Significance of $G$ being reductive

Question

Let $G$ be an algebraic group. What is the significance of $G$ being reductive?

By reductive, I mean that: Let $R(G)$ be the largest subgroup such that $R(G)$ is a connected, solvable, normal subgroup of $G$, and let $R_u(G)$ be the unipotent elements in $R(G)$, then $G$ is reductive if $R_u(G)=\{e\}$.

How should I think about the property of being reductive?

Answer 1

Over an algebraically closed field reductive groups are classified by their root datums. They are generally the classic matrix groups like $\mathrm{GL}_n(k)$, $\mathrm{SL}_n(k)$, $\mathrm{PGL}_n(k)$, $\mathrm{Sp}_n(k)$, $\mathrm{SO}_n(k)$. There are some more complicated ones like the spin groups and depending on how general you want to be there are also twists, but personally just I think of them as being classical matrix groups.

Answer 2

If you take for granted that semisimple algebraic groups are important, a connected, reductive group is nothing more than a fattening of a semisimple group by a torus. Specifically, $G$ is connected reductive if and only if it is isomorphic to a product $(H \times S)/N$, where $H$ is a semisimple algebraic group, $S$ is a torus, and $N$ is a finite normal subgroup of $H \times S$. One can then choose $N$ so that $H$ embeds into $G$ as its derived group as $h \mapsto (h,1)N$, and $S$ similarly embeds into $G$ as the connected component of the center. Then $G$ is generated by $H$ and $S$, which have finite intersection.

If you don't take semisimple groups for granted, I will write an answer from a different point of view tomorrow.

Answer 3

As an extended comment, it should also be noted that over a field of characteristic zero (where I spend most my days anyway), being reductive is equivalent to being geometrically reductive, which might be a more intuitively interesting property: Being geometrically reductive means that every (finite-dimensional) rational $G$-module $M$ decomposes as a direct sum of irreducible $G$-modules. Without the ability to decompose representations into irreducibles, representation theory becomes much harder.

Edit: Regarding the question why such groups are called geometrically reductive: I do not know the answer for certain and would welcome any comments, but I would speculate that this is historical in the following sense: What we call geometrically reductive today was studied very early in invariant theory (which classically happens over $\Bbb C$), and I dare say maybe they called it "reductive". It turns out that being geometrically reductive is a not-too-narrow, sufficient condition for admitting a Reynolds operator, which is a cruicial tool for constructing quotients in algebraic geometry. I suspect that our current notion of reductive was discovered later and deemed the better definition because the notions coincide in characteristic zero, but the modern one is more correct in the sense that it gives the same sufficient criterion for constructing quotients, regardless of the characteristic of the base field. Since the older of the two notions was rooted in Geometry, its new name was chosen to be "geometrically reductive". Again, this is speculation.

Answer 4

If $G$ is a connected, reductive group, you should think of the structure of $G$ in terms of its roots, and the facts about parabolic and Levi subgroups that the roots entail. Here is part of the story. Let $B$ be Borel subgroup of $G$, and let $T$ be a maximal torus of $G$. For an example, you may keep in mind $G = \textrm{GL}_n, B$ the upper triangulars, and $T$ the diagonal matrices. There is a unique Borel subgroup $B^-$ (think lower triangular) containing $T$ such that $B \cap B^- = T$. Let $U, U^-$ be the unipotent radicals of $B, B^-$. Then $B = TU, B^- = TU^-$ (semidirect products).

Let $\mathfrak g$ be the Lie algebra of $G$ (think all $n$ by $n$ matrices). The adjoint action $\textrm{Ad}: T \rightarrow \textrm{GL}(\mathfrak g)$ (think conjugation by a diagonal matrix) is a rational representation of $T$. Then $\mathfrak g$ decomposes into a direct sum of one dimensional spaces

