In the review of cosmological perturbations by Mukhanov et al linked:


on page 212 of the pdf, they introduce a symmetric, trace-free, rank-2 tensor perturbation $h_{ij}$ with constraints $h^i_i = 0$ and $\nabla^jh_{ij} = 0$. He claims that the rationale behind these constraints is that this does away with the scalar and vector components, thus making it truly tensorial.

I do not understand how the latter condition reconciles with the 'vector' component of rank-2 tensors usually being interpreted as the anti-symmetric component.

Any rank-2 tensor under the action of rotations (as is our interest in cosmology) can be decomposed into three irreducible components: an anti-symmetric part, and a symmetric part that is further broken down into a trace-free and scalar (trace) part:

$$X_{ab} = X_{[ab]} + \frac{1}{n}X\delta_{ab} + (X_{(ab)} - \frac{1}{n}X\delta_{ab}).$$

The trace transforms as a scalar under rotations; while the 3 independent components of the anti-symmetric part transform like a vector. Finally, the five independent components of the symmetric, trace-free part transform like a tensor.

How does setting $\nabla^jh_{ij} = 0$ reconcile with this interpretation?

In response to the answer by knzhou:

This makes sense, in that you're decomposing the symmetric perturbation tensor into its scalar, purely vector and purely tensor components. The first term accounts for the trace; the remaining are trace-free. The claim is then that the trace-free symmetric tensor can be broken down into components that transform like a vector and a tensor. The former is then broken down into the gradient of a scalar and a pure vector (the second and third terms respectively), with derivatives taken appropriately to tally indices. I just don't see how this exclusion of vector components happens via the derivative condition as imposed by Mukhanov et al.


1 Answer 1


Mukhanov is talking about the SVT decomposition, $$h_{ij} = 2 C \delta_{ij} + 2 \left( \partial_i \partial_j - \frac13 \delta_{ij} \nabla^2 \right) E + 2 \partial_{(i} \hat{E}_{j)} + 2 \hat{E}_{ij}$$ where $C$ and $E$ are scalars, $\hat{E}$ is a vector, and $\hat{E}_{ij}$ is a tensor. As you can see, all of the terms in the expansion are symmetric tensors, but they're written in terms of derivatives or $\delta_{ij}$ times scalars, vectors, and tensors. Mukhanov's conditions pick out the $\hat{E}_{ij}$ term. This decomposition is useful because each of these quantities evolves independently under the linearized Einstein equations.

  • $\begingroup$ I've modified my question to further clarify my misunderstanding; could you comment on it? It was a bit too long to write as a comment. $\endgroup$ Aug 31, 2021 at 21:15

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