Harmonic morphisms and shear-free perfect fluids coupled with gravity

Abstract

We employ the technique used in the classification of harmonic morphisms with one-dimensional fibers on four-dimensional Einstein manifolds (Pantilie in Commun Anal Geom 10:779–814, 2002) to give a simpler proof of the fact that the shear-free perfect fluids coupled to gravity are either irrotational or expansion free in the case when the equation of state is $\rho =-3p$.

1 Introduction

Harmonic (or wave) maps [4] are critical points of the Dirichlet’s energy

$$\begin{aligned} \mathcal{E }(\varphi )=\frac{1}{2}\int \limits _M \vert \text{ d}\varphi \vert ^2 \text{ vol}_g \end{aligned}$$

where $\varphi : (M, g) \rightarrow (N, h)$ denotes a smooth map between two (semi-) Riemannian manifolds $M$ and $N$. Harmonic morphisms form a subclass of harmonic maps characterized by an additional condition of transversal conformality. Both notions enjoy an extensive study and a deep mathematical understanding, cf. [1–3]. Among their various generalizations, we distinguish the $r$-harmonic maps and morphisms [8], defined in a similar way with respect to the energy $\mathcal E _r(\varphi )=\tfrac{1}{r}\int _M \vert \text{ d}\varphi \vert ^r \text{ vol}_g, r>0$. These developments make harmonic maps and morphisms a very appealing tool in applications, as shown by the successfully concept of sigma model in theoretical physics. In [15], we showed that they can also play a rôle in general relativity. This rôle is due to the local duality between $r$-harmonic morphisms with one-dimensional timelike fibers defined on a space-time and shear-free relativistic perfect fluids $(U, p = \frac{1}{3}(r - 3)\lambda ^{r}, \rho = \lambda ^{r})$ on that space-time, where $U$ is the unitary vector field tangent to the fibers and $\lambda$ is the dilation of the harmonic morphism [15, Prop. 3.2]. This duality is known to geometers through the notion of foliations that produce ($r$-)harmonic morphisms [11].

Harmonic morphisms (and, to a lesser extent, $r$-harmonic morphisms) with one dimensional fibres have been classified under various curvature restrictions on the domain metric, see e.g. [9, 12] and for a review [1, 13]. In this note we want to point out that the local part of Pantilie’s classification result [12] for harmonic morphisms on Riemannian four-manifolds satisfying Einstein (vacuum) condition extends to Lorentzian four-manifolds solving the Einstein field equations (with fluid source). In the dual perspective of relativistic fluids, the counterpart of this result is the positive answer to the following

Conjecture ([16]). In general relativity, if the velocity vector field of a barotropic perfect fluid ($p + \rho \ne 0$ and $\rho = \rho (p)$) is shear free, then either the expansion or the rotation of the fluid vanishes.

in the case when $\rho =-3p$. This is already known to be true [7, 18], but in the present approach, the proof is simpler and is written in coordinate-free differential geometric language and can provide a new insight into this conjecture still not established despite a long history of confirmations in particular cases.

For a recent discussion about the conjecture and the progress that have been made, see [5, 19].

In the following section, we review the basic definitions and results needed thereafter; in Sect. 3, we give the proof of the main result, and we end with the proofs of some simple technical Lemmas.

Throughout the paper, $(M,g)$ will be a Lorentzian manifold and $\mathrm Ric$, $\mathrm Scal$ will denote its Ricci and scalar curvature, respectively. All the considerations are local.

2 Basic material

2.1 Perfect fluids coupled with gravity

Definition 1

[10, 17] Let $(M,g)$ be a four-dimensional space-time. A triple $(U, p, \rho )$ is called (relativistic) perfect fluid if

• (i) $U$ is a timelike future-pointing unit vector field on $M$, called the flow vector field,

• (ii) $\rho , p: M\rightarrow \mathbb{R }$ are the real functions called mass (energy) density and pressure, respectively,

• (iii) the stress-energy tensor of the fluid is conserved:

  $$\begin{aligned} \mathrm div \left(p \, g + (p + \rho )U^{\flat } \otimes U^{\flat }\right)=0. \end{aligned}$$

If instead of $(iii)$, we impose the stronger condition

$(iii)^\prime$ the Einstein equations are satisfied

$$\begin{aligned} \mathrm Ric - \frac{1}{2}\mathrm Scal \cdot g = p \, g + (p + \rho )U^{\flat } \otimes U^{\flat }, \end{aligned}$$

(1)

then $(U, p, \rho )$ is called (relativistic) perfect fluid coupled with gravity.

