A canonical structure on the tangent bundle of a pseudo- or para-Kähler manifold

Abstract

It is a classical fact that the cotangent bundle $T^* {\mathcal {M}}$ of a differentiable manifold ${\mathcal {M}}$ enjoys a canonical symplectic form $\Omega ^*$. If $({\mathcal {M}},\mathrm{J} ,g,\omega )$ is a pseudo-Kähler or para-Kähler $2n$-dimensional manifold, we prove that the tangent bundle $T{\mathcal {M}}$ also enjoys a natural pseudo-Kähler or para-Kähler structure $({\tilde{\hbox {J}}},\tilde{g},\Omega )$, where $\Omega$ is the pull-back by $g$ of $\Omega ^*$ and $\tilde{g}$ is a pseudo-Riemannian metric with neutral signature $(2n,2n)$. We investigate the curvature properties of the pair $({\tilde{\hbox {J}}},\tilde{g})$ and prove that: $\tilde{g}$ is scalar-flat, is not Einstein unless $g$ is flat, has nonpositive (resp. nonnegative) Ricci curvature if and only if $g$ has nonpositive (resp. nonnegative) Ricci curvature as well, and is locally conformally flat if and only if $n=1$ and $g$ has constant curvature, or $n>2$ and $g$ is flat. We also check that (i) the holomorphic sectional curvature of $({\tilde{\hbox {J}}},\tilde{g})$ is not constant unless $g$ is flat, and (ii) in $n=1$ case, that $\tilde{g}$ is never anti-self-dual, unless conformally flat.

Appendix: the Weyl tensor in the pseudo-Kähler or para-Kähler cases

The Riemann curvature tensor ${\hbox {Rm}}$ of a pseudo-Riemannian manifold ${\mathcal {N}}$ may be seen as a symmetric form $\text {R}$ on bivectors of $\Lambda ^2 T{\mathcal {N}}$ (see [3] for references). Splitting $\text {R}$ along the eigenspaces $\Lambda ^+ \oplus \Lambda ^-$ of the Hodge operator $*$ on $\Lambda ^2 T{\mathcal {N}}$, yields the following block decomposition

$$\begin{aligned} \text {R}= \left( \begin{array}{cc} \text {W}^+ + \frac{\text {Scal}}{12} I &{} \text {Z}\\ \text {Z}^* &{} \text {W}^- + \frac{\text {Scal}}{12} I \end{array}\right) \end{aligned}$$

where $Z^*$ denotes the adjoint w.r.t. the induced metric on $\Lambda ^2 T{\mathcal {N}}$, so that $\text {W}= \text {W}^+ \oplus \text {W}^-$, the Weyl tensor seen as a 2-form on $\Lambda ^2 T{\mathcal {N}}$, is the traceless, Hodge-commuting part of the Riemann curvature operator $\text {R}$. Hence the following formula

$$\begin{aligned} \text {W}= {\hbox {Rm}}- \frac{1}{2} \text {Ric}\bigcirc \!\!\!\!\!\!\wedge \,g + \frac{\text {Scal}}{12} g \bigcirc \!\!\!\!\!\!\wedge \,g \, . \end{aligned}$$

If, additionally, ${\mathcal {N}}$ is a four dimensional Kähler manifold, then

Theorem 7.1

W$^+$ can be written as a multiple of the scalar curvature by a parallel non-trivial 2-form on $\Lambda ^2 T{\mathcal {N}}$.

See Prop. 2 in [5] for a proof and the explicit formula for the tensor involved. We do not need it explicitly since we are only interested in the following

Corollary 6

$({\mathcal {N}}, g, \mathrm{J} )$ is anti-self-dual ($\text {W}^+ = 0$) if and only if the scalar curvature vanishes.

The result extends to the two cases considered in this article: (1) neutral pseudo-Kähler manifolds and (2) para-Kähler manifolds, with a slight twist: $\text {W}^+$ is replaced by $\text {W}^-$. Precisely:

Theorem 7.2

Let $({\mathcal {N}}, g, \mathrm{J} )$ be a four dimensional manifold endowed with a pseudo-Kähler neutral metric (respectively a para-Kähler metric, necessarily neutral). Then the Weyl tensor W commutes with the Hodge operator and ${\mathcal {N}}$ is self-dual (W$^- = 0$) if and only if the scalar curvature vanishes.

The result for neutral pseudo-Kähler manifolds is probably known and relates to representation theory (see [3] for introduction and references), but since we could not find an explicit proof in the literature⁵, we will give a simple one below. To our knowledge, the proof for the para-Kähler case is new (albeit similar).

