# Notes on the Weyl, E/B, and 3+1 decompositions of Riemann, and curvature invariants

More notes I had that needed to live somewhere. I hope you find them useful.

Update 2018-03-26: I’m making public a companion xAct notebook which contains most of the calculations on this web page. You can also get this notebook from the xAct-contrib/examples repository.

# Weyl decomposition

The most basic ingredient here is splitting the Riemann tensor into trace and tracefree pieces. This should be covered in every standard GR textbook (see e.g. Wald1). The decomposition is, in $n$ dimensions,

\begin{align} \label{eq:R-decomp-1} R_{abcd} = C_{abcd} + \frac{2}{n-2}( g_{a[c} R_{d]b} - g_{b[c} R_{d]a}) - \frac{2}{(n-1)(n-2)}R g_{a[c} g_{d]b} \,, \end{align}

where $C_{abcd}$ is the tracefree Weyl tensor, and as usual, the Ricci tensor is $R_{bd} = R^a{}_{bad}$ and the Ricci scalar is $R = g^{ab}R_{ab}$.

Now, this is not a very pleasant expression to remember. It can be brought into a slightly nicer form by defining an ‘impose index symmetry’ operation with some group $G$, which is a subgroup of the signed permutation group (signed symmetric group / hyperoctahedral group), which is just the symmetric group $S_n$ where each permutation is further given a sign, $\pm 1$. The ‘Riemann symmetric’ group $r[abcd]$ is generated by the elements $\{ -(ab), -(cd), +(ac)(bd) \}$. That is, the Riemann tensor $R_{abcd}$ is separately antisymmetric on $ab$, separately antisymmetric on $cd$, and symmetric under the simultaneous interchange of $ab\leftrightarrow cd$. You can check that $r[abcd]$ is a group with 8 elements.

The ‘impose index symmetry’ operation on a tensor expression $T_{i\cdots}$ is given by

\begin{align} \label{eq:impose-sym-def} (S_G T)_{i\cdots} = \frac{1}{|G|} \sum_{g\in G} z(g) T_{g\circ i\cdots} \, \end{align}

where $|G|$ is the order of the group, $z(g)$ is the sign (+1 or -1) associated with the element $g$, and $g\circ i\cdots$ means to permute indices $i\cdots$ as specified by the permutation $g$. This is a generalization of the symmetrization $(ab)$ and antisymmetrization $[abc]$ operations that are commonly used, corresponding to respectively the whole symmetric group with all plus signs, or the whole symmetric group where signs $z(g)=sgn(g)$ are determined by the parity of the permutation $g$. This general operation is implemented in xAct as ImposeSymmetry[].

Having defined this operation, we can (slightly) improve Eq. \eqref{eq:R-decomp-1} via

\begin{align} \label{eq:R-decomp-proj} R_{abcd} = S_{r[abcd]} \left[ C_{abcd} + \frac{4}{n-2} R_{ac}g_{bd} - \frac{2}{(n-1)(n-2)} R g_{ac}g_{bd} \right] \,. \end{align}

# Curvature invariants 1

Each of the Weyl/Ricci tensor/Ricci scalar pieces of the decomposition are orthogonal to each other in the following way. Construct the Kretschmann invariant, $K\equiv R_{abcd}R^{abcd}$. In terms of the Weyl decomposition, it is given by

\begin{align} \label{eq:K-Weyl-decomp} K = C_{abcd}C^{abcd} + \frac{4}{n-2} R_{ab}R^{ab} - \frac{2}{(n-1)(n-2)}R^2 \,. \end{align}

Note: the coefficients in Eq. \eqref{eq:K-Weyl-decomp} are the same as those appearing in Eq. \eqref{eq:R-decomp-proj}. This has something to do with the impose-symmetry operation being a properly normalized idempotent projection operation. In this case, it is actually a Young projector for the standard Young tableau filled with {{1,3},{2,4}}.

