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

Henri Anciaux, Pascal Romon

► To cite this version:

Henri Anciaux, Pascal Romon. A canonical structure on the tangent bundle of a pseudo- or para-Kähler manifold. Monatshefte für Mathematik, 2014, 174 (3), pp.329-355. 10.1007/s00605-014-0630-6. hal-00778411v3

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

Henri Anciaux,\* Pascal Romon†

## Abstract

It is a classical fact that the cotangent bundle $T^*M$ of a differentiable manifold $M$ enjoys a canonical symplectic form $\Omega^*$. If $(M, J, g, \omega)$ is a pseudo-Kähler or para-Kähler $2n$-dimensional manifold, we prove that the tangent bundle $TM$ also enjoys a natural pseudo-Kähler or para-Kähler structure $(\tilde{J}, \tilde{g}, \Omega)$, 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 $(\tilde{J}, \tilde{g})$ 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 $(\tilde{J}, \tilde{g})$ is not constant unless $g$ is flat, and (ii) in $n = 1$ case, that $\tilde{g}$ is never anti-self-dual, unless conformally flat.

2010 MSC: 32Q15, 53D05

**Keywords:** tangent bundle, pseudo-Kähler geometry, para-Kähler geometry, self-duality and anti-self-duality.

## Introduction

It is a classical fact that given any differentiable manifold $M$, its cotangent bundle $T^*M$ enjoys a canonical symplectic structure $\Omega^*$.

Moreover, given a linear connection $\nabla$ on a manifold $M$, (e.g. the Levi-Civita connection of a Riemannian metric), the bundle $TTM$ splits into a direct sum of two subbundles $HM$ and $VM$, both isomorphic to $TM$. This allows to define an almost complex structure $J$ by setting $J(X_h, X_v) := (-X_v, X_h)$, where, for $X \in TTM = HM \oplus VM$, we write $X \simeq (X_h, X_v) \in TM \times TM$. Analogously, one may introduce a natural almost para-complex (or bi-Lagrangian) structure, setting $J'(X_h, X_v) := (X_v, X_h)$.

<footnote>\*Universidade de São Paulo; supported by CNPq (PQ 306154/2011-0) and Fapesp (2011/21362-2), henri.anciaux@gmail.com</footnote>

<footnote>†Université Paris-Est Marne-la-Vallée, pascal.romon@univ-mlv.fr</footnote>
It is also well known that the tangent bundle of a Riemannian manifold $(M, g)$ can be given a natural Riemannian structure, called Sasaki metric. A simple way to understand this construction, which extends verbatim to the case of a pseudo-Riemannian metric $g$ with signature $(p, m-p)$, is as follows: using the splitting $TM = HM \oplus VM$, we set:

$$G((X_h, X_v), (Y_h, Y_v)) := g(X_h, Y_h) + g(X_v, Y_v).$$

This metric has signature $(2p, 2(m-p))$ and is well behaved with respect to $J$ in two ways: (i) $G$ is compatible with $J$, i.e. $G(.,.) = G(J, J)$, and (ii) the symplectic form $\Omega := G(J,.)$ is nothing but the pull-back of $\Omega^*$ by the musical isomorphism between $TM \simeq_g T^*M$. In other words, the triple $(J, G, \Omega)$ defines an “almost pseudo-Kähler” structure<fnref n="1" /> on $TM$.

Unfortunately, this construction suffers two flaws: $J$ is not integrable unless $\nabla$ is flat and the metric $G$ is somewhat “rigid”: for example, if $G$ has constant scalar curvature, then $g$ is flat (see [15]). We refer to [5, 16] and the survey [9] for more detail on the Sasaki metric.

Another construction can be made in the case where $M$ is complex (resp. para-complex): in this case both $TM$ and $T^*M$ enjoy a canonical complex (resp. para-complex) structure which are defined as follows: given a family of holomorphic (resp. para-holomorphic<fnref n="2" />) local charts $\varphi: M \to U \subset \mathbb{R}^{2n}$ on $M$, we define holomorphic (resp. para-holomorphic) local charts $\bar{\varphi}: TM \to U \times \mathbb{R}^{2n}$ by $\bar{\varphi}(x, V) = (\varphi(x), d\varphi_x(V))$, $\forall (x, V) \in TM$ for the tangent bundle, and $\bar{\varphi}: T^*M \to U \times \mathbb{R}^{2n}$ by $\bar{\varphi}(x, \xi) = (\varphi(x), ((d\varphi_x)^t)^{-1}(\xi))$, $\forall (x, \xi) \in T^*M$ for the cotangent bundle. In the first section, we shall see that if $M$ is merely almost complex (resp. almost para-complex), then a more subtle argument allows to define again a canonical almost complex structure (resp. almost para-complex structure) on $TM$. On the other hand, we shall prove in the second section that if $M$ is pseudo- or para-Kähler, the corresponding structure on $TM$ can also be constructed using the splitting $HM \oplus VM$ induced by the Levi-Civita connection of the Kählerian metric.

Combining the canonical symplectic structure $\Omega^*$ of $T^*M$ with the canonical complex (resp. para-complex) structure $\tilde{J}^*$ just defined, it is natural to introduce a 2-tensor $\tilde{g}^*$ by the formula

$$\tilde{g}^* := \Omega^*(., \tilde{J}^*).$$

However, it turns out that $\Omega^*$ is not compatible with $\tilde{J}^*$, since it turns out that $\Omega^*(\tilde{J}^*, \tilde{J}^*) = -\varepsilon\Omega^*$ instead of the required formula $\Omega^*(\tilde{J}^*, \tilde{J}^*) = \varepsilon\Omega^*$ (here and in the following, in order to deal simultaneously with the complex

$$G'((X_h, X_v), (Y_h, Y_v)) := g(X_h, Y_h) - g(X_v, Y_v).$$

<footnote n="1">¹We might also define an “almost para-Kähler” structure on $TM$ by introducing the *para-Sasaki metric*</footnote>

<footnote>This metric has neutral signature $(m, m)$ ($m$ being the dimension of $M$), is compatible with $J'$ and verifies $\Omega := -G'(J',.)$.</footnote>

<footnote n="2">²The terminology *split-holomorphic* is sometimes used.</footnote>
and para-complex cases, we define $\epsilon$ to be such that $(\tilde{J}^*)^2 = -\epsilon\mathrm{Id}$, i.e. $\epsilon = 1$ in the complex case and $\epsilon = -1$ in the para-complex case). It follows that the tensor $\tilde{g}^*$ is not symmetric and therefore we failed in constructing a canonical pseudo-Riemannian structure on $T^*M$.

On the other hand, the same idea works well if one considers, instead of the cotangent bundle, the tangent bundle of a pseudo- or para-Kähler manifold $(M, J, g)$, thus obtaining a canonical pseudo- or para-Kähler structure. The purpose of this note is to investigate in detail this construction and to study its curvature properties. The results are summarized in the following:

**Main Theorem** Let $(M, J, g, \omega)$ be a pseudo- or para-Kähler manifold. Then $TM$ enjoys a natural pseudo- or para-Kähler structure $(\tilde{J}, \tilde{g}, \Omega)$ with the following properties:

- $\tilde{J}$ is the canonical complex or para-complex structure of $TM$ induced from that of $M$;

- $\Omega$ is the pull-back of $\Omega^*$ by the metric isomorphism $TM \simeq_g T^*M$;

- The pseudo-Riemannian metric $\tilde{g}$ can be recovered from $\tilde{J}$ and $\Omega$ by the equation $\tilde{g}(.,.) := \Omega(., \tilde{J}.)$;

- According to the splitting $TTM - H\mathcal{M} \oplus V\mathcal{M}$ induced by the Levi-Civita connection of $g$, the triple $(\tilde{J}, \tilde{g}, \Omega)$ takes the following expression:

$$ \tilde{J}(X_h, X_v) := (JX_h, JX_v) $$

$$ \tilde{g}((X_h, X_v), (Y_h, Y_v)) := g(X_v, JY_h) - g(X_h, JY_v) $$

$$ \Omega((X_h, X_v), (Y_h, Y_v)) := g(X_v, Y_h) - g(X_h, Y_v); $$

- The pseudo-Riemannian metric $\tilde{g}$ has the following properties:

(i) $\tilde{g}$ has neutral signature $(2n, 2n)$ and is scalar flat;

(ii) $(TM, \tilde{g})$ is Einstein if and only if $(M, g)$ is flat, and therefore $(TM, \tilde{g})$ is flat as well;

(iii) the Ricci curvature $\tilde{Ric}$ of $\tilde{g}$ has the same sign as the Ricci curvature $\tilde{Ric}$ of $g$;

(iv) $(TM, \tilde{g})$ is locally conformally flat if and only if $n = 1$ and $g$ has constant curvature, or $n > 2$ and $g$ is flat; if $n = 1$, $\tilde{g}$ is always self-dual, so anti-self-duality is equivalent to conformal flatness;

(v) the pair $(\tilde{J}, \tilde{g})$ has constant holomorphic curvature if and only if $g$ is flat.

**Remark 1.** We use in (iv) the general property that four-dimensional neutral pseudo-Kähler or para-Kähler manifolds are self-dual if and only if their scalar curvature vanishes. This is analogous to the case of Kähler four-dimensional manifolds, except that self-duality is exchanged with anti-self-duality. A proof of this statement is given in Theorem A.2 in the appendix.
This result is a generalization of previous work on the tangent bundle of a Riemannian surface (see [10], [11], [3]). The authors wish to thank Brendan Guilfoyle for his valuable suggestions and comments.

# 1 Almost complex and para-complex structures on the tangent bundle

Given a manifold $\mathcal{M}$ endowed with an almost complex or almost para-complex structure J, it is only natural to ask whether its tangent or cotangent bundle inherit such a structure. The answer is positive:

**Proposition 1.** Let $(\mathcal{M}, J)$ be an almost complex (resp. para-complex) manifold. Then its tangent bundle admits a canonical almost complex (resp. para-complex) structure $\tilde{J}$. Furthermore, if $J$ is complex (resp. para-complex), so is $\tilde{J}$.

**Remark 2.** Such a result has been proven already by Lempert & Szöke [14] for the tangent bundle in the almost complex case. Their construction uses the jets over $\mathcal{M}$ and is quite a bit more technical than our proof. However it gives an interesting interpretation of the meaning of $\tilde{J}$. We shall see below in Proposition 2 a different and simpler way of defining and understanding $\tilde{J}$, provided $\mathcal{M}$ is a pseudo- or para-Kähler manifold.

