Maximal tori in Lie vs algebraic groups

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 $\mathbb{C}^*$.

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 a 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 $GL(N, {\mathbb C})$ for some $N$ (Zariski closed means a subgroup defined via a system of polynomial equations), see here.

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 compact 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 $K$.