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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.

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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.

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

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    $\begingroup$ See arxiv.org/pdf/math/0101017.pdf p. 5 for details. $\endgroup$
    – Ben McKay
    Commented Apr 10, 2017 at 15:55
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    $\begingroup$ But the maximal compact subgroup is $PSO(4)=SO(3) \times SO(3)$, preserving the splitting, if I remember correctly. $\endgroup$
    – Ben McKay
    Commented Apr 10, 2017 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
    Commented Apr 10, 2017 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
    Commented Apr 10, 2017 at 18:07
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    $\begingroup$ 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
    Commented Apr 10, 2017 at 18:31

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