. For the variety $ X$ and a point $ x\in X$, we define the tangent space at $ x$ to be the affine space $ T(X)_x$ of $ k^n$ as the variety defined by $ d_xf_1,\dots,d_xf_n$. This definition for instance really does give the tangent line to a point on a smooth curve, although it makes sense at any point, regardless of whether it is "smooth".

Given any $ f\in k[X]$, the linear polynomial $ d_xf$ becomes a $ k$-valued function on the tangent space. In order to work with the tangent space as a vector space, it is useful to introduce the change of coordinates so that the point $ x$ becomes the origin. Once we make this change of coordinates, $ d_xf$ becomes a linear functional on the tangent space, and in these new coordinates, $ d_x(f) = \sum_{i=1}^n \partial f/\partial T_i|_xT_i$.

Let $ M_x$ be the ideal of $ k[X]$ of functions that vanish at $ x$. As vector spaces, $ k[X] = k\oplus M_x$. Indeed, if $ f\in k[X]$ then $ f(t) – f(x)$ vanishes at $ x$ so $ f(t) = f(t) – f(x) + f(x)$. Hence we can define a map $ \Phi:k[X]\to k\oplus M_x$ by $ \Phi(f) = (f(x),f(t) – f(x))$. This map is well-defined because if $ g$ is represented of the class of $ f$, then $ f(x) = g(x)$, and by inspection it is bijective.

Since $ d_x$ applied to any constant is zero, the map $ d_x$ by restriction becomes a surjective map $ d_x:M_x\to \rm{Hom}_k(T(X)_x,k) = T(X)_x^*$. Furthermore, the kernel of this map is $ M_x^2$. Thus, we have a vector space isomorphism

Taking duals again (i.e. applying the $ \rm{Hom}_k(-,k)$) functor gives us a map

Here, a functional $ \varphi$ is mapped to $ \varphi\circ d_x$. Now, this is really useful, because in the case when $ X=G$ is an algebraic group, the Lie algebra of $ G$ is isomorphic to the differentiation of the coordinate ring $ k[G]$. Recall that these differentiations are the $ k$-linear maps $ d:k[G]\to k$ that satisfy $ d(fg) = d(f)\epsilon(g) + d(g)\epsilon(f)$ where $ \epsilon:k[G]\to k$ is the counit morphism. Hence $ \epsilon(f)$ is the same thing as evaluation of $ f$ at the identity, so that $ \varphi\circ d_e$ is easily verified to be a differentiation, this time considered as a $ k$-linear function $ k[G]\to k$—this makes sense because a differentiation is completely determined by what it does to $ M_e/M_e^2$. It is an exercise to verify that this correspondence does give an isomorphism between differentiations and the dual space of $ M_e/M_e^2$. This shows that the Lie algebra of an algebraic group over a field is a finite dimensional vector space.

Thinking of the Lie algebra as differentiations is nice because it is easy to see directly how morphisms of algebraic groups translate into morphisms of Lie algebras.

### Matrix Groups

Now, suppose that we have a given algebraic group $ G$ realised as a subgroup of $ \rm{GL}_n(k)$. This means that we have written the coordinate ring of $ G$ as the quotient of $ k[T_{11},\dots,T_{nn},\rm{det}^{-1}]$, which is the $ k$-algebra that represents the algebraic group scheme $ \rm{GL}_n$ over $ k$. In this notation $ T_{ij}$ corresponds to the $ i,j$ entry of a matrix so that a matrix with entries in $ k$ corresponds to a $ k$-point of this group scheme.

Given this quotient $ k$-algebra, it is straightforward to calculate the Lie algebra $ \mathfrak{g}$ of $ G$ in matrix form using the original definition of the tangent space.

For instance, in the case of $ SL_2(k)$, we can write its representing Hopf algebra as $ k[T_{11},T_{12},T_{21},T_{22},\rm{det}^{-1}]/(T_{11}T_{22} – T_{12}T_{21} – 1)$. By computing the tangent space and changing coordinates so that the identity matrix becomes the origin, we see that the Lie algebra of $ SL_2(k)$ is $ \mathfrak{sl}_2(k)$, the matrices of zero trace.

