Invertible disformal transformations with higher derivatives

We first review the case of the conventional disformal transformation that contains up to the first derivative of the scalar field to clarify the reason why it is possible to construct the inverse transformation in this case. Let us consider a class of metric transformations of the form (1), which we recapitulate here for convenience:

$$\bar{g}_{\mu\nu}=F_{0}(\phi,X)g_{\mu\nu}+F_{1}(\phi,X)\phi_{\mu}\phi_{\nu},$$

(3)

where we have introduced $\phi_{\mu}\coloneqq\partial_{\mu}\phi$ and then $X\coloneqq\phi_{\mu}\phi^{\mu}$. Note that the following results hold in general $D$-dimensional spacetime.

A remarkable feature of this class of transformations is that it is equipped with two binary operations and hence forms a group under each of the two operations. One of the two operations is the matrix product of two disformal metrics, while the other is the functional composition of two sequential disformal transformations. In what follows, we demonstrate that the class of conventional disformal transformations is indeed closed under the two operations mentioned above.

1. [A]

   Closedness under the matrix product. We consider two independent transformations of the form (3),

   $$\bar{g}_{\mu\nu}=F_{0}(\phi,X)g_{\mu\nu}+F_{1}(\phi,X)\phi_{\mu}\phi_{\nu},\qquad\tilde{g}_{\mu\nu}=f_{0}(\phi,X)g_{\mu\nu}+f_{1}(\phi,X)\phi_{\mu}\phi_{\nu}.$$

   (4)

   By contracting $\bar{g}_{\mu\nu}$ and $\tilde{g}_{\mu\nu}$ with the unbarred metric, one can construct another disformal metric, which we call the matrix product of two disformal metrics. Written explicitly, the matrix product is computed as

   $$g^{\alpha\beta}\bar{g}_{\mu\alpha}\tilde{g}_{\beta\nu}=F_{0}f_{0}g_{\mu\nu}+\left[(F_{0}+XF_{1})f_{1}+F_{1}f_{0}\right]\phi_{\mu}\phi_{\nu},$$

   (5)

   which is again of the form (3).

1. [B]

   Closedness under the functional composition. We consider two sequential transformations of the form (3),

   $$\bar{g}_{\mu\nu}=F_{0}(\phi,X)g_{\mu\nu}+F_{1}(\phi,X)\phi_{\mu}\phi_{\nu},\qquad\hat{g}_{\mu\nu}=\bar{F}_{0}(\phi,\bar{X})\bar{g}_{\mu\nu}+\bar{F}_{1}(\phi,\bar{X})\phi_{\mu}\phi_{\nu}.$$

   (8)

   Here, $\bar{X}$ denotes the kinetic term of the scalar field contracted by $\bar{g}^{\mu\nu}$, which is computed by use of (7) as

   $$\bar{X}\coloneqq\bar{g}^{\mu\nu}\phi_{\mu}\phi_{\nu}=\frac{X}{F_{0}+XF_{1}}.$$

   (9)

   Then, the composition of the two transformations is given by

   $$\hat{g}_{\mu\nu}=F_{0}(\phi,X)\bar{F}_{0}(\phi,\bar{X}(\phi,X))g_{\mu\nu}+\left[\bar{F}_{1}(\phi,\bar{X}(\phi,X))+F_{1}(\phi,X)\bar{F}_{0}(\phi,\bar{X}(\phi,X))\right]\phi_{\mu}\phi_{\nu},$$

   (10)

   which is again of the form (3).

The above analysis demonstrates that the two properties [A] and [B] play an essential role in the systematic construction of the inverse metric and the inverse transformation. Note that for the existence of the inverse transformation, the existence of the inverse metric is necessary. As we saw above, when computing the functional composition of the two disformal transformations in (8), one needs to express $\bar{X}$ in terms of the unbarred quantities, in which the inverse metric $\bar{g}^{\mu\nu}$ shows up. In the next subsection, we study a generalized disformal transformation with higher derivatives satisfying the above two properties and explicitly construct its inverse transformation.

