I know that for a rank 2 tensor $A_{\mu \nu}$ I can write $$ A_{\mu \nu} = A_{(\mu \nu)} + A_{[\mu \nu]}, $$ where $A_{(\mu \nu)}$ is the symmetrization of $A$ and $A_{[\mu \nu]}$ is the anti-symmetrization. For rank $3$ tensors this obviously does not work, but suppose we are working on a 2d-manifold with a metric $g$ that induces a volume element $\varepsilon_{ab}$. Then I remember seeing the following decomposition of $A$: $$ A_{\mu \nu \lambda } \overset{?}{=} A_{(\mu \nu \lambda)} + \varepsilon_{\mu \nu} \varepsilon^{ab} A_{ab \lambda} + \varepsilon_{\mu \lambda} \varepsilon^{ab} A_{a \nu b} + \varepsilon_{\nu \lambda } \varepsilon^{ab} A_{\mu ab}. $$ I do not remember the exact decomposition, but I recall it was like "projecting" $A$ onto the volume element on the manifold. I have two doubts:

  1. What was the correct decomposition?
  2. Does this decomposition generalize to tensors of any rank and Riemannian manifolds of any dimension?
  • $\begingroup$ I have no idea what projecting a 3-tensor onto a skew-symmetric 2-tensor should mean. You’ve accounted for completely symmetric and skew-symmetric in the three pairs; what about those that are symmetric in one pair? $\endgroup$ Jun 23, 2023 at 23:52


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