Is the Eguchi-Hanson metric a metric on $\mathbb{R}\times S^3$ or tangent bundle of $S^2$?

In the paper "Asymptotically flat self-dual solutions to euclidean gravity" by T.Eguchi and A.J. Hanson, the author constructed a Ricci flat metric on $\mathbb{R}\times S^3$. Let $$\sigma_x=\frac12(-\cos \psi d\theta-\sin \theta \sin \psi d\phi),$$ $$\sigma_y=\frac12(\sin \psi d\theta-\sin \theta\cos \psi d\phi),$$ $$\sigma_z=\frac12(-d\psi-\cos \theta d\phi).$$ They obey the structure equations of the exterior algebra, $$d\sigma_x=2\sigma_y \Lambda \sigma_z, \quad etc.$$ Here $\theta, \psi$ and $\phi$ are Euler angles on $S^3$ with ranges $0\leqslant \theta \leqslant \pi, 0\leqslant \phi \leqslant 2\pi, 0\leqslant \psi \leqslant 4\pi$. Let me list one of the two types of metrics having axial symmetry in this paper: $$ds^2=f^2(r)dr^2+r^2(\sigma_x^2+\sigma_y^2)+r^2 g^2(r)\sigma_z^2,$$ where $$g(r)=f^{-1}(r)=[1-(\frac{a}{r})^4]^{\frac12},$$ $a$ is an integration constant.

The sectional curvature are one of $\pm \frac{2a^4}{r^6}, \pm \frac{4a^4}{r^6}$. The curvatures are regular everywhere for $r\geqslant a$. Although the metrics have an apparent singularity at $r=a$, it can be eliminated by a change of variable.

Question 1: I find in some other references, Eguchi-Hanson metric is a metric on the tangent bundle of $S^2$, not $\mathbb{R} \times S^3$, why?

Question 2: In the metric we should consider $r\geqslant a$ or $r\geqslant 0$? Is the tangent cone at infinity $(\eta M, p)\to (\mathbb{R}^4/Z_2,0)$ as $\eta \to 0$?

• It's a metric on the tangent bundle of $S^2$. The manifold $TS^2$ has a dense open subset (the complement of the zero section) which is diffeomorphic to $(S^3\times \mathbb{R}^+)/\mathbb{Z}_2\cong (S^3/\mathbb{Z}_2)\times (a,\infty)$, so a metric on $TS^2$ can be specified by giving a metric with certain boundary conditions on $S^3\times (a,\infty)$. You should look closely at the construction of the change of variable which "eliminates the singularity". – macbeth Nov 10 '17 at 15:22
• So at the zero section, the sectional curvature is one of $\pm 2, \pm 4$ times $\frac{a^2}{a^6}=\frac{1}{a^4}$? I compute curvature by formula for warped product in Peter Petersen's book "Riemannian Geometry". Let $\partial_r$ be the radial direction, $e_1,e_2,e_3$ be the tangent directions. Consider the warped product $$dt^2=dr^2+\frac{r^2}{f^2}(\sigma_x^2+\sigma_y^2)+\frac{r^2g^2}{f^2}\sigma_z^2=dr^2+\varphi_1^2(\sigma_x^2+\sigma_y^2)+\varphi_2^2\sigma_z^2.$$ Then $sec(\partial_r, e_3)=\frac{1}{f^2(r)} [-\frac{\varphi_2''}{\varphi_2}]=\frac{12r^2}{(r^4-a^4)^2}$, so where do I make a mistake? – mathmetricgeometry Nov 15 '17 at 4:06
• @macbeth: So at the zero section, the sectional curvature is one of $\pm 2, \pm 4$ times $\frac{a^2}{a^6}=\frac{1}{a^4}$? Compute curvature by the the formula for warped products in Peter Petersen's book "Riemannian Geometry". Let $\partial_r$ be the radial direction, $e_1,e_2,e_3$ be the tangent directions. Consider$$dt^2=dr^2+\frac{r^2}{f^2}(\sigma_x^2+\sigma_y^2)+\frac{r^2g^2}{f^2}\sigma_z^2=dr^2+\varphi_1^2(\sigma_x^2+\sigma_y^2)+\varphi_2^2\sigma_z^2$$. Then $sec(\partial_r, e_3)=\frac{1}{f^2(r)} [-\frac{\varphi_2''}{\varphi_2}]=\frac{12r^2}{(r^4-a^4)^2}$, so where do I make a mistake? – mathmetricgeometry Nov 15 '17 at 10:44
• You were using $ds^2=f^2(r)dr^2+r^2(\sigma_x^2+\sigma_y^2)+r^2 g^2(r)\sigma_z^2$ for the metric before, $dt^2=dr^2+\frac{r^2}{f^2}(\sigma_x^2+\sigma_y^2)+\frac{r^2g^2}{f^2}\sigma_z^2$ now, they differ by the conformal factor $f^2(r)$ which blows up at the zero section! – macbeth Nov 15 '17 at 23:48
• @macbeth: I have multiply the conformal factor, then I get $sec(\partial_r, e_3)=\frac{12r^2}{(r^4-a^4)^2}$. As I listed in the question, the curvature should be $\pm 2,\pm 4$ times $\frac{a^4}{r^6}$. So I don't know where I am wrong. – mathmetricgeometry Nov 16 '17 at 3:16