Canonical variational completion and 4D Gauss–Bonnet gravity

Abstract

Recently, a proposal to obtain a finite contribution of second derivative order to the gravitational field equations in $D = 4$ dimensions from a renormalized Gauss–Bonnet term in the action has received a wave of attention. It triggered a discussion whether the employed renormalization procedure yields a well-defined theory. One of the main criticisms is based on the fact that the resulting field equations cannot be obtained as the Euler–Lagrange equations from a diffeomorphism invariant action. In this work, we use techniques from the inverse calculus of variations to point out that the renormalized truncated Gauss–Bonnet equations cannot be obtained from any action at all (either diffeomorphism invariant or not), in any dimension. Then, we employ canonical variational completion, based on the notion of Vainberg–Tonti Lagrangian—which consists in adding a canonically defined correction term to a given system of equations, so as to make them derivable from an action. To apply this technique to the suggested 4D renormalized Gauss–Bonnet equations, we extend the variational completion algorithm to some classes of PDE systems for which the usual integral providing the Vainberg–Tonti Lagrangian diverges. We discover that in $D>4$ the suggested field equations can be variationally completed, choosing either the metric or its inverse as field variables; both approaches yield consistently the same Lagrangian, whose variation leads to fourth-order field equations. In $D=4$, the Lagrangian of the variationally completed theory diverges in both cases.

Appendices

Properties of the field equations from the $A^\mu {}_\mu$ term

We claimed in item 3 of Sect. 4 that the variationally completed field equations of the truncated Einstein–Gauss Bonnet field equations (20) contain higher than second-order derivatives in any dimension. This is true due to the following line of argument.

The trace of $A_{\mu \nu }$ contains non-trivial terms which are quadratic in the second derivatives of the metric and do not factor in a way that Bianchi identities cancel these terms. Hence, also $A_{\mu \nu }$ itself contains such terms. This can be explicitly realized by introducing a counting parameter $\epsilon$ and replacing every term $\partial _\mu \partial _\nu g_{\rho \sigma }$ by $\epsilon \partial _\mu \partial _\nu g_{\rho \sigma }$. Doing so, we can express $A^\mu {}_{\mu }$ as a polynomial in $\epsilon$ and find, with help of the computer algebra program xAct for Mathematica [98],

$$\begin{aligned} A^\mu {}_\mu&= \epsilon ^2 \frac{D-2}{2 (D-1)} g^{\mu \sigma } g^{\lambda \zeta } g^{\rho \omega } g^{\tau \nu } \nonumber \\&\quad \bigg ( D ( - \partial _{\zeta }\partial _{\sigma }g_{\mu \lambda } + 2 \partial _{\zeta }\partial _{\lambda }g_{\mu \sigma } ) \partial _{\nu }\partial _{\omega }g_{\rho \tau } \nonumber \\&\quad +\, ( D - 1)\partial _{\omega }\partial _{\rho }g_{\mu \lambda } \partial _{\nu }\partial _{\tau }g_{\sigma \zeta } - D \partial _{\zeta }\partial _{\lambda }g_{\mu \sigma } \partial _{\nu }\partial _{\tau }g_{\rho \omega } \nonumber \\&\quad +\, ( D - 1) \partial _{\rho }\partial _{\lambda }g_{\mu \sigma } (\partial _{\omega }\partial _{\zeta }g_{\tau \nu } - 4 \partial _{\nu }\partial _{\omega }g_{\zeta \tau } + 2 \partial _{\nu }\partial _{\tau }g_{\zeta \omega })\nonumber \\&\quad +\, 2 ( D - 1) \partial _{\rho }\partial _{\sigma }g_{\mu \lambda } (\partial _{\nu }\partial _{\omega }g_{\zeta \tau } - 2 \partial _{\nu }\partial _{\tau }g_{\zeta \omega } + \partial _{\nu }\partial _{\zeta }g_{\omega \tau })\bigg ) \nonumber \\&\quad +\, {\text {lower order terms in }} \epsilon . \end{aligned}$$

(35)

Hence, also the untraced tensor $A_{\mu \nu }$, which is part of the truncated field equations, must contain terms of the form $\partial _\mu \partial _\nu g_{\rho \sigma } \partial _\lambda \partial _\tau g_{\zeta \omega }$.