Having clarified the reason why the conventional disformal transformation (3) is invertible, we now consider transformations with higher derivatives. The main difficulty here is that the higher covariant derivatives depend on the Christoffel symbol, i.e., the derivative of the metric. This generically spoils the property [B] since a functional composition of two transformations generically yields unwanted extra terms with higher derivatives that are not contained in the original transformation law. In order to make a transformation invertible, one has to tune it so that such extra terms do not show up. As a general example of transformations for which this tuning is possible, let us consider a metric transformation in $D$-dimensional spacetime defined by

$$\bar{g}_{\mu\nu}=F_{0}g_{\mu\nu}+F_{1}\phi_{\mu}\phi_{\nu}+2F_{2}\phi_{(\mu}X_{\nu)}+F_{3}X_{\mu}X_{\nu}.$$

(15)

Here, $X_{\mu}\coloneqq\partial_{\mu}X=2\phi_{\alpha}\phi^{\alpha}_{\mu}$ with $\phi_{\mu\nu}\coloneqq\nabla_{\mu}\nabla_{\nu}\phi$ and $T_{({\mu\nu})}\coloneqq(T_{\mu\nu}+T_{\nu\mu})/2$. Also, $F_{i}$’s ($i=0,1,2,3$) are functions of $(\phi,X,Y,Z)$, where $Y$ and $Z$ are defined as follows:

$$Y\coloneqq\phi_{\mu}X^{\mu},\qquad Z\coloneqq X_{\mu}X^{\mu}.$$

(16)

Note that the conventional disformal transformation (3) is included as a special case with $F_{0}=F_{0}(\phi,X)$, $F_{1}=F_{1}(\phi,X)$, and $F_{2}=F_{3}=0$. In what follows, we explicitly construct the inverse transformation for (15) with a particular focus on the properties [A] and [B].

We first examine the property [A], i.e., the closedness under the matrix product. To this end, we consider two independent transformations of the form (15),

$$\begin{split}\bar{g}_{\mu\nu}&=F_{0}g_{\mu\nu}+F_{1}\phi_{\mu}\phi_{\nu}+2F_{2}\phi_{(\mu}X_{\nu)}+F_{3}X_{\mu}X_{\nu},\\ \tilde{g}_{\mu\nu}&=f_{0}g_{\mu\nu}+f_{1}\phi_{\mu}\phi_{\nu}+2f_{2}\phi_{(\mu}X_{\nu)}+f_{3}X_{\mu}X_{\nu}.\end{split}$$

(17)

The matrix product of $\bar{g}_{\mu\nu}$ and $\tilde{g}_{\mu\nu}$ is calculated as follows:

	$\displaystyle g^{\alpha\beta}\bar{g}_{\mu\alpha}\tilde{g}_{\beta\nu}=\;$	$\displaystyle F_{0}f_{0}g_{\mu\nu}+\left[F_{0}f_{1}+F_{1}\left(f_{0}+Xf_{1}+Yf_{2}\right)+F_{2}\left(Yf_{1}+Zf_{2}\right)\right]\phi_{\mu}\phi_{\nu}$
		$\displaystyle+\left[F_{0}f_{2}+F_{1}\left(Xf_{2}+Yf_{3}\right)+F_{2}\left(f_{0}+Yf_{2}+Zf_{3}\right)\right]\phi_{\mu}X_{\nu}$
		$\displaystyle+\left[F_{0}f_{2}+F_{2}\left(f_{0}+Xf_{1}+Yf_{2}\right)+F_{3}\left(Yf_{1}+Zf_{2}\right)\right]X_{\mu}\phi_{\nu}$
		$\displaystyle+\left[F_{0}f_{3}+F_{2}\left(Xf_{2}+Yf_{3}\right)+F_{3}\left(f_{0}+Yf_{2}+Zf_{3}\right)\right]X_{\mu}X_{\nu},$		(18)

