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

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.