A Lie algebra with trivial center and commutative radical

algebraic-groupslie-algebraslie-groupssemisimple-lie-algebras

Let $\mathfrak g$ be a complex linear Lie algebra. Assume that the center $\mathfrak z$ of $\mathfrak g$ is trivial

Let $\mathfrak r$ be the radical of $\mathfrak g$. If $\mathfrak r$ is abelian, then $\mathfrak g$ is semisimple?

What if $\mathfrak g$ is the Lie algebra of an algebraic complex linear group?

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Your question whether $\mathfrak{g}$ is semisimple is equivalent to whether necessarily $\mathfrak{r} = 0$. The answer is no. A counterexample is given by

$$\mathfrak{g} = \left\{ \pmatrix{a & b & d\\ c& -a& e\\ 0&0&0} : a,b,c,d,e \in \Bbb C\right\}$$

where the radical $$\mathfrak{r} = \left\{ \pmatrix{0 & 0 & d\\ 0& 0& e\\ 0&0&0} : d,e \in \Bbb C\right\}$$ is two-dimensional.

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Theorem of Highest Weight for Reductive vs Semisimple complex algebraic group

For a split reductive group there is exactly the same kind of story. I will describe the setup as precisely as possible so that you can see the connection. This may not be an answer to the question but it should go some way to describing what is happening.

Fix a field $k$, and let $T \subseteq B \subseteq G$ be a "pinning" of the split reductive group $G$, i.e. $B$ is a Borel subgroup containing a maximal torus $T$. The torus $T$ determines a weight lattice $X$ and coweight lattice $X^\vee$, a pair of free $\mathbb{Z}$-modules together in a perfect pairing $$ \langle-, - \rangle: X^\vee \times X \to \mathbb{Z}.$$ These $\mathbb{Z}$-modules are defined as hom-spaces in the category of algebraic groups: $$ X^\vee = \operatorname{Hom}_{\mathsf{AlgGrp}}(\mathbb{G}_m, T), \quad X = \operatorname{Hom}_{\mathsf{AlgGrp}}(T, \mathbb{G}_m),$$ and the perfect pairing is simply composing the homs to get an endomorphism of $\mathbb{G}_m$, which are classified by integers.

From $G$ we get sets of roots $\Phi \subseteq X$ and coroots $\Phi^\vee \subseteq X^\vee$, equipped with a bijection $\alpha \mapsto \alpha^\vee$. The Borel subgroup $B$ determines the set $\Phi^+$ of positive roots and the set $S \subseteq \Phi^+$ of simple roots.

Important note: The combinatorial data of $(X^\vee, X, \langle -, - \rangle, \Phi, \Phi^\vee, (-)^\vee: \Phi \to \Phi^\vee)$ completely determines the split reductive group $G$ up to isomorphism. This is called the root datum of the group.

The dominant weights $X_+ \subseteq X$ are those weights which pair with the simple coroots positively: $$ X_+ = \{\lambda \in X \mid \langle \alpha^\vee, \lambda \rangle \geq 0 \}.$$ Then the theorem is that the irreducible representations of $G$ are classified by the dominant weights $X_+$.

Here are some examples of root data.

