# A Metric that satisfies the gravitational monopole equations-II

Gourab Bhattacharya

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Gourab Bhattacharya. A Metric that satisfies the gravitational monopole equations-II. 2023. hal-04070293

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# A METRIC THAT SATISFIES THE GRAVITATIONAL MONOPOLE EQUATIONS-II

GOURAB BHATTACHARYA

ABSTRACT. Using Mumford's Geometric invariant theoretic techniques of poly-stability condition, we showed the existence of a Gravitational Monopole metric in the Theorem (8.2). It is a Kähler metric of constant negative scalar curvature.

## 1. INTRODUCTION

Mumford stability is used via the Narasimhan-Seshadri theorem to construct a Gravitational Monopole metric which is Kähler metric of constant negative scalar curvature. The proof is given in §(8), the Theorem (8.2).

## 2. THE GRAVITATIONAL MONOPOLE EQUATIONS

In [cf.1] the Gravitational Monopole equations were introduced in the following sense. Let $(X^4, g)$ be a Riemannian spin 4-manifold. Then the Clifford algebra bundle $Cl(X^4)$ is a vector bundle over $X^4$ with fibre at $x \in X^4$ is the Clifford algebra $Cl(T_x X)$. With respect to the metric $g$, one identifies (isomorphism) $Cl(T_x X)$ with $Cl(T_x^* X)$. Therefore, as a vector space, this is isomorphic to $\wedge T_x^* X$. Let us also assume $E \to X$ is a Clifford module bundle with a covariant derivative $\nabla^E$. Then for each $x \in X$ there is a Clifford action $c : T_x^* X \otimes E_x \to E_x$ via $c(\alpha \otimes s) = c(\alpha)s$.

**Definition 2.1.** The twisted Dirac operator associated to $(E, \nabla^E)$ is the operator,

$$ (2.1) \qquad \nabla^E := c \circ \nabla^E : C^\infty(X, E) \to C^\infty(X, E). $$

The equations we wish to consider are (sometimes we omit the mapping $c$ and the dimension 4 for the convenience of computations),

$$ (2.2) \qquad \begin{aligned} \nabla^E \psi &= (d + d^*) \psi = 0, \\ c(W_g^+) &= \frac{1}{4} \langle e_i \cdot e_j \psi, \psi \rangle e^i \wedge e^j, \\ \text{or, } \quad c((W_g^+)^{ijkl}) &= \frac{1}{4} \langle e_k \cdot e_l \psi, \psi \rangle. \end{aligned} $$

The Weitzenböck's formula [5]: the decomposition of the Laplace-Beltami operator as a generalized Laplacian is, with the left-Clifford multiplication $\epsilon_l = \text{ext}_l - \text{int}_l$, and corresponding right Clifford action $\epsilon_r$,

$$ (2.3) \qquad (d + d^*)^2 = \Delta^{\wedge T^* X} - \sum_{ijkl} R_{ijkl} \epsilon_l^i \epsilon_l^j \epsilon_r^k \epsilon_r^l + \frac{1}{4} s. $$

**Proposition 1.** For $k \neq l$, $\langle e_k e_l \psi, \psi \rangle$ is purely imaginary.

**Proposition 2.**

$$ (2.4) \qquad |\langle e_k e_l \psi, \psi \rangle|^2 = 2|\psi|^4. $$

We keep in mind the Seiberg-Witten analysis, and analogously define and get the following definition of Gravitational-Monopole functional,
**Definition 2.2.** The Gravitational-Monopole functional of a pair $(\psi, g)$ is given by,

$$ (2.5) \qquad S(g, \psi) = \int_X \left( |\nabla \psi|^2 + |W^+ - \frac{1}{4} \langle e_i \cdot e_j \psi, \psi \rangle e^i \wedge e^j|^2 \right) d(\mathrm{vol})_g \\ \text{i.e., } S(g, \psi) = \int_X \left( |(d + d^*)\psi|^2 + |W^+ - \frac{1}{4} \langle e_i \cdot e_j \psi, \psi \rangle e^i \wedge e^j|^2 \right) d(\mathrm{vol})_g. $$