1.1 A.1 The pseudo-Kähler case

We will write explicitly the Weyl tensor in a given positively oriented orthonormal frame, denoted by $( e_1, e_{1'}, e_2, e_{2'})$, where $e_{1'} = \mathrm{J} e_1, e_{2'} = \mathrm{J} e_2, g ( e_1) = g ( e_{1'}) = - 1$ and $g ( e_2) = g ( e_{2'}) = + 1$. (For brevity, $g(X)$ denotes the norm $g(X,X)$.) The pseudo-metric $g$ extends to bivectors, has signature $( 2, 4)$, and will be again denoted by $g$: $g ( e_a \wedge e_b) = g ( e_a) g ( e_b) - g ( e_a, e_b)^2 = g ( e_a) g ( e_b)$, so that $\mathcal {B}= ( e_1 \wedge e_{1'}, e_1 \wedge e_2, e_1 \wedge e_{2'}, e_{1'} \wedge e_2, e_{1'} \wedge e_{2'}, e_2 \wedge e_{2'})$ is an orthonormal frame of $\Lambda ^2$, with $g ( e_a \wedge e_b) = - 1$, except for $g ( e_1 \wedge e_{1'}) = g ( e_2 \wedge e_{2'}) = + 1$. (Note that the other convention, taking $- g$ does not change the induced metric on $\Lambda ^2$.)

Since the volume $e_1 \wedge e_{1'} \wedge e_2 \wedge e_{2'}$ is positively oriented, we construct an orthonormal eigenbasis for the Hodge star on $\Lambda ^2 T{\mathcal {N}}$:

$$\begin{aligned} \left\{ \begin{array}{l} E_1^{\pm } = \frac{\sqrt{2}}{2} ( e_1 \wedge e_{1'} \pm e_2 \wedge e_{2'})\\ E_2^{\pm } = \frac{\sqrt{2}}{2} ( e_1 \wedge e_2 \pm e_{1'} \wedge e_{2'})\\ E_3^{\pm } = \frac{\sqrt{2}}{2} ( e_1 \wedge e_{2'} \mp e_{1'} \wedge e_2) \end{array} \right. \end{aligned}$$

so that $\Lambda ^{\pm }$ is generated by $E_1^{\pm }, E_2^{\pm }, E_3^{\pm }$.

The Kähler condition implies

$$\begin{aligned} {\hbox {Rm}}( \mathrm{J} X, \mathrm{J} Y, Z, T) = {\hbox {Rm}}( X, Y, Z, T) = {\hbox {Rm}}( X, Y, \mathrm{J} Z, \mathrm{J} T), \end{aligned}$$

because $\mathrm{J}$ is isometric and parallel. The matrix of the symmetric 2-form $\text {R}$ in the orthonormal frame $\mathcal {B}$ is

	$e_{11'}$	$e_{12}$	$e_{12'}$	$e_{1'2}$	$e_{1'2'}$	$e_{22'}$
$e_{11'}$	$\text {R}_{11' 11'}$	$\text {R}_{11' 12}$	$\text {R}_{11' 12'}$	$\text {R}_{11' 1' 2} = - \text {R}_{11' 12'}$	$\text {R}_{11' 1' 2'} = \text {R}_{11' 12}$	$\text {R}_{11' 22'}$
$e_{12}$		$\text {R}_{1212}$	$\text {R}_{1212'}$	$\text {R}_{121' 2}= - \text {R}_{1212'}$	$\text {R}_{131' 2'} = \text {R}_{1212}$	$\text {R}_{1222'}$
$e_{12'}$			$\text {R}_{12' 12'}$	$\text {R}_{12' 1' 2} = - \text {R}_{12' 12'}$	$\text {R}_{12' 1' 2'} = \text {R}_{1212'}$	$\text {R}_{12' 22'}$
$e_{1'2}$				$\text {R}_{1' 21' 2} = \text {R}_{12' 12'}$	$\text {R}_{1' 21' 2'} = - \text {R}_{1212'}$	$\text {R}_{1' 222'} = - \text {R}_{12' 22'}$
$e_{1'2'}$					$\text {R}_{1' 2' 1' 2'} = \text {R}_{1212}$	$\text {R}_{1' 2'22'} = \text {R}_{1222'}$
$e_{22'}$						$\text {R}_{22' 22'}$

where $e_{ab}$ stands for $e_a \wedge e_b$, for greater legibility. We have written the matrix as a table for clarity and to make symmetries more obvious, and because $\text {R}$ is symmetric we need only write half the matrix. We have used the internal symmetries of $\text {R}$, to choose among equivalent coefficients the ones lowest in the lexicographic order of the indices.