which is again of the form (15), and hence the property [A] is satisfied.^*3*3*3Precisely speaking, the right-hand side of (18) is not of the form (15) as it is not symmetric in $\mu$ and $\nu$ in general. Therefore, for the closedness under the matrix product, the underlying set of transformations should be enlarged to include such asymmetric terms. Nevertheless, as mentioned in the main text, the inverse element of (15) can be found in the symmetric subset. This is reminiscent of the fact that a product of two symmetric matrices is not necessarily symmetric, while the inverse of a symmetric matrix is symmetric. The inverse matrix for $\bar{g}_{\mu\nu}$ is obtained by putting $g^{\alpha\beta}\bar{g}_{\mu\alpha}\tilde{g}_{\beta\nu}=g_{\mu\nu}$ in (18). While this requirement apparently yields five equations for four unknown functions $f_{0},f_{1},f_{2},f_{3}$, only four of them are independent, and hence the system is not overdetermined. Thus, the coefficient functions in $\tilde{g}_{\mu\nu}$ are fixed as

$$\begin{split}&f_{0}=\frac{1}{F_{0}},\qquad f_{1}=-\frac{F_{0}F_{1}-Z(F_{2}^{2}-F_{1}F_{3})}{F_{0}\mathcal{F}},\\ &f_{2}=-\frac{F_{0}F_{2}+Y(F_{2}^{2}-F_{1}F_{3})}{F_{0}\mathcal{F}},\qquad f_{3}=-\frac{F_{0}F_{3}-X(F_{2}^{2}-F_{1}F_{3})}{F_{0}\mathcal{F}}.\end{split}$$

(19)

Here, we assumed $F_{0}\neq 0$ and defined the following quantity:

$$\mathcal{F}\coloneqq F_{0}^{2}+F_{0}(XF_{1}+2YF_{2}+ZF_{3})+(F_{2}^{2}-F_{1}F_{3})(Y^{2}-XZ),$$

(20)

which was also assumed to be nonvanishing. As a result, the inverse metric $\bar{g}^{\mu\nu}$ is given by

$$\bar{g}^{\mu\nu}=\frac{1}{F_{0}}\bigg{[}g^{\mu\nu}-\frac{F_{0}F_{1}-Z(F_{2}^{2}-F_{1}F_{3})}{\mathcal{F}}\phi^{\mu}\phi^{\nu}-2\frac{F_{0}F_{2}+Y(F_{2}^{2}-F_{1}F_{3})}{\mathcal{F}}\phi^{(\mu}X^{\nu)}-\frac{F_{0}F_{3}-X(F_{2}^{2}-F_{1}F_{3})}{\mathcal{F}}X^{\mu}X^{\nu}\bigg{]}.$$

(21)

Next, let us study under which conditions the transformation (15) can satisfy the property [B], i.e., the closedness under the functional composition. We consider sequential transformations with

$$\begin{split}\bar{g}_{\mu\nu}&=F_{0}g_{\mu\nu}+F_{1}\phi_{\mu}\phi_{\nu}+2F_{2}\phi_{(\mu}X_{\nu)}+F_{3}X_{\mu}X_{\nu},\\ \hat{g}_{\mu\nu}&=\bar{F}_{0}\bar{g}_{\mu\nu}+\bar{F}_{1}\phi_{\mu}\phi_{\nu}+2\bar{F}_{2}\phi_{(\mu}\bar{X}_{\nu)}+\bar{F}_{3}\bar{X}_{\mu}\bar{X}_{\nu},\end{split}$$

(22)

with $\bar{F}_{i}$’s ($i=0,1,2,3$) being functions of $(\phi,\bar{X},\bar{Y},\bar{Z})$. In the case of the first-order disformal transformations studied in §II.1, the invertibility is guaranteed if $X$ can be locally expressed in terms of $\bar{X}$. In the present case with higher derivatives, we have

$$\bar{X}=\bar{g}^{\mu\nu}\phi_{\mu}\phi_{\nu}=\frac{XF_{0}-F_{3}(Y^{2}-XZ)}{\mathcal{F}},$$