But, any variational PDE system which is of second order must be linear in the second-order derivatives acting on the fundamental dynamical variable [96, p. 147]. Hence, the truncated field equations cannot be variational and the variation of $A^\mu {}_\mu$ cannot be of second order only, but must contain higher derivatives.

Necessity of densitysing in variational completion

In Sect. 4, we applied the variational completion algorithm to the original and to the truncated Einstein Gauss–Bonnet gravity field equations in any dimension.

An important first step in applying the algorithm was to define the densitized field equations in Eq. (21). In the following, we are going to prove that, if the expressions ${\mathcal {E}}^{\mu \nu }=-\frac{1}{2}E^{\mu \nu }\sqrt{-g}$ are the Euler–Lagrange expressions of a Lagrangian $\lambda ={\mathcal {L}}\mathrm {d}^{n}x,$ then the expressions $E^{\mu \nu }$ cannot arise as the Euler–Lagrange expressions of any Lagrangian (either coordinate-invariant or not).

Mathematically more precise, variationality is generally discussed for certain differential forms ${\mathcal {E}}$ on a jet bundle of a fibered manifold, rather than for PDE’s. These differential forms are called source forms and their local coefficients ${\mathcal {E}}_A$ are the left-hand sides of the given PDE’s. Multiplying a PDE system by a positive factor (such as $\sqrt{-g}$) will inevitably lead to a different source form; thus, this factor does not affect the set of solutions of them PDE system, but does affect its variationality.

To fix the notation, let $(Y\overset{\pi }{\rightarrow }M,F)$ be a fiber bundle over M, with a local coordinate system $(x^{\mu },y^{A})$ adapted to the fibration. Sections (physically interpreted as fields) are maps $\gamma :U\rightarrow Y$ (where $U\subset M$ is open), locally described as $\gamma :(x^{\mu })\mapsto (y^{A}(x^{\mu })).$ On the second-order jet bundle $J^{2}Y$, we denote the induced coordinates by $(x^{\mu },y^{A},y_{~\mu }^{A},y_{~\mu \nu }^{A}).$ On the jet bundle $J^{2}Y,$ the quantities $x^{\mu },y^{A},y_{~\mu }^{A},y_{~\mu \nu }^{A}$ are interpreted as coordinate functions (i.e., they are independent of one another); only when composed by (prolonged) sections, they provide the functions $(y^{A}(x^{\mu }))$ and their derivatives.

In [96, p. 147], it was shown that for a second order PDE system ${\mathcal {E}} _{A}={\mathcal {E}}_{A}(x^{\mu },y^{B},y_{~\mu }^{B},y_{~\mu \nu }^{B}),$ local variationality implies that the following Helmholtz conditions are identically satisfied by ${\mathcal {E}} _{A}$:

$$\begin{aligned} H_{AB}^{\mu \nu }({\mathcal {E}} )&:= \dfrac{\partial {\mathcal {E}} _{A}}{ \partial y_{~\mu \nu }^{B}}-\dfrac{\partial {\mathcal {E}} _{B}}{\partial y_{~\mu \nu }^{A}}=0 \end{aligned}$$

(36)

$$\begin{aligned} H_{~AB}^{\nu }({\mathcal {E}} )&:= \dfrac{\partial {\mathcal {E}} _{A}}{\partial y_{~\nu }^{B}}+\dfrac{\partial {\mathcal {E}} _{B}}{\partial y_{~\nu }^{A}} -\mathrm {d}_{\mu }\left( \dfrac{\partial {\mathcal {E}} _{A}}{\partial y_{~\mu \nu }^{B}}+ \dfrac{\partial {\mathcal {E}} _{B}}{\partial y_{~\mu \nu }^{A}}\right) =0 \end{aligned}$$