Now given a matrix $ M\in \mathfrak{sl}_2(k)$ of zero trace, how do we retrieve the corresponding differentiation on $ k[G]$? By our previous computation, it is $ \varphi_M\circ d_e$, where $ \varphi_M$ is the functional on functionals (= element of $ T(X)_x^{**}$) corresponding to the matrix $ M$. But from the natural isomorphism from a vector space to its double dual, we see that the differentiation corresponding to $ M$ is the map

where of course we keep in mind that $ d_ef$ is with respect to the coordinate change from $ e$ to $ 0$. Now how can we use this to determine the effect of an algebraic group homomorphism $ \alpha:G\to G$? Once we have a differentiation $ k[G]\to k$, we compose it with the comorphism $ \alpha^*:k[G]\to k[G]$. What does this mean for algebraic matrix groups?

The differentiation will always be of the form $ f\mapsto d_ef(M)$ where $ M$ is the corresponding matrix (=point in affine space) where $ M$ is in the Lie algebra. If $ \alpha:G\to G$ is a morphism of algebraic groups then $ d(\alpha)(M) = M'$ where $ d(\alpha)$ denotes the Lie functor applied to $ \alpha$. By our explicit formula, we must have $ d_x(\alpha^*(f))(M) = d_x(f)(M')$ for every $ f\in k[G]$. In particular, setting $ f = T_{ij}$ gives the right-hand side the $ i,j$-entry of $ M'$ and the left-hand side something computable in terms of $ M$, so that this gives $ M'$ explicitly.

### An Example: The Adjoint Representation

The above formalism is enough to completely work out how any endomorphism of an algebraic group $ G$ translates to a morphism of Lie algebras using the Lie functor. The same idea works for any morphism of algebraic groups. In fact, we don't even need our algebraic group written as a matrix group, although in matrix form multiplication is always the same so it's easier to do computations.

One of the most important examples is the case of $ x\in G$ acting on $ G$ via conjugation. This gives rise to the morphism $ \rm{Int}_x:G\to G$ given by $ y\mapsto xyx^{-1}$. The corresponding morphism of Lie algebras $ \rm{Ad}_x:\mathfrak{g}\to\mathfrak{g}$ is called the adjoint morphism and this gives an action of $ G$ on $ \mathfrak{g}$ called the adjoint representation.

If $ G$ is given in terms of a matrix group, then for a fixed $ x$, the morphism $ \rm{Int}_x$ is given by linear functions of the $ T_{ij}$ with coefficients depending on the matrix of $ x$. A good exercise is to write down the matrix and go through the method outlined above for calculating the corresponding adjoint action for some fixed algebraic group $ G$. Let us do so for $ \rm{SL}_n(k)$. In this case recall that we had the representing Hopf algebra as

As an algebraic group the $ k$-points are given by $ 2\times 2$ matrices of determinant one, like

Multiplication is given by matrix multiplication. If

is any element of $ SL_2(k)$ then (exercise):

Since these are linear functions of the $ T_{ij}$ as we have already remarked, the differential operator $ d_e$ will not change these linear polynomials once shifted from $ e$ to $ 0$ so that the formula

Shows that $ \rm{Ad}_x$ is also conjugation by $ x$; this of course will not happen in general if the corresponding comorphism is not given by linear polynomials as was above. Thus the adjoint representation of $ SL_2(k)$ on $ \mathfrak{sl}_2(k)$ is conjugation on the vector space of matrices of zero trace. Notice that if we put a zero-trace matrix in the above conjugation formula we again get a matrix of zero trace, which is to be expected, although we already know this *a priori* from linear algebra.

Of course, our formula for conjugation is rather messy. After all $ \mathfrak{sl}_2(k)$ is just a three-dimensional vector space, so we can see how this adjoint representation works more clearly by choosing a basis. For instance, a basis of $ \mathfrak{sl}_2(k)$ is given by

Then it is an exercise in linear algebra that with respect to this basis, the representation on $ k^3$ is given by

There you have it! The adjoint representation of $ \rm{SL}_2(k)$. Of course, there is nothing terribly special about the $ k$-points: this formula gives a Hopf algebra morphism $ k[\rm{GL}_3]\to k[\rm{SL}_2]$ and thus a morphism of group schemes $ \rm{SL}_2\to\rm{GL}_3$.

### Further Remarks

Notice how here we have not really used the Lie algebra structure of the Lie algebra at all. Rather, we just wanted to obtain a representation. In later posts, we will see how the Lie algebra structure comes into play and how we can prove some important properties of the adjoint representation.