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<\/p>\n\n\n\n
$$ F = *d_{\\omega} \\Phi $$<\/p>\n\n\n\n
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 $d \\Phi$.<\/p>\n\n\n\n
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.<\/p>\n\n\n\n
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?<\/p>\n\n\n\n
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.<\/p>\n\n\n\n
Write $q = a e^{i \\frac{\\psi}{2}}$ where $a$ is a pure imaginary quaternion and $\\psi \\in [0, 4\\pi)$. Differentiating, we get<\/p>\n\n\n\n
$$ dq = (da + \\frac{1}{2} d\\psi a i) e^{i \\frac{\\psi}{2}} $$<\/p>\n\n\n\n
Therefore<\/p>\n\n\n\n
\\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}<\/p>\n\n\n\n
Let $\\mathbf{r} = q i \\bar{q} = a i \\bar{a}$. We then obtain that<\/p>\n\n\n\n
$$ da d\\bar{a} – \\frac{1}{4r} (a i d\\bar{a} – da i \\bar{a})^2 = \\frac{1}{4r} d \\mathbf{r}^2 $$<\/p>\n\n\n\n
so that<\/p>\n\n\n\n
$$ g = \\frac{1}{4r} d \\mathbf{r}^2 + \\frac{r}{4} \\left( d\\psi + \\frac{1}{r}(ai d\\bar{a} – da i \\bar{a}) \\right)^2. $$<\/p>\n\n\n\n
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 $\\mathbb{R}^3$.<\/p>\n\n\n\n
We differentiate $\\omega$ and obtain<\/p>\n\n\n\n
$$ d \\omega = – \\frac{dr}{r^2} \\wedge (ai d\\bar{a} – da i \\bar{a}) + \\frac{2}{r} da i \\wedge d\\bar{a} .$$<\/p>\n\n\n\n
But $r^2 = – \\mathbf{r} \\mathbf{r}$, which yields, upon differentiation<\/p>\n\n\n\n
$$ 2 r dr = -(d\\mathbf{r} \\mathbf{r} + \\mathbf{r} d \\mathbf{r}). $$<\/p>\n\n\n\n
Hence<\/p>\n\n\n\n
\\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}<\/p>\n\n\n\n
But the last term on the last line vanishes, so we get<\/p>\n\n\n\n
$$ d \\omega = – \\frac{1}{2r^3} d\\mathbf{r} \\mathbf{r} \\wedge d \\mathbf{r}, $$<\/p>\n\n\n\n
which can be checked to be nothing but $*d ( 1 \/ r)$, as claimed.<\/p>\n\n\n\n
Many authors sweep this calculation under the rug, so I thought I’d write a post about it!<\/p>\n","protected":false},"excerpt":{"rendered":"
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…<\/p>\n