This post is about $M = \mathbb{H}$ with metric $g = dq d\bar{q}$. The Gibbons-Hawking ansatz is an ansatz for $4$-dimensional gravitational instantons with a $1$ parameter group of symmetries (think of $U(1)$ or $\mathbb{R}$). One may construct such an instanton from a connection $1$-form $\omega$ for a circle bundle $P$ over an open subset $B$ of $\mathbb{R}^3$ (assuming the group is $U(1)$) and from a smooth function $\Phi$ on $B$, thought of as a smooth section of the adjoint bundle of $P$, satisfying the so-called Bogomolny equation

$$ F = *d_{\omega} \Phi $$

where $F$ is the curvature of $\omega$, i.e. $F = d\omega$ (since the Lie group $U(1)$ is abelian), $*$ is the Hodge star with respect to the Euclidean metric on $B$ and $d_{\omega}$ is the exterior differential twisted with the connection $\omega$. In this special case, the Lie group $G = U(1)$ is abelian, so $d_{\omega} \Phi$ is just $d \Phi$.

There is also a converse to that construction, so that any gravitational instanton with $U(1)$ symmetry can be, under some assumptions, obtained from this construction.

When learning about the Gibbons-Hawking ansatz, it is natural to ask the question: how can one view the simplest $4$-dimensional gravitational instanton with $U(1)$ symmetry, $M = \mathbb{H}$, with metric $g = dq d\bar{q}$, where $\mathbb{H}$ denotes the quaternions, $q$ is a quaternionic coordinate on $M$, from the point of view of the Gibbons-Hawking ansatz?

I struggled a bit with the calculation and authors tend to completely omit the calculation, so I thought I would post about it here, hoping it will be useful to somebody else, or possibly myself at some point in the future.

Write $q = a e^{i \frac{\psi}{2}}$ where $a$ is a pure imaginary quaternion and $\psi \in [0, 4\pi)$. Differentiating, we get

$$ dq = (da + \frac{1}{2} d\psi a i) e^{i \frac{\psi}{2}} $$

Therefore

\begin{aligned} g &= (da + \frac{1}{2} d\psi a i) (d\bar{a} – \frac{1}{2} i \bar{a} d \psi) \\ &= da d\bar{a} + \frac{1}{2}( ai d\bar{a} – da i \bar{a}) d\psi + \frac{r}{4} d\psi^2 \\ &= da d\bar{a} + \frac{r}{4} \left( d\psi + \frac{1}{r}(ai d\bar{a} – da i \bar{a}) \right)^2 – \frac{1}{4r} (a i d\bar{a} – da i \bar{a})^2 \end{aligned}

Let $\mathbf{r} = q i \bar{q} = a i \bar{a}$. We then obtain that

$$ da d\bar{a} – \frac{1}{4r} (a i d\bar{a} – da i \bar{a})^2 = \frac{1}{4r} d \mathbf{r}^2 $$

so that

$$ g = \frac{1}{4r} d \mathbf{r}^2 + \frac{r}{4} \left( d\psi + \frac{1}{r}(ai d\bar{a} – da i \bar{a}) \right)^2. $$

Let $\omega = d\psi + \frac{1}{r}(ai d\bar{a} – da i \bar{a})$. I claim that $d \omega = * d(1/r)$. Note that $\Phi = 1/r$ is (up to a factor) the fundamental solution of the Laplacian on $\mathbb{R}^3$.

We differentiate $\omega$ and obtain

$$ d \omega = – \frac{dr}{r^2} \wedge (ai d\bar{a} – da i \bar{a}) + \frac{2}{r} da i \wedge d\bar{a} .$$

But $r^2 = – \mathbf{r} \mathbf{r}$, which yields, upon differentiation

$$ 2 r dr = -(d\mathbf{r} \mathbf{r} + \mathbf{r} d \mathbf{r}). $$

Hence

\begin{aligned} d \omega &= \frac{1}{2 r^3} d \mathbf{r} \mathbf{r} \wedge ( ai d\bar{a} – da i \bar{a}) – \frac{1}{2 r^3} ( ai d\bar{a} – da i \bar{a}) \mathbf{r} \wedge d \mathbf{r} – \frac{2}{r^3} da i \bar{a} \mathbf{r} a i d\bar{a} \\ &= – \frac{1}{r^3} a i d\bar{a} \mathbf{r} \wedge da i \bar{a} – \frac{1}{r^3} da i \bar{a} \mathbf{r} \wedge a i d\bar{a} \\ &= – \frac{1}{2r^3} d\mathbf{r} \mathbf{r} \wedge d \mathbf{r} + \frac{1}{2 r^3} (a i d\bar{a} – da i \bar{a}) \mathbf{r} \wedge (a i d\bar{a} – da i \bar{a}) \end{aligned}

But the last term on the last line vanishes, so we get

$$ d \omega = – \frac{1}{2r^3} d\mathbf{r} \mathbf{r} \wedge d \mathbf{r}, $$

which can be checked to be nothing but $*d ( 1 / r)$, as claimed.

Many authors sweep this calculation under the rug, so I thought I’d write a post about it!