*Proof.* We prove the result using coordinate charts, which amounts to showing that $\tilde{J}$ can be defined independently of any change of variable. Let $y = \varphi(x)$ be a local change of coordinates on $\mathbb{R}^n$ and write $\xi$ and $\eta$ respectively for the tangent coordinates induced by the charts (i.e. $\sum_i \xi^i \partial/\partial x^i = \sum_i \eta^i \partial/\partial y^i$). The change of tangent coordinates at $x$ is $\xi \mapsto \eta = d\varphi(x)\xi$, in other words $\varphi$ induces a chart $\Phi$ on $\mathbb{R}^{2n}$, $\Phi : (x, \xi) \mapsto (\varphi(x), d\varphi(x)\xi)$. The tangent coordinates at $(x, \xi)$ (resp. $(y, \eta)$) are denoted by $(X, \Xi)$ (resp. $(Y, H)$) and the change of (doubly) tangent coordinates is

$$d\Phi(x, \xi) : (X, \Xi) \mapsto (Y, H) = (d\varphi(x)X, d^2\varphi(x)(X, \xi) + d\varphi(x)\Xi).$$

Assume moreover that we have a (1, 1) tensor, which reads in the $x$ coordinate as the matrix $J(x)$ and in the $y$ coordinate as the matrix $J'(y) = J'(\varphi(x)) = d\varphi(x) \circ J(x) \circ (d\varphi(x))^{-1}$. Equivalently for any $X$ and $Y = d\varphi(x)X$, we have $J'(y)Y = J'(\varphi(x))d\varphi(x)X = d\varphi(x)J(x)X$. Differentiating this equality along $\xi$ yields

$$(D_{d\varphi(x)\xi}J')(\varphi(x))d\varphi(x)X + J'(\varphi(x))d^2\varphi(x)(X, \xi) = d\varphi(x)(D_{\xi}J)(x)X + d^2\varphi(x)(J(x)X, \xi), \quad (1)$$

where $(D_{\xi}J)(x)$ denotes in this proof the directional derivative of the matrix $J$ at $x$ in the direction $\xi$ (not a covariant derivative).
We now define the (1, 1) tensor $\tilde{J}$ in the $(x, \xi)$ coordinate by

$$ \tilde{J}(x, \xi) : (X, \Xi) \mapsto (J(x)X, J(x)\Xi + D_{\xi}J(x)X). $$

Let us prove that this definition is coordinate-independent (for greater readability we will often write $J, J'$ for $J(x), J'(y)$). Using (1) and the symmetry of the second order differential $d^2\varphi(x)$,

$$ \begin{aligned} d\Phi(x, \xi)(J(X, \Xi)) &= d\Phi(x, \xi)(JX, J\Xi + D_{\xi}J(x)X) \\ &= (d\varphi(x)JX, d^2\varphi(x)(JX, \xi) + d\varphi(x)(J\Xi + D_{\xi}J(x)X)) \\ &= (J'Y, J'd\varphi(x)\Xi \\ &\quad +(D_{d\varphi(x)\xi}J')(\varphi(x))d\varphi(x)X + J'd^2\varphi(x)(X, \xi)) \\ &= (J'Y, J'(d\varphi(x)\Xi + d^2\varphi(x)(X, \xi)) \\ &\quad +(D_{d\varphi(x)\xi}J')(\varphi(x))d\varphi(x)X) \\ &= (J'Y, J'H + D_{\eta}J'(y)Y) = \tilde{J}'(y, \eta)(Y, H), \end{aligned} $$

where $\tilde{J}'$ denotes the map corresponding to $\tilde{J}$ in the $(y, \eta)$ coordinates. Consequently the tensor on $\mathcal{M}$ extends naturally to $TM$.

We have so far defined a (1, 1) tensor on $TM$ without extra assumptions. Suppose now that $J$ is an almost complex (resp. para-complex) structure, so that $J^2 = -\varepsilon\mathrm{Id}$. Differentiating this property yields $J D_{\xi}J + (D_{\xi}J)J = 0$. Then

$$ \begin{aligned} \tilde{J}^2(X, \Xi) &= (J^2X, J(J\Xi + D_{\xi}JX) + D_{\xi}J(JX)) \\ &= (-\varepsilon X, -\varepsilon\Xi + J(dJ\xi)X + (dJ\xi)(JX)) = -\varepsilon(X, \Xi) \end{aligned} $$

so that $\tilde{J}$ is also an almost complex (resp. para-complex) structure.

Finally if $J$ is a complex (resp. para-complex) structure then we can use complex (resp. para-complex) coordinate charts, which amounts to saying that $J$ is a constant matrix. Then $\tilde{J}$ defined in the associated charts on $TM$ takes a simpler expression, and is also constant:

$$ \tilde{J}(x, \xi) : (X, \Xi) \mapsto (JX, J\Xi) $$

and that characterizes a complex (resp. para-complex) structure. $\square$

**Remark 3.** Finding a similar almost-complex structure on $T^*\mathcal{M}$ is much more difficult, and may not be true in all generality. The Reader will note that, whenever $\mathcal{M}$ is endowed with a pseudo-Riemannian metric, we have a musical correspondence between $TM$ and $T^*\mathcal{M}$, and $\tilde{J}$ induces a corresponding structure $\tilde{J}^*$ on $T^*\mathcal{M}$. However different metrics will yield different structures on $T^*\mathcal{M}$. There is one unambiguous case, which will be the setting in the remainder of this article, namely when $J$ is integrable.
## 2 The Kähler structure

Let $\mathcal{M}$ be a differentiable manifold. We denote by $\pi$ and $\pi^*$ the canonical projections $TM \to \mathcal{M}$ and $TM^* \to \mathcal{M}$. The subbundle $\ker(d\pi) := VM$ of $TTM$ (it is thus a bundle over $TM$) will be called the vertical bundle.

Assume now that $\mathcal{M}$ is equipped with a linear connection $\nabla$. The corresponding horizontal bundle is defined as follows: let $\bar{X}$ be a tangent vector to $TM$ at some point $(x_0, V_0)$. This implies that there exists a curve $\gamma(s) = (x(s), V(s))$ such that $(x(0), V(0)) = (x_0, V_0)$ and $\gamma'(0) = \bar{X}$. If $X \notin VM$ (which implies $x'(0) \neq 0$), we define the connection map (see [7], [3]) $K: TTM \to TM$ by $K\bar{X} = \nabla_{x'(0)}V(0)$, where $\nabla$ denotes the Levi-Civita connection of the metric $g$. If $X$ is vertical, we may assume that the curve $\gamma$ stays in a fiber so that $V(s)$ is a curve in a vector space. We then define $K\bar{X}$ to be simply $V'(0)$. The horizontal bundle is then $\ker(K)$ and we have a direct sum

$$ TTM = HM \oplus VM \approx TM \oplus TM \\ \bar{X} \approx (\Pi\bar{X}, K\bar{X}). \qquad (2) $$

Here and in the following, $\Pi$ is a shorthand notation for $d\pi$.

**Lemma 1.** [7] Given a vector field $X$ on $(\mathcal{M}, \nabla)$ there exists exactly one vector field $X^h$ and one vector field $X^v$ on $TM$ such that $(\Pi X^h, KX^h) = (X, 0)$ and $(\Pi X^v, KX^v) = (0, X)$. Moreover, given two vector fields $X$ and $Y$ on $(\mathcal{M}, \nabla)$, we have, at the point $(x, V)$:

$$ \begin{aligned} [X^v, Y^v] &= 0, \\ [X^h, Y^v] &= (\nabla_X Y)^v \approx (0, \nabla_X Y), \\ [X^h, Y^h] &\approx ([X, Y], -R(X, Y)V), \end{aligned} $$

where $R$ denotes the curvature of $\nabla$ and we use the direct sum notation (2).

The Reader should not confuse the horizontal lift $X^h$, which is a vector field on $TM$ constructed from a vector field $X \in \mathcal{X}(\mathcal{M})$, with the notation $\bar{X}_h = \Pi\bar{X}$ denoting the horizontal part of $\bar{X} \in \mathcal{X}(TM)$. Similarly, the vertical lift $X^v$ is not the vertical projection $\bar{X}_v = K\bar{X}$.

We say that a vector field $\bar{X}$ on $TM$ is projectable if it is constant on the fibres, i.e. $(\Pi\bar{X}, K\bar{X})(x, V) = (\Pi\bar{X}, K\bar{X})(x, V')$. According to the lemma above, it is equivalent to the fact that there exists two vector fields $X_1$ and $X_2$ on $\mathcal{M}$ such that $\bar{X} = (X_1)^h + (X_2)^v$.

Assume now that $\mathcal{M}$ is equipped with a pseudo-Riemannian metric $g$, i.e. a non-degenerate bilinear form. By the non-degeneracy assumption, we can identify $T^*\mathcal{M}$ with $TM$ by the following (musical) isomorphism:

$$ \iota: \begin{array}{ccc} TM & \to & TM^* \\ (x, V) & \mapsto & (x, \xi), \end{array} $$

where $\xi$ is defined by

$$ \xi(W) = g(V, W), \quad \forall W \in T_x\mathcal{M}. $$
The Liouville form $\alpha \in \Omega^1(T^*M)$ is the 1-form defined by $\alpha_{(x,\xi)}(\bar{X}) = \xi_x(d\pi^*(\bar{X}))$, where $\bar{X}$ is a tangent vector at the point $(x, \xi)$ of $T^*M$. The canonical symplectic form on $TM^*$ is defined to be $\Omega^* := -d\alpha$. There is an elegant, explicit formula for the symplectic form $\Omega := \iota^*(\Omega^*)$ in terms of the metric $g$ and the splitting induced by the Levi-Civita connection $\nabla$ (see [2], [13]):

**Lemma 2.** Let $\bar{X}$ and $\bar{Y}$ be two tangent vectors to $TM$; we have

$$ \Omega(\bar{X}, \bar{Y}) = g(K\bar{X}, \Pi\bar{Y}) - g(\Pi\bar{X}, K\bar{Y}). $$

**Proposition 2.** Let $(M, J, g)$ be a pseudo- or para-Kähler manifold. The canonical structure $\tilde{J}$ satisfies

$$ \tilde{J}\bar{X} \approx \tilde{J}(\Pi\bar{X}, K\bar{X}) = (J\Pi\bar{X}, JK\bar{X}). $$

**Corollary 1.** Let $(M, J, g)$ be a pseudo- or para-Kähler manifold. The 2-tensor $\tilde{g}(.,.) := \Omega(., \tilde{J})$ satisfies

$$ \tilde{g}(\bar{X}, \bar{Y}) = g(K\bar{X}, J\Pi\bar{Y}) - g(\Pi\bar{X}, JK\bar{Y}). $$

Moreover, $\tilde{g}$ is symmetric and therefore defines a pseudo-Riemannian metric on $TM$.