**Proposition 3.**

$$ (2.6) \qquad S(g, \psi) = \int_X \left( |\nabla \psi|^2 + |W^+|^2 + \frac{s}{4} |\psi|^2 + \frac{1}{8} |\psi|^4 \right) d(\mathrm{vol})_g $$

**Proposition 4.** As a direct consequence of (3), if the scalar curvature of $X$ is non-negative, all solutions of (2.2) have $\psi \equiv 0$.

### 3. ACTION OF SPECIAL ORTHOGONAL GROUP

Let $X$ be an oriented Riemannian manifold of even dimension $2l$ and we also assume $X$ is a spin manifold, that is the first and second Stiefel-Whitney classes vanish. We denote by $\wedge^p$ the bundle of exterior $p$-forms with $A^p = \Gamma(\wedge^p)$ its space of smooth sections. The Hodge star operator $*\wedge^p \to \wedge^{2l-p}$ is defined by,

$$ (3.1) \qquad \alpha \wedge *\beta = (\alpha, \beta)\omega \in \wedge^{2l} $$

where $\alpha, \beta \in \wedge^p$, $(\alpha, \beta)$ is the induced inner product on $p$-forms and $\omega$ is the volume form.

From now everything will be 4-dimensional unless otherwise stated. We start with the symmetry of the equations, namely the Lie algebras. The Lie algebra $so(4)$ of the special orthogonal group $SO(4)$ is not simple. It can be decomposed into the direct sum of two copies of the Lie algebra $so(3)$ of the group $SO(3)$:

$$ (3.2) \qquad \mathfrak{so}(4) \cong \mathfrak{so}(3) \oplus \mathfrak{so}(3). $$

In terms of the group theory, one understands the above decomposition corresponds to the fact that the universal covering group of $SO(4)$ is the product of the two copies of $SU(2)$. This fact in quantum mechanics corresponds to $\pm \frac{1}{2}$ spins of an electron for each factor $SU(2)$.

In terms of the geometry of the vector bundles, the decomposition $\mathfrak{so}(4) \cong \mathfrak{so}(3) \oplus \mathfrak{so}(3)$ induces the following decomposition (for a choice of $g$ on $X^4$) for the vector bundle $\wedge^2 T^*X \to X$,

$$ (3.3) \qquad \wedge^2 T^*X \cong \wedge^+ \oplus \wedge^-, $$

as a Whitney sum of two oriented 3-plane bundles. One can choose an oriented orthonormal frame for $T^*_U X$ for an open set $U \subset X$. One therefore has,

$$ (3.4) \qquad \wedge^{\pm} = \mathrm{Span}\{(e^1 \wedge e^2 \pm e^3 \wedge e^4), (e^2 \wedge e^3 \pm e^1 \wedge e^4), (e^3 \wedge e^1 \pm e^2 \wedge e^4)\}. $$

We now use the unique Levi-Civita connection $\nabla$ on $\wedge^2 T^*X$ to find a suitable decomposition of the curvature tensor under the action of $O(4)$. The first step towards it is to note that $\nabla g = 0$, this however means that $\nabla$ is covariantly constant, that is $\nabla$ maps sections of $\wedge^\pm$ into $\wedge^\pm \otimes T^*X$; there is no mixed term mapping $\wedge^+$ into $\wedge^- \otimes T^*X$. The curvature of the Levi-Civita connection defines a section of $\wedge^2 T^*X \otimes \wedge^2 T^*X$, correspondingly a decomposition of $\wedge^2 T^*X \otimes \wedge^2 T^*X$ into four matrix-blocks of size $3 \times 3$, more precisely, the Riemann curvature tensor defines, in general, a self-adjoint linear transformation $\mathcal{R}: \wedge^2 \to \wedge^2$ such that,

