In my lectures on groups and representations, we write the Clebsch-Gordan decomposition for addition of angular momenta $$r_{j_1}\otimes r_{j_2}=\bigoplus_{j=|j_1-j_2|}^{j_1+j_2}r_j\tag{1}$$ where $r_j$ is the irreducible representation of the $su(2)$ Lie algebra with angular momentum $j$. However, shouldn't the decomposition be of the Lie group tensor product representation rather than the algebra? It feels like $r_j$ should instead be the $R_j$ representation of the $SU(2)$ Lie group corresponding to the exponentiation of $r_j$.

Are the lectures just using sloppy notation, or is the decomposition really in the algebra?


2 Answers 2


The Clebsch-Gordan decomposition is for the underlying vector spaces

$$V_{j_1}\otimes V_{j_2}\cong\bigoplus_{j=|j_1-j_2|}^{j_1+j_2}V_j. \tag{1'}$$

Since the Lie group $G=SU(2)$ is simply connected, all its Lie algebra representations are also Lie group representations. [The opposite is trivial.] One can hence act$^1$ with both Lie algebra elements and Lie group elements on the underlying vector spaces (1').

$^1$ Be aware that a Lie algebra (group) element acts via Leibniz rule (simultaneously) on tensor product factors in (1'), respectively.

They both act simultaneously on direct summands in (1').

The Lie group and Lie algebra action are in this way compatible on (1') via the exponential map.

  • $\begingroup$ Is there also any rule for how the Lie algebra elements act on the direct sum vector space? $\endgroup$ Oct 8, 2023 at 14:53
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    $\begingroup$ Yes on $V= V_1\oplus V_2$ we have $J \mapsto J\oplus J$ and on the product $V_1\otimes V_2$ as $J\mapsto 1\otimes J+J\otimes 1$. $\endgroup$
    – mike stone
    Oct 8, 2023 at 14:58

The representation $r_j$ are representations of both the group and the algebra, so there is no difference.

The C-G decomposition,however, is a decomposition of representations and is a decomposition of neither the group nor the algebra.

  • $\begingroup$ Maybe I used sloppy wording myself. I know we are decomposing a tensor product representation, but it seems like whether this is a Lie group or Lie algebra representation should make a difference. For example, the direct sum of Lie algebra representations is not equivalent to the direct sum of the corresponding Lie group representations. $\endgroup$ Oct 8, 2023 at 14:28
  • $\begingroup$ Can you give an example where the direct sum of group reps differs from the direct sum of algebra reps? Obviously the co-product in the group ($\Delta g=g\otimes g$) differs from that of the algebra ($\Delta j={\mathbb I}\otimes j+ j\otimes {\mathbb I}$) but the vector spce decompositions coincide in all cases that I can think of. $\endgroup$
    – mike stone
    Oct 8, 2023 at 14:32
  • $\begingroup$ I'm sure you are right, but maybe I am just confused about the meaning of direct sum of algebra reps. I am imagining exponentiating the elements of the direct sum algebra rep to get the elements of the group rep, but this turns the direct sum into a direct product. Is it just a matter of notation, that the direct sum of algebra reps works differently to that of group reps? $\endgroup$ Oct 8, 2023 at 14:43
  • $\begingroup$ The elements of the representation spaces are vectors. You can't exponentiate them! But the algebra and the group act differently on the reps. I sugges that you look at the wiki article en.wikipedia.org/wiki/Coalgebra $\endgroup$
    – mike stone
    Oct 8, 2023 at 14:55
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    $\begingroup$ I think the problem has just been me misunderstanding what is meant by the direct product and sum, and it makes sense now to think in terms of the vector space as in the Qmechanic answer. Thanks for your help. $\endgroup$ Oct 8, 2023 at 15:55

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