5
$\begingroup$

Wikipedia1 defines a reductive group $G$ as an algebraic group with trivial unipotent radical. The radical is the connected component of identity in the maximal normal solvable subgroup of $G$. The unipotent radical is the set of the unipotent elements of the radical. An element $u$ is unipotent if $(u-1)$ is nilpotent.

Question 1 - This last line is where everything breaks down in my head. If we talk of unipotent then there is a clear difference between what $1$ is and what $0$ is. In this case we have to be in a ring. But here we are just in a group $(G)$. What am I missing?

EDIT -

Question 2 - In this book (chapter 5) I have come across the definition that an affine algebraic group $G$ is reductive if its maximal connected solvable subgroup is a torus. (This is what Tobias mentions in his comment). But I am not sure how the definitions are equivalent when $G$ is affine.

Thank you.

1Link to the revision of the Wikipedia article at the time of this post.

$\endgroup$
9
  • 1
    $\begingroup$ The very first line in the Wikipedia page you link to says "In mathematics, a reductive group is an algebraic group G over an algebraically closed field". Do you know what it means to talk of a group "over a field"? $\endgroup$
    – user247327
    Commented Jun 18, 2016 at 12:55
  • 2
    $\begingroup$ An algebraic group over a field is a group which is also a variety over that field (with the group operations being compatible with the variety structure in a particular way). Varieties are like manifolds but of a more algebraic nature. An affine variety over $\mathbb{F}$ is a subset of $\mathbb F^n$ definable as the set of roots of some polynomial. A general variety is a space made by stitching together affine varieties, i.e. it looks locally like an affine variety. $\endgroup$
    – Oscar Cunningham
    Commented Jun 18, 2016 at 15:39
  • 2
    $\begingroup$ @Rise For most algebraic groups you'll come across the underlying variety is affine. There's a theorem that such groups are isomorphic to a subset of some matrix group $\mathrm{GL}_n(\mathbb{F})$ (a subset defined as the roots of a polynomial in fact). In this case each element of the group can be seen as a matrix, so the expression "$u-1$" makes sense. $\endgroup$
    – Oscar Cunningham
    Commented Jun 18, 2016 at 15:46
  • 2
    $\begingroup$ I will just leave this as a comment as I feel it lacks too many details for a proper answer: Once you have shown that the concept of being unipotent actually makes sense (i.e. that it does not depend on a choice of embedding), one direction becomes easy as one just embeds the torus as the set of diagonal matrices. The other direction is a bit more involved. First, one shows (this is as far as I recall known as the Lie-Kolchin theorem) that a closed, conneded, solvable subgroup of $GL_n$ is isomorphic (in fact, conjugate) to a subgroup of the group of upper triangular matrices. (cont.) $\endgroup$
    – Tobias Kildetoft
    Commented Jun 18, 2016 at 18:43
  • 2
    $\begingroup$ Then one shows that if such a group of upper triangular matrices has no unipotents, then it consists of diagonal matrices (one thing to start with is noticing that all commutators are unipotent, so the group must at least be abelian). Then finally, one needs to use that a closed, connected subgroup of a torus is itself a torus. $\endgroup$
    – Tobias Kildetoft
    Commented Jun 18, 2016 at 18:44

1 Answer 1

Reset to default
3
$\begingroup$

I don't know how to answer for algebraic groups in general, only affine algebraic groups over an algebraically closed field $k$. Remember that an affine algebraic group is up to isomorphism the same thing as a closed subgroup of $\textrm{GL}_n$ for some $n$. However, an isomorphism of such a group into $\textrm{GL}_n$ is not canonical.

Question 1: For any matrix $x \in \textrm{GL}_n$, there exist unique $x_s, x_u \in \textrm{GL}_n$ such that $x = x_sx_u = x_ux_s$, $x_s$ is diagonalizable, and $x_u$ is unipotent. We call $x_u, x_s$ the unipotent and semisimple components of $x$, respectively. This is the multiplicative version of Jordan canonical form.

Let $G$ be such a LAG over $k$. If $x \in G$, there exist unique $x_s, x_u \in G$ such that for any isomorphism $\varphi$ of $G$ onto a closed subgroup of $\textrm{GL}_n$ for some $n$, $\varphi(x_u), \varphi(x_s)$ are the unipotent and semisimple components of $\varphi(x)$. This is proved in Chapter 2 of Springer, Linear Algebraic Groups. Thus an $x \in G$ being unipotent doesn't depend on how you regard $G$ as a closed subgroup of $\textrm{GL}_n$.

Let $R(G)$ be the radical of $G$: it is the unique maximal closed, connected, solvable, normal subgroup of $G$ (the notes you cited must have a typo; e.g. $\textrm{GL}_n$ is reductive, the upper triangular invertible matrices are a maximal connected closed solvable group, but that's not a torus).

Question 2: $R_u(G)$ is trivial if and only if $R(G)$ is a torus.

If $R(G)$ is a torus, then $R_u(G)$ (which is equal to the set of unipotent elements $R(G)_u$ of $R(G)$) is trivial, because it is a unipotent subgroup of a semisimple group.

Suppose $R_u(G) = R(G)_u$ is trivial. Now $R(G)$ is a connected solvable group. If $H$ is a connected solvable group, and $T$ is a maximal torus of $H$, then as a variety, $H$ is isomorphic to the product of $T$ and the set $H_u$ of unipotent elements of $H$ (6.3, Springer). If $H_u$ is trivial, that says that $T = H$. So a connected solvable group with no nontrivial unipotent elements is automatically a torus.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .