# Can a tensor of any rank be decomposed into symmetric/anti-symmetric parts?

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?
• 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? Jun 23, 2023 at 23:52