The Weyl tensor satisfies some of the $\mathrm{J}$-symmetries of $\text {R}$: indeed

$$\begin{aligned} \text {Ric}( \mathrm{J} X, \mathrm{J} Y)&= \sum _{i=1}^4 g ( e_i) {\hbox {Rm}}( \mathrm{J} X, e_i, \mathrm{J} Y, e_i) =\sum _{i=1}^4 g ( e_i) {\hbox {Rm}}( X, \mathrm{J} e_i, Y, \mathrm{J} e_i)\\&= \sum _{i=1}^4 g ( \mathrm{J} e_i) {\hbox {Rm}}( X, \mathrm{J} e_i, Y, \mathrm{J} e_i) = \text {Ric}( X, Y) \end{aligned}$$

because $( \mathrm{J} e_i)$ is again an orthonormal basis. In particular, this invariance implies $r_{11'} = \text {Ric}( e_1, e_{1'}) = r_{1' 1} = - r_{11'}$, so $r_{11'}$ vanish (and so does $r_{22'}$). For the Kulkarni–Nomizu product,

$$\begin{aligned} \text {Ric}\bigcirc \!\!\!\!\!\!\wedge \,g ( \mathrm{J} X, Y, Z, T)&= \text {Ric}(\mathrm{J} X, Z) g (Y, T) + \text {Ric}(Y, T) g (\mathrm{J} X, Z)\\&- \text {Ric}(\mathrm{J} X, T) g (Y, Z) - \text {Ric}(Y,Z) g (\mathrm{J} X, T)\\&= - \text {Ric}(X, \mathrm{J} Z) g (\mathrm{J} Y, \mathrm{J} T) - \text {Ric}(\mathrm{J} Y, \mathrm{J} T) g (X, \mathrm{J} Z)\\&+ \text {Ric}(X, \mathrm{J} T) g (\mathrm{J} Y, \mathrm{J} Z) + \text {Ric}(\mathrm{J} Y, \mathrm{J} Z) g (X, \mathrm{J} T)\\&= - \text {Ric}\bigcirc \!\!\!\!\!\!\wedge \,g ( X, \mathrm{J} Y, \mathrm{J} Z, \mathrm{J} T) \end{aligned}$$

so

$$\begin{aligned} \text {Ric}\bigcirc \!\!\!\!\!\!\wedge \,g ( \mathrm{J} X, \mathrm{J} Y, Z, T) = - \text {Ric}\bigcirc \!\!\!\!\!\!\wedge \,g ( X, \mathrm{J} ^2 Y, \mathrm{J} Z, \mathrm{J} T) = \text {Ric}\bigcirc \!\!\!\!\!\!\wedge \,g ( X, Y, \mathrm{J} Z, \mathrm{J} T) \, . \end{aligned}$$

Hence the following symmetries (fewer than for ${\hbox {Rm}}$) in the coefficients of $\text {Ric}\bigcirc \!\!\!\!\!\!\wedge \,g$, $g \bigcirc \!\!\!\!\!\!\wedge \,g$ and ${\hbox {Rm}}$, and therefore $\text {W}$:

	$e_{1 1'}$	$e_{1} \wedge e_{2}$	$e_{1 2'}$	$e_{1'} \wedge e_{2}$	$e_{1' 2'}$	$e_{2 2'}$
$e_{1 1'}$	$\text {W}_{11' 11'}$	$\text {W}_{11' 12}$	$\text {W}_{11' 12'}$	$\text {W}_{11' 1' 2} = - \text {W}_{11' 12'}$	$\text {W}_{11' 1' 2'} = \text {W}_{11' 12}$	$\text {W}_{11' 22'}$
$e_{1 2}$		$\text {W}_{1212}$	$\text {W}_{1212'}$	$\text {W}_{121' 2}$	$\text {W}_{121' 2'}$	$\text {W}_{1222'}$
$e_{1 2'}$			$\text {W}_{12' 12'}$	$\text {W}_{12' 1' 2}$	$\text {W}_{12' 1' 2'} = - \text {W}_{121' 2}$	$\text {W}_{12' 22'}$
$e_{1' 2}$				$\text {W}_{1' 21' 2} = \text {W}_{12' 12'}$	$\text {W}_{1' 21' 2'} = - \text {W}_{1212'}$	$\text {W}_{1' 222'} = - \text {W}_{12' 22'}$
$e_{1' 2'}$					$\text {W}_{1' 2' 1' 2'} = \text {W}_{1212}$	$\text {W}_{1' 2' 22'} = \text {W}_{1222'}$
$e_{2 2'}$						$\text {W}_{22' 22'}$

Expanding on the above eigenbasis of $\Lambda ^+ \oplus \Lambda ^-$ (which differs from the one in the positive definite case) yields the following Weyl tensor coefficients, which we have simplified using the symmetries above (up to a factor $1 / 2$ due to normalization):