$$ (3.5) \qquad \mathcal{R}(e^i \wedge e^j) = \frac{1}{2} \sum_{k,l} R_{ijkl} e^k \wedge e^l, $$

relative to the decomposition $\wedge^2 = \wedge^+ \otimes \wedge^-$, the operator $\mathcal{R}$ has the following form,

$$ (3.6) \qquad \mathcal{R} = \begin{bmatrix} A & B \\ B^t & C \end{bmatrix} $$
where, $B \in Hom(\Lambda^{-}, \Lambda^{+})$ (is the traceless Ricci curvature) Ric, and $A \in End(\Lambda^{+})$, that is A is symmetric about its diagonal, that is $A^t = A$, similarly for $C \in End(\Lambda^{-})$ we have $C^t = C$. This representation of the curvature tensor $\mathcal{R}$ gives us a complete decomposition of it into irreducible components, namely

$$ (3.7) \qquad \mathcal{R} \rightarrow (\operatorname{Tr} A, B, A - \frac{1}{3} \operatorname{Tr} A, C - \frac{1}{3} \operatorname{Tr} C) $$

$\operatorname{Tr} A = \operatorname{Tr} C = \frac{1}{4}s$ where s is the scalar curvature.

More elaborately, if the basis of $\Lambda^{\pm}$ in (3.4) is denoted by $\{x_{\pm}^{i}\}_{i=1}^{3}$, then the curvature tensor $\mathcal{R}$ has the following expansion

$$ (3.8) \qquad \mathcal{R} = W_{ij}^{+} x_{+}^{i} \otimes x_{+}^{j} + W_{ij}^{-} x_{-}^{i} \otimes x_{-}^{j} + B_{ij} x_{+}^{i} \otimes x_{-}^{j} + B_{ij}^{t} x_{-}^{i} \otimes x_{+}^{j} - \frac{s}{12} (x_{+}^{i} \otimes x_{-}^{j} + x_{-}^{i} \otimes x_{+}^{j}). $$

If we denote the projection operator by,

$$ (3.9) \qquad P_{\pm} := \frac{1}{2}(1 \pm *) : \Lambda \rightarrow \Lambda^{\pm}, $$

then,

$$ (3.10) \qquad W^{\pm} = P_{\pm} \circ Rm \circ P_{\pm} - \frac{s}{12} Id_{\pm} $$

### 4. TWISTOR CORRESPONDENCE FOR THE GRAVITATIONAL MONOPOLES

Let $t: Z \rightarrow M$ denote the twistor fibration and suppose $M$ is hermitian and self-dual. One concludes from the definition of the almost complex structure of $Z$: that $t$ is not a holomorphic map and the complex structure $J$ of $M$ defines a cross section $J: M \rightarrow Z$. Let us denote the image of $J$ by $\Sigma$. One can use the integrability of $J$, and can show $\Sigma$ is a complex hypersurface of $Z$ biholomorphic to $M$. Similarly $-J: M \rightarrow Z$ defines hypersurface $\bar{\Sigma}$. The "real structure" $\sigma$ of $Z$ switches the two hypersurfaces identifying one with the other in an antiholomorphic fashion [cf.9]. If $D$ denotes the divisor $\Sigma + \bar{\Sigma}$ in $Z$, we can consider the corresponding holomorphic line bundle $\mathcal{L} = [D]$; since $\sigma(D) = \sigma(\Sigma + \bar{\Sigma}) = \bar{\Sigma} + \Sigma = D$, so $\mathcal{L}$ is called a "real" bundle. Let us also assume $\Theta_{Z,D}$ is the sheaf of holomorphic vector fields on $Z$ which are tangent to $D$.

