# What is the purpose of emphasizing that an action is invariant under diffeomorphism?

When learning field theory and string theory, I always see physicists stress the fact that the action, which is an integral of the Lagrangian density $S(x)=\int L(x,\dot{x})dt$, is invariant under diffeomorphism. For example, in string theory, people always say that the Polyakov action is invariant under worldsheet diffeomorphisms.

I do not understand why this is important because as I see it, integrals are defined to be independent of the choice of coordinates and so actions are trivially invariant under diffeomorphism. Can any one show me an example of action which is not invariant under diffeomorphism?

• An action like $S=\int f d^4x$ in which $f$ is an invariant quantity under diffeomorphism is not invariant under diffeomorphism since the integral measure is a pseudo-tensor. Sep 29, 2016 at 8:26
• @Hosein By definition of integrals in calculus, isn't it proved in differential geometry that any integral of differential forms should be invariant under orientation-preserving diffeomorphisms? Suppose $f(x)=1$, which is constant, by definition of what an integral is, the number S is the same no matter in which coordinate! Sep 29, 2016 at 8:33
• @Hosein, Let us assume that the number $S$ is somehow dependent on the choice of coordinates, then we conclude that this integral, denoted by $S$ depends on a family of continuous variables labeling all possible choices of coordinates. As a result, the whole world does not make sense anymore. For example, the concept of length and area depends on how people measure. The distance between two fixed points are different in Cartesian coordinate and polar coordinate, which is absurd. Sep 29, 2016 at 8:39
• The point is that length and area are defined such that they remain unchanged under diffeomorphism, for example the volume is defined as $V=\int \sqrt{-g} d^4x$ for a space with a defined metric $g$ . And this quantity is invariant under diffeomorphism. Sep 29, 2016 at 8:44
• @Hosein, Yes the Riemannian volume form is just a special form. Isn't it a fact that integral of any differential form is invariant under arbitrary orientation-preserving diffeomorphism? Even in Euclidean space, integrals of a function is just a number. Isn't this a fact? Sep 29, 2016 at 8:47

After checking the book 'Differential Geometry and Lie Groups for Physicists' by Marian Fecko in section 16.4.1, I believe that I am close to understanding what physicists mean by invariance of an action under diffeomorphism. In what follows, I explain Marian Fecko's discussion on diffeomorphism invariance of an action.

Consider a classical field theory $\phi$ on a Riemannian manifold $(M,g)$, where $g_{ab}$ is its metric. We say the action natural with respect to diffeomorphism in the following sense.

Let us define such a differential form $\Omega[\phi,g]$ defined via the action $$S[\phi,g]=\int_{D}\Omega[\phi,g]=\int_{D}L(\phi,\nabla\phi,g)\omega_{g}$$ where $D\subset M$ is a submanifold and $\omega_{g}$ is the volume form on $D$ associated with the metric $g$.

Let us then assume that the variation of fields and the variation of spacetime coordinates vanish on the boundary $\partial D$. In mathematical terminology, this requirement corresponds to a flow $\Psi_{t}:M\rightarrow M$ that is arbitrary inside $D$ but vanishes on $\partial D$.

We say that the action is invariant (or natural) under the diffeomorphism $\Psi_{t}$ if the pull-back satisfies $$\Psi_{t}^{\ast}(\Omega[\phi,g])=\Omega[\Psi_{t}^{\ast}(\phi),\Psi_{t}^{\ast}(g)].$$