(23)

which is a function of $(\phi,X,Y,Z)$ in general. Let us consider to express $\hat{g}_{\mu\nu}$ as a functional of $\phi$ and $g_{\mu\nu}$. If $\bar{X}$ depends on $Y$ or $Z$ in a nontrivial manner, then the derivatives of $\bar{X}$ in $\hat{g}_{\mu\nu}$ yield derivatives of $Y$ or $Z$, which do not appear in the transformation law (15). On the other hand, so long as $\bar{X}$ has no dependence on either $Y$ or $Z$, then the composition of the two transformations is again of the form (15), meaning that the property [B] is satisfied. Therefore, we require

$$\bar{X}_{Y}=\bar{X}_{Z}=0,$$

(24)

where ${\bar{X}}_{Y}\coloneqq\partial{\bar{X}}/\partial Y$ and ${\bar{X}}_{Z}\coloneqq\partial{\bar{X}}/\partial Z$, so that $\bar{X}=\bar{X}(\phi,X)$. We also assume $\bar{X}_{X}\neq 0$ so that we can solve the relation $\bar{X}=\bar{X}(\phi,X)$ for $X$ to have $X=X(\phi,\bar{X})$. Then, we have $\bar{X}_{\mu}=\bar{X}_{X}X_{\mu}+\bar{X}_{\phi}\phi_{\mu}$ with ${\bar{X}}_{\phi}\coloneqq\partial{\bar{X}}/\partial\phi$, and hence

$$\begin{split}\bar{Y}&=\bar{g}^{\mu\nu}\phi_{\mu}\bar{X}_{\nu}=\bar{X}_{X}\frac{YF_{0}+F_{2}(Y^{2}-XZ)}{\mathcal{F}}+\bar{X}_{\phi}\bar{X},\\ \bar{Z}&=\bar{g}^{\mu\nu}\bar{X}_{\mu}\bar{X}_{\nu}=\bar{X}_{X}^{2}\frac{ZF_{0}-F_{1}(Y^{2}-XZ)}{\mathcal{F}}+2\bar{X}_{\phi}\bar{Y}-\bar{X}_{\phi}^{2}\bar{X}.\end{split}$$

(25)

Here, we require that these two equations can be solved for $Y$ and $Z$ to obtain $Y=Y(\phi,\bar{X},\bar{Y},\bar{Z})$ and $Z=Z(\phi,\bar{X},\bar{Y},\bar{Z})$, which is guaranteed if the Jacobian determinant $|\partial(\bar{Y},\bar{Z})/\partial(Y,Z)|$ is nonvanishing.

We are now ready to write down the expression for the inverse transformation for $\bar{g}_{\mu\nu}=\bar{g}_{\mu\nu}[g_{\alpha\beta},\phi]$. With the requirement $\bar{X}=\bar{X}(\phi,X)$, we can express $\hat{g}_{\mu\nu}$ in terms of the unbarred quantities as

$$\hat{g}_{\mu\nu}=F_{0}\bar{F}_{0}g_{\mu\nu}+(\bar{F}_{1}+F_{1}\bar{F}_{0}+2\bar{X}_{\phi}\bar{F}_{2}+\bar{X}_{\phi}^{2}\bar{F}_{3})\phi_{\mu}\phi_{\nu}+2(\bar{X}_{X}\bar{F}_{2}+F_{2}\bar{F}_{0}+\bar{X}_{\phi}\bar{X}_{X}\bar{F}_{3})\phi_{(\mu}X_{\nu)}+(\bar{X}_{X}^{2}\bar{F}_{3}+F_{3}\bar{F}_{0})X_{\mu}X_{\nu},$$

(26)

where the functions of $(\phi,\bar{X},\bar{Y},\bar{Z})$ in the right-hand side are regarded as functions of $(\phi,X,Y,Z)$ by (23) and (25). The inverse transformation can be obtained by putting $\hat{g}_{\mu\nu}=g_{\mu\nu}$, which fixes the coefficient functions in $\hat{g}_{\mu\nu}$ as

$$\bar{F}_{0}=\frac{1}{F_{0}},\qquad\bar{F}_{1}=-\frac{\bar{X}_{X}^{2}F_{1}-2\bar{X}_{\phi}\bar{X}_{X}F_{2}+\bar{X}_{\phi}^{2}F_{3}}{\bar{X}_{X}^{2}F_{0}},\qquad\bar{F}_{2}=-\frac{\bar{X}_{X}F_{2}-\bar{X}_{\phi}F_{3}}{\bar{X}_{X}^{2}F_{0}},\qquad\bar{F}_{3}=-\frac{F_{3}}{\bar{X}_{X}^{2}F_{0}}.$$