The following theorem is due to Penrose

**Theorem 4.1.**

$$ (4.1) \qquad \mathcal{L} \cong K_Z^{-1/2}, $$

if and only if the metric $h$ is conformal to a Kähler metric.

**Theorem 4.2.** If $M$ is not Ricci flat then,

$$ (4.2) \qquad H^0(Z, \Theta) \cong H^0(M, \Theta). $$

*Proof.* See [9].

□

**Proposition 5.** $H^0(Z, \Theta)$ is the complexification of the Lie algebra of parallel vector fields, and

$$ (4.3) \qquad h^0(M, \Theta) = h^0(M, \Omega^1) = \frac{1}{2} b_1(M). $$

*Proof.* See [9].

□

**Theorem 4.3.** Let $(M, \omega)$ be a compact almost Kähler surface with a positive definite signature $\sigma > 0$ and $c_1 \neq 0$, $c_1(L)^2 > 0$, we also assume $M$ is a Gravitational Monopole and let $Z$ be its twistor space, we also assume $H^2(Z, \Theta \otimes \mathcal{I}_D) = 0$ then $H^0(M, \Theta_M) = H^0(Z, \Theta_Z) = H^2(Z, \Theta_{Z,D}) = H^2(Z, \Theta_Z) = 0$.
*Proof.* The condition $H^0(M, \Theta_M) = H^0(Z, \Theta_Z) = 0$ follows from the non-existent of Killing vector fields, showed in [cf. Equation 3.17 of 3]. To prove the second part we use [cf.8]. Consider the following exact sequence of sheaves

$$ (4.4) \qquad 0 \to \Theta_{Z,D} \to \Theta_Z \to N_D \to 0, $$

and,

$$ (4.5) \qquad 0 \to \Theta_Z \otimes \mathcal{I}_D \to \Theta_{Z,D} \to \Theta_D \to 0. $$

From the short exact sequence (4.5), we get the long exact sequence

$$ (4.6) \qquad \dots \to H^2(Z, \Theta_Z \otimes \mathcal{I}_D) \to H^2(Z, \Theta_{Z,D}) \to H^2(D, \Theta_D) \to \dots, $$

since

$$ (4.7) \qquad H^2(Z, \Theta_Z \otimes \mathcal{I}_D) = 0 $$

we have

$$ (4.8) \qquad H^2(Z, \Theta_{Z,D}) = 0. $$

Now one uses the short exact sequence (4.4) to get the long exact sequence

$$ (4.9) \qquad \dots \to H^2(Z, \Theta_{Z,D}) \to H^2(Z, \Theta_Z) \to H^2(D, N_D) \to \dots $$

Now equation (4.8) says $H^2(Z, \Theta_{Z,D}) = 0$, therefore from the long exact sequence we get

$$ (4.10) \qquad H^2(Z, \Theta_Z) = 0. $$

This proves the theorem.

□

*Remark 4.4.* We can compare with [cf.8] to see the non-existence of holomorphic vector fields on Z also imply $H^2(Z, \Theta_Z \otimes \mathcal{I}_D) = 0$.

*Remark 4.5.* One can compare with [cf.9], it was proved there that when M is a scalar-flat Kähler surface but not Ricci-flat

$$ (4.11) \qquad H^0(M, \Theta_M) = H^0(Z, \Theta_Z) = 0 $$

holds true in contrast to our strictly negative scalar curvature treatment of Equation 3.17 in [cf.3] and corresponding vanishing of Killing fields.

The following theorem can be found in [cf.10], [cf.9],

**Theorem 4.6.** When obstruction vanishes, the deformation of scalar-flat Kähler metrics corresponds to the real slice of $H^1(Z, \Theta_{Z,D})$, and its obstruction lies in $H^2(Z, \Theta_{Z,D})$.

