Question about linear algebraic groups split vs isotropic

I am reading notes on linear algebraic groups and I'm getting confused with some definitions and I would appreciate any clarification.

They define $$G$$ to be split if there exists a maximal torus $$T$$ of $$G$$ that is split.

Then later on they say: If there is no split torus contained in $$G$$ then $$G$$ is said to be anisotropic. Otherwise $$G$$ is said to be isotropic. If $$G$$ is isotropic then there exists a maximal torus $$T$$ contained in $$G$$ unique up to conjugation.

1) With this definition, to me it looks like split and isotropic mean the same thing... What am I missing here?

2) When $$G$$ is split we have the decomposition of the Lie algebra of $$G$$ as $$\mathfrak{g} = \mathfrak{t} \oplus \oplus_{\alpha \in \Phi(G,T)} \mathfrak{g}_{\alpha}$$ where $$\mathfrak{t}$$ is the Lie algebra of $$T$$ (and the rest with usual notation of roots of $$T$$ in $$G$$ with root spaces).

But when isotropic we have $$\mathfrak{g} = \mathfrak{m} \oplus \oplus_{\alpha \in \Phi(G,T)} \mathfrak{g}_{\alpha}$$ where $$\mathfrak{m}$$ is the $$0$$ eigenspace. If someone could also explain (or provide me with some idea) me where this difference is coming from, I would greatly appreciate it.

Thank you.

PS Further clarification regarding the second question: when $$G$$ is split we have that the Lie algebra of $$G$$ has the decomposition $$\mathfrak{g} = \mathfrak{t} \oplus \oplus_{\alpha \in \Phi(G,T)} \mathfrak{g}_{\alpha}$$ where $$\mathfrak{t}$$ is the Lie algebra of $$T$$ and each $$\mathfrak{g}_{\alpha}$$ is $$1$$ dimensional. However in the situation between split and anisotropic, my understanding is that we don't have exactly the same situation. In the above notation we have $$\mathfrak{m}$$ is not (necessarily?) the Lie algebra of $$T$$ and each $$\mathfrak{g}_{\alpha}$$ is not (necessarily?) $$1$$ dimensional anymore.

I guess I was hoping I could get some idea on why this happens to be the case... (Even though maybe the only difference is that when there is a maximal split torus the situation is "nice" and not as nice otherwise)

The point is that being split is one extreme: $$G$$ contains a split maximal torus. Being anisotropic is the other extreme: $$G$$ contains no split torus. Being isotropic(which is not really a term I've ever heard one use, and I work with algebraic groups a lot) is just that it's somewhere between these two: $$G$$ contains a split torus, but perhaps no split maximal torus.

So, for example:

• The group $$\mathrm{GL}_n$$ over any field $$k$$ is split. Indeed it contains the diagonal split torus $$\mathbb{G}_m^n$$. Of course, be careful that this does not mean that every torus or even maximal torus in $$\mathrm{GL}_n$$ is split. In fact, the maximal tori in $$\mathrm{GL}_n$$ are of the form $$\displaystyle \prod_{i=1}^m \mathrm{Res}_{L_i/k}\mathbb{G}_{m,L_i}$$ where $$L_i/k$$ are finite separable extensions and $$\displaystyle \sum_i^m [L_i:k]=n$$.
• Let $$D$$ be a central division algebra over $$k$$. Consider the reductive group over $$k$$, usually denoted $$D^\times$$, given by sending a $$k$$-algebra $$R$$ to $$(R\otimes_k D)^\times$$. This group has a split connected center: $$Z(D^\times)\cong\mathbb{G}_m$$. The group $$D^\times/Z(D^\times)$$ is anisotropic.
• Let $$D$$ be as in the last example and and assume that $$\dim_k D>1$$. Then, $$D^\times$$ is isotropic but not split. Indeed, $$D^\times$$ is not anisotropic since it contains the split torus $$Z(D^\times)=\mathbb{G}_m$$ but it is not split (or even quasi-split). Indeed, if $$D^\times$$ were split then the same would be true of $$D^\times/Z(D^\times)$$ but the latter being anisotropic and split would imply that $$D^\times/Z(D^\times)$$ has a maximal torus of rank $$0$$. By reductiveness this implies that $$D^\times/Z(D^\times)$$ is trivial, which is absurd. More concretely, the dimension of a maximal split torus (i.e. the rank) in $$D^\times$$ is $$n$$ since it's a form of $$\mathrm{GL}_n$$, and so the rank of $$D^\times/Z(D^\times)$$ is $$n-1$$. Since $$\dim_k D>1$$ we see that $$D^\times/Z(D^\times)$$ is anisotropic of positive rank so not split.

As for your Lie algebra question, I'm not exactly sure what you have written there. Can you clarify?

• Thank you for this very clear answer! Now it makes sense! I have added some clarification regarding the Lie algebra part. I would greatly appreciate any comment regarding this, if you happen to have any, as well! – Takeshi Gouda Apr 16 at 13:19

The other answer is very good. I just add some more examples, on the Lie algebra level because I'm more familiar with it. I leave it to you to write down the corresponding linear algebraic groups.