(37)

$$\begin{aligned} H_{AB}({\mathcal {E}} )&:= \dfrac{\partial {\mathcal {E}} _{A}}{\partial y^{B}}- \dfrac{\partial {\mathcal {E}} _{B}}{\partial y^{A}}-\dfrac{1}{2}\mathrm {d}_{\nu }\left( \dfrac{\partial {\mathcal {E}} _{A}}{\partial y_{~\nu }^{B}}-\dfrac{\partial {\mathcal {E}} _{B}}{\partial y_{~\nu }^{A}}\right) =0 \end{aligned}$$

(38)

Here, $\mathrm {d}_{\mu }=\partial _{\mu }+y_{~\mu }^{A}\dfrac{\partial }{\partial y^{A}}+y_{~\mu \nu }^{A}\dfrac{\partial }{\partial y_{~\nu }^{A}}+y_{~\mu \nu \rho }^{A}\dfrac{\partial }{\partial y_{~\nu \rho }^{A}}$ is the total derivative operator (of order three) acting on functions $f:J^{2}Y\rightarrow {\mathbb {R}},$ $f=f(x^{\mu },y^{B},y_{~\mu }^{B},y_{~\mu \nu }^{B})$. In particular, $\mathrm {d}_{\mu }y^{A}=y_{~\mu }^{A}.$

Now, let us assume that ${\mathcal {E}} _{A}$ satisfies the Helmholtz conditions. Multiplying ${\mathcal {E}} _{A}$ by a factor $f=f(x^{\mu },y^{B})$, we obtain a new source form $f{\mathcal {E}},$ with local coefficients $f{\mathcal {E}}_{A}.$

The first Helmholtz condition (36) is, indeed, not affected by the rescaling. But, substituting $f{\mathcal {E}} _{A}$ instead of ${\mathcal {E}} _{A}$ into (37) gives:

$$\begin{aligned} H_{~AB}^{\nu }(f{\mathcal {E}} ):=fH_{~AB}^{\nu }({\mathcal {E}} )-(\mathrm {d}_{\mu }f)\left( \dfrac{\partial {\mathcal {E}} _{A}}{\partial y_{~\mu \nu }^{B}}+\dfrac{ \partial {\mathcal {E}} _{B}}{\partial y_{~\mu \nu }^{A}}\right) . \end{aligned}$$

The term $fH_{~AB}^{\nu }({\mathcal {E}} )$ vanishes by the variationality assumption on ${\mathcal {E}} _{A};$ using (36) in the remaining term, we get $\dfrac{\partial {\mathcal {E}} _{A}}{\partial y_{~\mu \nu }^{B}}+\dfrac{ \partial {\mathcal {E}} _{B}}{\partial y_{~\mu \nu }^{A}}=2\dfrac{\partial {\mathcal {E}} _{A}}{\partial y_{~\mu \nu }^{B}}$ and therefore:

$$\begin{aligned} H_{~AB}^{\nu }(f{\mathcal {E}} )=-2(\mathrm {d}_{\mu }f)\dfrac{\partial {\mathcal {E}} _{A}}{ \partial y_{~\mu \nu }^{B}}. \end{aligned}$$

(39)

In order to simplify calculations, we will contract the above equality by $y^{A}$, thus getting:

$$\begin{aligned} y^{A} H_{~AB}^{\nu }(f{\mathcal {E}} )=-2(\mathrm {d}_{\mu }f)\dfrac{\partial (y^{A}{\mathcal {E}} _{A})}{ \partial y_{~\mu \nu }^{B}}. \end{aligned}$$

(40)

A necessary (but not sufficient) condition for the variationality of the source form $f {\mathcal {E}}$ is then that the contracted expressions (40) vanish.

Now, in order to check the above condition for the expressions ${\mathcal {E}}^{\mu \nu }$ studied in Sect. 4, we will make the following substitutions: ${\mathcal {E}}_{A} \rightarrow {\mathcal {E}}^{\mu \nu },$ $y^{A}\rightarrow g_{\mu \nu },$ $f\rightarrow \dfrac{1}{\sqrt{-g}}$.

