3
$\begingroup$

Is a well known fact that $SO(3)$ acts transitively on $S^2$ and that the isotropy group of this action is $SO(2).$ In this case, $S^2$ has a natural structure of homogeneous space. In particular, I wonder that is true that $SO(3)\times SO(3)$ acts transitively on $S^2\times S^2$ and it has a natural structure of homogeneous space. If we consider in this space the natural $SO(3)-$invariant metric in $S^2$ then we have in $S^2\times S^2$ the product metric and in this metric there are several directions where the curvature is zero.

My question is the following, does there exists a homogeneous structure on $S^2\times S^2$ that is not similar to the product structure? I.e. such that $S^2\times S^2$ is not the quotient of a product of groups?

The point of this questions is that if it is the case, we can search for $G-$invariant metrics oh $S^2\times S^2$ using the possible reductive decomposition of the Lie algebra $\mathfrak{g}$, so it is a manner to search for eventual positively curved metrics.

$\endgroup$
1
$\begingroup$

This is just a comment to supply some details for Ben McKay's answer. The space $S=\Lambda^2(\mathbb{R}^4)$ has dimension $6$. There is an involution $\xi\colon S\to S$ defined as follows: given distinct indices $i,j\in\{1,2,3,4\}$ we let $k$ and $l$ denote the two remaining indices, and we let $s$ denote the signature of the permutation sending $1$, $2$, $3$ and $4$ to $i$, $j$, $k$ and $l$, and we put $\xi(e_i\wedge e_j) = s e_k\wedge e_l$. This satisfies $\xi^2=1$, so $S$ is the direct sum of the spaces $S_+=\ker(\xi-1)$ and $S_-=\ker(\xi+1)$. It is not hard to find bases and thus show that $\dim(S_+)=\dim(S_-)=3$. Now put $P_+=(S_+\setminus\{0\})/\mathbb{R}^+$ and similarly for $P_-$. These are both diffeomorphic to the sphere $S^2$. Given an oriented two-dimensional subspace $W<\mathbb{R}^4$ we can choose an oriented basis $u,v$ and then consider $(1+\xi)(u\wedge v)\in S_+$. One can check that this is nonzero, so it determines an element of $P_+$. Moreover, this does not depend on the choice of oriented basis, so we can call it $\phi_+(W)$. We can define $\phi_-(W)\in P_-$ in a similar way. It then turns out that the resulting map from the oriented Grassmannian to $P_+\times P_-$ is a diffeomorphism.

| cite | improve this answer | |
$\endgroup$
4
$\begingroup$

$S^2 \times S^2$ is the space of oriented 2-planes in $\mathbb{R}^4$, acted on by the projective linear group $PSL(4,\mathbb{R})$.

| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ See arxiv.org/pdf/math/0101017.pdf p. 5 for details. $\endgroup$ – Ben McKay Apr 10 '17 at 15:55
  • 1
    $\begingroup$ But the maximal compact subgroup is $PSO(4)=SO(3) \times SO(3)$, preserving the splitting, if I remember correctly. $\endgroup$ – Ben McKay Apr 10 '17 at 16:00
  • $\begingroup$ Could you please explain here how you construct a bijection between $S^2\times S^2$ and the set of oriented 2-planes in $\mathbb R^4$ ? $\endgroup$ – Mikhail Borovoi Apr 10 '17 at 16:37
  • $\begingroup$ @MikhailBorovoi: See math.stackexchange.com/questions/1828484/…, for example. You might find Qiaochu's answer easier to read than mine: math.stackexchange.com/questions/219100/… $\endgroup$ – Bombyx mori Apr 10 '17 at 18:07
  • 2
    $\begingroup$ You can use the compact version $SO(4)/S(O(2)\times O(2))$ (or something like that). people.sju.edu/~ktapp/so4.pdf gives a nice account on left-invariant metrics on $SO(4)$. On the other hand, I am pretty sure that homogeneous manifolds with positive curvature are already classified (please look at Ziller or Wilking survey on positively curved manifolds). $\endgroup$ – Llohann Apr 10 '17 at 18:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.