The expected dimension of the moduli space of scalar We have $N_D = K_D^{-1}$ [cf.9] and short exact sequence (4.4)

$$ (4.12) \qquad 0 \to \Theta_{Z,D} \to \Theta_Z \to N_D \to 0 $$

to give

$$ (4.13) \qquad \chi(Z, \Theta_Z) = \chi(Z, \Theta_{Z,D}) + 2\chi(M, K_M^{-1}). $$
## 5. THE CONSTRUCTION OF THE ELLIPTIC COMPLEX

To get an elliptic complex corresponding to the data of the Gravitational monopole equations, we modify the data a little bit. So, we assume, $A$ is a spin connection on the smooth 4-manifold $X$. $\psi$ is a section of $S^+$, so we relace the Dirac data $\psi \in \ker(d + d^*)$ to $\psi \in \ker D_A$. In this way, the Dirac operator becomes dependent on $A$ and we get a more general context. Let $\hat{A}$ be the induced connection on the line bundle $L \to X$. We denote by $F_{\hat{A}}$ the curvature corresponding to the connection $\hat{A}$. We put no restriction on $F_{\hat{A}}$. All the restrictions are on the self-dual part of the Weyl tensor of $X$.

In the case of a line bundle, the gauge group $\mathcal{G} = \mathcal{M}(X, U(1))$ as a space of maps is well-defined as it is only dependent on the transition functions. The action of $\mathcal{G}$ on the pair $(A, \psi)$ is given in the following way:

$$ (5.1) \qquad \lambda : (A, \psi) \to (A - \lambda^{-1}d\lambda, \lambda\psi), $$

and on $\hat{A}$ by

$$ (5.2) \qquad \hat{A} - 2i\lambda^{-1}d\lambda. $$

We verify the following:

$$ (5.3) \qquad D_{A-\lambda^{-1}d\lambda}(\lambda\psi) = \lambda D_A\psi + d\lambda \cdot \psi - d\lambda \cdot \psi. $$

The self-dual part $F_{\hat{A}}^+$ of the curvature tensor $F_{\hat{A}}$ also remains invariant

$$ (5.4) \qquad F_{\hat{A}-2\lambda^{-1}d\lambda}^{+} = F_{\hat{A}}^{+} - 2d^{+}(\lambda^{-1}d\lambda) = F_{\hat{A}}^{+}. $$

Since the metric $g_{ij} \mapsto g_{AA'BB'} = \epsilon_{AB}\epsilon_{A'B'}$, corresponding $U(1)$-action on the metric is just multiplication by $|\lambda|^2 = 1$, therefore $W^+$ remains invariant, on the other hand

$$ (5.5) \qquad \langle e_i e_j \lambda \psi, \lambda \psi \rangle = |\lambda|^2 \langle e_i e_j \psi, \psi \rangle = \langle e_i e_j \psi, \psi \rangle, $$

as $|\lambda| = 1$. Now we study the kernel of the linearized operator $T_g W^+ : T_g \mathcal{M} \to \Gamma(\mathrm{Sym}_0^2 \wedge_+^2)$. The deformations we shall consider will be represented by the first cohomology of the complex

$$ (5.6) \qquad \Gamma(TX) \oplus \Gamma(\mathbb{R}) \xrightarrow{\tau^*} \Gamma(\mathrm{Sym}_0^2 \wedge_+^2) \xrightarrow{TW^+} \Gamma(\mathrm{Sym}_0^2 \wedge_+^2), $$

where $\mathbb{R}$ represents trivial $\mathbb{R}$-bundle over $X$. $\Gamma(\mathrm{Sym}^2 T^* X) = T_g \mathcal{M} = \mathrm{Hom}(\wedge_+^2, \wedge_-^2) \oplus \mathbb{R}$. But $\mathrm{Hom}(\wedge_+^2, \wedge_-^2) \cong \wedge_+^2 \otimes \wedge_-^2$. If we want to mod out the trivial line bundle $\mathbb{R}$ then we must work with $\mathcal{M}/\mathcal{C}_+^\infty$, and the section space $\Gamma(\mathbb{R})$ is replaced by the orbit space $\mathcal{C}_+^\infty(g)$.