$SL_2$: $X^\vee = X = \mathbb{Z}$ with the perfect pairing being the dot product. The simple root is $\alpha = 2$, the simple coroot is $\alpha^\vee = 1$. Note that the simple coroot is a basis for $X^\vee$, but the root is not a basis for $X$. There is a fundamental weight $\varpi = 1$ defined by the condition $\langle \alpha^\vee, \varpi \rangle = 1$. The dominant weights are $0, 1, 2, \ldots$.
$PGL_2$: $X^\vee = X = \mathbb{Z}$ with the perfect pairing being the dot product. The simple root is $\alpha = 1$, the simple coroot is $\alpha^\vee = 2$. There is no fundamental weight $\varpi$ defined by the condition $\langle \alpha^\vee, \varpi \rangle = 1$, since we would have to have $\varpi = 1/2$ which is not a member of $X$. The dominant weights are $0, 1, 2, \ldots$.
$GL_n$: $X^\vee = X = \mathbb{Z}^n$ with the dot product. The simple roots and coroots are $\alpha_1 = (1, -1, 0, \ldots, 0)$, $\alpha_2 = (0, 1, -1, 0, \ldots, 0)$, $\ldots$, $\alpha_{n-1} = (0, \ldots, 0, 1, -1)$. The fundamental weight $\varpi_i$ should satisfy $\langle \alpha_j^\vee, \varpi_i \rangle = \delta_{ij}$, however since there are only $n-1$ simple coroots and we have a rank $n$ space, there is no unique choice of $\varpi_i$. We could choose $\varpi_1 = (1, 0, \ldots, 0)$, $\varpi_2 = (1, 1, 0, \ldots, 0)$ and so on up to $\varpi_{n-1} = (1, \ldots, 1, 1, 0)$, then also set $\det = (1, \ldots, 1)$, then the vectors $(\varpi_1, \ldots, \varpi_{n-1}, \det)$ would form a basis for $X$. Note that this is a non-canonical choice however. We could add or subtract multiples of $\det$ from any of our $\varpi_i$ and still have vectors satisfying the pairing requirement above.

So how is the case of $G$ any different to that of $\mathfrak{g} = \operatorname{Lie}(G)$? The key point is integrality: for reductive groups it is automatic and canonical, for reductive Lie algebras it is non-canonical and needs to be imposed.

Passing to $\mathfrak{g}$ loses some data even for semisimple algebraic groups, for instance $SL_n$ and $PGL_n$ are nonisomorphic as algebraic groups, but their Lie algebras are isomorphic. However, the root data sees the difference, which has to do with exactly how the roots and coroots sit inside their lattices.

Passing to $\mathfrak{g}$ for reductive groups which are not semisimple has a larger problem: we do not know what to call an "integral" weight for the centre, since we have thrown away the integrality that used to be there. What's worse is that there is no canonical way of getting it back.

And in fact, if you simply start with the reductive Lie algebra $\mathfrak{g}$, then anything goes for the centre! Consider for example the one-dimensional reductive Lie algebra $\mathfrak{g} = \mathfrak{gl}_1$: then irreducible finite-dimensional representations of this are classified by $\mathbb{C}$, not by any $\mathbb{Z}$-lattice. Similarly, irreducible finite-dimensional representations of $\mathfrak{gl}_n$ are classified by an irrep of $\mathfrak{sl}_n$ and a complex number $\mathbb{C}$.

To wrap this rambling post up, I would say that there is a connection between the classification of irreps in a reductive group $G$ and its reductive Lie algebra $\mathfrak{g} = \operatorname{Lie}(G)$, but the best way to uniformly state it is for reductive groups, and then extend it to the Lie algebra.

From the pinning $T \subseteq B \subseteq G$ of the group you get the Lie algebra $\mathfrak{g} = \operatorname{Lie}(G)$, a choice of Borel subalgebra $\mathfrak{b} = \operatorname{Lie}(B)$, and Cartan subalgebra $\mathfrak{h} = \operatorname{Lie}(T)$. Furthermore, every weight $\lambda \in X$ is a group homomorphism $\lambda: T \to \mathbb{G}_m$ which can be differentiated to a Lie algebra homomorphism $d\lambda: \mathfrak{h} \to \mathbb{C}$, i.e. an element of the dual $\mathfrak{h}^*$. So you get a canonical set of integral weights $X \subseteq \mathfrak{h}^*$ containing a root system and so on, and now you can define what it means to be an "integral" representation of $\mathfrak{g}$. And from here the classification will go through.

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Theorem of Highest Weight for Reductive vs Semisimple complex algebraic group

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