*Proof of Proposition 2.* Let us write the splitting of $TTM$ in a local coordinate $x$ as in the proof of Proposition 1 (³). The Levi-Civita connection is expressed through its connection form $\mu$: $\nabla_X Y = dY(X) + \mu(X)Y$. Consequently, if $(X, \Xi) \in T_{(x,\xi)}TM$, $\Pi(X, \Xi) = X$ and $K(X, \Xi) = \Xi + \mu(X)\xi$. Thus

$$ \Pi(\tilde{J}(X, \Xi)) = JX \text{ and } K(\tilde{J}(X, \Xi)) = J(x)\Xi + (dJ(x)\xi)X + \mu(J(x)X)\xi. $$

Because $J$ is integrable, we may choose $x$ to be a complex coordinate, so that $J$ is a constant endomorphism, and $dJ(x)\xi$ vanishes. Because $M$ is Kähler, we know that $\mu(X)$ commutes with $J$. However, $\nabla$ being without torsion, $\mu(X)Y = \mu(Y)X$, so

$$ K(\tilde{J}(X, \Xi)) = J\Xi + J\mu(X)\xi = JK(X, \Xi). \quad \square $$

**Corollary 2.** The symplectic form $\Omega$ is compatible with the complex or para-complex structure $\tilde{J}$.

*Proof.* Using Lemma 2, we compute

$$ \begin{align*} \Omega(\tilde{J}\bar{X}, \tilde{J}\bar{Y}) &= g(K\tilde{J}\bar{X}, \Pi\tilde{J}\bar{Y}) - g(\Pi\tilde{J}\bar{X}, K\tilde{J}\bar{Y}) \\ &= g(JK\bar{X}, J\Pi\bar{Y}) - g(J\Pi\bar{X}, JK\bar{Y}) \\ &= \varepsilon g(K\bar{X}, \Pi\bar{Y}) - \varepsilon g(\Pi\bar{X}, K\bar{Y}) \\ &= \varepsilon \Omega(\bar{X}, \bar{Y}). \end{align*} $$

<figure><img src="image_2.png" /></figure>

<footnote n="3">³ The Reader should be aware of the conflicting notation: the splitting of $TTM \simeq \mathbb{R}^{4n}$ as $\mathbb{R}^{2n} \oplus \mathbb{R}^{2n}$ induced by the coordinate charts (e.g. $\bar{X} \simeq ((x, \xi), (X, \Xi)))$ differs a priori from the connection-induced splitting $\bar{X} \simeq (\Pi\bar{X}, K\bar{X})$.</footnote>
### 3 The Levi-Civita connection of $\tilde{g}$

The following lemma describes the Levi-Civita connection $\tilde{\nabla}$ of $\tilde{g}$ in terms of the direct decomposition of $TTM$, the Levi-Civita connection $\nabla$ of $g$ and its curvature tensor $R$.

**Lemma 3.** Let $\bar{X}$ and $\bar{Y}$ be two vector fields on $TM$ and assume that $\bar{Y}$ is projectable, then at the point $(x, V)$ we have

$$ (\tilde{\nabla}_{\bar{X}} \bar{Y})|_V = (\nabla_{\Pi\bar{X}} \Pi\bar{Y}, \nabla_{\Pi\bar{X}} K\bar{Y} - T_1(\Pi\bar{X}, \Pi\bar{Y}, V)) $$

where

$$ T_1(X, Y, V) = \frac{1}{2} (R(X, Y)V - \varepsilon R(V, JX)JY - \varepsilon R(V, JY)JX) $$

Moreover, if $M$ is a pseudo-Riemannian surface with Gaussian curvature $c$, we have

$$ T_1(X, Y, V) = \begin{cases} -2cg(V, X)Y & \text{in the Kähler case,} \\ +2cg(V, Y)X & \text{in the para-Kähler case.} \end{cases} $$

*Proof.* We use Lemma 1 together with the Koszul formula:

$$ \begin{aligned} 2\tilde{g}(\tilde{\nabla}_{\bar{X}}\bar{Y}, \bar{Z}) &= \bar{X}\tilde{g}(\bar{Y}, \bar{Z}) + \bar{Y}\tilde{g}(\bar{X}, \bar{Z}) - \bar{Z}\tilde{g}(\bar{X}, \bar{Y}) + \tilde{g}([\bar{X}, \bar{Y}], \bar{Z}) \\ &\quad - \tilde{g}([\bar{X}, \bar{Z}], \bar{Y}) - \tilde{g}([\bar{Y}, \bar{Z}], \bar{X}), \end{aligned} $$

where $X, Y$ and $Z$ are three vector fields on $TM$. From the fact that $[X^v, Y^v]$ and $\tilde{g}(X^v, Y^v)$ vanish we have:

$$ \begin{aligned} 2\tilde{g}(\tilde{\nabla}_{X^v} Y^v, Z^v) &= X^v \tilde{g}(Y^v, Z^v) + Y^v \tilde{g}(X^v, Z^v) - Z^v \tilde{g}(X^v, Y^v) \\ &\quad + \tilde{g}([X^v, Y^v], Z^v) - \tilde{g}([X^v, Z^v], Y^v) - \tilde{g}([Y^v, Z^v], X^v) \\ &= 0. \end{aligned} $$

Moreover, taking into account that $\tilde{g}(Y^v, Z^h)$ and similar quantities are constant on the fibres, we obtain

$$ \begin{aligned} 2\tilde{g}(\tilde{\nabla}_{X^v} Y^v, Z^h) &= X^v \tilde{g}(Y^v, Z^h) + Y^v \tilde{g}(X^v, Z^h) - Z^h \tilde{g}(X^v, Y^v) \\ &\quad + \tilde{g}([X^v, Y^v], Z^h) - \tilde{g}([X^v, Z^h], Y^v) - \tilde{g}([Y^v, Z^h], X^v) \\ &= -\tilde{g}(-\nabla_Z X)^v, Y^v) - \tilde{g}(-\nabla_Z Y)^v, X^v) \\ &= 0. \end{aligned} $$

From these last two equations we deduce that $\tilde{\nabla}_{X^v} Y^v$ vanishes. Analogous computations show that $\tilde{\nabla}_{X^v} Y^h$ vanishes as well. From Lemma 1 and the formula $[\bar{X}, \bar{Y}] = \tilde{\nabla}_{\bar{X}}\bar{Y} - \tilde{\nabla}_{\bar{Y}}\bar{X}$, we deduce that

$$ \tilde{\nabla}_{X^h} Y^v \simeq (0, \nabla_X Y). \qquad (3) $$

Finally, introducing

$$ T_1(X, Y, V) := \frac{1}{2} (R(X, Y)V - \varepsilon R(V, JY)JX - \varepsilon R(V, JX)JY), $$
we compute that

$$
\begin{align*}
2\tilde{g}(\tilde{\nabla}_{X^h} Y^h, Z^h) &= -g(R(X, Y)V, JZ) + g(R(X, Z)V, JY) + g(R(Y, Z)V, JX) \\
&= -g(R(X, Y)V, JZ) + g(R(V, JY)X, Z) + g(R(V, JX)Y, Z) \\
&= -g(R(X, Y)V, JZ) + \varepsilon g(R(V, JY)JX, JZ) + \varepsilon g(R(V, JX)JY, JZ) \\
&= -g(2T_1(X, Y, V), JZ)
\end{align*}
$$

and

$$
\tilde{g}(\tilde{\nabla}_{X^h} Y^h, Z^v) = -g(\nabla_X Y, JZ),
$$

from which we deduce that

$$
\tilde{\nabla}_{X^h} Y^h (V) = (\nabla_X Y, -T_1(X, Y, V)). \quad (4)
$$

From (3) and (4) we deduce the required formula for $\tilde{\nabla}_X \tilde{Y}$.

If $n=1$, we have $R(X, Y)Z = c(g(Y, Z)X - g(X, Z)Y)$, hence the tensor $T_1$
becomes:

$$
\begin{align*}
2T_1(X, Y, V) &= R(X, Y)V + \varepsilon JR(V, JX)Y + \varepsilon JR(V, JY)X \\
&= c(g(Y, V)X - g(X, V)Y \\
&\quad - \varepsilon J(g(JX, Y)V - g(V, Y)JX + g(JY, X)V - g(V, X)JY)) \\
&= c(g(Y, V)X - g(X, V)Y \\
&\quad - \varepsilon(g(JX, Y)JV + g(V, Y)X + g(JY, X)JV + g(V, X)Y)) \\
&= c((1 - \varepsilon)g(V, Y)X - (1 + \varepsilon)g(V, X)Y).
\end{align*}
$$

<figure><img src="image_3.png" /></figure>

**Remark 4.** It should be noted that covariant derivatives with respect to a pro-
jectable vertical field $X^v$ always vanish.

**Proposition 3.** The structure $\tilde{J}$ is parallel with respect to $\tilde{\nabla}$.

Proof. It can be seen as a trivial consequence of the fact that $\tilde{J}$ is complex (resp. para-complex) and $\Omega$ is closed, but can also be checked directly, using the equivariance properties of $J$ w.r.t. the connection $\nabla$ and the curvature tensor $R$. Using the definition of $\tilde{J}$ and Lemma 3, $\tilde{\nabla}_{\tilde{X}} \tilde{J} \tilde{Y}$ is obvious provided $T_1(X, JY, V) = J T_1(X, Y, V)$. That is indeed the case since

$$
\begin{align*}
& 2(T_1(X, JY, V) - J T_1(X, Y, V)) \\
&= R(X, JY) V + R(V, JX) Y + R(V, Y) JX \\
&\quad - R(X, Y) JV - R(V, JX) Y - R(V, JY) X \\
&= R(X, JY) V + R(JY, V) X + J(R(V, Y) X + R(Y, X) V) \\
&= R(V, X) JY + JR(X, V) Y = 0,
\end{align*}
$$

where we have used Bianchi's identity.

<figure><img src="image_4.png" /></figure>
# 4 Curvature properties of ($\tilde{J}$, $\tilde{g}$)

## 4.1 The Riemannian curvature tensor of $\tilde{g}$

**Proposition 4.** The curvature tensor $\tilde{Rm} := -\tilde{g}(\tilde{R}, \cdot)$ of $\tilde{g}$ at $(x, V)$ is given by the formula

$$ \tilde{Rm}(\bar{X}, \bar{Y}, \bar{Z}, \bar{W}) = g(T_2(\Pi\bar{X}, \Pi\bar{Y}, \Pi\bar{Z}, V), J\Pi\bar{W}) \\ -Rm(\Pi\bar{X}, \Pi\bar{Y}, \Pi\bar{Z}, JK\bar{W}) - Rm(\Pi\bar{X}, \Pi\bar{Y}, JK\bar{Z}, \Pi\bar{W}) \\ -Rm(\Pi\bar{X}, JK\bar{Y}, \Pi\bar{Z}, \Pi\bar{W}) + Rm(JK\bar{X}, \Pi\bar{Y}, \Pi\bar{Z}, \Pi\bar{W}), $$

where

$$ T_2(X, Y, Z, V) := (\nabla_X T_1)(Y, Z, V) - (\nabla_Y T_1)(X, Z, V). $$

Moreover, $(TM, \tilde{g})$ is scalar flat and the Ricci tensor of $\tilde{g}$ is

$$ \tilde{Ric}(\bar{X}, \bar{Y}) = 2\tilde{Ric}(\Pi\bar{X}, \Pi\bar{Y}). $$

**Corollary 3.** $(TM, \tilde{g})$ is Einstein if and only if $(M, g)$ is flat. Moreover $(TM, \tilde{g})$ has nonnegative (resp. nonpositive) Ricci curvature if and only if $(M, g)$ has nonnegative (resp. nonpositive) Ricci curvature as well.