From here forward we focus on dimension $n=4$. Now we can also define the (left) dual Riemann tensor, via

\begin{align} {}^*\!R^{ab}{}_{cd} \equiv \frac{1}{2}\epsilon^{abef}R_{efcd} \,. \end{align}

With this dual, we can define the Pontryagin invariant,

\begin{align} P \equiv {}^*\!R^{abcd} R_{abcd} ={}^*\!C^{abcd} C_{abcd} \,, \end{align}

where ${}^*\!C^{abcd}$ is the (left) dual Weyl tensor, defined in the same way as the dual to Riemann. Interestingly, the Pontryagin is completely insensitive to the Ricci tensor.

For future purposes, we will also define a complex tensor where the real part is Weyl and the imaginary part is the (left) dual of Weyl,

\begin{align} \tilde{C}_{abcd} \equiv C_{abcd} + i {}^*\! C_{abcd} \,. \end{align}

The square of this tensor contains both Kretschmann (minus the Ricci parts) and Pontryagin:

\begin{align} \tilde{C}_{abcd} \tilde{C}^{abcd} = 2 ( C_{abcd} C^{abcd} + i {}^*\!C_{abcd} C^{abcd} ) \,. \end{align}

This quantity is related to the so-called $I$ invariant which is usually defined in spinor language (see e.g. Stephani et al.2). The completely symmetric ‘Weyl spinor’ $\Psi_{(ABCD)}$ related to the tensor $\tilde{C}$ is

\begin{align} \label{eq:tilde-C-to-spinor} \tilde{C}_{abcd} = 2\Psi_{ABCD} \varepsilon_{\dot{A}\dot{B}} \varepsilon_{\dot{C}\dot{D}} \,, \end{align}

where tensor indices $a, b, c, d$ are related to spinor index pairs $A\dot{A}, B\dot{B}$ and so on. In other words, the soldering form (or van der Waerden symbols) $\sigma^a{}_{A\dot{A}}\sigma^b{}_{B\dot{B}}\cdots$ are to be inferred on the LHS of Eq. \eqref{eq:tilde-C-to-spinor}.

The aforementioned $I$ invariant is given by

\begin{align} \label{eq:I-inv-def} I &\equiv \frac{1}{2} \Psi_{ABCD} \Psi^{ABCD} = \frac{1}{32} \tilde{C}_{abcd} \tilde{C}^{abcd} \\ I &= \frac{1}{2}( \Psi_0 \Psi_4 - 4 \Psi_1 \Psi_3 + 6 \Psi_2^2 - 4 \Psi_1 \Psi_3 + \Psi_0 \Psi_4 ) \,, \end{align}

where the Weyl scalars in some spinor dyad basis $\{ o^A, \iota^A \}$ are defined as usual (the subscript counts the number of $\iota$’s contracted into the spinor),

\begin{align} \Psi_0 &= \Psi_{oooo} \,, & \Psi_1 &= \Psi_{ooo\iota} \,, \\ \Psi_2 &= \Psi_{oo\iota\iota} \,, & \Psi_3 &= \Psi_{o\iota\iota\iota} \,, \\ \Psi_4 &= \Psi_{\iota\iota\iota\iota} \,. \end{align}

Similarly, for the $J$ invariant,

\begin{align} J &\equiv \frac{1}{6} \Psi_{ABCD} \Psi^{CDEF} \Psi_{EF}{}^{AB} = \frac{1}{384} \tilde{C}_{abcd} \tilde{C}^{cdef} \tilde{C}_{ef}{}^{ab} \,, \\ J &= \Psi_0 \Psi_2 \Psi_4 + 2 \Psi_1 \Psi_2 \Psi_3 - \Psi_0 \Psi_3^2 - \Psi_1^2 \Psi_4 - \Psi_2^3 \,. \end{align}

A mnemonic for remembering $J$ in terms of the Weyl scalars is given by the determinant expression

\begin{align} J = \left| \begin{matrix} \Psi_0 & \Psi_1 & \Psi_2 \\ \Psi_1 & \Psi_2 & \Psi_3 \\ \Psi_2 & \Psi_3 & \Psi_4 \end{matrix} \right| \,. \end{align}
$\newcommand{\lie}{\mathcal{L}} \newcommand{\Rf}{ {}^{(4)}\!R} \newcommand{\Rt}{ {}^{(3)}\!R} \newcommand{\cd}{\nabla}$