These are functions on the jet bundle $J^{2}{{\,\mathrm{Met}\,}}(M)$, where ${{\,\mathrm{Met}\,}}(M)$ is the fiber bundle of symmetric and nondegenerate tensors of type (0, 2) over the spacetime manifold M, [96, p. 172]. On this bundle, a system of fibered coordinate functions has the form $(x^{\mu };g_{\mu \nu };g_{\mu \nu ,\rho };g_{\mu \nu ,\rho \tau })$.

A brief direct computation, using $\dfrac{\partial g}{\partial g_{\nu \rho }}=g^{\nu \rho }g$, gives:

$$\begin{aligned} \mathrm {d}_{\mu }f= \dfrac{1}{2}(-g)^{-1/2}g^{\nu \rho }g_{\nu \rho ,\mu }= \dfrac{1}{ \sqrt{-g}}\varGamma _{~\mu \nu }^{\nu }, \end{aligned}$$

where the $\varGamma _{~\mu \nu }^{\nu }$ are formal Christoffel symbols, i.e., in their expressions, $x^{\mu },$ $g_{\mu \nu }$ and $g_{\mu \nu ,\rho }$ are all regarded as independent variables (it is only along given sections that we can state that $g_{\mu \nu }=g_{\mu \nu }(x^{\rho })$). In particular, we cannot tune the coordinates $x^{\mu }$ in such a way as to have $\varGamma _{~\mu \nu }^{\nu }=0$ even at a single point (let alone having this equality identically satisfied).

The second factor $\dfrac{\partial (y^{A}{\mathcal {E}} _{A})}{\partial y_{~\mu \nu }^{B}}$ in (40) becomes, in our case: $\dfrac{\partial (g_{\alpha \beta }{\mathcal {E}}^{\alpha \beta })}{\partial g_{\gamma \delta ,\mu \nu }}$. Using (10) and (28), we find:

$$\begin{aligned} g_{\alpha \beta }{\mathcal {E}}^{\alpha \beta } = \left[ M_{\text {P}}^2\left( 1-\dfrac{D}{2}\right) R+\varLambda _0 D- \dfrac{\alpha }{2}{\mathcal {G}}\right] \sqrt{-g}. \end{aligned}$$

(41)

The expressions (40) are then of the form:

$$\begin{aligned} 2M_{\text {P}}^2\left( 1-\dfrac{D}{2}\right) \varGamma _{~\mu \tau }^{\tau }\dfrac{\partial R}{\partial g_{\gamma \delta ,\mu \nu }}+ \cdots , \end{aligned}$$

(42)

where the dots stand for terms which, after differentiation, will still contain curvature components; but, using the identity: $\dfrac{\partial R}{\partial g_{\gamma \delta ,\mu \nu }}=g^{\gamma \nu }g^{\delta \mu }-g^{\mu \nu }g^{\gamma \delta }$, the explicitly listed term above is, up to multiplication by a constant: $(\varGamma ^{\tau \delta }{}_\tau g^{\gamma \nu }-\varGamma ^{\tau \nu }{}_\tau g^{\gamma \delta })\not =0$.

Therefore, there is no chance that the full Helmholtz expressions $H^{\nu (\alpha \beta )(\gamma \delta )}(f{\mathcal {E}} )$ (which also involve nontrivial curvature terms) would identically vanish, which means that the functions $E^{\mu \nu }=-\dfrac{2}{\sqrt{ -g}}{\mathcal {E}}^{\mu \nu }$ cannot be the Euler–Lagrange expressions of any Lagrangian (either coordinate invariant or not).

Extending the Vainberg–Tonti Lagrangian

Here, we prove Theorem 1 of Sect. 3.2, which extends the definition of the Vainberg–Tonti Lagrangian to cases when the domain of definition of the functions ${\mathcal {E}}_{A}$ is not vertically star-shaped with center 0.

