Is the Clebsch-Gordan decomposition of Lie algebra or Lie group representations?

Question

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?

Answer 1

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').

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$^1$ 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.

Answer 2

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.