# 3+1 and electric/magnetic decomposition

Now let us suppose we have a 3+1 decomposition of the spacetime, i.e. that (a region of) the spacetime is foliated by spacelike hypersurfaces (see standard references on this topic, e.g. Baumgarte and Shapiro3). We have a timelike unit normal $n^a$, with $n^a n_a = -1$. This induces a splitting of the tangent spaces into time components (parallel to $n$) and ‘spatial’ components (orthogonal to $n$). We have the spatial metric induced on the spatial hypersurface $\gamma_{ab}$, with $\gamma_{ab}n^b = 0$, given by

\begin{align} g_{ab} = \gamma_{ab} - n_a n_b \,. \end{align}

The metric-compatible derivative of $g_{ab}$ is $\cd_a$, the metric-compatible derivative of $\gamma_{ab}$ is $D_a$, the extrinsic curvature is $K_{ab} = -\frac{1}{2}\lie_n \gamma_{ab}$.

To avoid confusion, we will now label the 4-Riemann (and 4-Ricci) with $\Rf$, and the 3-Riemann (and 3-Ricci, a.k.a. “thricci”) with $\Rt$.

We can now decompose all 4-dimensional objects using this splitting of the tangent bundle. We will need to split the 4-dimensional volume form, via (Warning: conventions may vary!)

\begin{align} \epsilon_{abcd} &= -4 n_{[a} \epsilon_{bcd]} \,, & \epsilon_{abc} &= n^d \epsilon_{dabc} \,. \end{align}

There is no need for 3/4 superscripts because just reading the number of indices tells you whether this is the 3-volume or 4-volume form.

Now we proceed to splitting the Weyl tensor into space- and time pieces. This proceeds by analogy to the Maxwell tensor. The “electric” part of Weyl is defined via

\begin{align} \label{eq:E-def} E_{ab} \equiv C_{acbd} n^c n^d \,, \end{align}

while the “magnetic” part of Weyl is defined via (Warning: conventions may vary!)

\begin{align} \label{eq:B-def} B_{ab} \equiv {}^*\!C_{acbd} n^c n^d \,. \end{align}

From these two it is obvious that

\begin{align} \tilde{C}_{acbd} n^c n^d = E_{ab} + i B_{ab} \,. \end{align}

Each of $E_{ab}$ and $B_{ab}$ are symmetric and trace-free spatial tensors. This is as far as you can decompose Weyl without imposing additional structure (like some preferred direction).

We can re-express Weyl in terms of $E_{ab}$, $B_{ab}$, the timelike unit vector and spatial metric, and 3-epsilon tensor. I don’t know if there is a clever way to do this other than to take the below expression and verify that it’s correct. Let me know if you have some better insight. Regardless, the most compact way of writing this is again using the above ‘impose index symmetry’ operation $S_G$ in Eq. \eqref{eq:impose-sym-def} with $G=r[abcd]$, the index symmetries of Riemann. The expression is (thanks to Alfonso García-Parrado Gómez-Lobo)

\begin{align} \label{eq:weyl-in-E-B} C_{abcd} = S_{r[abcd]} \left[ 4 E_{ac} (\gamma_{bd} + n_b n_d) + \epsilon_{ab}{}^e n_d B_{ce} \right] \,. \end{align}

# Curvature invariants 2

Obviously we would like to express curvature invariants in terms of the electric and magnetic Weyl tensors. This is a snap if you use xTensor (otherwise there is a lot of algebra). The relationship is

\begin{align} I &= \frac{1}{2} (E_{ab}E^{ab}-B_{ab}B^{ab}) + i E_{ab}B^{ab} \,, \\ I &= \frac{1}{2} (E_{ab} + i B_{ab})( E^{ab} + i B^{ab} ) \,. \end{align}

Now it is easy to see that Weyl squared and Pontryagin are simply

\begin{align} \label{eq:Weyl-squared-in-E-B} C_{abcd} C^{abcd} &= 8 (E_{ab}E^{ab}-B_{ab}B^{ab}) \,, \\ {}^*C_{abcd} C^{abcd} &= 16 E_{ab} B^{ab} \,. \end{align}

Kretshmann of course also adds the 4-Ricci terms to Eq. \eqref{eq:Weyl-squared-in-E-B}, via the relationship in Eq. \eqref{eq:K-Weyl-decomp}.