(27)

Thus, we have obtained the inverse transformation in the following form:

$$g_{\mu\nu}=\frac{1}{F_{0}}\left(\bar{g}_{\mu\nu}-\frac{\bar{X}_{X}^{2}F_{1}-2\bar{X}_{\phi}\bar{X}_{X}F_{2}+\bar{X}_{\phi}^{2}F_{3}}{\bar{X}_{X}^{2}}\phi_{\mu}\phi_{\nu}-2\frac{\bar{X}_{X}F_{2}-\bar{X}_{\phi}F_{3}}{\bar{X}_{X}^{2}}\phi_{(\mu}\bar{X}_{\nu)}-\frac{F_{3}}{\bar{X}_{X}^{2}}\bar{X}_{\mu}\bar{X}_{\nu}\right),$$

(28)

where the functions of $(\phi,X,Y,Z)$ in the right-hand side can be translated back into functions of $(\phi,\bar{X},\bar{Y},\bar{Z})$ by use of (23) and (25).^*4*4*4The above analysis shows that the transformation $\bar{D}\colon(\bar{g}_{\mu\nu},\phi)\mapsto(g_{\mu\nu},\phi)$ defined by (28) is the left inverse of the transformation $D\colon(g_{\mu\nu},\phi)\mapsto(\bar{g}_{\mu\nu},\phi)$ defined by (15), i.e., $\bar{D}\circ D(g_{\mu\nu},\phi)=(g_{\mu\nu},\phi)$. Since the left inverse of a group element is also its right inverse, it follows that $D\circ\bar{D}(\bar{g}_{\mu\nu},\phi)=(\bar{g}_{\mu\nu},\phi)$.

To summarize, we have obtained a set of sufficient conditions for the generalized disformal transformation (15) to be invertible. The conditions are summarized as

$$F_{0}\neq 0,\qquad\mathcal{F}\neq 0,\qquad\bar{X}_{X}\neq 0,\qquad\bar{X}_{Y}=\bar{X}_{Z}=0,\qquad\left|\frac{\partial(\bar{Y},\bar{Z})}{\partial(Y,Z)}\right|\neq 0.$$

(29)

This set of conditions can be used not only as a simple criterion for the invertibility of a given transformation of the form (15) but also as a useful tool to construct invertible generalized disformal transformations as we shall see below. In order for the condition $\bar{X}_{Y}=\bar{X}_{Z}=0$ and $\bar{X}_{X}\neq 0$ to be satisfied, let us take $\bar{X}=\bar{X}_{0}(\phi,X)$ as an input, with $\bar{X}_{0}$ being an arbitrary function of $(\phi,X)$ such that $\bar{X}_{0X}\neq 0$. Then, by use of (23), e.g., the function $F_{3}$ is written in terms of $\bar{X}_{0}$, $F_{0}$, $F_{1}$, and $F_{2}$ as

$$F_{3}=\frac{XF_{0}-\bar{X}_{0}(\phi,X)\left[F_{0}(F_{0}+XF_{1}+2YF_{2})+F_{2}^{2}(Y^{2}-XZ)\right]}{Y^{2}-XZ+\bar{X}_{0}(\phi,X)\left[ZF_{0}-F_{1}(Y^{2}-XZ)\right]}.$$

(30)

Therefore, we obtain invertible transformations by choosing the functions $\bar{X}_{0}$, $F_{0}$, $F_{1}$, and $F_{2}$ so that they satisfy the remaining conditions in (29), i.e., $F_{0}\neq 0$, $\mathcal{F}\neq 0$, and $|\partial(\bar{Y},\bar{Z})/\partial(Y,Z)|\neq 0$. In particular, for the above $F_{3}$, the condition $\mathcal{F}\neq 0$ yields

$$\mathcal{F}\propto\left[YF_{0}+F_{2}(Y^{2}-XZ)\right]^{2}\neq 0.$$

(31)

We shall use this strategy to construct a nontrivial example of invertible transformations of the form (15) in §III.1.

