Stationary Harrison one-form and metric

PDF p. 6: §1.4; PDF p. 8: §2.1; (2.2)–(2.4)

Let $(M,g,N,T)$ be stationary Lorentzian data, with $T^{\mathrm{♭}}=g(T,\cdot )$ and $f=-g(T,T)$. On an open set $U$ carrying the supplied normalized Ernst potential $\sigma =f+i\psi$, let $F_{ab}={\mathrm{\nabla }}_a{T}_b$ and let ${\star }_g$ be the oriented metric Hodge star. Fix $q\in [0,1]$, and put $D_q=\mathrm{\mid }q(1+\sigma )+(1-\sigma ){\mathrm{\mid }}^2$. Suppose a supplied point $p\in U$ satisfies $T_p=0$, set $d_q=D_q(p)$, and restrict to an open subset $V\subseteq U$ on which $D_q>0$.

A Harrison one-form for these data on $V$ is a supplied smooth real one-form $W_q$ satisfying ${\mathcal{L}}_T{W}_q=0$, $d{W}_q=4(1-q)(((1-q)f-(1+q))F+(1-q)\psi {\star }_{g}F)$, and $W_q(T)=d_q-D_q$. Here $d$ is the exterior derivative and ${\mathcal{L}}_T{W}_q=0$ means that the one-form is stationary under the local flow of $T$. The conditions specify a choice; they do not assert that such a one-form exists or is unique.

For such a choice define the covariant two-tensor $g_q=\frac{1}{4}(D_{q}g+T^{\mathrm{♭}}\otimes W_q+W_q\otimes T^{\mathrm{♭}}-f{D}_q^{-1}{W}_q\otimes W_q)$ and the bundle map $A_{q}v=v+D_q^{-1}{W}_q(v)T$. Its fiberwise pullback is $(A_q^{\ast }g)(u,v)=g(A_{q}u,A_{q}v)$. Then $g_q=(D_q\mathrm{/}4)A_q^{\ast }g$; in particular $g_q$ is a smooth Lorentzian metric on $V$.

Stationary Lorentzian data and tensor conventions

Killing two-form and normalized Ernst notation

Complex self-dual forms and metric-dependent contraction

Real and imaginary parts, complex conjugation, and modulus

A smooth differential k-form

The exterior derivative by the invariant vector-field formula

The Lie derivative of a differential form

**Given:**The data, parameter, positivity restriction and supplied one-form in the Definition.

Goal: Justify the displayed tensor, pullback identity and Lorentzian signature.

1.1 Writing $q(1+\sigma )+(1-\sigma )=(1+q)-(1-q)f-i(1-q)\psi$, the modulus convention gives $D_q=((1+q)-(1-q)f{)}^2+(1-q{)}^2{\psi }^2$. At $p$, $T_p=0$ gives $f(p)=0$, so $d_q=(1+q{)}^2+(1-q{)}^2\psi (p{)}^2>0$. On $V$, the assumed $D_q>0$ makes all displayed divisions smooth. The contraction condition gives $A_{q}T=(d_q\mathrm{/}{D}_q)T$. Direct composition shows that $v\mapsto v-d_q^{-1}{W}_q(v)T$ is the inverse of $A_q$: the coefficient of $T$ cancels using $W_q(T)=d_q-D_q$. Thus $A_q$ is a smooth fiberwise invertible bundle map. [given, F1, F3, F5, algebra]

2.1 Expanding $g(A_{q}u,A_{q}v)$ gives $g(u,v)+D_q^{-1}(T^{\mathrm{♭}}(u)W_q(v)+W_q(u)T^{\mathrm{♭}}(v))-f{D}_q^{-2}{W}_q(u)W_q(v)$. Multiplication by $D_q\mathrm{/}4$ is exactly the displayed formula for $g_q$. The pullback has the signature of $g$, and multiplication by the positive scalar $D_q\mathrm{/}4$ preserves it; hence $g_q$ is smooth and Lorentzian. The differential and stationarity equations remain conditions on the supplied one-form, not conclusions of this calculation. [given, F1, F2, F4, step 1.1, algebra] ∎

The cited article attributes the Harrison formulas. The justification uses the supplied definitions and pointwise algebra; it does not establish the existence or construction of $W_q$.

The cited article attributes the Harrison formulas. The justification uses the supplied definitions and pointwise algebra; it does not establish the existence or construction of $W_q$.