*Proof of Proposition 4.* We will compute the curvature tensor for projectable vector fields, and need only do so for the following six cases, due to the symmetries of $\tilde{Rm}$. Remark 4 simplifies computations greatly, since most vertical derivatives vanish, except when the derived vector field is not projectable. In particular $\tilde{R}(X^v, Y^v)$ vanishes as endomorphism, hence:

$$ \tilde{Rm}(X^v, Y^v, Z^v, W^v) = 0 $$

$$ \tilde{Rm}(X^v, Y^v, Z^v, W^h) = 0 $$

$$ \tilde{Rm}(X^v, Y^v, Z^h, W^v) = 0 $$

To obtain the last three combinations, let us first derive $\tilde{R}(X^h, Y^h)Z^h$. This is more delicate since we have to covariantly differentiate non-projectable quantities. Indeed

$$ \begin{align*} \tilde{R}(X^h, Y^h)Z^h &= \tilde{\nabla}_{X^h} \tilde{\nabla}_{Y^h} Z^h - \tilde{\nabla}_{Y^h} \tilde{\nabla}_{X^h} Z^h - \tilde{\nabla}_{[X^h, Y^h]} Z^h \\ &= \tilde{\nabla}_{X^h} (\nabla_Y Z, -T_1(Y, Z, V)) - \tilde{\nabla}_{Y^h} (\nabla_X Z, -T_1(X, Z, V)) \\ &\quad - \tilde{\nabla}_{[X,Y], -R(X,Y)V} Z^h \\ &= (\nabla_X \nabla_Y Z, -T_1(X, \nabla_Y Z, V)) - D_{X^h}(0, T_1(Y, Z, V)) \\ &\quad -(\nabla_Y \nabla_X Z, -T_1(Y, \nabla_X Z, V)) + D_{Y^h}(0, T_1(X, Z, V)) \\ &\quad -(\nabla_{[X,Y]} Z, -T_1([X, Y], Z, V)) \\ &= (R(X, Y)Z, 0) \\ &\quad -(0, T_1(X, \nabla_Y Z, V) - T_1(Y, \nabla_X Z, V) - T_1([X, Y], Z, V)) \\ &\quad -\tilde{\nabla}_{X^h}(0, T_1(Y, Z, V)) + \tilde{\nabla}_{Y^h}(0, T_1(X, Z, V)) \end{align*} $$
Recalling the lemma<fnref n="4" /> in [12], there exists a vector field $U$ on $M$ such that $U(x) = V$ and $(\nabla_X U)(x) = 0$. Then the vertical lift of $T_1(Y, Z, U)$ is seen to agree to first order with

$$ (x, V) \mapsto (0, T_1(X(x), Z(x), V)) $$

thus allowing us to use the formula in Lemma 3:

$$ \begin{aligned} \tilde{\nabla}_{X^h}(0, T_1(Y, Z, \cdot)) &= \tilde{\nabla}_{X^h}(T_1(Y, Z, U)^v) \\ &= (0, \nabla_X(T_1(Y, Z, U))) \\ &= (0, (\nabla_X T_1)(Y, Z, U) + T_1(\nabla_X Y, Z, U) \\ & \qquad + T_1(Y, \nabla_X Z, U) + T_1(Y, Z, \nabla_X U)) \end{aligned} $$

which, evaluated at $(x, V)$, yields

$$ \tilde{\nabla}_{X^h}(0, T_1(Y, Z, \cdot))|_{(x,V)} = (0, (\nabla_X T_1)(Y, Z, V) + T_1(\nabla_X Y, Z, V) + T_1(Y, \nabla_X Z, V)). $$

Summing up,

$$ \begin{aligned} \tilde{R}(X^h, Y^h)Z^h|_{(x,V)} &= (R(X, Y)Z, \\ & \quad -T_1(X, \nabla_Y Z, V) + T_1(Y, \nabla_X Z, V) \\ & \quad +T_1([X, Y], Z, V) - (\nabla_X T_1)(Y, Z, V) \\ & \quad -T_1(\nabla_X Y, Z, V) - T_1(Y, \nabla_X Z, V) \\ & \quad +(\nabla_Y T_1)(X, Z, V) + T_1(\nabla_Y X, Z, V) \\ & \quad +T_1(X, \nabla_Y Z, V)) \\ &= (R(X, Y)Z, -(\nabla_X T_1)(Y, Z, V) + (\nabla_Y T_1)(X, Z, V)) \\ &= (R(X, Y)Z, -T_2(X, Y, Z, V)). \end{aligned} $$

From that we deduce directly

$$ \begin{aligned} \widetilde{\mathrm{Rm}}(X^h, Y^h, Z^h, W^v) &= -\mathrm{Rm}(X, Y, Z, JW) \\ \widetilde{\mathrm{Rm}}(X^h, Y^h, Z^h, W^h)|_{(x,V)} &= g(T_2(X, Y, Z, V), JW). \end{aligned} $$

On the other hand, using repeatedly Remark 4,

$$ \begin{aligned} \widetilde{\mathrm{Rm}}(X^h, Y^v, Z^h, W^v) &= \tilde{g}(\tilde{\nabla}_{X^h} \tilde{\nabla}_{Y^v} W^v - \tilde{\nabla}_{Y^v} \tilde{\nabla}_{X^h} W^v - \tilde{\nabla}_{[X^h, Y^v]} W^v, Z^h) \\ &= \tilde{g}(-\tilde{\nabla}_{Y^v}(0, \nabla_X W), Z^h) = \tilde{g}(0, Z^h) = 0. \end{aligned} $$

The claimed formula is easily deduced using the symmetries of the curvature tensor.

<footnote n="4">⁴Note that computations in [12] are done for the Sasaki metric, hence direct results do not apply.</footnote>
In order to calculate the Ricci curvature of $\tilde{g}$, we consider a Hermitian pseudo-orthonormal basis $(e_1, \dots, e_{2n})$ of $T_{x,M}$, i.e. $g(e_a, e_b) = \varepsilon_a \delta_{ab}$, where $\varepsilon_a = \pm 1$, and $e_{n+a} = Je_a$. In particular, $\varepsilon_{n+a} = \varepsilon\varepsilon_a$. This gives a (non-orthonormal) basis of $T_{(x,V)}TM$:

$$ \bar{e}_a := (e_a)_h \qquad \bar{e}_{2n+a} := (e_a)^v. $$

A calculation using Corollary 1 shows that the expression of $\tilde{g}$ in this basis is:

$$ [\tilde{g}_{\mu\nu}]_{1\le\mu,\nu\le 4n} := \begin{pmatrix} 0 & 0 & 0 & \Delta \\ 0 & 0 & -\Delta & 0 \\ 0 & -\Delta & 0 & 0 \\ \Delta & 0 & 0 & 0 \end{pmatrix}, $$

where $\Delta = \varepsilon \mathrm{diag}(\varepsilon_1, \dots, \varepsilon_n) = \mathrm{diag}(\varepsilon_{n+1}, \dots, \varepsilon_{2n})$. It follows that $\widetilde{\mathrm{Ric}}(X^v, Y^v)$ and $\widetilde{\mathrm{Ric}}(X^h, Y^h)$ vanish.

Moreover, noting that $\tilde{g}^{\mu\nu} = \tilde{g}_{\mu\nu}$,

$$
\begin{align*}
\widetilde{\mathrm{Ric}}(X^h, Y^h) &= \sum_{\mu,\nu=1}^{4n} \tilde{g}^{\mu\nu} \widetilde{\mathrm{Rm}}(X^h, \bar{e}_\mu, Y^h, \bar{e}_\nu) \\
&= \sum_{a=1}^{n} \varepsilon \varepsilon_a \left( \widetilde{\mathrm{Rm}}(X^h, (e_a)^h, Y^h, (Je_a)^v) - \widetilde{\mathrm{Rm}}(X^h, (Je_a)^h, Y^h, (e_a)^v) \right. \\
&\qquad \left. - \widetilde{\mathrm{Rm}}(X^h, (e_a)^v, Y^h, (Je_a)^h) + \widetilde{\mathrm{Rm}}(X^h, (Je_a)^v, Y^h, (e_a)^h) \right) \\
&= \sum_{a=1}^{n} \varepsilon \varepsilon_a \left( - \mathrm{Rm}(X, e_a, Y, J^2 e_a) + \mathrm{Rm}(X, Je_a, Y, Je_a) \right. \\
&\qquad \left. + \mathrm{Rm}(Y, Je_a, X, Je_a) - \mathrm{Rm}(Y, e_a, X, J^2 e_a) \right) \\
&= 2 \sum_{a=1}^{n} \left( \varepsilon_a \mathrm{Rm}(X, e_a, Y, e_a) + \varepsilon_{a+n} \mathrm{Rm}(X, e_{a+n}, Y, e_{a+n}) \right) \\
&= 2 \sum_{k=1}^{2n} \varepsilon_k \mathrm{Rm}(X, e_k, Y, e_k) = 2 \mathrm{Ric}(X, Y).
\end{align*}
$$

We see easily that $\widetilde{\mathrm{Ric}}$ vanishes whenever one of the vectors is along the vertical fiber, thus the expected formula.

Finally the scalar curvature

$$ \widetilde{\mathrm{Scal}} = \sum_{\mu,\nu=1}^{4} \tilde{g}^{\mu\nu} \widetilde{\mathrm{Ric}}(\bar{e}_{\mu}, \bar{e}_{\nu}) = 0, $$

since $\tilde{g}^{\mu\nu}$ vanishes as soon as both $\bar{e}_\mu, \bar{e}_\nu$ are both horizontal. $\square$

<figure><img src="image_5.png" /></figure>
## 4.2 The Weyl curvature tensor of $\tilde{g}$

**Proposition 5.** The Weyl tensor $\tilde{W}$ at $(x, V)$ is given by

$$ \tilde{W}(\bar{X}, \bar{Y}, \bar{Z}, \bar{W}) = \tilde{\mathrm{Rm}}(\bar{X}, \bar{Y}, \bar{Z}, \bar{W}) - \frac{1}{2n-1} (\mathrm{Ric}(\Pi\bar{X}, \Pi\bar{Z})\tilde{g}(\bar{Y}, \bar{W}) + \mathrm{Ric}(\Pi\bar{Y}, \Pi\bar{W})\tilde{g}(\bar{Y}, \bar{W}) - \mathrm{Ric}(\Pi\bar{X}, \Pi\bar{W})\tilde{g}(\bar{Y}, \bar{Z}) - \mathrm{Ric}(\Pi\bar{Y}, \Pi\bar{Z})\tilde{g}(\bar{X}, \bar{W})) $$

In particular, if $n = 1$,

$$ \tilde{W}(\bar{X}, \bar{Y}, \bar{Z}, \bar{W}) = g(T_2(\Pi\bar{X}, \Pi\bar{Y}, \Pi\bar{Z}, V), J\Pi\bar{W}). $$

**Corollary 4.** $(TM, \tilde{g})$ is locally conformally flat if and only if $n = 1$ and $g$ has constant curvature, or $n \ge 2$ and $g$ is flat.