Condition $(iii)$ of the definition decomposes into the fluid’s equations:

$$\begin{aligned} (\rho + p)\mathrm div U + U(\rho )&= 0 \quad (\text{ the} \text{ conservation} \text{ of} \text{ energy} \text{ along} \text{ the} \text{ flow}),\nonumber \\ (\rho +p)\nabla _{U}U + \mathrm grad ^{\mathcal{H }}p&= 0 \quad (\text{ the} \text{ Euler} \text{ equations}), \end{aligned}$$

(2)

where $\mathrm grad ^{\mathcal{H }}p$ is the spatial pressure gradient, that is, the component orthogonal to $U$. If moreover the perfect fluid is coupled with gravity, we have the following block-diagonal structure of the Ricci tensor of $(M,g)$:

$$\begin{aligned} \mathrm Ric (U,U)&= \frac{1}{2}(\rho + 3p), \nonumber \\ \mathrm Ric (X,U)&= 0, \quad \forall X \perp U, \nonumber \\ \mathrm Ric (X,Y)&= \frac{1}{2}(\rho - p)g(X,Y) , \quad \forall X,Y \perp U. \end{aligned}$$

(3)

Note that $\mathrm Scal = \rho -3p$.

In this paper, we are concerned with perfect fluids coupled with gravity satisfying $\rho = -3 p$. This equation of state is supposed to represent a hypothetical form of matter called quintessence characterized by negative pressure and $-1 < p/\rho \le 0$. Moreover, as shown by Vilenkin [20], a randomly oriented distribution of (infinitely thin) straight strings averaged over all directions behaves like a perfect fluid with $\rho =-3p$.

2.2 Harmonic morphisms and curvature restrictions

Recall [1, 6] that a harmonic morphism $\varphi : (M,g) \rightarrow (N,h)$ between (semi-) Riemannian manifolds can be characterized as a harmonic map (i.e., $\mathrm trace \nabla \text{ d}\varphi =0$) which is moreover horizontally weakly conformal of dilation $\lambda : M \rightarrow \mathbb{R }_+$ (i.e., $\varphi ^* h = \lambda ^2 g \vert _{(\mathrm Ker \text{ d}\varphi )^\perp \times (\mathrm Ker \text{ d}\varphi )^\perp }$).

Let $\varphi$ be a harmonic morphism defined on a four-dimensional space-time taking values in a three-dimensional Riemannian manifold, denote by $U$ the unit vertical timelike vector that spans $\mathcal{V }= \mathrm Ker \text{ d}\varphi$ and by $\mathcal{H }= \mathcal{V }^\perp$ the horizontal distribution associated with $\varphi$. Then it is known that

$$\begin{aligned} -\nabla _U U + \mathrm grad ^{\mathcal{H }}(\ln \lambda )&= 0,\end{aligned}$$

(4)

$$\begin{aligned} \mathrm div U + 3U(\ln \lambda )&= 0,\end{aligned}$$

(5)

$$\begin{aligned} (\mathcal L _{U}g)(X,Y)-\frac{2}{3}\mathrm div U \cdot g(X,Y)&= 0, \quad \forall X,Y \in \mathcal{H }, \end{aligned}$$

(6)

where (4) is the fundamental equation of the harmonic morphism $\varphi$, (5) is a geometric identity and (6) is equivalent to the horizontal conformality of $\varphi$.

Recall [1, p.59] also that a section $\sigma$ of $(\otimes ^r \mathcal{H }) \otimes (\otimes ^s \mathcal{H }^*)$ is called basic if $(\mathcal L _V \sigma )^\mathcal{H }=0$ for all $V \in \Gamma (\mathcal{V })$. In particular, a function $f$ on $M$ is basic if $V(f)=0$ and a horizontal vector field $X$ on $M$ is basic if $[V, X]^\mathcal{H }=0$, for all $V \in \Gamma (\mathcal{V })$; the latter condition means that $X$ is projectable on $N$.

