Posted by Jason Polak on 23. August 2014 · Write a comment · Categories: algebraic-geometry, group-theory · Tags:

Suppose one day you run into the following algebraic group, defined on $\mathbb{Z}$-algebras $R$ by
$$G(R) = \left\{ \begin{pmatrix}
a_{11} & a_{12} & a_{13} & a_{14} \\
a_{12} & a_{11} & -a_{14} & -a_{13} \\
a_{13} & -a_{14} & a_{11} & -a_{12} \\
a_{14} & -a_{13} & -a_{12} & a_{11}
\end{pmatrix} \in\mathrm{GL}_4(R) \right\}$$
Can you figure out what this group is? I actually did run into this group one day, but luckily I discovered its true identity. Today we’ll see one way to do so.

One thing to try to identify unknown algebraic groups in practice is to check out their matrix multiplication. If $A = (a_{ij}), B = (b_{ij})$ have the above form then $C = AB$ has first row
$$
c_{11} = a_{11} b_{11} + a_{12} b_{12} + a_{13} b_{13} + a_{14} b_{14} \\c_{12}= a_{12} b_{11} + a_{11} b_{12} – a_{14} b_{13} – a_{13} b_{14} \\c_{13}= a_{13} b_{11} – a_{14} b_{12} + a_{11} b_{13} – a_{12} b_{14} \\c_{14}= a_{14} b_{11} – a_{13} b_{12} – a_{12} b_{13} + a_{11} b_{14}
$$
(We only need to specify the first row to get the whole matrix in this group). Can you tell what the group is yet? At first, I couldn’t, so let’s try something else. One can check now that this group is actually commutative, and this would be enough to determine what it is given the context of where it came from, but let’s assume we don’t know that. Instead, let’s take the determinant:
$$
\mathrm{det}(A) = (a_{11} + a_{12} + a_{13} – a_{14})(a_{11} + a_{12} – a_{13} + a_{14})\\\times(a_{11} – a_{12} + a_{13} + a_{14})(a_{11} – a_{12} – a_{13} – a_{14})$$
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Here is an old classic from linear algebra: given an $ n\times n$ matrix $ A = (a_{ij})$, the determinant of $ A$ can be calculated using the permuation formula for the determinant:

$ \det(A) = \sum_{\sigma\in S_n} (-1)^\sigma a_{1\sigma(1)}\cdots a_{n\sigma(n)}$.

Here $ S_n$ denotes the permutation group on $ n$ symbols and $ (-1)^\sigma$ denotes the sign of the permutation $ \sigma$. The computation of the determinant is much more easily done via the ‘minor expansion method’, but this just reduces to the above formula. Now, typically we would never use the above formula to calculate the determinant, but that doesn’t mean it isn’t useful!

The Lie Algebra of an Algebraic Group

In fact, computating the Lie algebra of $ \mathrm{SL}_n$ over a fixed commutative ring $ k$ happens to be a quick and interesting application of the above determinant formula. First, let us recall the definition of the Lie algebra, or at least one of the many equivalent definitions. Consider the algebra of dual numbers, $ k[\tau]/(\tau^2)$. Each element in this algebra can be written uniquely as $ a + b\tau$: this gives a $ k$-algebra map $ C: k[\tau]/(\tau^2)\to k$ by sending $ a + b\tau$ to $ a$.
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Posted by Jason Polak on 03. August 2013 · Write a comment · Categories: algebraic-geometry, group-theory · Tags: ,

Conventions: $ G$ is an algebraic group over an algebraically closed field $ k$ and we identify $ G$ with $ G(k)$.

Consider the algebraic groups $ \mathbb{G}_a$ and $ \mathbb{G}_m$. They are the only one-dimensional connected groups and they are both solvable. What about two-dimensional connected groups? It turns out that if $ \mathrm{dim} G\leq 2$ then $ G$ is solvable.

For $ \mathrm{dim} G= 3$, this is no longer true, for instance $ \mathrm{SL}_2$ is $ 3$-dimensional but not solvable since it is perfect, i.e. equal to its commutator subgroup. So let’s prove our theorem:

Theorem. Let $ G$ be a connected algebraic group over $ k$. If $ \mathrm{dim} G = 2$ then $ G$ is solvable.

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Here is a classic problem of geometric invariant theory: let $ G$ be a reductive linear algebraic group such as $ \mathrm{GL}_n$ and let $ \mathfrak{g}$ be its Lie algebra. Determine the invariant functions $ k[\mathfrak{g}]^G$, where $ G$ acts on $ \mathfrak{g}$ via the adjoint action. This problem is motivated by the search for quotients: What is the quotient $ \mathfrak{g}/G$? Here, the action of $ G$ on $ \mathfrak{g}$ is given by the adjoint action. More explicitly, an element $ g\in G$ acts via the differentiation of $ \mathrm{Int}_g$, where $ \mathrm{Int}_g$ is conjugation by $ g$ on $ G$.

For simplicity, we will stay in the realm of varieties over an algebraically closed field $ k$ of characteristic zero.

First, we should ask:

What should $ \mathfrak{g}/G$ even mean?

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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.
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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.
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