**Remark 5.** This result has been proved in the case $n = 1$ and $\varepsilon = 1$ in [10].

*Proof of Proposition 5.* Since the scalar curvature vanishes, we have

$$ \tilde{W} = \tilde{\mathrm{Rm}} - \frac{1}{4n-2} \mathrm{Ric} \otimes \tilde{g}, $$

where $\otimes$ denotes the Kulkarni–Nomizu product. Recall that $\mathrm{Ric}(\bar{X}, \bar{Y}) = 0$ if one of the two vectors $\bar{X}$ and $\bar{Y}$ is vertical. Consequently

$$ \tilde{\mathrm{Ric}} \otimes \tilde{g}(\bar{X}, \bar{Y}, \bar{Z}, \bar{W}) = 2 (\mathrm{Ric}(\Pi\bar{X}, \Pi\bar{Z})\tilde{g}(\bar{Y}, \bar{W}) + \mathrm{Ric}(\Pi\bar{Y}, \Pi\bar{W})\tilde{g}(\bar{Y}, \bar{W}) - \mathrm{Ric}(\Pi\bar{X}, \Pi\bar{W})\tilde{g}(\bar{Y}, \bar{Z}) - \mathrm{Ric}(\Pi\bar{Y}, \Pi\bar{Z})\tilde{g}(\bar{X}, \bar{W})) $$

The expression of the Weyl tensor follows easily.

In the case $n = 1$ of a surface with Gaussian curvature $c$, we have $\mathrm{Ric}(X, Y) = cg(X, Y)$ and $\mathrm{Rm}(X, Y, Z, W) = c(g(X, Z)g(Y, W) - g(X, W)g(Y, Z))$. Hence using Proposition 4, the expression of Weyl tensor simplifies and we get the claimed formula. $\square$

*Proof of Corollary 4.* We first deal with the case $n = 1$. Lemma 3 implies that $T_1(X, Y, Z) = -2cg(Z, X)Y$ when $\varepsilon = 1$ (resp. $2cg(Z, Y)X$ when $\varepsilon = -1$). Therefore, if $\varepsilon = 1$,

$$ \begin{aligned} T_2(X, Y, Z, W) &= \nabla_X T_1(Y, Z, W) - \nabla_Y T_1(X, Z, W) \\ &= -2(X.c)g(W, Y)Z + 2(Y.c)g(W, X)Z \\ &= 2g((Y.c)X - (X.c)Y, W)Z, \end{aligned} $$

which vanishes if and only if $(X.c)Y = (Y.c)X$ for all vectors $X, Y$, i.e. the curvature $c$ is constant. Analogously, if $\varepsilon = -1$,

$$ \begin{aligned} T_2(X, Y, Z, W) &= \nabla_X T_1(Y, Z, W) - \nabla_Y T_1(X, Z, W) \\ &= 2(X.c)g(W, Z)Y - 2(Y.c)g(W, Z)X \\ &= 2((X.c)Y - (Y.c)X)g(W, Z), \end{aligned} $$
which again vanishes if and only if the curvature c is constant.

Assume now that $(TM, \tilde{g})$ is conformally flat with $n \ge 2$. Thus in particular

$$ \begin{aligned} \tilde{W}(X^h, Y^h, Z^h, W^v) &= -\mathrm{Rm}(X, Y, Z, JW) \\ &= -\frac{1}{2n-1}\left(-\mathrm{Ric}(X, Z)g(Y, JW) + \mathrm{Ric}(Y, Z)g(X, JW)\right) \end{aligned} $$

vanishes, so

$$ \mathrm{Rm}(X, Y, Z, JW) = \frac{1}{2n-1} \left( \mathrm{Ric}(X, Z)g(Y, JW) - \mathrm{Ric}(Y, Z)g(X, JW) \right). $$

(Observe that this equation always holds if $\mathcal{M}$ is a surface.) Let us apply the symmetry property of the curvature tensor to this equation with $Z = X$ and $JW = Y$, assuming furthermore that $X$ and $Y$ are two non-null vectors:

$$ \begin{aligned} 0 &= (2n-1)(\mathrm{Rm}(X, Y, X, Y) - \mathrm{Rm}(Y, X, Y, X)) \\ &= \mathrm{Ric}(X, X)g(Y, Y) - \mathrm{Ric}(Y, X)g(X, Y) \\ &\quad -\mathrm{Ric}(Y, Y)g(X, X) + \mathrm{Ric}(X, Y)g(Y, X) \\ &= \mathrm{Ric}(X, X)g(Y, Y) - \mathrm{Ric}(Y, Y)g(X, X). \end{aligned} $$

Hence

$$ \frac{\mathrm{Ric}(X, X)}{g(X, X)} = \frac{\mathrm{Ric}(Y, Y)}{g(Y, Y)}. $$

The set of non null vectors being dense in $TM$, it follows by continuity that $g$ is Einstein. We deduce that

$$ \begin{aligned} \mathrm{Rm}(X, Y, X, Y) &= \frac{1}{2n-1} \left( \mathrm{Ric}(X, X)g(Y, Y) - \mathrm{Ric}(Y, X)g(X, Y) \right) \\ &= c \left( g(X, X)g(Y, Y) - g(X, Y)g(X, Y) \right), \end{aligned} $$

so $g$ has constant curvature. But since $\mathcal{M}$ is Kähler and has dimension $2n \ge 4$, it must be flat. $\square$

Finally, we recall the general result linking the Weyl tensor to the scalar curvature in dimension four: for a neutral pseudo-Kähler or para-Kähler metric, self-duality is equivalent to scalar flatness (see Theorem A.2 in annex). We can therefore conclude

**Corollary 5.** In dimension four ($n = 1$), the metric $\tilde{g}$ is anti-self-dual if and only the curvature $c$ of $g$ is constant.

*Proof.* Thanks to proposition 4, we know that $\tilde{g}$ is scalar flat, hence self-dual ($W^-$ vanishes identically). In order for $\tilde{g}$ to be also anti-self-dual, the Weyl tensor has to vanish completely, which amounts, following corollary 4, to having constant (sectional) curvature $c$ on $\mathcal{M}$. $\square$
## 4.3 The holomorphic sectional curvature of $(\tilde{J}, \tilde{g})$

**Proposition 6.** $(\tilde{J}, \tilde{g})$ has constant holomorphic sectional curvature if and only if $g$ is flat.

*Proof.* Define the holomorphic sectional curvature tensor of $\tilde{g}$ by $\widetilde{\mathrm{Hol}}(\tilde{X}) := \widetilde{\mathrm{Rm}}(\tilde{X}, \tilde{J}\tilde{X}, \tilde{X}, \tilde{J}\tilde{X})$. Writing any doubly tangent vector $\tilde{X}$ as the sum of a horizontal and a vertical factor, we will compute $\widetilde{\mathrm{Hol}}(X^h + Y^v)$. We deduce from Proposition 4 that $\widetilde{\mathrm{Rm}}$ vanishes whenever two or more entries are vertical. Hence, using the antisymmetric properties of the Riemann tensor w.r.t. the complex or para-complex structure,

$$ \begin{aligned} \widetilde{\mathrm{Hol}}(X^h + Y^v) &= \widetilde{\mathrm{Rm}}(X^h, JX^h, X^h, JX^h) \\ &\quad + \widetilde{\mathrm{Rm}}(X^h, JX^h, X^h, JY^v) + \widetilde{\mathrm{Rm}}(X^h, JX^h, Y^v, JX^h) \\ &\quad + \widetilde{\mathrm{Rm}}(X^h, JY^v, X^h, JX^h) + \widetilde{\mathrm{Rm}}(Y^v, JX^h, X^h, JX^h) \\ &= \widetilde{\mathrm{Rm}}(X^h, JX^h, X^h, JX^h) + 4\widetilde{\mathrm{Rm}}(X^h, JX^h, X^h, JY^v) \\ &= g(T_2(X, JX, X, V) - 4\varepsilon\mathrm{R}(X, Y)X, JX). \end{aligned} $$

In particular,

$$ \begin{aligned} \widetilde{\mathrm{Hol}}(X^v) &= 0 \\ \widetilde{\mathrm{Hol}}(X^h + X^v) &= g(T_2(X, JX, X, V), JX) \\ \widetilde{\mathrm{Hol}}(X^h + (JX)^v) &= g(T_2(X, JX, X, V), JX) + 4\varepsilon\mathrm{Hol}(X). \end{aligned} $$

It follows from the first equation that if $\widetilde{\mathrm{Hol}}$ is constant, it must be zero. Hence, from the second and third equation we deduce that Hol must vanish, i.e. $g$ is flat. $\square$

# 5 Examples

The simplest examples where we may apply the construction above is where $(M, J, g, \omega)$ is the plane $\mathbb{R}^2$ equipped with the flat metric $g := dq_1^2 + \varepsilon dq_2^2$ and the complex or para-complex structure $J$ defined by $J(\partial_{q_1}, \partial_{q_2}) = (-\varepsilon\partial_{q_2}, \partial_{q_1})$. In other words, $\mathbb{R}^2$ is identified with the complex plane $\mathbb{C}$ or the para-complex plane $\mathbb{D}$. We recall that $\mathbb{D}$, called the algebra of double numbers, is the two-dimensional real vector space $\mathbb{R}^2$ endowed with the commutative algebra structure whose product rule is given by

$$ (u, v).(u', v') = (uu' + vv', uv' + u'v). $$

The number $(0, 1)$, whose square is $(1, 0)$ and not $(-1, 0)$, will be denoted by $\tau$.

We claim that in the complex case $\varepsilon = 1$, the structure $(\tilde{J}, \tilde{g}, \Omega)$ just constructed on $\mathbb{T}\mathbb{C}$ is equivalent to that of the standard complex pseudo-Euclidean
plane $(\mathbb{C}^2, \bar{J}, \langle . , . \rangle_2, \omega_1)$, where $\bar{J}$ is the canonical complex structure, ($z_1 = x_1 + iy_1, z_2 = x_2 + iy_2$) are the canonical coordinates and

$$ \langle . , . \rangle_2 := -dx_1^2 - dx_2^2 + dx_2^2 + dy_2^2 \\ \omega_1 := -dx_1 \wedge dy_1 + dx_2 \wedge dy_2. $$

To see this, it is sufficient to consider the following complex change of coordinates

$$ \left\{ \begin{array}{l} z_1 := \frac{\sqrt{2}}{2}((p_1 + ip_2) + i(q_1 + iq_2)) \\ z_2 := \frac{\sqrt{2}}{2}(p_1 + ip_2 - i(q_1 + iq_2)), \end{array} \right. $$

which preserves the symplectic form, since we have

$$ \omega_1 := -dx_1 \wedge dy_1 + dx_2 \wedge dy_2 = dq_1 \wedge dp_1 + dq_2 \wedge dp_2 = \Omega, $$

where $\Omega$ is the canonical symplectic form of $T^*\mathbb{C} \simeq_g T\mathbb{C}$. The metric of a pseudo-Kähler structure being determined by the complex structure and the symplectic form through the formula $\tilde{g} = \Omega(., \bar{J}.)$, we have the required identification.