With the notations in the previous appendix, we consider arbitrary second order PDE systems ${\mathcal {E}}_{A}(x^{\mu },y^{B},y_{~\mu }^{B},y_{~\mu \nu }^{B})=0$. Any such PDE system defines a source form $\varepsilon ={\mathcal {E}}_{A}\omega ^{A}\wedge \mathrm {d}^{n}x$, where $\omega ^{A}=\mathrm {d}y^{A}-y_{~i}^{A}\mathrm {d}x^{i}$, on some fibered chart $(V^{2},\psi ^{2})$ of $J^{2}Y$. In the following, we will consider that the fibered chart domain $V^{2}$ is completely arbitrary, i.e., its image $\psi ^2(V^2)$ through the coordinate homeomorphism $\psi ^2$ is not necessarily vertically star-shaped.

Lemma 1

Let ${\mathcal {E}}_{A} = 0$ be an arbitrary second order PDE system, and a, b two arbitrary real numbers. Define, at each point in the domain of definition of ${\mathcal {E}}_{A}$:

$$\begin{aligned} {{\mathcal {L}}}_{{\mathcal {E}}}(x^{\mu },y^{B},y_{~\mu }^{B},y_{~\mu \nu }^{B}):=y^{A}\overset{b}{\underset{a}{\int }} {\mathcal {E}}_{A}(x^{\mu },ty^{B},ty_{~\mu }^{B},ty_{~\mu \nu }^{B})\mathrm {d}t. \end{aligned}$$

(43)

If the equations ${\mathcal {E}}_{A} = 0$ are locally variational and the above integrals exist and are finite, then, the Euler–Lagrange expressions (13) associated with ${{\mathcal {L}}}_{{\mathcal {E}}}$ are:

$$\begin{aligned} \tilde{{\mathcal {E}}}_{B}=b {\mathcal {E}}_{B}(x^{\mu } b y^{A},by_{~\mu }^{A},by_{~\mu \nu }^{A})-a {\mathcal {E}}_{B}(x^{\mu },ay^{A},ay_{~\mu }^{A},ay_{~\mu \nu }^{A}). \end{aligned}$$

(44)

Proof

Since ${\mathcal {E}}_{A} = 0$ are assumed to be variational, the Helmholtz conditions (36)–(38) hold. Further, let us calculate the Euler–Lagrange expressions $\tilde{{\mathcal {E}}}_{B}({\mathcal {L}}_{{\mathcal {E}}}).$ Denoting by $\chi _{t}:V^{2} \rightarrow V^{2}$ the fiber homothety $(x^{\mu },y^{B},y_{~\mu }^{B},y_{~\mu \nu }^{B})\mapsto (x^{\mu },ty^{B},ty_{~\mu }^{B},ty_{~\mu \nu }^{B})$ (defined for t such that $\chi _{t}(V^{2}) \subset V^{2}$, where $V^{2}$ is the domain of definition of ${\mathcal {E}}_{A}$), we can write (43) in a more compact way as:

$$\begin{aligned} \mathcal {{L}}_{{\mathcal {E}}}:=y^{A}\overset{b}{\underset{a}{\int }} {\mathcal {E}} _{A}\circ \chi _{t}\mathrm {d}t. \end{aligned}$$

(45)

Further,

$$\begin{aligned} \dfrac{\partial \mathcal {{L}}_{{\mathcal {E}} }}{\partial y^{B}}=\overset{ b}{\underset{a}{\int }}{\mathcal {E}}_{B}\circ \chi _{t}\mathrm {d}t+y^{A}\overset{b}{ \underset{a}{\int }}t\dfrac{\partial {\mathcal {E}}_{A}}{\partial y^{B}}\circ \chi _{t}\mathrm {d}t. \end{aligned}$$

