I have a compact connected Lie group $G$ with Lie algebra $\mathfrak{g}$ and a codimension one closed subgroup $H$ with Lie algebra $\mathfrak{h}$. Using an inner product on $\mathfrak{g}$ one can then choose some complementary Lie subalgebra $\mathfrak{k}$ and exponeniate that to get a Lie subgroup $K$. Now, as there are a lot of inner products, there is presumably a lot of ways to choose a complementary Lie subalgebra. However, there are clearly some choices that are better than others e.g. I would like to choose $\mathfrak{k}$ so that $\exp(t\mathfrak{k})$ is closed and so the associated Lie group is isomorphic to $S^1$. Then I would like to choose it so that it's the most "basic" $S^1$ possible, which I guess is asking that it represents a generator of $H_1(G)$... Is there a way to choose the inner product so that one chooses the "correct" complementary Lie algebra?
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$\begingroup$ I understand "I would like to choose $\mathfrak{k}$ so that $\exp(t\mathfrak{k})$ is closed and so the associated Lie group is isomorphic to $S^1$" but not "I would like to choose it so that it's the most "basic" $S^1$ possible, which I guess is asking that it represents a generator of $H_1(G)$" $\endgroup$– Praphulla KoushikCommented Apr 7, 2020 at 16:02
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$\begingroup$ It won't generate $H_1(G)$ unless $H_1(G)$ is rank 1. $\endgroup$– Ben McKayCommented Apr 7, 2020 at 16:09
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1$\begingroup$ A remark is that $H$ is necessarily normal and contains $[G,G]$. If $Z$ is the connected center of $G$, we have $G=HZ$, and the question boils down to the same question within $Z$ and its codimension-1 subgroup $H\cap Z$, i.e., boils down to the abelian case. Also note that because of this remark, if the conjugacy-invariant scalar product is chosen to be rational on $Z$ i.e., take rational values on the kernel of the exponential map $\mathfrak{z}\to G$), then the exponential of the orthogonal is automatically closed. $\endgroup$– YCorCommented Apr 7, 2020 at 16:52
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$\begingroup$ You don't need an inner product to construct a complementary one-dimensional subspace $\frak k$ to a Lie suabalgebra $\frak h$ of $\frak g$ of codimension 1. You just choose any one-dimensional subspace $\frak k$ not contained in $\frak h$. However, as you write, some choices are better than others. $\endgroup$– Mikhail BorovoiCommented Apr 8, 2020 at 13:36
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$\begingroup$ As @YCor has noticed, $\frak h$ contains $[\frak g,\frak g]$. It would be wise to choose $\frak k$ inside the center $\frak z(\frak g)$. $\endgroup$– Mikhail BorovoiCommented Apr 8, 2020 at 13:39
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