	$E_1^+$	$E_2^+$	$E_3^+$
$E_1^+$	$\text {W}_{11' 11'}$ + $\text {W}_{22' 22'}$ + $2 \text {W}_{11' 22'}$	$2 ( \text {W}_{11' 12} + \text {W}_{1222'})$	$2 ( \text {W}_{11' 12'} + \text {W}_{12' 22'})$
$E_2^+$		$2 ( \text {W}_{1212} + \text {W}_{121' 2'})$	$2 ( \text {W}_{1212'} - \text {W}_{121' 2})$
$E_3^+$			$2 ( \text {W}_{12' 12'} - \text {W}_{12' 1' 2})$
$E_1^-$
$E_2^-$
$E_3^-$

	$E_1^-$	$E_2^-$	$E_3^-$
$E_1^+$	$\text {W}_{11' 11'} - \text {W}_{22' 22'}$	0	0
$E_2^+$	$2 ( \text {W}_{11' 12} - \text {W}_{1222'})$	0	0
$E_3^+$	$2 ( \text {W}_{11' 12'} - \text {W}_{12' 22'})$	0	0
$E_1^-$	$\text {W}_{11' 11'} + \text {W}_{22' 22'} - 2 \text {W}_{11' 22'}$	0	0
$E_2^-$		$2 ( \text {W}_{1212} - \text {W}_{121' 2'})$	$2 ( \text {W}_{1212'} + \text {W}_{121' 2})$
$E_3^-$			$2 ( \text {W}_{12' 12'} + \text {W}_{12' 1' 2})$

(Again only half the coefficients are written down.) Further simplifications come from computing $\text {W}$, and using

$$\begin{aligned} \text {Scal}&= - r_{11} - r_{1' 1'} + r_{22} + r_{2' 2'} = 2 ( r_{22} -r_{11})\\&= 2 ( - ( - \text {R}_{11' 11'} + \text {R}_{1212} + \text {R}_{12' 12'}) + ( - \text {R}_{1212} - \text {R}_{1' 21' 2} + \text {R}_{22' 22'}))\\&= 2 ( \text {R}_{11' 11'} - 2 ( \text {R}_{1212} + \text {R}_{12' 12'}) + \text {R}_{22' 22'}) \, . \end{aligned}$$

First prove that the Hodge star commutes with $\text {W}$ by considering $\text {W}( \Lambda ^+, \Lambda ^-)$:

$$\begin{aligned} \text {W}_{11' 11'}&= \text {R}_{11' 11'} + \frac{1}{2} ( r_{11} + r_{1' 1'}) + \frac{\text {Scal}}{6} = \text {R}_{11' 11'} + r_{11} + \frac{\text {Scal}}{6}\\&= \text {R}_{1212} + \text {R}_{12' 12'} + \frac{\text {Scal}}{6}\\ \text {W}_{22' 22'}&= \text {R}_{22' 22'} - \frac{1}{2} ( r_{22} + r_{2' 2'}) + \frac{\text {Scal}}{6} = \text {R}_{22' 22'} - r_{22} + \frac{\text {Scal}}{6}\\&= \text {R}_{1212} + \text {R}_{12' 12'} + \frac{\text {Scal}}{6} \end{aligned}$$

so that $\text {W}_{11' 11'} - \text {W}_{22' 22'} = 0$. Similarly

$$\begin{aligned} \text {W}_{11' 12} = \text {R}_{11' 12} + \frac{r_{1' 2}}{2}, \text {W}_{1222'} = \text {R}_{1222'} + \frac{r_{12'}}{2} = \text {R}_{1222'} - \frac{r_{1' 2}}{2} \end{aligned}$$

so

$$\begin{aligned}&\text {W}_{11' 12} - \text {W}_{1222'} = \text {R}_{11' 12} - \text {R}_{1222'} + r_{1' 2} = 0\\&\text {W}_{11' 12'} = \text {R}_{11' 12'} + \frac{r_{1' 2'}}{2} = \text {R}_{11' 12'} + \frac{r_{12}}{2}, \text {W}_{12' 22'} = \text {R}_{12' 22'} - \frac{r_{12}}{2},\\&\text {W}_{11' 12'} - \text {W}_{12' 22'} = \text {R}_{11' 12'} - \text {R}_{12' 22'} + r_{12} = 0 . \end{aligned}$$

That proves that $\text {W}$ is block-diagonal.