Analogously, in the para-complex case $\varepsilon = -1$, the structure $(\tilde{J}, \tilde{g}, \Omega)$ constructed on $T\mathbb{D}$ is equivalent to that of the standard para-complex plane $(\mathbb{D}^2, \bar{J}, \langle . , . \rangle_*, \omega_*)$, where $\bar{J}$ is the canonical para-complex structure, ($w_1 = u_1 + \tau u_1, w_2 = u_2 + \tau y_2$) are the canonical coordinates and

$$ \langle . , . \rangle_* := du_1^2 - dv_1^2 + du_2^2 - dv_2^2 \\ \omega_* := du_1 \wedge dv_1 + du_2 \wedge dv_2. $$

Here we have to be careful with the identification of $T^*\mathbb{D}$ with $T\mathbb{D}$: since the metric $g$ is $dq_1^2 - dq_2^2$, we have $q_1 := dp_1 \simeq_g \partial_{p_1}$ and $q_2 := dp_2 \simeq_g -\partial_{q_2}$. Hence $\Omega^* = dq_1 \wedge dp_1 + dq_2 \wedge dp_2$ and $\Omega = dq_1 \wedge dp_1 - dq_2 \wedge dp_2$. Introducing the change of para-complex coordinates

$$ \left\{ \begin{array}{l} w_1 := \frac{\sqrt{2}}{2}((p_1 + \tau p_2) - \tau(q_1 + \tau q_2)) \\ w_2 := \frac{\sqrt{2}}{2}(\tau(p_1 + \tau p_2) + (q_1 + \tau q_2)), \end{array} \right. $$

we check that

$$ \omega_* = du_1 \wedge dv_1 + du_2 \wedge dv_2 = dq_1 \wedge dp_1 - dq_2 \wedge dp_2 = \Omega, $$

hence we obtain the identification between $(T\mathbb{D}, \tilde{J}, \tilde{g}, \Omega)$ and $(\mathbb{D}^2, \bar{J}, \langle . , . \rangle_*, \omega_*)$. Of course the metrics considered in these two examples are flat.

The next simplest examples of pseudo-Riemannian surfaces are the two-dimensional space forms, namely the sphere $\mathbb{S}^2$, the hyperbolic plane $\mathbb{H}^2 := \{x_1^2 + x_2^2 - x_3^2 = -1\}$ and the de Sitter surface $d\mathbb{S}^2 := \{x_1^2 + x_2^2 - x_3^2 = 1\}$. Their tangent bundles enjoy a interesting geometric interpretation (see [10]):
the tangent bundle $TS^2$ is canonically identified with the set of oriented lines of Euclidean three-space:

$$L(\mathbb{R}^3) \ni \{V + tx | t \in \mathbb{R}\} \simeq (x, V - \langle V, x \rangle_0 x) \in TS^2.$$

Analogously, the tangent bundle $TH^2$ is canonically identified with the set of oriented negative (timelike) lines of three-space endowed with the metric $\langle . , . \rangle_1 := dx_1^2 + dx_2^2 - dx_3^2$:

$$L_{1,-}^3 \ni \{V + tx | t \in \mathbb{R}\} \simeq (x, V - \langle V, x \rangle_1 x) \in TH^2,$$

Finally, the tangent bundle $TdS^2$ is canonically identified with the set of oriented positive (spacelike) lines of three-space endowed with the metric $\langle . , . \rangle_1$:

$$L_{1,+}^3 \ni \{V + tx | t \in \mathbb{R}\} \simeq (x, V - \langle V, x \rangle_1 x) \in TdS^2.$$

Observe that the metric constructed on $TS^2$ (resp. $TH^2$) has non-negative (resp. non-positive) Ricci curvature.

# A The Weyl tensor in the pseudo-Kähler or para-Kähler cases

The Riemann curvature tensor $Rm$ of a pseudo-Riemannian manifold $N$ may be seen as a symmetric form $R$ on bivectors of $\Lambda^2TN$ (see [4] for references). Splitting $R$ along the eigenspaces $\Lambda^+ \oplus \Lambda^-$ of the Hodge operator $*$ on $\Lambda^2TN$, yields the following block decomposition

$$R = \begin{pmatrix} W^+ + \frac{\text{Scal}}{12} I & Z \\ Z^* & W^- + \frac{\text{Scal}}{12} I \end{pmatrix}$$

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

$$W = Rm - \frac{1}{2} \text{Ric} \otimes g + \frac{\text{Scal}}{12} g \otimes g.$$

If, additionally, $N$ is a four dimensional Kähler manifold, then

**Theorem A.1.** $W^+$ can be written as a multiple of the scalar curvature by a parallel non-trivial 2-form on $\Lambda^2TN$.

See Prop. 2 in [6] 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.** $(N, g, J)$ is anti-self-dual ($W^+ = 0$) 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: $W^+$ is replaced by $W^-$. Precisely:

**Theorem A.2.** Let $(\mathcal{N}, g, J)$ 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 [4] for introduction and references), but since we could not find an explicit proof in the literature<fnref n="5" />, we will give a simple one below. To our knowledge, the proof for the para-Kähler case is new (albeit similar).

## 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 $e_{1'} = Je_1$, $e_{2'} = Je_2$, $g(e_1) = g(e_{1'}) = -1$ 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 $B = (e_1 \wedge e_{1'}, e_1 \wedge e_2, e_1 \wedge e_{2'}, e_{1'} \wedge e_2, e_{1'} \wedge e_{2'}, e_2 \wedge e_{2'})$ 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 $\Lambda^2 T\mathcal{N}$:

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

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

The Kähler condition implies

$$ \mathrm{Rm}(JX, JY, Z, T) = \mathrm{Rm}(X, Y, Z, T) = \mathrm{Rm}(X, Y, JZ, JT), $$

because $J$ is isometric and parallel. The matrix of the symmetric 2-form $R$ in

<footnote n="5">⁵On the contrary, some authors seem to imply that scalar flatness is equivalent to anti-self-duality, see [8]). However this contradiction could possibly come from a different choice of orientation, which would exchange self-dual with anti-self-dual.</footnote>
the orthonormal frame $\mathcal{B}$ is

<table><thead><tr><th></th><th>e<sub>11'</sub></th><th>e<sub>12</sub></th><th>e<sub>12'</sub></th><th>e<sub>1</sub>'<sub>2</sub></th><th>e<sub>1</sub>'<sub>2'</sub></th><th>e<sub>22'</sub></th></tr></thead><tbody><tr><th>e<sub>11'</sub></th><td>R<sub>11'11'</sub></td><td>R<sub>11'12</sub></td><td>R<sub>11'12'</sub></td><td>R<sub>11'1'2</sub> =<br/>-R<sub>11'12'</sub></td><td>R<sub>11'1'2'</sub> =<br/>R<sub>11'12</sub></td><td>R<sub>11'22'</sub></td></tr><tr><th>e<sub>12</sub></th><td></td><td>R<sub>1212</sub></td><td>R<sub>1212'</sub></td><td>R<sub>121'2</sub> =<br/>-R<sub>1212'</sub></td><td>R<sub>131'2'</sub> =<br/>R<sub>1212</sub></td><td>R<sub>1222'</sub></td></tr><tr><th>e<sub>12'</sub></th><td></td><td></td><td>R<sub>12'12'</sub></td><td>R<sub>12'1'2</sub> =<br/>-R<sub>12'12'</sub></td><td>R<sub>12'1'2'</sub> =<br/>R<sub>1212'</sub></td><td>R<sub>12'22'</sub></td></tr><tr><th>e<sub>1</sub>'<sub>2</sub></th><td></td><td></td><td></td><td>R<sub>1'21'2</sub> =<br/>R<sub>12'12'</sub></td><td>R<sub>1'21'2'</sub> =<br/>-R<sub>1212'</sub></td><td>R<sub>1'222'</sub> =<br/>-R<sub>12'22'</sub></td></tr><tr><th>e<sub>1</sub>'<sub>2'</sub></th><td></td><td></td><td></td><td></td><td>R<sub>1'2'1'2'</sub> =<br/>R<sub>1212</sub></td><td>R<sub>1'2'22'</sub> =<br/>R<sub>1222'</sub></td></tr><tr><th>e<sub>22'</sub></th><td></td><td></td><td></td><td></td><td></td><td>R<sub>22'22'</sub></td></tr></tbody></table>

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 $R$ is symmetric we need only write half the matrix. We have used the internal symmetries of $R$, to choose among equivalent coefficients the ones lowest in the lexicographic order of the indices.

The Weyl tensor satisfies some of the J-symmetries of $R$: indeed

$$ \begin{aligned} \mathrm{Ric}(JX, JY) &= \sum_{i=1}^{4} g(e_i) \mathrm{Rm}(JX, e_i, JY, e_i) = \sum_{i=1}^{4} g(e_i) \mathrm{Rm}(X, Je_i, Y, Je_i) \\ &= \sum_{i=1}^{4} g(Je_i) \mathrm{Rm}(X, Je_i, Y, Je_i) = \mathrm{Ric}(X, Y) \end{aligned} $$

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

$$ \begin{aligned} \mathrm{Ric} \otimes g(JX, Y, Z, T) &= \mathrm{Ric}(JX, Z)g(Y, T) + \mathrm{Ric}(Y, T)g(JX, Z) \\ &\quad - \mathrm{Ric}(JX, T)g(Y, Z) - \mathrm{Ric}(Y, Z)g(JX, T) \\ &= - \mathrm{Ric}(X, JZ)g(JY, JT) - \mathrm{Ric}(JY, JT)g(X, JZ) \\ &\quad + \mathrm{Ric}(X, JT)g(JY, JZ) + \mathrm{Ric}(JY, JZ)g(X, JT) \\ &= - \mathrm{Ric} \otimes g(X, JY, JZ, JT) \end{aligned} $$

so

$$ \mathrm{Ric} \otimes g(JX, JY, Z, T) = -\mathrm{Ric} \otimes g(X, J^2Y, JZ, JT) = \mathrm{Ric} \otimes g(X, Y, JZ, JT). $$

Hence the following symmetries (fewer than for $Rm$) in the coefficients of $\mathrm{Ric} \otimes g$,
$g \otimes g$ and Rm, and therefore W:

<table><thead><tr><th></th><th>e<sub>11'</sub></th><th>e<sub>1</sub> &and; e<sub>2</sub></th><th>e<sub>12'</sub></th><th>e<sub>1'</sub> &and; e<sub>2</sub></th><th>e<sub>1'2'</sub></th><th>e<sub>22'</sub></th></tr></thead><tbody><tr><th>e<sub>11'</sub></th><td>W<sub>11'11'</sub></td><td>W<sub>11'12'</sub></td><td>W<sub>11'12'</sub></td><td>W<sub>11'1'2'</sub> =<br/>-W<sub>11'12'</sub></td><td>W<sub>11'1'2'</sub> = W<sub>11'12</sub></td><td>W<sub>11'22'</sub></td></tr><tr><th>e<sub>12</sub></th><td></td><td>W<sub>1212'</sub></td><td>W<sub>1212'</sub></td><td>W<sub>121'2</sub></td><td>W<sub>121'2'</sub></td><td>W<sub>1222'</sub></td></tr><tr><th>e<sub>12'</sub></th><td></td><td></td><td>W<sub>12'12'</sub></td><td>W<sub>12'1'2</sub></td><td>W<sub>12'1'2'</sub> =<br/>-W<sub>121'2</sub></td><td>W<sub>12'22'</sub></td></tr><tr><th>e<sub>1'2</sub></th><td></td><td></td><td></td><td>W<sub>1'21'2</sub><br/>= W<sub>12'12'</sub></td><td>W<sub>1'21'2'</sub> =<br/>-W<sub>1212'</sub></td><td>W<sub>1'222'</sub> =<br/>-W<sub>12'22'</sub></td></tr><tr><th>e<sub>1'2'</sub></th><td></td><td></td><td></td><td></td><td>W<sub>1'2'1'2'</sub><br/>= W<sub>1212</sub></td><td>W<sub>1'2'22'</sub><br/>= W<sub>1222'</sub></td></tr><tr><th>e<sub>22'</sub></th><td></td><td></td><td></td><td></td><td></td><td>W<sub>22'22'</sub></td></tr></tbody></table>