Performing integration by parts in the first term, this becomes, after a brief computation:

$$\begin{aligned} \dfrac{\partial \mathcal {{L}}_{{\mathcal {E}} }}{\partial y^{B}}&=t{\mathcal {E}} _{B}(x^{\mu },ty^{A},ty_{~\mu }^{A},ty_{~\mu \nu }^{A})\mid _{a^{{}}}^{b_{{}}} \\&\quad +\overset{b}{\underset{a}{\int }}t\left[ y^{A}\left( \dfrac{\partial {\mathcal {E}}_{A}}{ \partial y^{B}} -\dfrac{\partial {\mathcal {E}}_{B}}{\partial y^{A}}\right) \circ \chi _{t} - y_{~\mu }^{A} \dfrac{ \partial {\mathcal {E}}_{B}}{\partial y_{\mu ~}^{A}}\circ \chi _{t}\right. \\&\quad \left. -\,y_{~\mu \nu }^{A} \dfrac{\partial {\mathcal {E}}_{B}}{\partial y_{\mu \nu ~}^{A}}\circ \chi _{t}\right] \mathrm {d}t. \end{aligned}$$

The other derivatives appearing in $\tilde{{\mathcal {E}}}_{B}(\mathcal {{L}}_{{\mathcal {E}}})$ are:

$$\begin{aligned} \dfrac{\partial \mathcal {{L}}_{{\mathcal {E}} }}{\partial y_{~\mu }^{B}}&=y^{A}\overset{b}{\underset{a}{\int }}t\dfrac{\partial {\mathcal {E}} _{A}}{ \partial y_{\mu }^{B}}\circ \chi _{t}\mathrm {d}t \\ \Rightarrow \ \mathrm {d}_{\mu }\left( \dfrac{\partial \mathcal {\tilde{L}}_{{\mathcal {E}} }}{\partial y_{~\mu }^{B}}\right)&=y_{~\mu }^{A}\overset{b}{\underset{a}{\int }}t\dfrac{\partial {\mathcal {E}} _{A}}{\partial y_{\mu }^{B}}\circ \chi _{t}\mathrm {d}t +y^{A}\overset{b}{\underset{a}{\int }}t \mathrm {d}_{\mu }\left( \dfrac{\partial {\mathcal {E}} _{A}}{ \partial y_{\mu }^{B}}\circ \chi _{t}\right) \mathrm {d}t; \\ \dfrac{\partial \mathcal {\tilde{L}}_{{\mathcal {E}} }}{\partial y_{~\mu \nu }^{B}}&=y^{A}\overset{b}{\underset{a}{\int }}t\dfrac{\partial {\mathcal {E}} _{A}}{\partial y_{\mu \nu }^{B}}\circ \chi _{t}\mathrm {d}t \\ \Rightarrow \mathrm {d}_{\mu }\mathrm {d}_{\nu }\left( \dfrac{\partial \mathcal {\tilde{L}}_{{\mathcal {E}} }}{ \partial y_{~\mu \nu }^{B}}\right)&=y_{~\mu \nu }^{A}\overset{b}{\underset{a}{ \int }}t\dfrac{\partial {\mathcal {E}} _{A}}{\partial y_{\mu \nu }^{B}}\circ \chi _{t} \mathrm {d}t +2y_{~\mu }^{A}\overset{b}{\underset{a}{\int }}t \mathrm {d}_{\nu }(\dfrac{\partial {\mathcal {E}} _{A}}{\partial y_{\mu \nu }^{B}}\circ \chi _{t})\mathrm {d}t \\&\quad +\,y^{A}\overset{b}{\underset{a}{\int }}t \mathrm {d}_{\mu } \mathrm {d}_{\nu }\left( \dfrac{\partial {\mathcal {E}} _{A}}{\partial y_{\mu \nu }^{B}}\circ \chi _{t}\right) \mathrm {d}t. \end{aligned}$$