The $\text {W}^-$ term satisfies

$$\begin{aligned} \text {W}_{11' 11'} + \text {W}_{22' 22'} - 2 \text {W}_{11' 22'}&= \text {R}_{11' 11'} + r_{11} + \text {R}_{22' 22'} - r_{22} + \frac{\text {Scal}}{3}\\&- 2 \text {R}_{11' 22'}\\&= \text {R}_{11' 11'} + \text {R}_{22' 22'} - 2 \text {R}_{11' 22'} - \frac{\text {Scal}}{6}\\&= \text {R}_{11' 11'} + \text {R}_{22' 22'} - 2 ( \text {R}_{1212} + \text {R}_{12' 12'})\\&- \frac{\text {Scal}}{6}\\&= \frac{\text {Scal}}{2} - \frac{\text {Scal}}{6} = \frac{\text {Scal}}{3} \end{aligned}$$

using the first Bianchi identity (and the invariance of ${\hbox {Rm}}$):

$$\begin{aligned} \text {R}_{11' 22'}&= - \text {R}_{1' 212'} - \text {R}_{211' 2'} = \text {R}_{12' 12'} + \text {R}_{1212} .\\ \text {W}_{1212} - \text {W}_{121' 2'}&= \text {R}_{1212} + \frac{r_{22} - r_{11}}{2} - \frac{\text {Scal}}{6} - \text {R}_{121' 2'} = \frac{\text {Scal}}{4} - \frac{\text {Scal}}{6}\\&= \frac{\text {Scal}}{12}\\ \text {W}_{12' 12'} + \text {W}_{12' 1' 2}&= \text {R}_{12' 12'} + \frac{\text {Scal}}{4} - \frac{\text {Scal}}{6} + \text {R}_{12' 1' 2} = \frac{\text {Scal}}{12}\\ \text {W}_{1212'} + \text {W}_{121' 2}&= \text {R}_{1212'} + \frac{r_{22'}}{2} + \text {R}_{121' 2} - \frac{r_{11'}}{2} = \frac{1}{2} ( r_{22'} - r_{11'}) = 0 . \end{aligned}$$

Finally,

$$\begin{aligned} \text {W}^- = \text {Scal} \left( \begin{array}{ccc} 1 / 3 &{} &{} \\ &{} 1 / 6 &{} \\ &{} &{} 1 / 6 \end{array}\right) = \frac{\text {Scal}}{6} \text {Id} + \frac{\text {Scal}}{6} E^-_1 \otimes E^-_1 \end{aligned}$$

(and indeed this matrix is traceless w.r.t. the pseudo-metric $g$). One should note that the above expression differs from the Riemannian case, where

$$\begin{aligned} \text {W}^+ = \text {Scal} \left( \begin{array}{ccc} 1 / 3 &{} &{} \\ &{} - 1 / 6 &{} \\ &{} &{} - 1 / 6 \end{array}\right) = - \frac{\text {Scal}}{6} \text {Id} + \frac{\text {Scal}}{3} E_1^+ \otimes E_1^+ . \end{aligned}$$

We let the Reader check that in the neutral case, the $\text {W}^+$ part is not a multiple of the scalar curvature, which completes the proof of Theorem 7.2.

1.2 A.2 The para-Kähler case

The computations are almost identical, but the results differ from the pseudo-Kähler setup, because the para-complex structure $\mathrm{J}$ is now an anti-isometry: $\text {R}( \mathrm{J} X, \mathrm{J} Y)Z = - \text {R}( X, Y) Z$. We pick an orthonormal basis $( e_1, e_{1'}, e_2, e_{2'})$ with $e_{1'} = \mathrm{J} e_1, e_{2'} = \mathrm{J} e_2$, and $g(e_1) = g ( e_2) = + 1$, $g ( e_{1'}) = g ( e_{2'}) = - 1$. The frame $\mathcal {B}= ( e_1 \wedge e_{1'}, e_1 \wedge e_2, e_1 \wedge e_{2'}, e_{1'} \wedge e_2, e_{1'} \wedge e_{2'}, e_2 \wedge e_{2'})$ of $\Lambda ^2 T{\mathcal {N}}$ is also orthonormal w.r.t. the induced metric on $\Lambda ^2$, again denoted by $g$, which has signature $( 2, 4)$: $g ( e_a \wedge e_b) = g ( e_a) g ( e_b) = - 1$, except for $g ( e_1 \wedge e_2) = g ( e_{1'} \wedge e_{2'}) = + 1$.

An orthonormal eigenbasis for the Hodge operator is the following:

$$\begin{aligned} \left\{ \begin{array}{l} E_1^{\pm } = \frac{\sqrt{2}}{2} ( e_1 \wedge e_{1'} \mp e_2 \wedge e_{2'})\\ E_2^{\pm } = \frac{\sqrt{2}}{2} ( e_1 \wedge e_2 \mp e_{1'} \wedge e_{2'})\\ E_3^{\pm } = \frac{\sqrt{2}}{2} ( e_1 \wedge e_{2'} \mp e_{1'} \wedge e_2) \end{array} \right. \end{aligned}$$

where the $E_a^+$ (resp. $E_a^-$) span $\Lambda ^+$ (resp. $\Lambda ^-$). (Note the sign differences w.r.t. the pseudo-Kähler case.)