The fundamental vector field of the map $\varphi$ is $V := \lambda U$. Analogously to [1, Lemma 11.7.2], the importance of this vector field is given by the fact that it allows us to rewrite Eq. (4) as follows:

$$\begin{aligned}{}[V, X]=0, \quad \forall X \ \text{ basic} \text{ vector} \text{ field} \text{ on}\ M. \end{aligned}$$

(7)

Consider the 1-form dual to $V$, defined by $\vartheta (X) := -\lambda ^{-2}g(X, V)$, $\forall X$, and the 2-form $\Omega :=\text{ d}\vartheta$. We notice that $\Omega (X,Y)=\lambda ^{-2}g([X,Y],V)$ for all $X, Y \in \Gamma (\mathcal{H })$.

Lemma 1

Let $\varphi :(M^4,g) \rightarrow (N^3,h)$ be a harmonic morphism defined on a four-dimensional space-time. Then,

• (i) $i_W \Omega =0$, for all $W \in \Gamma (\mathcal{V })$ ($\Omega$ is a horizontal $2$-form)

• (ii) $\mathcal{H }$ is integrable if and only if $\Omega =0$ ($\Omega$ is the integrability $2$-form of $\mathcal{H }$);

• (iii) $\mathcal L _W \Omega =0$ or, equivalently, $W(\Omega (X,Y))=0$, for all $X, Y$ basic vectors and for all $W \in \Gamma (\mathcal{V })$ ($\Omega$ is a basic 2-form).

Analogously to [12], [1, p. 343], we can prove the following

Proposition 1

Let $(M,g)$ be four-dimensional space-time, $(N,h)$ a Riemannian 3-manifold and $\varphi :(M,g) \rightarrow (N,h)$ a harmonic morphism with one-dimensional timelike fibers tangent to the unit vector $U$. Let $\lambda : M \rightarrow \mathbb{R }_+$ denote the dilation of $\varphi$. Then, we have the identities:

$$\begin{aligned} \mathrm Ric (U,U)&= \Delta \ln \lambda + 4 U(U(\ln \lambda )) - 6 [U(\ln \lambda )]^2 + \frac{\lambda ^{2}}{2} \vert \Omega \vert ^2,\end{aligned}$$

(8)

$$\begin{aligned} \mathrm Ric (X,U)&= 2 X(U(\ln \lambda )) - \frac{\lambda }{2} \{ \delta \Omega (X) + 2\Omega (X, \mathrm grad \ln \lambda )\},\end{aligned}$$

(9)

$$\begin{aligned} \mathrm Ric (X,Y)&= (\varphi ^* \mathrm Ric ^N) (X, Y) + g(X, Y) \Delta \ln \lambda - 2 X (\ln \lambda ) Y(\ln \lambda ) + \frac{\lambda ^{2}}{2} \langle i_X \Omega , i_Y \Omega \rangle ,\nonumber \\ \end{aligned}$$

(10)

where $\mathrm Ric ^N$ is the Ricci curvature of $(N,h)$ and $X,Y$ are horizontal vectors.

Let us recall [1, 12] also that a $\varphi$ is of Killing type iff $\mathrm grad ^{\mathcal{V }}\lambda = 0$, of warped product type iff it has totally geodesic fibers (i.e., $\nabla _U U=0$) and $\mathcal{H }$ is integrable and of type three (T) iff $\vert \mathrm grad ^{\mathcal{V }}\lambda \vert$ is a nonzero function of $\lambda$. The local part of Pantilie’s classification result [12] states that if $(M,g)$ is an orientable (Rie-mannian) Einstein four-manifold, then $\varphi$ belongs to one of these types.

2.3 The duality

In the setting of the previous subsection, from a dual perspective, Eqs. (4), (5) can be identified respectively with the Euler equations and the energy conservation along the flow (2) for the relativistic fluid $(U, p = -\frac{1}{3}\lambda ^2, \rho = \lambda ^2)$, while (6) tells that the fluid is shear free. In this case, the additional assumption (3) for the fluid to be coupled with gravity becomes

$$\begin{aligned} \mathrm Ric (U,U)&= \mathrm Ric (X,U)= 0 \nonumber \\ \mathrm Ric (X,Y)&= \frac{2}{3}\lambda ^2 g(X,Y), \quad \forall X,Y \perp U. \end{aligned}$$

(11)

Conversely, given a perfect fluid $(U, p, \rho = -3p)$, we can always integrate it locally to obtain a (local) harmonic morphism from $M$ into some Riemannian 3-manifold $(N,h)$ such that $\text{ d}\varphi (U)=0$ and $\rho$ equal the squared dilation of $\varphi$. For more details see [15].