A caveat here is that the transformation law could be ill defined for some particular configuration of $(g_{\mu\nu},\phi)$. Nevertheless, it is still possible to perform the invertible disformal transformation $\bar{g}_{\mu\nu}=\bar{g}_{\mu\nu}[g_{\alpha\beta},\phi]$ on some seed action of scalar-tensor theories $\bar{S}[\bar{g}_{\mu\nu},\phi]$ to generate a new action $S[g_{\mu\nu},\phi]$. For instance, provided that $\bar{X}_{0}$, $F_{0}$, and $F_{1}$ are regular functions, the denominator of (30) vanishes for configurations with $Y=Z=0$, which happens when $X={\rm const}$. This means that, even if the new action $S[g_{\mu\nu},\phi]$ admits a solution with $X={\rm const}$, one cannot map the solution via the disformal transformation to generate a solution in the original frame. On the other hand, so long as we consider configurations for which the transformation law is well defined, there is one-to-one correspondence between the configuration space in the new frame and the one in the original frame.

One may think that arbitrary tensors of the form $\Phi_{\mu\nu}^{n}\coloneqq\phi_{\mu}^{\alpha_{1}}\phi_{\alpha_{1}}^{\alpha_{2}}\cdots\phi_{\alpha_{n-1}\nu}$ (e.g., $\Phi_{\mu\nu}^{1}=\phi_{\mu\nu}$ and $\Phi_{\mu\nu}^{2}=\phi_{\mu}^{\alpha}\phi_{\alpha\nu}$) and/or scalar quantities constructed from $g_{\mu\nu}$, $\phi_{\mu}$, and $\Phi_{\mu\nu}^{n}$ (e.g., $\Box\phi$ and $\phi_{\alpha}^{\beta}\phi_{\beta}^{\alpha}$) can be included in the transformation law (15). For instance, one could consider transformations of the form (2), in which a term with $\phi_{\mu\nu}$ is present. In this case, one can make use of the Cayley-Hamilton theorem, which allows us to write any $\Phi_{\mu\nu}^{n}$ with $n\geq D$ in terms of $g_{\mu\nu},\phi_{\mu\nu},\Phi_{\mu\nu}^{2},\cdots,\Phi_{\mu\nu}^{D-1}$. Therefore, considering a transformation composed of $g_{\mu\nu},\phi_{\mu\nu},\Phi_{\mu\nu}^{2},\cdots,\Phi_{\mu\nu}^{D-1}$, the property [A] may be satisfied, which allows us to systematically construct the inverse metric. However, as mentioned earlier, a composition of such transformations generates various terms with the third-order derivative of the scalar field as well as the second-order derivative of the metric through the Christoffel symbol [see (51) and (54)], which are not contained in the original transformation law. Hence, it is practically difficult to remove all such terms, implying that the property [B] cannot be satisfied in general. This explains why transformations of the form (2) are noninvertible, as shown in Babichev et al. (2021). The reason why we could obtain a concise invertibility condition for the transformation (15) is that there is only a single function $\bar{X}(\phi,X,Y,Z)$ that controls whether or not the class of transformations is closed under the functional composition. The point is that, so long as the conditions in (29) are satisfied, the Christoffel symbols are encapsulated in two sets of scalar quantities $(Y,Z)$ and $(\bar{Y},\bar{Z})$, between which the invertibility is manifest.

As a final remark, the above discussion can be extended to more general transformations containing the third derivative of $\phi$,

	$\displaystyle\bar{g}_{\mu\nu}=\;$	$\displaystyle F_{0}g_{\mu\nu}+F_{1}\phi_{\mu}\phi_{\nu}+2F_{2}\phi_{(\mu}X_{\nu)}+F_{3}X_{\mu}X_{\nu}+2F_{4}\phi_{(\mu}Y_{\nu)}+2F_{5}\phi_{(\mu}Z_{\nu)}+2F_{6}X_{(\mu}Y_{\nu)}+2F_{7}X_{(\mu}Z_{\nu)}$
		$\displaystyle+F_{8}Y_{\mu}Y_{\nu}+2F_{9}Y_{(\mu}Z_{\nu)}+F_{10}Z_{\mu}Z_{\nu},$		(32)

where $Y_{\mu}\coloneqq\partial_{\mu}Y$, $Z_{\mu}\coloneqq\partial_{\mu}Z$, and here

$$F_{i}=F_{i}(\phi,X,Y,Z,\phi_{\mu}Y^{\mu},\phi_{\mu}Z^{\mu},X_{\mu}Y^{\mu},X_{\mu}Z^{\mu},Y_{\mu}Y^{\mu},Y_{\mu}Z^{\mu},Z_{\mu}Z^{\mu}).$$

