Suppose that $T$ is a closed subtorus (compact, connected, abelian subgroup) of $\mathbb{T}^d:=\mathbb{R}^d/\mathbb{Z}^d$. Is there a closed subtorus $T^\perp$ of $\mathbb{T}^d$ such that $\mathbb{T}^d=T^\perp \oplus T$?
To elaborate more, I know that $T$ would then be isomorphic as a Lie group to $\mathbb{T}^k$ for some $k$, hence is a divisible group. Therefore, I know that $T$ is a summand in $\mathbb{T}^d$. So my question is whether or not the complementary summand is also a closed Lie subgroup.
Thanks in advance for the help!