$$\mathfrak g = \bigoplus\limits_{\alpha \in X(T)} \mathfrak g_{\alpha}$$

where $\alpha: T \rightarrow \mathbb{G}_m$ is a rational homomorphism of $T$ which acts on $\mathfrak g_{\alpha}$ as $\textrm{Ad}(t)X_{\alpha} = \alpha(t) X_{\alpha}$. Then $\mathfrak g_0$ is the Lie algebra of $T$ (think all diagonal matrices), and it is a consequence of reductivity of $G$ that for $0 \neq \alpha \in X(T)$, $\mathfrak g_{\alpha}$ is either one or zero dimensional. Those $0 \neq \alpha$ for which $\mathfrak g_{\alpha}$ is one dimensional are called the roots of $T$ in $G$ (in our example, the $\mathfrak g_{\alpha}$ will be the one dimensional spaces spanned by the elementary matrices $E_{ij}, i \neq j$). Let $\Phi$ be the set of roots.

Let $\mathfrak u, \mathfrak u^-$ be the Lie algebras of $U, U^-$ (for $\mathfrak u$, think upper triangular matrices with zeroes on the diagonal, similarly $\mathfrak u^-$). Then $\mathfrak g = \mathfrak u^- \oplus \mathfrak g_0 \oplus \mathfrak u$, and each of $\mathfrak u, \mathfrak u^-$ is a direct sum of the various one dimensional spaces $\mathfrak g_{\alpha} : \alpha \in \Phi$. Together they exhaust all the roots. Those $\alpha$ for which $\mathfrak g_{\alpha} \subseteq \mathfrak u$ will be called positive, and the others negative. Let $\Phi^+$ be the set of positive roots, and $\Phi^-$ the set of negative roots. Then $\Phi^- = - \Phi^+$.

Let $\Delta$ be the set of positive roots which cannot be expressed as a sum of two other positive roots (if $e_1, ... , e_n$ is the usual basis for $X(T)$, then $\Delta = \{e_1 - e_2, ... , e_{n-1} - e_n \}$, and $\Phi^+ = \{ e_i - e_j : i < j \}$). For each root $\alpha$, there is a unique connected one dimensional algebraic group $U_{\alpha}$, isomorphic to $\mathbb{G}_a$, which is normalized by $T$ and whose Lie algebra is $\mathfrak g_{\alpha}$ (if $\alpha = e_i - e_j$, then $\mathfrak g_{\alpha}$ has basis $E_{ij}$, and $U_{\alpha} = \{ I + xE{ij} : x \in \mathbb G_a \}$).

The standard Levi subgroups of $G$ are in bijection with the subsets of $\Delta$. Specifically, if $\theta$ is any subset of $\Delta$, let

$$T_{\theta} = (\bigcap\limits_{\alpha \in \theta} \textrm{Ker}(\alpha))^0$$

and define the Levi subgroup $M_{\theta} = Z_G(T_{\theta})$. For example, in $\textrm{GL}_4$, let $\theta = \{e_1 - e_2, e_3 - e_4\}$. Then $T_{\theta} = \textrm{diag}(x,x,y,y)$, and $Z_{\theta}$ is the block diagonal sum of two copies of $\textrm{GL}_2$. Then $Z_{\theta}$ is connected reductive, with maximal torus $T$ and Borel subgroup $Z_{\theta} \cap B$, with respect to which it has its own set of roots, namely those roots of $\Phi$ obtained by linear combination of $\theta$.

The standard parabolic subgroups of $G$ are also in bijection with the subsets of $\theta$. Every positive root is a unique positive, integer-linear combination of elements of $\Delta$. As $\alpha$ ranges through those positive roots which, when written as a combination of things in $\Delta$, has a nonzero coefficient belonging to something outside $\theta$, the subgroup $N_{\theta}$ generated by these $U_{\alpha}$ is a connected, unipotent group. The parabolic subgroup $P_{\theta}$ is $M_{\theta}N_{\theta}$, which is a semidirect product, with $M_{\theta}$ normalizing $N_{\theta}$. For example, in the example I mentioned with $\textrm{GL}_4$,

$$N_{\theta} = \begin{pmatrix} 1 & 0 & a & b \\ 0 & 1 & c & d \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$$

The standard parabolic subgroups $P_{\theta} : \theta \subseteq \Delta$ are exactly those subgroups of $G$ which contain $B$. Note that $P_{\Delta} = G, P_{\emptyset} = B$.