Similarly, the $J$ invariant turns out to be

\begin{align} J = \frac{-1}{6} (E^a{}_b + i B^a{}_b)(E^b{}_c + i B^b{}_c)(E^c{}_a + i B^c{}_a) \,. \end{align}

# Computing from Gauss-Codazzi relations

Let’s suppose you have at your disposal a numerical relativity codebase (such as SpEC) and thus you have access to the 3-metric $\gamma_{ab}$, extrinsic curvature $K_{ab}$, the 3-Ricci $\Rt_{ab}$ and 4-Ricci $\Rf_{ab}$; and you would like to compute $E_{ab}$ and $B_{ab}$. Certainly you should not attempt to construct the full Riemann or Weyl tensors, because they have 256 numbers per grid point but there are only 10 algebraically independent components in $E_{ab}$ and $B_{ab}$!

Therefore we would like an expression for how to compute $E_{ab}$ and $B_{ab}$ in terms of the available quantities, without all the overhead of computing Riemann or Weyl. To do so requires using the Gauss-Codazzi equations (see e.g. Baumgarte and Shapiro3 for a didactic exposition). The Gauss and Codazzi equations relate the orthogonal decomposition of the 4-curvature to the intrinsic and extrinsic curvature of the spatial submanifold. These are implemented in xAct with the command GaussCodazzi[].

The Gauss equation relates the completely spatial projection,

\begin{align} \label{eq:Gauss-eq} \gamma_a^p \gamma_b^q \gamma_c^r \gamma_d^s \Rf_{pqrs} = \Rt_{abcd} + K_{ac}K_{bd} - K_{ad}K_{bc} \,. \end{align}

The recipe for finding a formula for $E_{ab}$ is as follows:

1. Use the 4-metric $g^{ac}$ to trace over two indices
2. Use the Weyl decomposition \eqref{eq:R-decomp-1} for $\Rf$ on the left hand side
3. Insert the E/B decomposition \eqref{eq:weyl-in-E-B} for the Weyl tensor on the left hand side
4. You should now have $E_{bd}$ with free indices and no prefactor

I highly recommend using xAct for this calculation, to avoid errors (see the companion notebook). The result is

\begin{align} E_{ab} =& K_{ab} K^c_c - K_a^c K_{bc} + \Rt_{ab} \nonumber \\ & - \frac{1}{2} \gamma_a^c \gamma_b^d \Rf_{cd} - \frac{1}{2} \gamma_{ab} \gamma^{cd} \Rf_{cd} + \frac{1}{3} \gamma_{ab} \Rf \,. \end{align}

The three terms on the second line all vanish when in a 4-Ricci-flat manifold, e.g. in vacuum in GR.

To find an expression for the magnetic part $B_{ab}$ involves the Codazzi equation. This is the decomposition of 4-Riemann when one index is contracted with $n^a$ and the remaining three are projected onto spatial directions. The Codazzi equation is

\begin{align} \gamma_a^p \gamma_b^q \gamma_c^r n^s \Rf_{pqrs} = D_b K_{ac} - D_a K_{bc} \,. \end{align}

The recipe for finding a formula for $B_{ab}$ is as follows:

1. Contract with the 3-epsilon $\epsilon_d{}^{bc}$
2. Use the Weyl decomposition \eqref{eq:R-decomp-1} for $\Rf$ on the left hand side
3. Insert the E/B decomposition \eqref{eq:weyl-in-E-B} for the Weyl tensor on the left hand side
4. Symmetrize on indices $(a,d)$
5. You should now have $B_{ad}$ with free indices and no prefactor

Once again you should use xAct (see the companion notebook). The result is simply

\begin{align} B_{ab} = \epsilon_{cd(a} D^c K_{b)}{}^d \,. \end{align}

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