Let $G$ be a connected complex semisimple Lie group and $S$, $T$ two maximal tori in $G$. Is there a known upper bound on the number of connected components of $S\cap T$? For example, is it bounded by the cardinality of the centre $Z\_G$:
$$|\pi\_0(S\cap T)|\leq|Z\_G|?$$



Number of connected components of the intersection of two maximal tori
**Summary:** Let $X = \mathrm{Hom}(T,\mathbb{G}\_m)$ be the weight lattice, $\Phi \subset X$ the root system. Define a sublattice $L$ of $X$ to be a "root sublattice" if $L$ is generated as an abelian group by $L \cap \Phi$. Then the possible component groups of $S \cap T$ are the torsion subgroups of $X/L$, as $L$ ranges over root sublattices. 


This will follow from: 


**Theorem:** Let $Z$ be a subgroup of $T$. Then there exists a maximal torus $S$ with $S \cap T = Z$ if and only if there is a connected subgroup $H$ with $T \subseteq H \subseteq G$ such that $Z = Z(H)$.


We first make some comments about connected groups $H$ with $T \subseteq H \subseteq G$. Write $\mathfrak{g} = \mathfrak{t} \oplus \bigoplus\_{\beta \in \Phi} \mathfrak{g}\_{\beta}$, where $\mathfrak{g}$ and $\mathfrak{t}$ are the Lie algebras of $G$ and $T$ and $\mathfrak{g}\_{\beta}$ are the root spaces. 


Connected subgroups are determined by their Lie subalgebras, and a subalgebra containing $\mathfrak{t}$ must be of the form $\mathfrak{t} \oplus \bigoplus\_{\beta \in I} \mathfrak{g}\_{\beta}$ for some subset $I$ of $\Phi$. More specifically, $\mathfrak{t} \oplus \bigoplus\_{\beta \in \Phi} \mathfrak{g}\_{\beta}$ will be a Lie-sub-algebra if and only if, for $\beta\_1$ and $\beta\_2 \in I$, if $\beta\_1+\beta\_2 \in \Phi$ then $\beta\_1 + \beta\_2 \in I$. For such an $I$, we write $H\_I$ for the corresponding connected subgroup. We note that there are only finitely many $H\_I$, since there are only finitely many subsets of $\Phi$.


For any such $I$, set $J = I \cap (-I)$. Then $H\_J$ is a reductive subgroup of $G$, and we have a short exact sequence $0 \to N\_{I \setminus J} \to H\_I \to H\_J \to 0$ where $N\_{I \setminus J}$ is the unipotent group corresponding to $\bigoplus\_{\beta \in I \setminus J} \mathfrak{g}\_{\beta}$. This sequence is semidirect.


**Lemma:** The centralizer of $T$ in any $H\_I$ is $T$.


**Proof:** Let $J = I \cap (-I)$ and consider the above semidirect sequence $0 \to N\_{I \setminus J} \to H\_I \to H\_J \to 0$. Let $\pi$ be the map $H\_I \to H\_J$. Let $Z\_{H\_I}(T)$ be the centralizer of $T$ in $H\_I$. Then $\pi(Z\_{H\_I}(T)) \subseteq Z\_{H\_J}(T) = T$, where the latter inequality is standard. On the other hand, $T$ clearly does centralize $T$. So we have $T \subseteq Z\_{H\_I}(T) \subseteq \pi^{-1}(T)$ and thus we have a short exact sequence $0 \to N\_{I \setminus J} \cap Z\_{H\_I}(T) \to Z\_{H\_I}(T) \to T \to 0$. But the Lie algebra of $N\_{I \setminus J}$ is a direct sum of weight spaces for $T$ with nonzero character, so no element of $N\_{I \setminus J}$ centralizes $T$. We deduce that $N\_{I \setminus J} \cap Z\_{H\_I}(T)$ is trivial, so $Z\_{H\_I}(T) = T$. $\square$


**Corollary:** The center of $H\_I$ is contained in $T$.


**Proof:** Clearly, the center of $H\_I$ centralizes $T$. $\square$.


We can now show that $S \cap T$, for any maximal torus $S$, is of the form $Z(H\_I)$ for some $I$. Let $H$ be the Lie-sub-group generated by $S$ and $T$. Clearly, $T \subseteq H$
and $H$ is connected, so $H$ is of the form $H\_I$ for some $I$. By the corollary, $Z(H\_I) \subseteq T$ and similarly $Z(H\_I) \subseteq T$. This shows $Z(H\_I) \subseteq S \cap T$. On the other hand, $S$ and $T$ commute with every element of $S \cap T$, so $H\_I$ commutes with every element of $S \cap T$ and we have $S \cap T \subseteq Z(H\_I)$. We have proven both containments.


