Background. (Can be skipped if you already know what is the Eguchi-Hanson metric.) The Eguchi-Hanson metric $g$ is a complete Ricci-flat Riemannian metric on the cotangent bundle of the 2-sphere, $T^*S^2$. By removing the zero-section, we can identify it with $S^3/\mathbb{Z}_2 \times (0, \infty) = (\mathbb{R}^4\setminus\{0\}) / \mathbb{Z}_2$, where $\mathbb{Z}_2$ acts by antipodal reflections. It then has the explicit description
$$g = \frac{r^2}{\sqrt{1+r^4}}(dr^2 + r^2 \alpha_1^2) + \sqrt{1 + r^4}(\alpha_2^2+\alpha_3^2),$$
where $r^2 = x_0^2 + x_1^2 + x_2^2 + x_3^2$,
$$\alpha_1 = \frac{1}{r^2} (x^0 dx^1 – x^1 dx^0 + x^2 dx^3 – x^3 dx^2)$$
and $\alpha_2, \alpha_3$ are defined with the same formula by cyclic permutations of $(1, 2, 3)$. It has also been described by Calabi as a Kähler metric on $T^*\mathbb{CP}^1$, where the Kähler form is $\pi^*\omega_{FS} + i\partial\bar{\partial} (u \circ t)$, where $\omega_{FS}$ is the Fubini-Study Kähler form on $\mathbb{CP}^1$, $t : T^*\mathbb{CP}^1 \to \mathbb{R}$ is the squared-norm function with respect to the Fubini-Study metric, and $u(t) = 4\sqrt{1+t} – 4\log(1+\sqrt{1+t})$.
Question. Is the bundle map $\pi : T^*S^2 \to S^2$ a Riemannian submersion?
In other words, does $\pi$ restrict to isometries $d\pi : (T_\xi(T^*S^2))^{\mathrm{horizontal}} \to T_xS^2$? I tried to work this out explicitly using the above coordinate expressions, but it is very messy, and I couldn't do it. I was wondering if there is another argument, or perhaps a way to see that it's not a submersion using asymptotic properties of the metric.
Best Answer
No. The reason is basically the same why if you take flat $\mathbb R^4\setminus 0$ and quotient by the standard Hopf $\mathbb S^1$ action you get a punctured cone over $\mathbb S^2$ but the projection to that $\mathbb S^2$ is not a Riemannian submersion because the horizontal spaces scale with $r$. The Euguchi-Hanson metric is asymptotic to this one and the same exact thing happens.
What IS true for the Euguchi-Hansen metric is that if you fix $r$ then the projection from $\mathbb S^3/\mathbb Z_2$ to $\mathbb S^2$ is a Riemannian submersion up to scaling. But the scaling changes with $r$. This can be seen as follows.
The forms $\alpha_1,\alpha_2,\alpha_3$ form the standard left invariant orthornormal basis in the round binivariant metric on $\mathbb S^3$ (and its factor $SO(3)=\mathbb S^3/\mathbb Z_2$). Here $\alpha_1$ can be thought of as the inner product (with respect to the binivariant metric) with the Hopf vector field $X$ generating a free isometric $\mathbb S^1$ action on $SO(3)$. For any fixed $r$ the difference between the round metric on $SO(3)$ and the Euguchi Hanson one is that this circle is scaled by a function depending on $r$. But this does not affect the quotient metric at all and the projections $SO(3)\to \mathbb S^2$ are Riemannian submersions (up to scaling) with respect to both metrics on $SO(3)$. However, as I said you get a scaling depending on $r$ so the global map $TS^2\to S^2$ is not a Riemannian submersion.