Grouping terms, we immediately find:

$$\begin{aligned} \tilde{{\mathcal {E}}}_{B}(\mathcal {{L}}_{{\mathcal {E}}})&= -\overset{b}{\underset{a}{\int }} t[y^{A}(H_{BA}\circ \chi _{t})+y_{~\mu }^{A}(H_{BA}^{\mu }\circ \chi _{t})+y_{~\mu \nu }^{A}(H_{BA}^{\mu }\circ \chi _{t})] \mathrm {d}t, \end{aligned}$$

where $H_{AB}, H_{AB}^{\mu }, H_{AB}^{\mu }$ are the components of the Helmholtz form, defined in (36)–(38). By virtue of the Helmholtz conditions, these vanish identically and therefore:

$$\begin{aligned} {\mathcal {E}}_{B}(\mathcal {\tilde{L}}_{{\mathcal {E}} })=\left( t{\mathcal {E}} _{B}(x^{\mu },ty^{A},ty_{~\mu }^{A},ty_{~\mu \nu }^{A})\right) \mid _{a^{{}}}^{b_{{}}}, \end{aligned}$$

which is just (44). $\square$

The standard Vainberg–Tonti Lagrangian was determined by a similar reasoning, choosing $b=1$ and $a=0,$ see [96]. From the above Lemma, we immediately obtain the desired theorem:

Theorem 2

Let ${\mathcal {E}}_{A} = 0$ be an arbitrary second order PDE system and $a \in {\mathbb {R}} \cup \{\pm \infty \}$ such that

$$\begin{aligned} \underset{t\rightarrow a}{\lim }\left( t{\mathcal {E}} _{B}(x^{\mu },ty^{A},ty_{~\mu }^{A},ty_{~\mu \nu }^{A})\right) =0, \end{aligned}$$

at all $(x^{\mu },y^{B},y_{~\mu }^{B},y_{~\mu \nu }^{B})$ in the domain of ${\mathcal {E}}_{A}$. Define, at these points, the extended Vainberg–Tonti Lagrangian $\lambda = {\mathcal {L}} \mathrm {d}^{n}x$, by the rule:

$$\begin{aligned} {\mathcal {L}}_{{\mathcal {E}} }(x^{\mu },y^{B},y_{~\mu }^{B},y_{~\mu \nu }^{B}):=y^{A}\overset{1}{\underset{a}{\int }}{\mathcal {E}} _{A}(x^{\mu },ty^{B},ty_{~\mu }^{B},ty_{~\mu \nu }^{B})\mathrm {d}t. \end{aligned}$$

(46)

If the above integrals exist and are finite at all points in the given domain, then:

• 1. If the equations ${\mathcal {E}}_{A} = 0$ are variational, then $\lambda$ is a (locally defined) Lagrangian for these, i.e., the Euler–Lagrange expressions of (46) are precisely ${\mathcal {E}}_{A}$:

  $$\begin{aligned} \tilde{{\mathcal {E}}}_{A} ={\mathcal {E}} _{A}. \end{aligned}$$

  (47)

• 2. If ${\mathcal {E}}_{A} = 0$ are not variational, then the Euler–Lagrange expressions of (46) are their canonical variational completion; the correction terms are expressed in terms of the coefficients of the Helmholtz form:

  $$\begin{aligned} \tilde{{\mathcal {E}}}_{A} = {\mathcal {E}} _{A} + H_A, \end{aligned}$$

  (48)

  where

  $$\begin{aligned} H_A = -\overset{b}{\underset{a}{\int }} t[y^{B}(H_{AB}\circ \chi _{t})+y_{~\mu }^{B}(H_{AB}^{\mu }\circ \chi _{t})+y_{~\mu \nu }^{B}(H_{AB}^{\mu }\circ \chi _{t})] \mathrm {d}t. \end{aligned}$$

  (49)

The above result is extremely useful in the case when ${\mathcal {E}} _{A}$ is homogeneous of negative degree $k < -1$ in $y^{A}.$ In this case, the integral $y^{A} {\int \nolimits _{a}^{1} }{\mathcal {E}} _{A}(x^{\mu },ty^{B},ty_{~\mu }^{B},ty_{~\mu \nu }^{B})\mathrm {d}t$ diverges for $a=0$, but it can be replaced with an integral from $a=\infty$ to 1.

A special case. The case when ${\mathcal {E}}_{A}$ are homogeneous functions of degree $-1$ in the fiber variables, is a degenerate one. In this case, there is no integration endpoint a which satisfies the hypothesis of the above Theorem. Therefore, in this case, we cannot define the Vainberg–Tonti Lagrangian (46).