(33)

Likewise, it is straightforward to include arbitrarily higher-order derivatives of $\phi$ in the transformation. Rather than presenting a general discussion, we shall provide an example of invertible disformal transformations with the third derivative of $\phi$ in §III.2.

As an example of invertible disformal transformations of the form (15), let us consider the case with $F_{0}=1$, $F_{1}=F_{2}=0$, and $F_{3}=F_{3}(\phi,X,Y,Z)\neq 0$, i.e.,

$$\bar{g}_{\mu\nu}=g_{\mu\nu}+F_{3}(\phi,X,Y,Z)X_{\mu}X_{\nu}.$$

(34)

In this case, we have

$$\bar{X}(\phi,X,Y,Z)=\frac{X-F_{3}(Y^{2}-XZ)}{1+ZF_{3}},$$

(35)

which we require to be a function only of $(\phi,X)$. Assuming that $\bar{X}=X+P(\phi,X)$ with $P\neq 0$ and $P_{X}\neq-1$, from (30) we have

$$F_{3}=-\frac{P(\phi,X)}{Y^{2}+ZP(\phi,X)},$$

(36)

for which the transformation law of the metric is explicitly written as

$$\bar{g}_{\mu\nu}=g_{\mu\nu}-\frac{P(\phi,X)}{Y^{2}+ZP(\phi,X)}X_{\mu}X_{\nu}.$$

(37)

Then, the inverse metric is given by

$$\bar{g}^{\mu\nu}=g^{\mu\nu}+\frac{P(\phi,X)}{Y^{2}}X^{\mu}X^{\nu}.$$

(38)

We also have

$$\begin{split}\bar{Y}&=(1+P_{X})\left(Y+\frac{ZP}{Y}\right)+P_{\phi}(X+P),\\ \bar{Z}&=(1+P_{X})\left(Y+\frac{ZP}{Y}\right)\left[\frac{Z}{Y}(1+P_{X})+2P_{\phi}\right]+P_{\phi}^{2}(X+P),\end{split}$$

(39)

which can be inverted as

$$\begin{split}Y&=\frac{\bar{Y}^{2}-\bar{Z}P-P_{\phi}(\bar{X}-P)(2\bar{Y}-\bar{X}P_{\phi})}{(1+P_{X})(\bar{Y}-\bar{X}P_{\phi})},\\ Z&=\frac{\bar{Z}-P_{\phi}(2\bar{Y}-\bar{X}P_{\phi})}{(1+P_{X})(\bar{Y}-\bar{X}P_{\phi})}Y.\end{split}$$

(40)

The inverse disformal transformation takes the form,

$$g_{\mu\nu}=\bar{g}_{\mu\nu}+\frac{P}{\bar{Y}^{2}-\bar{Z}P-P_{\phi}(\bar{X}-P)(2\bar{Y}-\bar{X}P_{\phi})}\left(P^{2}_{\phi}\phi_{\mu}\phi_{\nu}-2P_{\phi}\phi_{(\mu}\bar{X}_{\nu)}+\bar{X}_{\mu}\bar{X}_{\nu}\right),$$

