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.



Homogenous structure on $S^2\times S^2$ and its geometry
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.


