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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)

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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?

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  • $\begingroup$ 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! $\endgroup$ – Takeshi Gouda Apr 16 at 13:19
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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.

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

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