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It is well-known that 3-dimensional real projective space $\mathbb{R}P^3$ is diffeomorphic to $T^1S^2$, the unit tangent bundle of the 2-sphere. However, I could not find any reference to whether these spaces are also isometric as Riemannian manifolds, where $\mathbb{R}P^3$ is given its canonical metric of constant curvature 1 (coming from its double cover by $S^3$), and $T^1 S^2$ is given the Sasaki metric. I feel like this should be true, but I am not sure. Can anyone help? Any reference would be appreciated.

TilBe
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  • What is the curvature tensor of the Sasaki metric on $T^1S^2$? Is it also constant curvature? Is it Einstein? (This isn't a rhetorical question; I have no idea what is meant by "the Sasaki metric on $T^1S^2$".) – Jesse Madnick Aug 11 '21 at 10:20
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    The Sasaki metric on a tangent bundle of some manifold $M$ is defined using the natural splitting of the double tangent bundle $TTM$ into a horizontal and vertical part (using the Levi-Civita connection), both of which are isomorphic to $TM$. Using such isomorphisms, one can pull back $g$ and declare the horizontal and vertical part to be orthogonal. By "Sasaki metric on $T^1S^2$ I mean the restriction of the Sasaki metric of $TS^2$ to $T^1S^2$. – TilBe Aug 11 '21 at 10:48
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    Both metrics are equivariant for the action of $\mathrm{SO}(3)$. So you only need to check it at a point. – Kapil Aug 11 '21 at 10:53
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    The Sasaki metric on $T_1\mathbb{M}^2_\kappa$ is of constant curvature only for $\kappa=0,1$, in which case you get $K=0$ and $K=1/4$ respectively. See Klingenberg & Sasaki (1975) for the original proof of $\kappa=1$ case. – user10354138 Aug 11 '21 at 10:57
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    @user10354138: Theorem 1 in the cited article gives the precise answer to TilBe's question - maybe you want to summarise it in an answer? – Jan Bohr Aug 11 '21 at 11:06
  • Indeed, it does. Thank you! – TilBe Aug 11 '21 at 11:13

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