Remark 1

If the associated harmonic morphism $\varphi$ is of Killing type, then the fluid is expansion free (i.e., $\mathrm div U=0$), while if it is of warped product type, then the perfect fluid is irrotational (i.e., $\Omega =0$).

3 The main result

Following the same lines in [1, 12], we prove the following

Theorem 1

Let $(M,g)$ be four-dimensional space-time and $\varphi :(M,g) \rightarrow (N,h)$ be a harmonic morphism into a Riemannian manifold, with one-dimensional timelike fibers tangent to the unit vector $U$. Let $\lambda$ denote its dilation. If $\mathrm Ric (X, U)=0$ and $\mathrm Ric (X, Y)= \nu g(X,Y)$ for all $X, Y$ orthogonal to $U$ and for some function $\nu : M \rightarrow \mathbb{R }$, then locally $\varphi$ is either of Killing type or of warped product type.

Proof

Let $(M,g)$ be a four-dimensional space-time and $\varphi :(M, g)\rightarrow (N,h)$ be a harmonic morphism into a Riemannian manifold.

Choose a local orthogonal frame $\{X, Y, Z\}$ of basic horizontal vector fields around a regular point $x$ in $M$; we may suppose that their lengths satisfy $\vert X \vert = \vert Y \vert = \vert Z \vert = 1/\lambda$ and that $Z$ satisfies $i_Z\Omega = 0$ ¹. Then, the contractions $i_X\Omega$ and $i_Y\Omega$ are both basic and orthogonal and $\vert i_X\Omega \vert ^2=\vert i_Y\Omega \vert ^2=\lambda ^2\Omega (X,Y)^2$.

Since according to the hypothesis, $\mathrm Ric \vert _{\mathcal{H }\times \mathcal{H }}$ must be proportional to $g^{\mathcal{H }}$, Eq. (10) applied to all pairs of vectors from $\{X, Y, Z\}$ gives us: $(\varphi ^* \mathrm Ric ^N) (X, Y) - 2 X (\ln \lambda ) Y(\ln \lambda )=0$, $(\varphi ^* \mathrm Ric ^N) (Y, Z) - 2 Y (\ln \lambda ) Z(\ln \lambda )=0$, and $(\varphi ^* \mathrm Ric ^N) (X, Z) - 2 X (\ln \lambda ) Z(\ln \lambda )=0$. In particular, $X(\ln \lambda )Y(\ln \lambda )$, $Y(\ln \lambda )Z(\ln \lambda )$, and $Z(\ln \lambda )X (\ln \lambda )$ are all basic functions. Let $W$ be the domain of the frame $\{X, Y, Z, V\}$. Consider the following closed subset in $W$:

$$\begin{aligned} S = \{x \in W : X_x(\ln \lambda ) = Y_x(\ln \lambda ) = 0\}. \end{aligned}$$

Denote its complement in $W$ by $S^c$—this will be an open subset of $M$.

Case I - on the open set $S^c$. Suppose that $X_x(\ln \lambda ) \ne 0$. By applying Equation (10) to the pair $\{X + Y, X - Y\}$, we deduce that $X(\ln \lambda )^2 - Y(\ln \lambda )^2$ is basic; moreover, since $X(\ln \lambda )Y(\ln \lambda )$ is basic too, $X (\ln \lambda )$ and $Y(\ln \lambda )$ have to be both basic. Then also $Z(\ln \lambda )$ is basic (since $Z(\ln \lambda )X (\ln \lambda )$ is basic). Hence $\mathrm grad (V(\ln \lambda ))$ is vertical.

Assume that at some point, $\mathrm grad (V(\ln \lambda )) \ne 0$ so by continuity this holds in a neighborhood.

• As $V(\ln \lambda )$ is non-constant and its level surfaces are horizontal, then $\mathcal{H }$ is integrable (so $\Omega =0$). Applying Equation (9) combined with the hypothesis $\mathrm Ric (X,U)=0$, we obtain that $X(U(\ln \lambda ))=0$, or equivalently $X(\ln \lambda )V(\ln \lambda )=0$. As we are on $S^c$, this would imply that $V(\ln \lambda )=0$, a contradiction.

