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LSpice
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Let$\DeclareMathOperator\rank{rank}$Let $G$ be a compact connected Lie group and $H$, $K$ be two closed connected subgroups. By the answerMikhail Borovoi's answer to the question asked here

In a compact lie group, can two closed connected subgroups generate a non-closed subgroup?

 , $L=\langle K, H\rangle$ (the subgroup generated by $K$, $H$) is a closed connected subgroup of $G$ (hence a Lie subgroup). Now assume additionally that $$r=rank~(H\cap K)=rank~H=rank~K.$$$$r=\rank(H\cap K)=\rank H=\rank K.$$ My question is if we can conclude that $$rank~L=r.$$$$\rank L=r.$$

Let $G$ be a compact connected Lie group and $H$, $K$ be two closed connected subgroups. By the answer to the question asked here

In a compact lie group, can two closed connected subgroups generate a non-closed subgroup?

 $L=\langle K, H\rangle$ (the subgroup generated by $K$, $H$) is a closed connected subgroup of $G$ (hence a Lie subgroup). Now assume additionally that $$r=rank~(H\cap K)=rank~H=rank~K.$$ My question is if we can conclude that $$rank~L=r.$$

$\DeclareMathOperator\rank{rank}$Let $G$ be a compact connected Lie group and $H$, $K$ be two closed connected subgroups. By Mikhail Borovoi's answer to In a compact lie group, can two closed connected subgroups generate a non-closed subgroup?, $L=\langle K, H\rangle$ (the subgroup generated by $K$, $H$) is a closed connected subgroup of $G$ (hence a Lie subgroup). Now assume additionally that $$r=\rank(H\cap K)=\rank H=\rank K.$$ My question is if we can conclude that $$\rank L=r.$$

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Jayq
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Rank of a Lie subgroup generated by two Lie subgroups

Let $G$ be a compact connected Lie group and $H$, $K$ be two closed connected subgroups. By the answer to the question asked here

In a compact lie group, can two closed connected subgroups generate a non-closed subgroup?

$L=\langle K, H\rangle$ (the subgroup generated by $K$, $H$) is a closed connected subgroup of $G$ (hence a Lie subgroup). Now assume additionally that $$r=rank~(H\cap K)=rank~H=rank~K.$$ My question is if we can conclude that $$rank~L=r.$$