1
$\begingroup$

I am reading course notes by Tamás Szamuely called "lectures on linear algebraic groups". The definition he gives for an element of the general linear group to be unipotent is not entirely clear to me. I will present my understanding of it, any remarks whether my understanding is correct or wrong is appreciated.

Szamuely writes: "An endomorphism $h \in \text{End}(V)$ is unipotent if $h-\text{id}_V$ is nilpotent. (...) Let $V$ be a finite-dimensional vector space, $g\in \text{GL}(V)$. There exist uniquely determined elements $g_s,g_u \in \text{GL}(V)$ with $g_s$ semisimple, $g_u$ unipotent (...).".

I understand the definition of being unipotent in rings, such as the ring of endomorphisms $\text{End}(V)$. However, in the context of groups, such as the general linear group $\text{GL}(V)$, it confuses me to make use of the operation of addition and the neutral element $0$ with respect to this addition. Both of these do not exist in the general linear group. Do we, in some sense, embed the general linear group into the ring of endomorphisms in order to be able to talk about addition and zero? And if so, am I correct in saying that it is a crucial fact that any affine algebraic group is isomorphic to a Zariski-closed subgroup of the general linear group, because this allows us to define unipotency of elements in any affine algebraic group via the above mentioned embedding, which is not possible for every abstract group?

$\endgroup$
  • $\begingroup$ At the end of your question you ask about abstract groups. This is absolutely a concept specific to algebraic and Lie groups, and not abstract groups. My answer below explains how to extend the concept more generally to group schemes, but I don't think there's an analogous concept, for example, in finitely generated groups. $\endgroup$ – Jackson 2 hours ago
0
$\begingroup$

You're correct initially—there needs to be some notion of (linear) endomorphism ring to define unipotency. It is straightforward to view $\mathrm{GL}(V)$ as sitting inside $\mathrm{End}(V)$ to make this definition. And it is true that every affine algebraic group over a field is a closed subgroup of some $\mathbf{GL}_n$.

That said, in light of Kolchin's fixed point theorem (taken from Waterhouse, ch. 8):

Let $G$ be a group consisting of unipotent matrices. Then in some basis all elements of $G$ are strictly upper triangular.

We can define unipotency more broadly for an affine group scheme $G$ in terms of its linear representations. $G$ is a unipotent group scheme if for every nonzero linear representation $V$ of $G$ there is a nonzero fixed vector $v$ (i.e., a nonzero subspace $kv \subset V$ on which $G$ acts trivially). Then an element $g \in G(R)$ is unipotent if the closed subgroup it topologically generates is a unipotent group scheme.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.