Since $\mathrm{J}$ is anti-isometric and parallel,

$$\begin{aligned} {\hbox {Rm}}( \mathrm{J} X, \mathrm{J} Y, Z, T) = - {\hbox {Rm}}( X,Y, Z, T) = {\hbox {Rm}}( X, Y, \mathrm{J} Z, \mathrm{J} T) \, . \end{aligned}$$

Hence the following symmetries of the Riemannian curvature operator $\text {R}$, expressed in the frame $\mathcal {B}$ (for symmetry reasons and greater legibility, lower left coefficients are not written in this and the subsequent matrices):

	$e_{1 1'}$	$e_{12}$	$e_{1 2'}$	$e_{1'} \wedge e_{2}$	$e_{1' 2'}$	$e_{2 2'}$
$e_{1 1'}$	$\text {R}_{11' 11'}$	$\text {R}_{11' 12}$	$\text {R}_{11' 12'}$	$\text {R}_{11' 1' 2} = - \text {R}_{11' 12'}$	$\text {R}_{11' 1' 2'} = - \text {R}_{11' 12}$	$\text {R}_{11' 22'}$
$e_{1 2}$		$\text {R}_{1212}$	$\text {R}_{1212'}$	$\text {R}_{121' 2} = - \text {R}_{1212'}$	$\text {R}_{121' 2'} = - \text {R}_{1212}$	$\text {R}_{1222'}$
$e_{1 2'}$			$\text {R}_{12' 12'}$	$\text {R}_{12' 1' 2} = - \text {R}_{12' 12'}$	$\text {R}_{12' 1' 2'} = - \text {R}_{1212'}$	$\text {R}_{12' 22'}$
$e_{1' 2}$				$\text {R}_{1' 21' 2} = \text {R}_{12' 12'}$	$\text {R}_{1' 21' 2'} = \text {R}_{1212'}$	$\text {R}_{1' 222'} = - \text {R}_{12' 22'}$
$e_{1' 2'}$					$\text {R}_{1' 2' 1' 2'} = \text {R}_{1212}$	$\text {R}_{1' 2'22'} = - \text {R}_{1222'}$
$e_{2 2'}$						$\text {R}_{22' 22'}$

(Note again the similarity with the pseudo-Kähler case: only a few signs change.)

The Weyl tensor satisfies some of the $\mathrm{J}$-symmetries of ${\hbox {Rm}}$ since

$$\begin{aligned} \text {Ric}( \mathrm{J} X, \mathrm{J} Y)&= \sum _{i=1}^4 g(e_i) {\hbox {Rm}}( \mathrm{J} X, e_i, \mathrm{J} Y, e_i) = \sum _{i=1}^4 g ( e_i) {\hbox {Rm}}( X, \mathrm{J} e_i, Y, \mathrm{J} e_i)\\&= - \sum _{i=1}^4 g ( \mathrm{J} e_i) {\hbox {Rm}}( X, \mathrm{J} e_i, Y, \mathrm{J} e_i) = - \text {Ric}( X, Y) \end{aligned}$$

since $( \mathrm{J} e_i)$ is also an orthonormal basis. In particular this invariance implies $r_{1' 1} = r_{11'} = - r_{1' 1}$, so $r_{11'}$ vanishes (and so does $r_{22'}$). Finally,

$$\begin{aligned} \frac{\text {Scal}}{2} = r_{11} + r_{22} = - \text {R}_{11' 11'} + 2 ( \text {R}_{1212} - \text {R}_{12' 12'}) - \text {R}_{22' 22'} . \end{aligned}$$

The Kulkarni–Nomizu product $\text {Ric}\bigcirc \!\!\!\!\!\!\wedge \,g$ satisfies

$$\begin{aligned} \text {Ric}\bigcirc \!\!\!\!\!\!\wedge \,g ( \mathrm{J} X, Y, Z, T)&= \text {Ric}(\mathrm{J} X, Z) g (Y, T) + \text {Ric}(Y, T) g (\mathrm{J} X, Z) \\&- \text {Ric}(\mathrm{J} X, T) g (Y, Z) - \text {Ric}(Y,Z) g (\mathrm{J} X, T) \\&= \text {Ric}(X, \mathrm{J} Z) g (\mathrm{J} Y, \mathrm{J} T) + \text {Ric}(\mathrm{J} Y, \mathrm{J} T) g (X, \mathrm{J} Z)\\&- \text {Ric}(X, \mathrm{J} T) g (\mathrm{J} Y, \mathrm{J} Z) - \text {Ric}(\mathrm{J} Y, \mathrm{J} Z) g (X, \mathrm{J} T)\\&= \text {Ric}\bigcirc \!\!\!\!\!\!\wedge \,g ( X, \mathrm{J} Y, \mathrm{J} Z, \mathrm{J} T) \end{aligned}$$

