I read that any rank-2 tensor can be decomposed into the sum of a traceless symmetric tensor, an anti-symmetric tensor and a unit tensor, all closed under $SO(3)$. The three form an irreducible representation of $SO(3)$. The same is said to be possible for any types of tensors.

My question is: how a rank-3 tensor (and beyond) can be decomposed into parts that are closed under $SO(3)$?

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    $\begingroup$ A rank $n$ tensor in your definition is an element of $\mathbb{R}^3 \otimes \cdots \otimes \mathbb{R}^3$, the tensor product of $n$ copies of $\mathbb{R}^3$? $\endgroup$ – Vincent Jul 6 '20 at 20:38
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    $\begingroup$ This is explained quite well in the book by Fulton and Harris: Representation Theory, a first course $\endgroup$ – Vincent Jul 6 '20 at 20:39
  • $\begingroup$ Yes, by rank-n tensor I mean something with n indices and transform like n vectors. $\endgroup$ – Xiaowen Shan Jul 6 '20 at 21:52
  • $\begingroup$ Fulton and Harris is beyond the comprehension of a non-mathematician like me. ;-) Could you point me to some simpler answers, e.g., the rank-3 equivalence to the decomposition of $5\oplus 3\oplus 1$ for rank-2? Many thanks! $\endgroup$ – Xiaowen Shan Jul 6 '20 at 22:07
  • $\begingroup$ You say you are not a mathematician, but are you by any chance a physicist? In that case you might enjoy Wikipedia's take on this: en.wikipedia.org/wiki/Clebsch%E2%80%93Gordan_coefficients. As a non-physicist I find it hard to read but I am pretty sure that ultimately it boils down to the same thing I wrote in the answer $\endgroup$ – Vincent Jul 7 '20 at 16:34

You know the decomposition $3 \otimes 3 = 5 \oplus 3 \oplus 1$. The more general form is $n \otimes n = (2n - 1) \oplus (2n - 3) \oplus \ldots \oplus 3 \oplus 1$. So this pretty picture about how the sum of consecutive odd numbers is a square you can find elsewhere on MSE has a surprising interpretation in terms of $SO(3)$-representations.

Now there is an even more general form. If I remember correctly it is:

$m \otimes n = (m + n - 1) \oplus (m + n - 3) \oplus \ldots \oplus (m -n + 3) \oplus (m -n + 1)$, assuming that $m \geq n$.

Now we can use this to compute $3 \otimes 3 \otimes 3$:

$$3 \otimes 3 \otimes 3 = 3 \otimes (5 \oplus 3 \oplus 1) = 5 \otimes 3 \oplus 3 \otimes 3 \oplus 3 \otimes 1 = (7 \oplus 5 \oplus 3) \oplus (5 \oplus 3 \oplus 1) \oplus 3 \\ = 7 \oplus 5 \oplus 5 \oplus 3 \oplus 3 \oplus 3 \oplus 1$$

Quick check: do these numbers add up to $27$? Yes. Ok, good.

  • $\begingroup$ Here the numbers are the dimensions of irreducible representations of $SO(3)$: for every odd number $n$ there is one $n$-dimensional space closed under the action of $SO(3)$ such that no subspace is mapped into itself by the $SO(3)$-action. $\endgroup$ – Vincent Jul 7 '20 at 16:31
  • $\begingroup$ And here is the picture I was referring to: math.stackexchange.com/a/639079/101420 $\endgroup$ – Vincent Jul 7 '20 at 16:37
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    $\begingroup$ Nice! One step forward, for the case of rank-2, the 5, 3 and 1 stand for the spaces of traceless symmetric, anti-symmetric and unit tensors. What are the seven subspaces? I guess the 1 is the Levi-Civita symbol, and one of the 3 is $a_i\delta_{jk}$? $\endgroup$ – Xiaowen Shan Jul 8 '20 at 19:36
  • $\begingroup$ Yes this is a good question, I don't know the answer from the top of my head. Maybe you can make it into a seperate question here on MSE? $\endgroup$ – Vincent Jul 9 '20 at 20:17

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