The problem is not underdetermined as the Gravitational monopole equation (2.2) gives additional restriction on the scalar curvature, that is $s = s_g < 0$ [cf.1], hence a linearization at $g$ gives the following complex

$$ (5.7) \qquad \Gamma(TX) \xrightarrow{\delta^*} \Gamma(\mathrm{Sym}^2 T^* X) \xrightarrow{TW_+ \oplus T_g} \Gamma(\mathrm{Sym}_0^2 \wedge_+^2) \oplus \Gamma(\mathbb{R}). $$

## 6. A COMPUTATION FOR A LINE BUNDLE.

The Riemann-Roch theorem, we get for any holomorphic line bundle $L$ on $M$,

$$ (6.1) \qquad \chi(L) = \chi(\mathcal{O}_M) + \int_M \frac{c_1(L)(c_1(L) + c_1(K^{-1}))}{2}. $$

In particular for $L = K^{-1}$ the following holds

$$ (6.2) \qquad \chi(K^{-1}) = \chi(\mathcal{O}_M) + c_1(K^{-1}) \cdot c_1(K^{-1}) = \chi(\mathcal{O}_M) + c_1^2(M). $$

As

$$ (6.3) \qquad \chi(\mathcal{O}_M) = \frac{1}{12}(c_1^2 + c_2) $$

we have,

$$ (6.4) \qquad \chi(K^{-1}) = \frac{1}{12}(c_1^2 + c_2) + c_1^2. $$
Now,

$$ (6.5) \qquad c_1^2(M) = 2\chi + 3\sigma, \quad c_2(M) = \chi $$

we get,

$$ (6.6) \qquad \begin{aligned} \chi(M, K_M^{-1}) &= \frac{1}{12}(2\chi + 3\sigma + \chi) + 2\chi + 3\sigma \\ &= \frac{1}{4}\chi + \frac{1}{4}\sigma + 2\chi + 3\sigma. \end{aligned} $$

Therefore,

$$ (6.7) \qquad \begin{aligned} \chi(Z, \Theta_Z) &= \chi(Z, \Theta_{Z,D}) + 2\chi(M, K_M^{-1}) \\ &= 3\chi + 8\sigma + \frac{1}{2}\chi + \frac{1}{2}\sigma + 4\chi + 3\sigma \end{aligned} \\ \text{i.e., } \chi(Z, \Theta_Z) = \frac{1}{2}(15\chi + 29\sigma). $$

Therefore the dimension of the moduli space $\mathcal{M}$ of self-dual metrics are given by

$$ (6.8) \qquad \dim \mathcal{M} = -\frac{1}{2}(15\chi + 29\sigma). $$

## 7. THE GENERAL CASE OF A VECTOR BUNDLE

**Theorem 7.1.** The complex (5.7) is elliptic with index equals to

$$ (7.1) \qquad \frac{1}{2}(29|\sigma(X)| - 15\chi(X)), $$

where $\chi(X)$ is the Euler characteristic of X and $\sigma(X)$ are the signature of X.

By the index theorem for the twisted Dirac operator

$$ (7.2) \qquad \mathrm{Index}(D_A) = - \int_X \mathrm{ch}(\sqrt{L}) \hat{A}(X). $$

The Chern character is known to be

$$ (7.3) \qquad \mathrm{ch}(\sqrt{L}) = 2(1 + c_1(L)^2 + \dots) $$

the $\hat{A}$-genus is

$$ (7.4) \qquad \hat{A}(X) = 1 - \frac{1}{24}p_1(X) + \dots, $$

where $p_1(X)$ is the first-Pontrjagin class. The top degree form of $\mathrm{ch}(\sqrt{L})\hat{A}(X)$ is $\frac{1}{12}p_1(X) + c_1(L)^2$, but $\frac{1}{3}p_1(X) = \sigma$, therefore we have the result

$$ (7.5) \qquad \mathrm{Index}(D_A) = c_1(\sqrt{L})^2 - \frac{\sigma}{4}. $$

We find the following [cf.2])