• So $\mathrm grad (V(\ln \lambda )) = 0$ and then $V(\ln \lambda )=c$ a constant.

• Assume that $c \ne 0$.

• Replacing $V(\ln \lambda )=c$ in Equation (9), we obtain:

  $$\begin{aligned} 0= - 2c X(\ln \lambda ) - \frac{\lambda ^2}{2} \{ \delta \Omega (X) + 2\Omega (X, \mathrm grad \ln \lambda )\}. \end{aligned}$$

  (12)

• Notice that $\Omega (X, \mathrm grad \ln \lambda )\!=\!\lambda ^2 Y(\ln \lambda )\Omega (X, Y)$ and this implies $V(\lambda ^{-2}\Omega (X, \mathrm grad \ln \lambda ))=0$. A more elaborate computation leads to: So taking the derivative of (12) along $V$ will give

  $$\begin{aligned} 0= V(\lambda ^4) \lambda ^{-2}\{ \delta \Omega (X) + 2\Omega (X, \mathrm grad \ln \lambda )\}. \end{aligned}$$

  Therefore, the term inside the brackets must vanish; substituting this back into (12) gives us $X(\ln \lambda )=0$, contradiction.

Lemma 2

On $S^c$, we have $V\left(\lambda ^{-2}\delta \Omega (X)\right)=0$.

Therefore, the constant $c$ must be zero; this implies $U(\ln \lambda )=0$. We have proved that on $S^c$ (so on $\overline{S^c}$), $\varphi$ is of Killing type.

Case II: on the open set $\text{ int}S$. Let

$$\begin{aligned} A = \{x \in \mathrm int S : Z_x(\ln \lambda ) = 0\}. \end{aligned}$$

• $(II_a)$ On the subset $A$, we have $\mathrm grad ^{\mathcal{H }}\ln \lambda =0$, so $\mathrm grad \ln \lambda$ is vertical. Thus, around a point where $\lambda$ is non-constant, its level surfaces are horizontal so $\mathcal{H }$ is integrable and $\varphi$ is of warped product type (if $\lambda$ is constant, $\mathrm div U =0$ and $\varphi$ is of Killing type).

• $(II_b)$ On the open subset $A^c \cap \mathrm int S$, we have $X(\ln \lambda ) = 0$, $Y(\ln \lambda ) = 0$ and $Z(\ln \lambda ) \ne 0$. The identity (10) and $\text{ Ric}(X,X)=\mathrm Ric (Z,Z)$ (consequence of our hypothesis) imply

  $$\begin{aligned} 4 Z(\ln \lambda )^2 + \lambda ^4 \Omega (X,Y)^2 = 2\left((\varphi ^* \mathrm Ric ^N) (Z, Z)-(\varphi ^* \mathrm Ric ^N) (X, X)\right). \end{aligned}$$

Since the right-hand term is a basic function, the derivative along $V$ of the left-hand term must vanish. This gives us immediately, by using Lemma 7:

$$\begin{aligned} -\lambda ^2 \Omega (X,Y)^2 \dfrac{V(\ln \lambda )}{Z(\ln \lambda ) } = 2\lambda ^{-2}Z(V(\ln \lambda )). \end{aligned}$$

(13)

Combine this relation with the following

Lemma 3

On $A^c \cap \mathrm int S$, we have

$$\begin{aligned} \delta \Omega (Z)= -\lambda ^2 \Omega (X,Y)^2 \dfrac{V(\ln \lambda )}{Z(\ln \lambda ) }. \end{aligned}$$

(14)

to obtain

$$\begin{aligned} \delta \Omega (Z)= 2\lambda ^{-2}Z(V(\ln \lambda )). \end{aligned}$$

(15)

Now Equation (9) with $\mathrm Ric (Z,U)=0$ implies

$$\begin{aligned} 4[Z(V(\ln \lambda ))-Z(\ln \lambda )V(\ln \lambda )]-\lambda ^2 \delta \Omega (Z)=0. \end{aligned}$$

Substituting (14) in the above relation, we get

$$\begin{aligned} Z(V(\ln \lambda ))-2Z(\ln \lambda )V(\ln \lambda )=0. \end{aligned}$$

(16)