Under such a requirement, since the flow does not move points on the boundary, under any infinitesimal amount of variation $\Psi_{\epsilon}$ generated by a vector field $V$ in $M$, we have $\Psi_{\epsilon}(D)=D$. It follows that $$S[\phi,g]=\int_{D}\Omega[\phi,g]=\int_{\Psi_{\epsilon}(D)}\Omega[\phi,g]=\int_{D}\Psi_{\epsilon}^{\ast}(\Omega[\phi,g])$$ $$=\int_{\Psi_{\epsilon}(D)}\Psi_{\epsilon}^{\ast}(\Omega[\phi,g])=\Psi_{\epsilon}^{\ast}(S[\phi,g])$$ $$=\int_{D}\Omega[\Psi_{\epsilon}^{\ast}(\phi),\Psi_{\epsilon}^{\ast}(g)]=\int_{D}\Omega[\phi+\epsilon\mathcal{L}_{V}\phi,g+\epsilon\mathcal{L}_{V}g]+o(\epsilon),$$ where $\mathcal{L}_{V}$ is the Lie derivative along the flow, and we have used integration by substitution in the first line. Since classical fields must be on-shell, $\phi$ extremizes the action $S[\phi,g]$. We then have $$\int_{D}\Omega[\Psi_{\epsilon}^{\ast}(\phi),\Psi_{\epsilon}^{\ast}(g)]=\int_{D}\Omega[\phi+\epsilon\mathcal{L}_{V}\phi,g+\epsilon\mathcal{L}_{V}g]=\int_{D}\Omega[\phi,g+\epsilon\mathcal{L}_{V}g].$$ and so $$S[\phi,g]=\int_{D}\Omega[\phi,g]=\Psi_{\epsilon}^{\ast}(S[\phi,g])=\int_{D}\Omega[\phi,g+\epsilon\mathcal{L}_{V}g]+o(\epsilon).$$ Therefore, by definition of energy-momentum tensor, we have $$S[\phi,g]=S[\phi,g]-\epsilon\int_{D}\frac{1}{2}(\mathcal{L}_{V}g)_{ab}T^{ab}\omega_{g}+o(\epsilon),$$ and so $$\int_{D}(\mathcal{L}_{V}g)_{ab}T^{ab}\omega_{g}=0,$$ for arbitrary variation $\delta g$, we conclude that the energy-momentum is conserved i.e. $\nabla_{a}T^{ab}=0$.

In summary, the diffeomorphism invariance of an action is a crucial condition for the conservation of energy-momentum tensor of the system.

I hope this can contribute the understanding of those who once were also confused by it. Welcome to clarify any mistakes and any misunderstandings I have.

Perhaps a simple example is in order: The action for a non-relativistic free particle

$$S[x]~=~\int \! dt ~L, \qquad L ~=~ \frac{m}{2}\dot{x}^2,$$

is not form invariant under time-reparametrizations

$$t\quad\longrightarrow \quad t^{\prime} ~=~f(t).$$

In contrast, modern fundamental physics (such as, e.g. string theory) is believed to be geometric, and the action formulation are expected to be reparametrization and diffeomorphism invariant.

• Thank you Qmechanic but I still do not understand why such an action is not invariant under time-diffeomorphism. Any action is just an integral of some function, which is just a number that does not depend on variables. Sep 29, 2016 at 8:59
• I believe that now I can understand what you meant. I think GR is a diffeomorphism invariant theory, but non-relativistic particle is not. Oct 2, 2016 at 7:44

In a sense you're completely right: integrals are invariant under change of variables. But in physics there's an essential point, not often emphasized, that whatever function you're integrating is given by a formula that should work in all coordinate systems.

In other words, to do a change of variables properly you need to include a Jacobian. But from a physical point of view, to include a Jacobian is to know what coordinate system you're in; otherwise how do you know that you should include a Jacobian? So in our integrals we want the Jacobian to be unity; in a general manifold this is done by sticking the factor of $\sqrt{-g}$.

Let me give you an example. Suppose we have the integral $\int_0^1\int_0^1\ dx\ dy$, and let's say its result is physically meaningful. Let's do a change of coordinates $x = x'^2$: according to the change of variables theorem, our integral is now written as $\int_0^1\int_0^1 2x'\ dx'\ dy$. Even though the result is the same, the integral is not invariant, because I need to know what coordinates I'm using to know whether I should include the $2x'$. The original formula is supposed to work the same in all coordinate systems.

The function itself should also be invariant, and this is another place where physicists and mathematicians use the same word for different things. To us a scalar is not just a number; it's supposed to be the same in all coordinate systems. You might complain that a number such as $4$ is the same in all coordinate systems, but again: in physics our functions are defined by formulas, and the same formula should work for everyone no matter what their coordinates are. For example, if you have a 2-dimensional manifold with coordinates $(x,y)$ and you have the function $f(x,y)=x$, then this function is not a scalar! Calculating it in different coordinates is going to give different results.

• Thanks Javier. I will think of your answer. Different integrands may lead to the same EOM. I am not sure if this may be linked with your statement. Sep 29, 2016 at 10:46