**Theorem 7.2.** The moduli space of solutions of the Gravitational monopole equation has dimension (assuming the index $\sigma(X) \ge 0$)

$$ (7.6) \qquad \begin{aligned} d := \frac{1}{2}(29\sigma(X) - 15\chi(X)) + c_1(\sqrt{L})^2 - \frac{\sigma}{4} \\ = (c_1(\sqrt{L}))^2 + \frac{57\sigma(X) - 30\chi(X)}{4} \end{aligned} $$
If $\sqrt{L}$ is globally defined, then,

$$ (7.7) \qquad \begin{aligned} d_{\text{Global}} \sqrt{L} &:= (c_1(\sqrt{L}))^2 + \frac{57\sigma(X) - 30\chi(X)}{4} \\ &= \frac{1}{4}(c_1(L))^2 + \frac{57\sigma(X) - 30\chi(X)}{4} \\ &= \frac{1}{4}[(c_1(L))^2 + 57\sigma(X) - 30\chi(X)] \end{aligned} $$

## 8. THE PROOF

Let $M$ be the total space of an oriented 2-sphere bundle $M \to \Sigma$ over a compact oriented Riemann surface. Let $M$ has a Kähler metrics of constant negative scalar curvature $s$ with Kähler form $\omega$. We impose the negativity assumption

$$ (8.1) \qquad \underbrace{\int sd\mu = 4\pi c_1 \cdot [\omega]}_{\text{true for any Kähler metric of } [\omega]} < 0 $$

to restrict ourselves to Kähler metrics of constant negative scalar curvature only.

**Proposition 6.** Kähler classes with $c_1 \cdot [\omega] < 0$ will exist on the ruled surface $M \to \Sigma$ if and only if and only if the curve $\Sigma$ has genus $g \ge 2$.

Since the genus of $\Sigma$ is $\ge 2$, it becomes the ideal ground to apply a theorem of Narasimhan-Seshadri [cf.15] that says any complex vector $E$ bundle over $\Sigma$ with $g \ge 2$ is classified by the projective representations of $SU(2)$. (For $g = 1$ we have Birkhoff-Grothendieck theorem, and the vector bundle $E$ becomes a finite some of lines $CP^1$.) The classification comes with a concept called *polystability*, a vector bundle $E$ is called polystable if it is a semi-stable bundle of the form $E = \bigoplus_{i=1}^n E_i$ with each $E_i$'s are stable.

For bundles over a projective algebraic curve, the polystabilty is equivalent to the existence of a flat projective unitary connection on $\mathbb{P}(E)$. If the bundle $E$ is polystable, we have

$$ (8.2) \qquad M = \Sigma \times_{\rho} CP^1 $$

for a representation $\rho : \pi_1(\Sigma) \to \mathbb{P}(SU(2)) = SO(3)$ of the fundamental group which is unique up to conjugation in $SO(3)$. In this case the local products of constant curvature Kähler metrics on $\Sigma$ and $CP^1$ inform us with an existence of a constant scalar curvature Kähler metric in each Kähler class. One therefore wish to show that any Kähler metric on $\mathbb{P}(E)$ with $s = \text{const} < 0$ is necessarily of this form.

LeBrun in [cf.14] proved the following theorem without the assumption on the universal cover but using the Seiberg-Witten equations that gives equivalent criteria. For our seek of necessity we added the assumption that *universal cover of* $(M, g)$ *is isometric to the product* $S^2 \times \mathcal{H}^2$. The rest of the proof is as given in [cf.14].