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):

<table><thead><tr><th></th><th>E<sub>1</sub><sup>+</sup></th><th>E<sub>2</sub><sup>+</sup></th><th>E<sub>3</sub><sup>+</sup></th></tr></thead><tbody><tr><th>E<sub>1</sub><sup>+</sup></th><td>W<sub>11'11'</sub> + W<sub>22'22'</sub> + 2W<sub>11'22'</sub></td><td>2(W<sub>11'12</sub> + W<sub>1222'</sub>)</td><td>2(W<sub>11'12'</sub> + W<sub>12'22'</sub>)</td></tr><tr><th>E<sub>2</sub><sup>+</sup></th><td></td><td>2(W<sub>1212</sub> + W<sub>121'2'</sub>)</td><td>2(W<sub>1212'</sub> - W<sub>121'2</sub>)</td></tr><tr><th>E<sub>3</sub><sup>+</sup></th><td></td><td></td><td>2(W<sub>12'12'</sub> - W<sub>12'1'2</sub>)</td></tr><tr><th>E<sub>1</sub></th><td></td><td></td><td></td></tr><tr><th>E<sub>2</sub><sup>-</sup></th><td></td><td></td><td></td></tr><tr><th>E<sub>3</sub><sup>-</sup></th><td></td><td></td><td></td></tr></tbody></table>

<table><thead><tr><th></th><th>E<sub>1</sub><sup>-</sup></th><th>E<sub>2</sub><sup>-</sup></th><th>E<sub>3</sub><sup>-</sup></th></tr></thead><tbody><tr><th>E<sub>1</sub><sup>+</sup></th><td>W<sub>11'11'</sub> - W<sub>22'22'</sub></td><td>0</td><td>0</td></tr><tr><th>E<sub>2</sub><sup>+</sup></th><td>2(W<sub>11'12</sub> - W<sub>1222'</sub>)</td><td>0</td><td>0</td></tr><tr><th>E<sub>3</sub><sup>+</sup></th><td>2(W<sub>11'12'</sub> - W<sub>12'22'</sub>)</td><td>0</td><td>0</td></tr><tr><th>E<sub>1</sub><sup>-</sup></th><td>W<sub>11'11'</sub> + W<sub>22'22'</sub> - 2W<sub>11'22'</sub></td><td>0</td><td>0</td></tr><tr><th>E<sub>2</sub><sup>-</sup></th><td></td><td>2(W<sub>1212</sub> - W<sub>121'2'</sub>)</td><td>2(W<sub>1212'</sub> + W<sub>121'2</sub>)</td></tr><tr><th>E<sub>3</sub><sup>-</sup></th><td></td><td></td><td>2(W<sub>12'12'</sub> + W<sub>12'1'2</sub>)</td></tr></tbody></table>

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

$$
\begin{align*}
\text{Scal} &= -r_{11} - r_{1'1'} + r_{22} + r_{2'2'} = 2(r_{22} - r_{11}) \\
&= 2(-(-R_{11'11'} + R_{1212} + R_{12'12'}) + (-R_{1212} - R_{1'21'2} + R_{22'22'})) \\
&= 2(R_{11'11'} - 2(R_{1212} + R_{12'12'}) + R_{22'22'}).
\end{align*}
$$

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

$$
\begin{align*}
W_{11'11'} &= R_{11'11'} + \frac{1}{2}(r_{11} + r_{1'1'}) + \frac{\text{Scal}}{6} = R_{11'11'} + r_{11} + \frac{\text{Scal}}{6} \\
&= R_{1212} + R_{12'12'} + \frac{\text{Scal}}{6}
\end{align*}
$$
$$
\begin{align*}
W_{22'22'} &= R_{22'22'} - \frac{1}{2}(r_{22} + r_{2'2'}) + \frac{\text{Scal}}{6} = R_{22'22'} - r_{22} + \frac{\text{Scal}}{6} \\
&= R_{1212} + R_{12'12'} + \frac{\text{Scal}}{6}
\end{align*}
$$

so that $W_{11'11'} - W_{22'22'} = 0$. Similarly

$$
W_{11'12} = R_{11'12} + \frac{r_{1'2}}{2}, \quad W_{1222'} = R_{1222'} + \frac{r_{12'}}{2} = R_{1222'} - \frac{r_{1'2}}{2}
$$

so

$$
W_{11'12} - W_{1222'} = R_{11'12} - R_{1222'} + r_{1'2} = 0
$$

$$
W_{11'12'} = R_{11'12'} + \frac{r_{1'2}}{2} = R_{11'12'} + \frac{r_{12}}{2}, \quad W_{12'22'} = R_{12'22'} - \frac{r_{12}}{2},
$$

$$
W_{11'12'} - W_{12'22'} = R_{11'12'} - R_{12'22'} + r_{12} = 0.
$$

That proves that W is block-diagonal.

The $W^-$ term satisfies

$$
\begin{align*}
W_{11'11'} + W_{22'22'} - 2W_{11'22'} &= R_{11'11'} + r_{11} + R_{22'22'} - r_{22} + \frac{\text{Scal}}{3} \\
&\quad -2R_{11'22'} \\
&= R_{11'11'} + R_{22'22'} - 2R_{11'22'} - \frac{\text{Scal}}{6} \\
&= R_{11'11'} + R_{22'22'} - 2(R_{1212} + R_{12'12'}) \\
&\quad - \frac{\text{Scal}}{6} \\
&= \frac{\text{Scal}}{2} - \frac{\text{Scal}}{6} = \frac{\text{Scal}}{3}
\end{align*}
$$

using the first Bianchi identity (and the invariance of Rm):

$$
R_{11'22'} = -R_{1'212'} - R_{211'2'} = R_{12'12'} + R_{1212}.
$$

$$
\begin{align*}
W_{1212} - W_{121'2'} &= R_{1212} + \frac{r_{22} - r_{11}}{2} - \frac{\text{Scal}}{6} - R_{121'2'} = \frac{\text{Scal}}{4} - \frac{\text{Scal}}{6} \\
&= \frac{\text{Scal}}{12}
\end{align*}
$$

$$
W_{12'12'} + W_{12'1'2} = R_{12'12'} + \frac{\text{Scal}}{4} - \frac{\text{Scal}}{6} + R_{12'1'2} = \frac{\text{Scal}}{12}
$$

$$
W_{1212'} + W_{121'2} = R_{1212'} + \frac{r_{22'}}{2} + R_{121'2} - \frac{r_{11'}}{2} = \frac{1}{2}(r_{22'} - r_{11'}) = 0.
$$

Finally,

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

$$W^+ = \mathrm{Scal} \begin{pmatrix} 1/3 & & \\ & -1/6 & \\ & & -1/6 \end{pmatrix} = -\frac{\mathrm{Scal}}{6} \mathrm{Id} + \frac{\mathrm{Scal}}{3} E_1^+ \otimes E_1^+.$$

We let the Reader check that in the neutral case, the $W^+$ part is not a multiple of the scalar curvature, which completes the proof of Theorem A.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 $J$ is now an anti-isometry: $R(JX, JY)Z = -R(X, Y)Z$. We pick an orthonormal basis $(e_1, e_{1'}, e_2, e_{2'})$ with $e_{1'} = Je_1, e_{2'} = Je_2$, 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:

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

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

Since $J$ is anti-isometric and parallel,

$$\mathrm{Rm}(JX, JY, Z, T) = -\mathrm{Rm}(X, Y, Z, T) = \mathrm{Rm}(X, Y, JZ, JT).$$

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

<table><thead><tr><th></th><th>e<sub>11'</sub></th><th>e<sub>12</sub></th><th>e<sub>12'</sub></th><th>e<sub>1'</sub> &and; e<sub>2</sub></th><th>e<sub>1'2'</sub></th><th>e<sub>22'</sub></th></tr></thead><tbody><tr><th>e<sub>11'</sub></th><td>R<sub>11'11'</sub></td><td>R<sub>11'12</sub></td><td>R<sub>11'12'</sub></td><td>R<sub>11'1'2</sub><br>= -R<sub>11'12'</sub></td><td>R<sub>11'1'2'</sub><br>= -R<sub>11'12</sub></td><td>R<sub>11'22'</sub></td></tr><tr><th>e<sub>12</sub></th><td></td><td>R<sub>1212</sub></td><td>R<sub>1212'</sub></td><td>R<sub>121'2</sub><br>= -R<sub>1212'</sub></td><td>R<sub>121'2'</sub><br>= -R<sub>1212</sub></td><td>R<sub>1222'</sub></td></tr><tr><th>e<sub>12'</sub></th><td></td><td></td><td>R<sub>12'12'</sub></td><td>R<sub>12'1'2</sub><br>= -R<sub>12'12'</sub></td><td>R<sub>12'1'2'</sub><br>= -R<sub>1212'</sub></td><td>R<sub>12'22'</sub></td></tr><tr><th>e<sub>1'2</sub></th><td></td><td></td><td></td><td>R<sub>1'21'2</sub><br>= R<sub>12'12'</sub></td><td>R<sub>1'21'2'</sub><br>= R<sub>1212'</sub></td><td>R<sub>1'222'</sub><br>= -R<sub>12'22'</sub></td></tr><tr><th>e<sub>1'2'</sub></th><td></td><td></td><td></td><td></td><td>R<sub>1'2'1'2'</sub><br>= R<sub>1212</sub></td><td>R<sub>1'2'22'</sub><br>= -R<sub>1222'</sub></td></tr><tr><th>e<sub>22'</sub></th><td></td><td></td><td></td><td></td><td></td><td>R<sub>22'22'</sub></td></tr></tbody></table>
(Note again the similarity with the pseudo-Kähler case: only a few signs change.)