We now know that all intersections are of the form $Z(H\_I)$. We want to show, in reverse, that any group of the form $Z(H\_I)$ occurs as $S \cap T$. Given $I$, let $L \subseteq X$ be the lattice generated by $I$. Let $K = L \cap \Phi$. Then $Z(H\_I)$ is the subgroup of $T$ on which the characters of $L$ vanish, and we thus deduce that $Z(H\_I) = Z(H\_K)$. So it is enough to show that $Z(H\_K)$ is of the form $S \cap T$. The group $H\_K$ is reductive, so all we need is


**Lemma:** Let $H\_K$ be as above. There is a maximal torus $S$ in $H\_K$ such that $S \cap T = Z(H\_K)$.


**Proof:** Let $Y = \bigcup\_{K' \subsetneq K} H\_{K'}$. Then $Y$ is a union of finitely many subgroups of lower dimension, so the complement of $Y$ is Zariski dense. Let $s$ be a regular element in $H\_K \setminus Y$, and let $S$ be the connected centralizer of $s$. We claim that $S \cap T = Z(H\_K)$. We know that every maximal torus in $H\_K$ contains $Z(H\_K)$. Suppose, for the sake of contradiction, that $t \in T \setminus Z(H\_K)$ and $t \in S$. Let $Z(t)$ be the centralizer of $t$ in $H\_K$; since $t$ is not central, $Z(t)$ is not $H\_K$. Let $Z(t)\_0$ be the connected component of the identity of $Z(t)$. So $Z(t)\_0$ is a connected subgroup of $H\_K$ containing $T$, and must be of the form $H\_{K'}$ for some $K' \subsetneq K$. Also, since $t \in S$, we have $S \subseteq Z(t)\_0$. So $s \in H\_{K'}$, contrary to the choice of $s$. We have obtained a contradiction, and deduce that $S \cap T = Z(H\_K)$.


We have now proven the theorem. As we noted above, $Z(H\_I)$ is the subgroup of $T$ where the characters in $I$ vanish. We deduce that $Z(H\_I)$ is the dual group to $X/\mathrm{Span}\_{\mathbb{Z}} I$ and the component group of $Z(H\_I)$ is the torsion subgroup of $X/\mathrm{Span}\_{\mathbb{Z}} I$.




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What remains is combinatorics. 


As nfdc23 suggests, it is convenient to work with the adjoint form of the group, in which case $X = \mathrm{Span}\_{\mathbb{Z}} \Phi$. For the general case, multiply all bounds by $|X / \mathrm{Span}\_{\mathbb{Z}} \Phi| = |Z(G)|$. I'll list the root sublattices and state the largest one. Proofs will be provided if requested. I've chosen


In $A\_n$, the root sublattices are $A\_{n\_1} \oplus A\_{n\_2} \oplus \cdots A\_{n\_r}$ for $\sum n\_i = n$ and $Z(H\_I)$ is trivial.


In $B\_n$, we obviously have $B\_{n\_1} \oplus B\_{n\_2} \oplus \cdots \oplus B\_{n\_r}$. Each of these $B\_m$'s, in turn, contain $D\_m$ and $A\_{m-1}$. The largest index comes from $D\_2^{\lfloor n/2 \rfloor}$ giving index $2^{\lfloor n/2 \rfloor}$. Here $\lfloor x \rfloor$ means $x$ rounded down, and $D\_2 = \{ \pm e\_1 \pm e\_2 \} \subset B\_2 = \{ \pm e\_1 \pm e\_2, \pm e\_1, \pm e\_2 \}$. 


In $C\_n$, we obviously have $C\_{n\_1} \oplus C\_{n\_2} \oplus \cdots \oplus C\_{n\_r}$ and we also have $A\_{m-1} \subset C\_m$. The largest index comes from $C\_1^{\oplus n}$, that is to say, from $\{ \pm 2 e\_i \}$ inside $C\_n = \{ \pm 2 e\_i, \pm e\_i \pm e\_j \}$, with index $2^n$.


In $D\_n$, we obviously have $D\_{n\_1} \oplus D\_{n\_2} \oplus \cdots \oplus D\_{n\_r}$ and we also have $A\_{m-1} \subset D\_m$. The largest index comes from $D\_2^{\lfloor n/2 \rfloor}$, giving $2^{\lfloor n/2 \rfloor -1}$.


The exceptional types seem like a pain, but they definitely harbor some surprises! Both $A\_8$ and $D\_8$ are root sublattices of $E\_8$, with index $3$ and $2$ respectively.