Reinserting (16) in the derived constraint (13), we obtain

$$\begin{aligned} V(\ln \lambda )\left[4Z(\ln \lambda )^2 + \lambda ^4 \Omega (X,Y)^2\right]=0. \end{aligned}$$

Since $Z(\ln \lambda )$ cannot vanish, we must have $V(\ln \lambda )=0$ so that $\varphi$ is of Killing type on the considered subset. $\square$

Taking Remark 1 into account and that coupling with gravity (11) is a special case of Theorem 1 hypothesis, we can state the following

Corollary 1

If the velocity vector field of a perfect fluid coupled with gravity satisfying $\rho =-3p$ is shear free, then either the expansion or the rotation of the fluid vanishes.

Notice that the condition $\mathrm Ric (U,U)=0$ from (3) was not employed.

Corollary 2

A harmonic morphism with one-dimensional timelike fibers from an Einstein space-time or from a vacuum space-time is either of Killing type or of warped product type.

The above Corollary is the Lorentzian analog of the result in [12] (see also [1, p. 380]). Note that the change of sign in the identity (10) simplifies radically the Case $(II_b)$ with respect to the Riemannian case so that type (T) morphisms are now excluded. Recall that type (T) harmonic morphisms allow both expansion and rotation (for instance on $\mathbb{R }^4$ with the Eguchi–Hanson-type metric [14]).

Remark 2

The relativistic fluids with $\rho = -3p$ (and harmonic morphisms) represent a genuine particular case of the shear-free conjecture. For an equation of state $\rho = w\, p$ with $w \ne -3$ (for $r$-harmonic morphisms with $r\ne 2$) in the Ricci identity (10), it will appear a second-order term in $\mathrm Hess (\ln \lambda )$, as we can check by performing a conformal change in metric to render the $r$-harmonic morphism a harmonic morphism. This will make inapplicable the entire argument used in the present case.

4 The Proofs of Lemmas

Proof of the Lemma 2

The basic ingredients are Lemma 1 and the following identity [1, p. 119]

$$\begin{aligned} \left(\nabla ^{\varphi }_{X}\text{ d}\varphi (Y) \right)^{\widehat{}} - \nabla _X^\mathcal{H } Y = X(\ln \lambda )Y + Y(\ln \lambda )X - g(X,Y) \mathrm grad ^\mathcal{H } \ln \lambda , \end{aligned}$$

where $\widehat{}$ stands for the horizontal lift operator. Notice that the first term in the above identity is a basic vector field. With this in hand, we can check by straightforward but lengthy calculation that

$$\begin{aligned} \delta \Omega (X)&= \lambda ^{2}\big [2Y(\ln \lambda )\Omega (X, Y)+Y(\Omega (X,Y)) + \Omega \left(\left(\nabla _Y^{\varphi }\text{ d}\varphi (Y)\right)^{\widehat{}}+ \left(\nabla _Z^{\varphi }\text{ d}\varphi (Z)\right)^{\widehat{}},X\right)\\&\quad +\Omega \left(Y, \left(\nabla _Y^{\varphi }\text{ d}\varphi (X)\right)^{\widehat{}}\,\,\right)\big ] \end{aligned}$$

and, since each term inside the brackets is basic, the conclusion follows. $\square$

Proof of the Lemma 3

An easy computation shows us that

$$\begin{aligned} \delta \Omega (Z)= -\lambda ^4 \Omega (X,Y) g([X,Y],Z). \end{aligned}$$

(17)

But $[X,Y](\ln \lambda )=0$ on $\mathrm int S$ implies

$$\begin{aligned}{}[X,Y]^{\mathcal{H }}(\ln \lambda ) - \Omega (X,Y) V(\ln \lambda )=0, \end{aligned}$$

where we have used $\Omega (X,Y)=\lambda ^{-2}g([X,Y],V)$.

As on $A^c \cap \mathrm int S$, we have $[X,Y]^{\mathcal{H }}(\ln \lambda )=\lambda ^{2}g([X,Y],Z)Z(\ln \lambda )$, it follows that

$$\begin{aligned} g([X,Y],Z)= \Omega (X,Y) \dfrac{V(\ln \lambda )}{\lambda ^2 Z(\ln \lambda ) }. \end{aligned}$$

Reinserting in (15) gives us the result. $\square$