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It is well known that one can decompose tensor product spaces into direct sums and that sometimes the components of the direct sums are not unique. Taking Wikipedia's example, the addition of three spin-$1/2$ particles yields a single spin-$3/2$ particle and two spin-$1/2$ particles: $$\mathbf{2}\otimes\mathbf{2}\otimes\mathbf{2}=\mathbf{4}\oplus\mathbf{2}\oplus\mathbf{2}.$$ Can this be written as $$\mathbf{2}\otimes\mathbf{2}\otimes\mathbf{2}=\mathbf{4}\oplus\mathbf{2}^{\oplus2}?$$ And, in general, can we write direct sums of the form $$\bigoplus_k \mathbf{k}^{\oplus d_k},$$ where $d_k$ is the multiplicity associated with dimension $k$? I have seen this written as $$\bigoplus_k d_k\mathbf{k},$$ for example on the same Wikipedia page and in the references cited there, but that seems more confusing, especially if we try to decompose spin operators as $$\mathbf{J}=\bigoplus_k d_k \mathbf{J}_k \quad \mathrm{versus}\quad \mathbf{J}=\bigoplus_k \mathbf{J}_k^{\oplus d_k}.$$

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    $\begingroup$ You "can" do whatever you want I guess, it's just that other people will not know what you mean. Also, in this case an exponent makes no sense, while a (direct) product has the right meaning. $\endgroup$
    – fqq
    Jun 15, 2021 at 23:38
  • $\begingroup$ I'm confused; how do you write $\mathbf{2}\otimes\mathbf{2}\otimes\mathbf{2}\otimes\mathbf{2}=$ $\mathbf{5}\oplus\mathbf{3 } \oplus\mathbf{3 } \oplus\mathbf{3 } \oplus\mathbf{1 } \oplus\mathbf{1 } $? $\endgroup$ Jun 16, 2021 at 0:01
  • $\begingroup$ what do you mean by decomposing $$\mathbf{J}=\bigoplus_k d_k \mathbf{J}_k \quad \mathrm{versus}\quad \mathbf{J}=\bigoplus_k \mathbf{J}_k^{\oplus d_k}$$ mean anyway? $\endgroup$ Jun 16, 2021 at 1:26
  • $\begingroup$ Recall in the conventional notation direct sums and products reduce to straight grade school arithmetic checks. Whereas in your vision…. $\endgroup$ Jun 16, 2021 at 2:20
  • $\begingroup$ @CosmasZachos the goal would be $\mathbf{5}\oplus \mathbf{3}^{\oplus 3}\oplus \mathbf{1}^{\oplus 2}$. This agrees with regular arithmetic: $a^4=aaaa$ means write $a$ next to itself 4 times with the standard multiplication operation; $a^{\otimes 4}=a\otimes a\otimes a \otimes a$ means write $a$ next to itself 4 times with a tensor product as the operation, so I could conclude that $a^{\oplus 4}=a\oplus a\oplus a\oplus a$ means compose $a$ next to itself 4 times with a direct sum as the composition operation. $\endgroup$ Jun 16, 2021 at 3:06

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You evidently missed my point about trivial arithmetic checks on the Kronecker product reduction, in the convention where representations are labelled by their dimensionality, used here for SU(2) (in contrast to the atomic physics convention) and flavor SU(3) and GUTs in particle physics.

In this convention,
$$\mathbf{2}\otimes\mathbf{2}\otimes\mathbf{2}\otimes\mathbf{2}=\mathbf{5}\oplus\mathbf{3}\oplus\mathbf{3}\oplus\mathbf{3}\oplus\mathbf{1}\oplus\mathbf{1}.$$ Converting irreps to numbers (their dimensionality!), and direct products into products and direct sums to sums, you get a numerical grade-school arithmetic check; in this case, you get 16=16.

That is, 16×16 matrices reduce to a 5×5 matrix block, three 3×3 blocks, and two 1×1 singlet entries/blocks, upon a Clebsch orthogonal transformation. The dimensionality of the reducible representation is then borne out by inspection!

