# Variation of action in terms of energy-momentum tensor

In Di Francesco, Mathieu, and Sénéchanl, Conformal Field theory section 4.2.2 it is stated that under an arbitrary diffeomorphism $$x\rightarrow x+\epsilon$$ the action transforms like $$\delta S=\int d^dx T^{\mu\nu}\partial_\mu \epsilon_\nu\tag{4.34}$$ even for fields that do not satisfy the equations of motion. I don't get this. I get that via Noether's theorem $$\delta S=\int d^dx \partial_\mu(T^{\mu\nu} \epsilon_\nu)$$ for all fields. Then, if the theory has translation symmetry and the field equations are satisfied, we obtain $$\partial_\mu T^{\mu\nu}=0$$, which immediately leads to the first equation. However, my argument requires the use of the equations of motion. I believe I am overlooking something extremely simple. I would appreciate if anyone can point this out to me.

The energy momenum tensor is defined by $$\delta S[\phi, g_{\mu\nu}]= \frac 12 \int d^dx \sqrt{g} T^{\mu\nu}\delta g_{\mu\nu}.$$ In this variation we vary the geometry but keep the fields $$\phi(x)$$ fixed. If we start in flat space where $$g_{\mu\nu}=\delta_{\mu\nu}$$ and make a diffeomorphism then $$\delta g_{\mu\nu}= \partial_\mu \epsilon_\nu+ \partial_\nu\epsilon_\mu$$ so $$\delta S[\phi, g] = \frac 12 \int d^dx T^{\mu\nu}(\partial_\mu \epsilon_\nu+ \partial_\nu\epsilon_\mu).$$ No equations of motion are needed, but again the fields are to be unchanged. What does need the fields to obey their EofM is conservation $$\partial_\mu T^{\mu\nu}=0.$$ This is because the action is unchanged a under a coordinate chage $$x\to x+\epsilon$$. A coordinate change requires that we change both $$g_{\mu\nu}$$ and $$\phi(x) \to \phi(x+\epsilon)$$. If the change in $$\phi$$ is to have no effect on $$S[\phi,g]$$ then we must require the field to satsify its EofM.
Strictly speaking, if one reads Ref. 1, then eq. (4.34) refers to eq. (2.191), which in turn refers to eq. (2.142), or better $$\delta S~=~ -\int \! d^d x~j^{\mu} \partial_{\mu}\omega_a \tag{2.140}.$$ Eq. (2.140) follows directly from Noether's first theorem, cf. e.g. this Phys.SE post, which means that the Noether current is the canonical SEM tensor. In footnotes 2 & 6 it is explained that we can improve the canonical SEM tensor into a symmetric SEM, cf. Belinfante et al, in a way so that eq. (4.34) still holds with the symmetric SEM tensor (up to possibly boundary terms).