so

$$\begin{aligned} \text {Ric}\bigcirc \!\!\!\!\!\!\wedge \,g ( \mathrm{J} X, \mathrm{J} Y, Z, T) = \text {Ric}\bigcirc \!\!\!\!\!\!\wedge \,g ( X, \mathrm{J} ^2 Y, \mathrm{J} Z, \mathrm{J} T) = \text {Ric}\bigcirc \!\!\!\!\!\!\wedge \,g ( X, Y, \mathrm{J} Z, \mathrm{J} T) \end{aligned}$$

and the same property holds for $g \bigcirc \!\!\!\!\!\!\wedge \,g$. Hence the following symmetries (fewer than for ${\hbox {Rm}}$) in the coefficients of $\text {Ric}\bigcirc \!\!\!\!\!\!\wedge \,g$, $g \bigcirc \!\!\!\!\!\!\wedge \,g$ and ${\hbox {Rm}}$, and therefore $\text {W}$:

	$e_{1 1'}$	$e_{1 2}$	$e_{1 2'}$	$e_{1' 2}$	$e_{1' 2'}$	$e_{2 2'}$
$e_{1 1'}$	$\text {W}_{11' 11'}$	$\text {W}_{11' 12}$	$\text {W}_{11' 12'}$	$\text {W}_{11' 1' 2} = - \text {W}_{11' 12'}$	$\text {W}_{11' 1' 2'} = - \text {W}_{11' 12}$	$\text {W}_{11' 22'}$
$e_{1 2}$		$\text {W}_{1212}$	$\text {W}_{1212'}$	$\text {W}_{121' 2}$	$\text {W}_{121' 2'}$	$\text {W}_{1222'}$
$e_{1 2'}$			$\text {W}_{12' 12'}$	$\text {W}_{12' 1' 2}$	$\text {W}_{12' 1' 2'} = \text {W}_{121' 2}$	$\text {W}_{12' 22'}$
$e_{1' 2}$				$\text {W}_{1' 21' 2} = \text {W}_{12' 12'}$	$\text {W}_{1' 21' 2'} = \text {W}_{1212'}$	$\text {W}_{1' 222'} = - \text {W}_{12' 22'}$
$e_{1' 2'}$					$\text {W}_{1' 2' 1' 2'} = \text {W}_{1212}$	$\text {W}_{1' 2'22'} = - \text {W}_{1222'}$
$e_{2 2'}$						$\text {W}_{22' 22'}$

Let us now express $\text {W}$ in the Hodge basis defined earlier, using the above symmetries (up to a factor $1 / 2$ due to normalization).

	$E_1^+$	$E_2^+$	$E_3^+$
$E_1^+$	$\text {W}_{11' 11'} + \text {W}_{22' 22'} - 2 \text {W}_{11' 22'}$	$2 ( \text {W}_{11' 12} - \text {W}_{1222'})$	$2 ( \text {W}_{11' 12'} - \text {W}_{12' 22'})$
$E_2^+$		$2 ( \text {W}_{1212} - \text {W}_{121' 2'})$	$2 ( \text {W}_{1212'} - \text {W}_{121' 2})$
$E_3^+$			$2 ( \text {W}_{12' 12'} - \text {W}_{12' 1' 2})$
$E_1^-$
$E_2^-$
$E_3^-$

	$E_1^-$	$E_2^-$	$E_3^-$
$E_1^+$	$\text {W}_{11' 11'} - \text {W}_{22' 22'}$	0	0
$E_2^+$	$2 ( \text {W}_{11' 12} + \text {W}_{1222'})$	0	0
$E_3^+$	$2 ( \text {W}_{11' 12'} + \text {W}_{12' 22'})$	0	0
$E_1^-$	$\text {W}_{11' 11'} + \text {W}_{22' 22'} + 2 \text {W}_{11' 22'}$	0	0
$E_2^-$		$2 ( \text {W}_{1212} + \text {W}_{121' 2'})$	$2 ( \text {W}_{1212'} + \text {W}_{121' 2})$
$E_3^-$			$2 ( \text {W}_{12' 12'} + \text {W}_{12' 1' 2})$

Only three terms in the off-block-diagonal part are not obviously zero.