**Theorem 8.1.** For some complex structure $J$, suppose that there is a Kähler metric $g$ of constant negative scalar curvature on $M$. We also assume that the universal cover of $(M, g)$ is isometric to the product $S^2 \times \mathcal{H}^2$ where the 2-sphere $S^2$ and hyperbolic 2-space $\mathcal{H}^2$ are endowed with appropriate constant multiples of their standard metrics. Let $E \to \Sigma$ be a rank-2 holomorphic vector bundle over a compact complex curve, and let $M := \mathbb{P}(E)$ be the total space of the associated $CP^1$-bundle. Let $[\omega]$ be a Kähler class on $M$ with $c_1 \cdot [\omega] < 0$. Then $[\omega]$ contains a Kähler metric of constant scalar curvature if and only if $E$ is a polystable vector bundle. Moreover, when such a metric exists, it is unique modulo biholomorphisms of $(M, J)$.

*Proof.* The following proof is taken from LeBrun with a little bit of different assumption but the philosophy remains the same [cf.14], we repeat it for the sake of completeness. Let us assume that $g$ is a Kähler metric of constant negative scalar curvature on $(M, J)$, and recall that by the assumption of the theorem that the universal cover of $(M, g)$ must be a product $S^2 \times \mathcal{H}^2$ of spaces of constant curvature. The product complex structure necessarily agrees with the lift of $J$ because the holonomy of $S^2 \times \mathcal{H}^2$ is $U(1) \times U(1)$. Now $P_1(M) = \pi_1(\Sigma)$ acts on $S^2 \times \mathcal{H}^2$ by holomorphic isometries, and thus sends any holomorphic curve $S^2 \times pt$
to another curve of this form. The induced action on the hyperbolic plane $\mathcal{H}^2$ is free and proper as $\mathbb{S}^2$ is compact and every rotation of $\mathbb{S}^2$ has a fixed point. Thus

$$ (8.3) \qquad \mathbb{P}(E) \cong (\mathbb{S}^2 \times_{\rho} \mathcal{H}^2)/\pi_1(\Sigma) $$

is biholomorphic to $\tilde{\Sigma} \times_{\rho} \mathbb{CP}^1$ for some compact Riemann surface $\tilde{\Sigma}$ and some representation $\rho: \pi_1(\Sigma) \to \mathbb{P}(SU(2)) = SO(3)$. But since the fibers of $\tilde{\Sigma} \times_{\rho} \mathbb{CP}^1 \to \Sigma$ are the only rational curves in $(M, J)$, there is a biholomorphism $\Sigma \to \Sigma$ such that the diagram

$$ (8.4) \qquad \begin{tikzcd} \mathbb{P}(E) \arrow[r, "\phi"] & \tilde{\Sigma} \times_{\rho} \mathbb{CP}^1 \arrow[d, "\pi"] \\ \Sigma \arrow[u, "\eta"'] \arrow[r, "\psi"] & \tilde{\Sigma} \arrow[d, "\psi"] \end{tikzcd} $$

commutes. Therefore $E$ has a flat unitary projective connection and we conclude $E$ is polystable. $\square$

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

The next claim is the following,

**Theorem 8.2.** The metric $\omega$ as constructed in the Theorem (8.1) necessarily solves the Gravitational Monopole equations.

*Proof.* The proof is by contradiction. Suppose $\omega$ does not satisfy the Gravitational Monopole equations, then $[\omega]$ violates equation (8.1), that is the basic negativity assumption of (8.1), for the sake of completeness we repeat the assumption in the equation described below, namely the inequality

$$ (8.5) \qquad \underbrace{\int sd\mu = 4\pi c_1 \cdot [\omega]}_{\text{true for any Kähler metric of } [\omega]} < 0. $$

Hence $\omega$ satisfies the Gravitational Monopole equations. $\square$

<figure><img src="image_4.png" /></figure>

In Theorem (8.2) we showed the existence of a Gravitational Monopole metric.

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Email address: bhattacharya@ihes.fr, gourabmath@gmail.com