The Weyl tensor satisfies some of the J-symmetries of Rm since

$$
\begin{align*}
\mathrm{Ric}(\mathrm{JX}, \mathrm{JY}) &= \sum_{i=1}^{4} g(e_i) \mathrm{Rm}(\mathrm{JX}, e_i, \mathrm{JY}, e_i) = \sum_{i=1}^{4} g(e_i) \mathrm{Rm}(\mathrm{X}, \mathrm{Je}_i, \mathrm{Y}, \mathrm{Je}_i) \\
&= -\sum_{i=1}^{4} g(\mathrm{Je}_i) \mathrm{Rm}(\mathrm{X}, \mathrm{Je}_i, \mathrm{Y}, \mathrm{Je}_i) = -\mathrm{Ric}(\mathrm{X}, \mathrm{Y})
\end{align*}
$$

since $(\mathrm{Je}_i)$ 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,

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

The Kulkarni–Nomizu product Ric ⊗ g satisfies

$$
\begin{align*}
\mathrm{Ric} \otimes g(\mathrm{JX}, \mathrm{Y}, \mathrm{Z}, \mathrm{T})
  &= \mathrm{Ric}(\mathrm{JX}, \mathrm{Z})g(\mathrm{Y}, \mathrm{T}) + \mathrm{Ric}(\mathrm{Y}, \mathrm{T})g(\mathrm{JX}, \mathrm{Z}) \\
  &\quad - \mathrm{Ric}(\mathrm{JX}, \mathrm{T})g(\mathrm{Y}, \mathrm{Z}) - \mathrm{Ric}(\mathrm{Y}, \mathrm{Z})g(\mathrm{JX}, \mathrm{T}) \\
  &= \mathrm{Ric}(\mathrm{X}, \mathrm{JZ})g(\mathrm{JY}, \mathrm{JT}) + \mathrm{Ric}(\mathrm{JY}, \mathrm{JT})g(\mathrm{X}, \mathrm{JZ}) \\
  &\quad - \mathrm{Ric}(\mathrm{X}, \mathrm{JT})g(\mathrm{JY}, \mathrm{JZ}) - \mathrm{Ric}(\mathrm{JY}, \mathrm{JZ})g(\mathrm{X}, \mathrm{JT}) \\
  &= \mathrm{Ric} \otimes g(\mathrm{X}, \mathrm{JY}, \mathrm{JZ}, \mathrm{JT})
\end{align*}
$$

so

$$
\mathrm{Ric} \otimes g(\mathrm{JX}, \mathrm{JY}, \mathrm{Z}, \mathrm{T}) = \mathrm{Ric} \otimes g(\mathrm{X}, \mathrm{J}^2\mathrm{Y}, \mathrm{JZ}, \mathrm{JT}) = \mathrm{Ric} \otimes g(\mathrm{X}, \mathrm{Y}, \mathrm{JZ}, \mathrm{JT})
$$

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

<table>
  <thead>
    <tr>
      <th></th>
      <th>e<sub>11'</sub></th>
      <th>e<sub>12</sub></th>
      <th>e<sub>12'</sub></th>
      <th>e<sub>1'2</sub></th>
      <th>e<sub>1'2'</sub></th>
      <th>e<sub>22'</sub></th>
    </tr>
  </thead>
  <tbody>
    <tr>
      <th>e<sub>11'</sub></th>
      <td>W<sub>11'11'</sub></td>
      <td>W<sub>11'12</sub></td>
      <td>W<sub>11'12'</sub></td>
      <td>W<sub>11'1'2</sub> =<br/>-W<sub>11'12'</sub></td>
      <td>W<sub>11'1'2'</sub> = -W<sub>11'12</sub></td>
      <td>W<sub>11'22'</sub></td>
    </tr>
    <tr>
      <th>e<sub>12</sub></th>
      <td></td>
      <td>W<sub>1212</sub></td>
      <td>W<sub>1212'</sub></td>
      <td>W<sub>121'2</sub></td>
      <td>W<sub>121'2'</sub></td>
      <td>W<sub>1222'</sub></td>
    </tr>
    <tr>
      <th>e<sub>12'</sub></th>
      <td></td>
      <td></td>
      <td>W<sub>12'12'</sub></td>
      <td>W<sub>12'1'2</sub></td>
      <td>W<sub>12'1'2'</sub> = W<sub>121'2</sub></td>
      <td>W<sub>12'22'</sub></td>
    </tr>
    <tr>
      <th>e<sub>1'2</sub></th>
      <td></td>
      <td></td>
      <td></td>
      <td>W<sub>1'21'2</sub> = W<sub>12'12'</sub></td>
      <td>W<sub>1'21'2'</sub> = W<sub>1212'</sub></td>
      <td>W<sub>1'222'</sub> = -W<sub>12'22'</sub></td>
    </tr>
    <tr>
      <th>e<sub>1'2'</sub></th>
      <td></td>
      <td></td>
      <td></td>
      <td></td>
      <td>W<sub>1'2'1'2'</sub> = W<sub>1212</sub></td>
      <td>W<sub>1'2'22'</sub> = -W<sub>1222'</sub></td>
    </tr>
    <tr>
      <th>e<sub>22'</sub></th>
      <td></td>
      <td></td>
      <td></td>
      <td></td>
      <td></td>
      <td>W<sub>22'22'</sub></td>
    </tr>
  </tbody>
</table>

Let us now express W in the Hodge basis defined earlier, using the above sym-
metries (up to a factor 1/2 due to normalization).

<table><thead><tr><th></th><th>E<sub>1</sub><sup>+</sup></th><th>E<sub>2</sub><sup>+</sup></th><th>E<sub>3</sub><sup>+</sup></th></tr></thead><tbody><tr><th>E<sub>1</sub><sup>+</sup></th><td>W<sub>11'11'</sub> + W<sub>22'22'</sub> - 2W<sub>11'22'</sub></td><td>2(W<sub>11'12</sub> - W<sub>1222'</sub>)</td><td>2(W<sub>11'12'</sub> - W<sub>12'22'</sub>)</td></tr><tr><th>E<sub>2</sub><sup>+</sup></th><td></td><td>2(W<sub>1212</sub> - W<sub>121'2'</sub>)</td><td>2(W<sub>1212'</sub> - W<sub>121'2</sub>)</td></tr><tr><th>E<sub>3</sub><sup>+</sup></th><td></td><td></td><td>2(W<sub>12'12'</sub> - W<sub>12'1'2</sub>)</td></tr><tr><th>E<sub>1</sub><sup>-</sup></th><td></td><td></td><td></td></tr><tr><th>E<sub>2</sub><sup>-</sup></th><td></td><td></td><td></td></tr><tr><th>E<sub>3</sub><sup>-</sup></th><td></td><td></td><td></td></tr></tbody></table>

<table><thead><tr><th></th><th>E<sub>1</sub><sup>-</sup></th><th>E<sub>2</sub><sup>-</sup></th><th>E<sub>3</sub><sup>-</sup></th></tr></thead><tbody><tr><th>E<sub>1</sub><sup>+</sup></th><td>W<sub>11'11'</sub> - W<sub>22'22'</sub></td><td>0</td><td>0</td></tr><tr><th>E<sub>2</sub><sup>+</sup></th><td>2(W<sub>11'12</sub> + W<sub>1222'</sub>)</td><td>0</td><td>0</td></tr><tr><th>E<sub>3</sub><sup>+</sup></th><td>2(W<sub>11'12'</sub> + W<sub>12'22'</sub>)</td><td>0</td><td>0</td></tr><tr><th>E<sub>1</sub><sup>-</sup></th><td>W<sub>11'11'</sub> + W<sub>22'22'</sub> + 2W<sub>11'22'</sub></td><td>0</td><td>0</td></tr><tr><th>E<sub>2</sub><sup>-</sup></th><td></td><td>2(W<sub>1212</sub> + W<sub>121'2'</sub>)</td><td>2(W<sub>1212'</sub> + W<sub>121'2</sub>)</td></tr><tr><th>E<sub>3</sub><sup>-</sup></th><td></td><td></td><td>2(W<sub>12'12'</sub> + W<sub>12'1'2</sub>)</td></tr></tbody></table>

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

$$W_{11'11'} = R_{11'11'} - \frac{1}{2}(-r_{11} + r_{1'1'}) - \frac{\text{Scal}}{6} = R_{11'11'} + r_{11} - \frac{\text{Scal}}{6}$$

$$W_{22'22'} = R_{22'22'} - \frac{1}{2}(-r_{22} + r_{2'2'}) - \frac{\text{Scal}}{6} = R_{22'22'} + r_{22} - \frac{\text{Scal}}{6}$$

but $r_{11} = -R_{11'11'} + R_{1212} - R_{12'12'}$ and $r_{22} = R_{2121} - R_{21'21'} - R_{22'22'} = R_{1212} - R_{12'12'} - R_{22'22'}$ so that

$$W_{11'11'} - W_{22'22'} = R_{11'11'} - R_{22'22'} + r_{11} - r_{22} = 0.$$

Similarly

$$W_{11'12} + W_{1222'} = R_{11'12} - \frac{r_{12}}{2} + R_{1222'} + \frac{r_{12'}}{2} = R_{11'12} + R_{1222'} - r_{1'2} = 0$$

$$W_{11'12'} + W_{12'22'} = R_{11'12'} - \frac{r_{12'}}{2} + R_{12'22'} + \frac{r_{12}}{2} = R_{11'12'} + R_{12'22'} + r_{12} = 0$$

which proves that W is block-diagonal, i.e. commutes with the Hodge operator.

Let us now look more closely at the W<sup>-</sup> term

$$\left( \begin{array}{cc} W_{11'11'} + W_{22'22'} + 2W_{11'22'} & 0 & 0 \\ 2(W_{1212} + W_{121'2'}) & 2(W_{1212'} + W_{121'2}) & 2(W_{12'12'} + W_{12'1'2}) \end{array} \right)$$
$$
\begin{align*}
W_{11'11'} + W_{22'22'} + 2W_{11'22'}
&= R_{11'11'} + r_{11} - \frac{\text{Scal}}{6} + R_{22'22'} + r_{22} - \frac{\text{Scal}}{6} + 2R_{11'22'} \\
&= R_{11'11'} + R_{22'22'} + 2R_{11'22'} + \frac{\text{Scal}}{2} - \frac{\text{Scal}}{3} \\
&= R_{11'11'} + R_{22'22'} + 2(-R_{1212} + R_{12'12'}) + \frac{\text{Scal}}{6} = -\frac{\text{Scal}}{3}
\end{align*}
$$

where we have used the first Bianchi identity (and the invariance of Rm)

$$
R_{11'22'} = -R_{1'212'} - R_{211'2'} = R_{12'12'} - R_{1212}.
$$

$$
\begin{align*}
W_{1212} + W_{121'2'} &= R_{1212} - \frac{r_{22} + r_{11}}{2} + \frac{\text{Scal}}{6} + R_{121'2'} \\
&= R_{1212} - \frac{\text{Scal}}{4} + \frac{\text{Scal}}{6} + R_{121'2'} = -\frac{\text{Scal}}{12}
\end{align*}
$$

$$
W_{12'12'} + W_{12'1'2} = R_{12'12'} + \frac{\text{Scal}}{4} - \frac{\text{Scal}}{6} + R_{12'1'2} = \frac{\text{Scal}}{12}
$$

$$
W_{1212'} + W_{121'2} = R_{1212'} - \frac{r_{22'}}{2} + R_{121'2} - \frac{r_{11'}}{2} = 0.
$$

Finally,

$$
W^{-} = \mathrm{Scal} \begin{pmatrix} -1/3 & & \\ & -1/6 & \\ & & 1/6 \end{pmatrix}
$$

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

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