If $G$ is a Lie group, we define a maximal [Lie] torus in $G$ to be a maximal connected compact abelian Lie subgroup of $G$. These guys correspond to Cartan subalgebras of $\mathfrak{g}=Lie(G)$.

If $G$ is a linear algebraic group, we define a maximal [algebraic] torus to be a maximal abelian semisimple subgroup isomorphic to a power of the multiplicative group of the base field.

Suppose now I take $G$ to be linear algebraic over $\mathbb{C}$, so it is also a complex Lie group. What is the relationship between maximal Lie tori and maximal algebraic tori? How do we see algebraic tori at the Lie algebra level?

The wiki page about algebraic tori says they were introduced in "analogy with the theory of tori in Lie group theory", but I have difficulties in understanding the explicit connection, beyond the "picture" of $S^1$ inside $\mathbb{C}^*$.

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    $\begingroup$ The given definition of a maximal torus in a Lie group is incorrect: The compactness assumption is wrong. Instead, assume that it is a maximal connected abelian subgroup whose adjoint representation is diagonalizable over the complex numbers. (You can also define the maximal split torus by requiring "diagonaliable over the real numbers".) Then everything becomes easy via the unitary trick: The maximal compact subtorus in the algebraic torus $(C^*)^n$ will be complex-diagonalizable and, hence, the entire algebraic torus. $\endgroup$ – Moishe Kohan Feb 27 '17 at 18:02

The question is quite muddled and is based on a misconception. In the theory of Lie groups, a "torus" (as a subgroup) need not be homeomorphic to the topological torus.

Let me work with complex semisimple Lie groups $G$, since the discussion is cleaner in this case and this is what the question was about. I will assume that $G$ has finitely many connected components. Then $G$ has complex algebraic structure meaning that $G$ embeds as a Zariski closed subgroup of $GL(N, {\mathbb C})$ (i.e. a subgroup defined via a system of polynomial equations).

A torus in $G$ is a connected abelian complex Lie subgroup $H< G$ whose (almost faithful) linear representation is diagonalizable. Which representation you take, does not matter, you can take the above representation $G\to GL(N, {\mathbb C})$ or you can take the adjoint representation of $G$. (Almost faithful means that the kernel is finite.) A maximal torus (usually denoted $T$ or $A$) is a maximal (with respect to inclusion) subgroup with this property. All maximal tori are conjugate to each other. (Their Lie algebras are Cartan subalgebras, i.e. commutative self-normalizing subalgebras of the Lie algebra of $G$.) Since $H$ is diagonalizable, it is isomorphic to a closed subgroup of $({\mathbb C}^\times)^N$, which implies that $H\cong ({\mathbb C}^\times)^n$ for some $n$.

Conversely, suppose that $H< G$ is a subgroup isomorphic to $({\mathbb C}^\times)^n$. I claim that $H$ is diagonalizable as a subgroup of $GL(N, {\mathbb C})$. Prove this by induction on $N$. The case $N=1$ is clear. Then verify that every connected abelian subgroup of $GL(N, {\mathbb C})$ has an invariant line $L\subset {\mathbb C}^N$ (actually, this is even true for solvable subgroups). How do we know this? Since we are working over the algebraically closed field, each nontrivial element of $H$ has an eigenvector in ${\mathbb C}^N$. By commutativity of $H$, the eigenspace decomposition is $H$-invariant. Now, use the induction hypothesis. In any case, we got an invariant line, so we have another linear representation of $H$ on the vector space $V={\mathbb C}^N/L$. We need to prove that the sequence of $H$-modules splits: $$ 0\to L\to {\mathbb C}^N \to V\to 0. $$ So far our proof used only the fact that $H$ is abelian. Now, I will use that $H$ contains the "compact subtorus" $K_H=(S^1)^n$. The next part of the proof is called the "unitary trick" (more precisely, this is a very special case of the unitary trick). Since the subtorus $K_T$ is compact, it preserves some hermitian inner product on ${\mathbb C}^N$. Thus, $K_T$ preserves $L^\perp$, the orthogonal complement to $L$ in ${\mathbb C}^N$ defined via this inner product. But $H$ is the complexification of $K_T$, hence, the entire $H$ preserves $L^\perp$. (This is a pleasant exercise in complex analysis: If $f$ is a holomorphic function on $H$ vanishing on $K_H$, then $f$ is identically zero. Now, convert this into the statement about invariance of $L^\perp$.) As a $K$-module, $L^\perp$ is isomorphic to $V$ of course. Therefore, the above sequence splits. By the induction hypothesis, $V$ is diagonalizable as an $H$-module (i.e. we have an isomorphism of $H$-modules $V\cong L_1\oplus ... \oplus L_{n-1}$ where each $L_i$ is 1-dimensional) therefore, adding the extra factor $L$ to this decomposition we obtain a diagonalization of $H$. qed

Few more remarks in the non-anlegbraically closed case. A split torus (over some field) in the language of algebraic groups, is a group isomorphic to $(G_m)^n$ for some $n$. Thus, if we treat $H=({\mathbb C}^\times)^n$ as a real algebraic group, then the split torus in $H$ is its set of real points, i.e. the subgroup $({\mathbb R}^\times)^n$. The complementary subgroup $(S^1)^n$ is sometimes called a "compact torus". This part indeed is a torus in the topological sense. But this torus will never be maximal in this situation. One more thing to note: Start with a complex Lie group $K$ (without abelian factors). Then tori in $K$ indeed are all of the form $T=(S^1)^n$. Once you complexify $K$, you obtain a complex semisimple Lie group $G$. The tori in $G$ will be (up to conjugation) of the form $T^{\mathbb C}$, complexifications of tori in $K$.

  • $\begingroup$ Thank you for this very clarifying answer! Just a few more doubts about "complementary compact tori"; in the notations you have at the end of your answer, let me take K to be a real Lie group that we can reasonably complexify (compact or real algebraic). Then let me consider compact tori in K. I understand compact tori are not maximal among compact subgroups, nor among abelian diagonalizable ones, but let T be a maximal compact torus. If I understand correctly, the complexification of T should be an algebraic torus in the complexification of K. Is this correct? $\endgroup$ – Lor Mar 2 '17 at 11:45
  • $\begingroup$ If this is correct, the next question would be: how do we see the compact torus in the Lie algebra of the complexification of K? My baby example is SL(2,R), for which I see the compact torus SO(2,R) somehow as (a conjugate of the exponential of) the intersection (inside sl(2,C)) between the compact form su(2) and the usual Cartan subalgebra of the split real form sl(2,R). But I do not know how to make this precise, nor how general this phenomenon is. $\endgroup$ – Lor Mar 2 '17 at 11:50
  • $\begingroup$ For your first question, the answer is yes: Each complex torus in an algebraic group is an algebraic subgroup. For the 2nd question, you can first identify the compact part ${\mathfrak k}$ of the complex Lie algebra ${\mathfrak g}_{\mathbb C}$, say, by looking at the real Killing form. Then find maximal Cartan subalgebras in ${\mathfrak k}$. $\endgroup$ – Moishe Kohan Mar 2 '17 at 19:03
  • $\begingroup$ @Lor: I assume that you already know that on a compact Lie algebra the Killing from is negative definite; thus, you will be looking for the maximal subspace with negative definite Killing form in order to identify ${\mathfrak k}$. $\endgroup$ – Moishe Kohan Mar 2 '17 at 21:47
  • $\begingroup$ Yes, this makes perfectly sense; thank you. $\endgroup$ – Lor Mar 3 '17 at 17:00

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