(41)

where $X$ in the arguments of $P$ and $P_{\phi}$ is regarded as a function of $(\phi,\bar{X})$ by solving $\bar{X}=X+P(\phi,X)$ for $X$. Note that, while the transformation (37) does not contain either $\phi_{\mu}\phi_{\nu}$ or $\phi_{(\mu}X_{\nu)}$, in general these terms show up in the inverse transformation (41). If $P=P(X)$, such terms vanish in (41). The simplest case would be $P(\phi,X)=c_{0}$ with $c_{0}$ being a nonvanishing constant, for which the disformal transformation (37) and its inverse (41) are explicitly written as

$$\bar{g}_{\mu\nu}=g_{\mu\nu}-\frac{c_{0}}{Y^{2}+c_{0}Z}X_{\mu}X_{\nu},\qquad g_{\mu\nu}=\bar{g}_{\mu\nu}+\frac{c_{0}}{\bar{Y}^{2}-c_{0}\bar{Z}}\bar{X}_{\mu}\bar{X}_{\nu}.$$

(42)

Let us now consider another example of invertible transformations of the following form:

$$\bar{g}_{\mu\nu}=\frac{V^{2}X-U^{2}Z}{V^{2}(X+c_{1})-U^{2}Z}\left[g_{\mu\nu}+\frac{c_{1}Z}{V^{2}X-U^{2}Z+c_{1}(V^{2}-WZ)}Z_{\mu}Z_{\nu}\right],$$

(43)

where $c_{1}$ is a nonvanishing constant and we have defined

$$U\coloneqq\phi_{\mu}Z^{\mu},\qquad V\coloneqq X_{\mu}Z^{\mu},\qquad W\coloneqq Z_{\mu}Z^{\mu}.$$

(44)

This transformation is of the form (32) and the third derivative of $\phi$ appears in $Z_{\mu}=8\phi_{\alpha}\phi^{\alpha}_{\beta}\nabla_{\mu}(\phi^{\gamma}\phi^{\beta}_{\gamma})$. The inverse metric takes the form

$$\bar{g}^{\mu\nu}=\frac{V^{2}(X+c_{1})-U^{2}Z}{V^{2}X-U^{2}Z}\left[g^{\mu\nu}-\frac{c_{1}Z}{V^{2}(X+c_{1})-U^{2}Z}Z^{\mu}Z^{\nu}\right].$$

(45)

Then, the relevant scalar quantities transform as follows:

$$\bar{X}=X+c_{1},\qquad\bar{Z}=Z,\qquad\frac{\bar{U}}{U}=\frac{\bar{V}}{V}=\frac{\bar{W}}{W}=1+\frac{c_{1}(V^{2}-WZ)}{V^{2}X-U^{2}Z}.$$

(46)

Note that the property [B] is guaranteed by $\bar{Z}_{U}=\bar{Z}_{V}=\bar{Z}_{W}=0$, which is a natural generalization of the condition $\bar{X}_{Y}=\bar{X}_{Z}=0$ in (29).^*5*5*5It should also be noted that $\bar{Y}$ takes the form, $$\bar{Y}=Y+\frac{c_{1}V(VY-UZ)}{V^{2}X-U^{2}Z},$$ and hence has a nontrivial dependence on $U$ and $V$, but this does not spoil the invertibility as the transformation (43) is independent of $Y$. On the other hand, if the transformation law had a nontrivial $Y$ dependence, then $\bar{Y}$ should satisfy $\bar{Y}_{U}=\bar{Y}_{V}=\bar{Y}_{W}=0$. The relations in (46) can be solved for the unbarred quantities as

$$X=\bar{X}-c_{1},\qquad Z=\bar{Z},\qquad\frac{U}{\bar{U}}=\frac{V}{\bar{V}}=\frac{W}{\bar{W}}=1-\frac{c_{1}(\bar{V}^{2}-\bar{W}\bar{Z})}{\bar{V}^{2}\bar{X}-\bar{U}^{2}\bar{Z}}.$$

(47)

Hence, the inverse transformation is obtained as follows:

$$g_{\mu\nu}=\frac{\bar{V}^{2}\bar{X}-\bar{U}^{2}\bar{Z}}{\bar{V}^{2}(\bar{X}-c_{1})-\bar{U}^{2}\bar{Z}}\left[\bar{g}_{\mu\nu}-\frac{c_{1}\bar{Z}}{\bar{V}^{2}\bar{X}-\bar{U}^{2}\bar{Z}-c_{1}(\bar{V}^{2}-\bar{W}\bar{Z})}\bar{Z}_{\mu}\bar{Z}_{\nu}\right].$$

(48)