$$\begin{aligned} \text {W}_{11' 11'}&= \text {R}_{11' 11'} - \frac{1}{2} ( - r_{11} + r_{1' 1'}) - \frac{\text {Scal}}{6} = \text {R}_{11' 11'} + r_{11} - \frac{\text {Scal}}{6}\\ \text {W}_{22' 22'}&= \text {R}_{22' 22'} - \frac{1}{2} ( - r_{22} + r_{2' 2'}) - \frac{\text {Scal}}{6} = \text {R}_{22' 22'} + r_{22} - \frac{\text {Scal}}{6} \end{aligned}$$

but $r_{11} = - \text {R}_{11' 11'} + \text {R}_{1212} - \text {R}_{12' 12'}$ and $r_{22} = \text {R}_{2121} - \text {R}_{21' 21'} - \text {R}_{22' 22'} = \text {R}_{1212} - \text {R}_{12' 12'} - \text {R}_{22' 22'}$ so that

$$\begin{aligned} \text {W}_{11' 11'} - \text {W}_{22' 22'} = \text {R}_{11' 11'} - \text {R}_{22' 22'} + r_{11} - r_{22} = 0 . \end{aligned}$$

Similarly

$$\begin{aligned}&\text {W}_{11' 12} + \text {W}_{1222'} = \text {R}_{11' 12} - \frac{r_{1' 2}}{2} + \text {R}_{1222'} + \frac{r_{12'}}{2} = \text {R}_{11' 12} + \text {R}_{1222'} - r_{1' 2} = 0\\&\text {W}_{11' 12'} + \text {W}_{12' 22'} = \text {R}_{11' 12'} - \frac{r_{1' 2'}}{2} + \text {R}_{12' 22'} + \frac{r_{12}}{2} = \text {R}_{11' 12'} + \text {R}_{12' 22'} + r_{12} = 0 \end{aligned}$$

which proves that $\text {W}$ is block-diagonal, i.e. commutes with the Hodge operator.

Let us now look more closely at the $\text {W}^-$ term

$$\begin{aligned}&\left( \begin{array}{ccc} \text {W}_{11' 11'} + \text {W}_{22' 22'} + 2 \text {W}_{11' 22'} &{} 0 &{} 0\\ &{} 2 ( \text {W}_{1212} + \text {W}_{121' 2'}) &{} 2 ( \text {W}_{1212'} + \text {W}_{121' 2})\\ &{} &{} 2 ( \text {W}_{12' 12'} + \text {W}_{12' 1' 2}) \end{array}\right) \\&\text {W}_{11' 11'} + \text {W}_{22' 22'} + 2 \text {W}_{11' 22'}\\&\qquad = \text {R}_{11' 11'} + r_{11} - \frac{\text {Scal}}{6} + \text {R}_{22' 22'} + r_{22} - \frac{\text {Scal}}{6} + 2 \text {R}_{11' 22'}\\&\qquad = \text {R}_{11' 11'} + \text {R}_{22' 22'} + 2 \text {R}_{11' 22'} + \frac{\text {Scal}}{2} - \frac{\text {Scal}}{3}\\&\qquad = \text {R}_{11' 11'} + \text {R}_{22' 22'} + 2 ( - \text {R}_{1212} + \text {R}_{12' 12'}) + \frac{\text {Scal}}{6} = - \frac{\text {Scal}}{3} \end{aligned}$$

where we have used the first Bianchi identity (and the invariance of ${\hbox {Rm}}$)

$$\begin{aligned} \text {R}_{11' 22'}&= - \text {R}_{1' 212'} - \text {R}_{211' 2'} = \text {R}_{12' 12'} - \text {R}_{1212} .\\ \text {W}_{1212} + \text {W}_{121' 2'}&= \text {R}_{1212} - \frac{r_{22} + r_{11}}{2} + \frac{\text {Scal}}{6} + \text {R}_{121' 2'}\\&= \text {R}_{1212} - \frac{\text {Scal}}{4} + \frac{\text {Scal}}{6} + \text {R}_{121' 2'} = - \frac{\text {Scal}}{12}\\ \text {W}_{12' 12'} + \text {W}_{12' 1' 2}&= \text {R}_{12' 12'} + \frac{\text {Scal}}{4} - \frac{\text {Scal}}{6} + \text {R}_{12' 1' 2} = \frac{\text {Scal}}{12}\\ \text {W}_{1212'} + \text {W}_{121' 2}&= \text {R}_{1212'} - \frac{r_{22'}}{2} + \text {R}_{121' 2} - \frac{r_{11'}}{2} = 0 . \end{aligned}$$

Finally,

$$\begin{aligned} \text {W}^- = \text {Scal} \left( \begin{array}{ccc} - 1 / 3 &{} &{} \\ &{} - 1 / 6 &{} \\ &{} &{} 1 / 6 \end{array}\right) \end{aligned}$$

vanishes if and only if $\text {Scal} = 0$. (The Reader will check that this matrix is indeed traceless w.r.t. the pseudo-metric $g$.)