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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 $SO(3)$?

Xiaowen Shan
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    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 $\mathbb{R}^3$? – Vincent Jul 06 '20 at 20:38
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    This is explained quite well in the book by Fulton and Harris: Representation Theory, a first course – Vincent Jul 06 '20 at 20:39
  • Yes, by rank-n tensor I mean something with n indices and transform like n vectors. – Xiaowen Shan Jul 06 '20 at 21:52
  • 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! – Xiaowen Shan Jul 06 '20 at 22:07
  • 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: https://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 – Vincent Jul 07 '20 at 16:34
  • You can also use spherical tensors, because they are irreducible by definition. Similarly to what @Vincent wrote on how many independent components each part in the expansion has, you can also count how many spherical tensors of a given rank $k$ you need to represent your rank-$3$ Cartesian tensor. A simple calculation will show that you need $1:k=3$, $2:k=2$, $3:k=1$, $1:k=0$. As the number of components for a sph. tensor is $2k+1$, you find $1(23+1)+2(22+1)+3(21+1)+1(2*0+1)=27=3^3$ :^) – Sl0wp0k3 Jun 02 '23 at 10:52

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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 that 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.

Vincent
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  • 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. – Vincent Jul 07 '20 at 16:31
  • And here is the picture I was referring to: https://math.stackexchange.com/a/639079/101420 – Vincent Jul 07 '20 at 16:37
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    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 $a_i\delta_{jk}$? – Xiaowen Shan Jul 08 '20 at 19:36
  • 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? – Vincent Jul 09 '20 at 20:17