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I am currently interested in quotients of Lie groups by submaximal tori.

$G = Sp(1) \times Sp(1)$ double-covers $SO(4)$, as noted at The Quaternions and $SO(4)$. Define a circle subgroup $T = \{1\} \times \{\cos \theta + i \sin \theta : \theta \in \mathbb{R}\}$ of $G$, where $i$ is a purely imaginary second coordinate for the quaternions. Then $T$ descends to a circle subgroup $S$ of $SO(4)$.

I think the quotient $SO(4)/S$ is a product $S^3 \times S^2$ of spheres, because the quotient $G/T$ is. (Right?)

Is $S$ conjugate to the block-diagonal $S' = SO(2) \times \{I\}$ subgroup of $SO(4)$? The corresponding quotient $SO(4)/SO(2)$ is the Stiefel manifold $V_{2,4}$).

Intuitively, an element of this manifold is given by taking a unit vector in $\mathbb{R}^4$ and then an orthogonal vector, so $V_{2,4}$ should be homeomorphic to the unit tangent bundle $T^1(S^3)$, which should be trivial by Tangent Bundle on S^3 (because $S^3$ is naturally a Lie group $Sp(1)$ and left-multiplication translation of a basis gives a trivialization of the tangent bundle of a Lie group).

But that's not a particular reason to think $S$ and $S'$ are conjugate. Are they? Should that be clear?

How different can quotients $SO(4)/SO(2)$ be?

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jdc
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To answer your first question, actually, $SO(4)/S$ is diffeomorphic to $\mathbb{R}P^3 \times S^3$.

To see this, recall that $SO(4) = G/\pm(1,1)$ is actually diffeomorphic to $\mathbb{R}P^3 \times S^3$. In fact, an explicit diffeomorphism is given by the map $f:G/\pm(1,1)\rightarrow \mathbb{R}P^3\times S^3$ given by $f[p,q] = ([p],pq)$. Notice that the $q$ factor (on which $T$ acts) only appears in the $S^3$ factor in $\mathbb{R}P^3\times S^3$.

Thus, using $f$ to transport the $S^1$ action to $\mathbb{R}P^3\times S^3,$ we have an $S^1$-equivariant diffeomorphism between the $S$ action on $SO(4)$ and the action of $S^1$ on $\mathbb{R}P^3\times S^3$ which is trivial on the $\mathbb{R}P^3$ factor and Hopf on the $S^3$ factor. Hence, $f$ induces a diffeomorphism between $SO(4)/S$ and $\mathbb{R}P^3\times S^2$.

In general, if you have a covering of Lie groups $\pi: G\rightarrow H$ with a subgroup $K\subseteq G$, then you get a covering $G/K\rightarrow H/\pi(K)$. The new cover can not have more sheets than the old one, but it can have fewer, depending on how $K$ intersects the center of $G$.

To answer your next question, no, $S$ and $S'$ are not conjugate. This actually follows by the answer above together with your observation that $SO(4)/S^1 = S^3 \times S^2$: $C_g:G\rightarrow G$, conjugation by $g\in G$ is a diffeomorphism, i.e., a single sheeted covering of $G$. It follows $C_g$ induces a covering of $\leq 1$ sheets (i.e., a diffeomorphism) $G/K\rightarrow G/C_g(K_)$. Since $SO(4)/S$ and $SO(4)/S'$ are not diffeomorphism (or even homotopy equivalent), $S$ and $S'$ are not conjugate.

But, there is a much easier way to see that $S$ and $S'$ are not conjugate. Namely, $S$ in the basis $\{1,i,j,k\}$ of $\mathbb{R}^4 = \mathbb{H}$, $S$ has the matrix form $\begin{bmatrix} \cos \theta & \sin\theta & & \\ -\sin\theta & \cos\theta & & \\ & & \cos \theta & -\sin\theta \\ & & \sin \theta & \cos \theta\end{bmatrix}.$

This has eigenvalues $e^{i\theta}$ and $e^{-i\theta}$ each with multiplicity $2$. On the other hand, matrices in $S'$ have eigenvalues $e^{i\theta}, e^{-i\theta},1,1$. Since conjugate matrices have the same eigenvalues, it follows that no non-identity element of $S$ is conjugate to any element of $S'$.

Finally, a fun fact, since I can't resist. The circle subgroups of $G$ are parametrized, up to conjugacy, by two relatively prime integers $a$ and $b$. They are of the form $S_{a,b} = (z^a, z^b)$ with $z\in S^1$. It turns out that no matter what $a$ and $b$ are, the quotient $G/S^1_{a,b}$ is always diffeomorphic to $S^3\times S^2$. (It follows that $SO(4)/T$, for any circle $T$ is diffeomorphic to $S^3\times S^2$ or double covered by it). The only proof I know uses the Barden-Smale classification of closed simply connected $5$-manifolds (well, just the Smale part). My advisor once said that if anyone actually cooks up an explicit diffeomorphism between $G/S_{a,b}$ and $S^3\times S^2$, he would take that person out to any restaurant of their choosing.

Jason DeVito - on hiatus
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  • Wow, thanks. Some of those things were pretty obvious, and I am kind of embarrassed, but the fun fact is pretty wild. I am tempted to drop what I am doing and try to work out a diffeomorphism, although I obviously would be unsuccessful. Is there an $S_{a,b}$ that corresponds to the other obvious double cover, $S^3 \times \mathbb{R}P^2$? – jdc Feb 19 '14 at 01:15
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    @jdc: They're only obvious in retrospect ;-). If you do solve it, I'd love to join you on the dinner. Incidentally, there are biquotients of the form $G // S^1$ which are NOT diffeomorphic to $S^2\times S^3$. Because $S^3\times \mathbb{R}P^2$ is not orientable, there is no homogeneous space $G/H = S^3\times \mathbb{R}P^2$ with $H$ connected, so there's no way to make it using an $S_{a,b}$. – Jason DeVito - on hiatus Feb 19 '14 at 01:43
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    This last fact follows from the observation that $H$ acts on $G/H$ by left multiplication. Since $H eH = eH$, this induces an $H$ action on $T_{eH} G/H$. By putting a $G$-invariant metric on $G/H$, $H$ acts by isometries, so this action on $T_{eH} G/H$ defines an embedding $H\rightarrow O(n)$. If $H$ is connected, then the image actually lies in $SO(n)$. Now, pick any nonvanishing top degree form on $T_{eh}G/H$ and average it via the $H$ action. Because $H$ is orientation preserving, this averaged form is non-zero, and $H$ invariant. Now use left multiplication by $G$ to move the form... – Jason DeVito - on hiatus Feb 19 '14 at 01:47
  • around to all of $G/H$. $H$-invariance implies this is well defined, so now you have a non-vanishing form on all of $G/H$, so $G/H$ is orientable. – Jason DeVito - on hiatus Feb 19 '14 at 01:48