As you see in your WP reference, the established condensed combinatoric convention for the above is, instead, $$ \mathbf{5}\oplus 3\times\mathbf{3} \oplus 2\times\mathbf{1}, $$ comporting with this arithmetic check, albeit clumsily; it is mostly useful to the Catalan triangle algorithm.

In your stillborn scheme, by contrast, $ \mathbf{5}\oplus\mathbf{3}^{\oplus 3} \oplus\mathbf{1}^{\oplus 2}$, you'd get 6+27=33, instead, ensuring confusion for your hapless reader. You are free to devise any conventions you like, but caveat lector.

NB Unfriendliness to the reader is routine in this field, so, instead of using the above SU(2) expression, in atomic physics they use irrep labels j for d=2j+1 above, $$\mathbf{1/2}\otimes\mathbf{1/2}\otimes\mathbf{1/2}\otimes\mathbf{1/2}=\mathbf{2}\oplus\mathbf{1}\oplus\mathbf{1}\oplus\mathbf{1}\oplus\mathbf{0}\oplus\mathbf{0},$$ which forfeits the trivial arithmetic check, straightforward to perform in your head, however.

I have the distinct impression you have not fully mastered the irrep coproduct (spin-composition in QM) which dictates this notation and makes the expressions so simple and natural, $$ \Delta(J_z)= J_z\otimes \mathbb{1}_2\otimes \mathbb{1}_2\otimes \mathbb{1}_2 +\mathbb{1}_2\otimes J_z\otimes \mathbb{1}_2\otimes \mathbb{1}_2 \\ + \mathbb{1}_2\otimes \mathbb{1}_2\otimes J_z\otimes \mathbb{1}_2+ \mathbb{1}_2\otimes \mathbb{1}_2\otimes \mathbb{1}_2 \otimes J_z.$$ It acts on each subspace simultaneously ("synchronized swimming") in the suitable irrep. The structure is additive, not multiplicative, so multiplexing is a multiplication, not an exponentiation.

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  • $\begingroup$ Evidently! As charged, I am more familiar with the atomic convention. We still have the bulky notation $\Delta (J_z)=J_z\oplus 1_3\oplus 1_3\oplus 1_3\oplus 1_1\oplus 1_1 + 1_5\oplus J_z\oplus 1_3\oplus 1_3\oplus 1_1\oplus 1_1+\cdots$, so it would be nice to simplify the notation using either $J_z\oplus 1_3\oplus 1_3\oplus 1_3\oplus 1_1\oplus 1_1 + 1_5\oplus 3 J_z \oplus 1_1\oplus 1_1+ 1_5\oplus 1_3\oplus 1_3\oplus 1_3\oplus 2J_z$ or $J_z\oplus 1_3\oplus 1_3\oplus 1_3\oplus 1_1\oplus 1_1 + 1_5\oplus J_z^{\oplus 3} \oplus 1_1\oplus 1_1+ 1_5\oplus 1_3\oplus 1_3\oplus 1_3\oplus J_z^{\oplus 2}$ $\endgroup$ Jun 16, 2021 at 14:36
  • $\begingroup$ But, of course, I can live without that (caveat auctor) $\endgroup$ Jun 16, 2021 at 14:39
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    $\begingroup$ No, the coproduct you wrote is unsound! It is a direct sum of the $J_z$s in the respective irrep, after Clebsching! Look at a simple example of two doublets composing to a singlet and triplet. $\endgroup$ Jun 16, 2021 at 14:40
  • $\begingroup$ Wait... yes, it would be the product of all of the cumbersome things I wrote. The question still becomes whether there is a better notation for $J_z^{(5)}\oplus J_z^{(3)}\oplus J_z^{(3)}\oplus J_z^{(3)}\oplus J_z^{(1)}\oplus J_z^{(1)}$ but I will rest. Thank you for the indulgence. $\endgroup$ Jun 16, 2021 at 14:56
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    $\begingroup$ Yes, the last expression here is right. The same operator repeats itself in disjoint matrix blocks... cf here, an additive structure. $\endgroup$ Jun 16, 2021 at 15:00

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