However, I think your root space decomposition for the non-split case is a little off. I would expect something like

$$\mathfrak{g} \simeq \mathfrak{z}(\mathfrak{a}) \oplus \bigoplus_{\lambda \in R(\mathfrak{a})} \mathfrak{g}_\lambda$$

where $$\mathfrak a$$ is a maximal split torus (I say "torus" for short instead of "toral subalgebra" even in the Lie algebra setting), and $$\mathfrak{z}(\mathfrak{a})$$ is its centraliser (= $$0$$-eigenspace), which probably is your $$\mathfrak{m}$$ (even though you should be aware that it's not the $$0$$-eigenspace of the entire maximal torus you call $$T$$). But then also the root system $$R(\mathfrak a)$$ is what is often called a "rational" root system, again not of the maximal torus $$T$$, but of the maximal split torus $$\mathfrak{a}$$. Such an $$R(\mathfrak a)$$ can be empty (in the anisotropic case) or non-reduced i.e. e.g. of type $$BC_n$$. And yes, the root spaces $$\mathfrak{g}_\lambda$$ can have dimension $$> 1$$.

Namely, let's look at the following three Lie algebras over $$\Bbb R$$:

$$\mathfrak{g}_1 = \mathfrak{sl}_3(\Bbb R) = \lbrace \begin{pmatrix} a & c & e\\ f & b & d\\ h & g & -a-b \end{pmatrix} : a, ..., h \in \mathbb{R} \rbrace$$;

$$\mathfrak{g}_2 = \mathfrak{su}_{1,2} := \lbrace \begin{pmatrix} a+bi & c+di & ei\\ f+gi & -2bi & -c+di\\ hi & -f+gi & -a+bi \end{pmatrix} : a, ..., h \in \mathbb{R} \rbrace$$;

$$\mathfrak{g}_3 = \mathfrak{su}_{3} := \lbrace \begin{pmatrix} ia & c+di & g+hi\\ -c+di & ib & e+fi\\ -g+hi & -e+fi & -ai-bi \end{pmatrix} : a, ..., h \in \mathbb{R} \rbrace$$.

They are all simple and $$8$$-dimensional -- indeed, they all have isomorphic complexification $$(\mathfrak{g}_i)_\Bbb C = \Bbb C \otimes_\Bbb R \mathfrak{g}_i \simeq \mathfrak{sl}_3(\Bbb C)$$, meaning that they are "real forms" of $$\mathfrak{sl}_3$$, or expressed with root systems, real forms of type $$A_2$$. (And over $$\Bbb R$$, they are the only such forms up to isomorphism.) Indeed, in each of them, the diagonal matrices would form a maximal torus (or rather, toral subalgebra in the Lie algebra setting), which is two-dimensional; and its roots in the complexification form a root system of type $$A_2$$; but only for the first one is this torus a split torus.

So $$\mathfrak{g}_1$$ is split. On the other extreme, $$\mathfrak{g}_3$$ is anisotropic, as it contains no split torus $$\neq 0$$ at all: The only maximal split torus is $$\mathfrak a = 0$$, and $$R(0) = \emptyset$$; hence, $$\mathfrak{g}_3 = \mathfrak m = \mathfrak z(0)$$.

These are the extreme cases.

Now $$\mathfrak{g}_2$$ lies between them, so it would be "isotropic" in your nomenclature. (Indeed, this one has a much more special property called quasi-split, vaguely meaning that it's closer to being split than to being anisotropic, but there are more complicated examples of something that is not quasi-split and not anisotropic either.) In this answer I mentioned the maximal split tori of it, which have dimension $$1$$; the most obvious one being $$\mathfrak{a} = \begin{pmatrix} a & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -a \end{pmatrix}.$$ Notice that its $$0$$-eigenspace = centraliser (what you seem to call $$\mathfrak{m}$$) is $$\mathfrak{a} \oplus \mathfrak{t}_0$$ where $$\mathfrak{t}_0 = \begin{pmatrix} bi & 0 & 0\\ 0 & -2bi & 0\\ 0 & 0 & bi \end{pmatrix}$$; in this case (this is a special feature of quasi-split Lie algebras though and not true for all "isotropic" cases), this sum happens to be exactly a maximal (but "only half split") torus and becomes the standard maximal split = split maximal torus in the complexification $$(\mathfrak{g}_{2})_\mathbb{C} \simeq \mathfrak{sl}_3(\mathbb{C})$$. (In general, this $$\mathfrak m$$ can be bigger, as shown in extreme in the anisotropic case.)

Now the rational root system $$R(\mathfrak a)$$ is indeed of the non-reduced type $$BC_1$$, as it consists of the four roots $$\pm \lambda, \pm 2 \lambda$$, where $$\lambda ( \begin{pmatrix} a & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -a \end{pmatrix})= a$$. You'll see that e.g. $$\mathfrak g_\lambda = \lbrace \begin{pmatrix} 0 & c+di & 0\\ 0 & 0 & -c+di\\ 0 & -0 & 0 \end{pmatrix}: c,d \in \Bbb R \rbrace$$ has dimension $$2$$, whereas $$\mathfrak g_{2\lambda} = \lbrace \begin{pmatrix} 0 & 0 & ei\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{pmatrix} : e \in \Bbb R \rbrace$$ has dimension $$1$$. This and more examples were discussed in greater generality in https://math.stackexchange.com/a/3133194/96384.

• Thank you very much for this! It clarified for me the second part of my question. Thank you! – Takeshi Gouda Apr 17 at 13:23