I think that a reformulation of my question is necessary:
An intertwiner $\iota:\; V_{j_{1}}\bigotimes V_{j_{2}}\rightarrow V_{j_{3}}$ is defined as:
$\forall g\in SU(2),\;\forall u_{i},v_{i},w_{i},…\in V_{j_{i}}:\;\iota((\rho_{j_{1}}\bigotimes\rho_{j_{2}})(g)\;[v_{1}\bigotimes v_{2}])=\rho_{j_{3}}(g)[\iota(v_{1}\bigotimes v_{2})]$
where $\rho_{j_{i}}$ is the representation map corresponding to the irrep of spin $j_i$ of $SU(2)$, and $V_{j_i}$ are the invariant spaces upon which acts the $\rho_{j_{i}}(g)$ for $g\in SU(2)$
I know that the Shur's lemma is: if
$\iota:\; V_{j}\rightarrow V_{k}$
is an interwtwiner, then is it either a scalar (if $j=k$) or zero ($j\not= k$)
Now, what I want to know, is if $V_{j_{1}}\bigotimes V_{j_{2}} = \dots\oplus V_{j_{2}} \oplus \dots$ take an example $V_{j_{1}}\bigotimes V_{j_{2}} = V_{j_{4}} \oplus V_{j_{2}} \oplus V_{j_{5}}$ I can write:
$\forall g\in SU(2),\;\forall u_{i},v_{i},w_{i},…\in V_{j_{i}}:\;\iota((\rho_{j_{4}}\oplus\rho_{j_{3}}\oplus\rho_{j_{5}})(g)\;[v_{1}\bigotimes v_{2}])=\rho_{j_{3}}(g)[\iota(v_{1}\bigotimes v_{2})]$
in this case how to prove that $\iota$ is a scalar? (by Shur's lemma)
Best Answer
I presume by $V_j$ you mean the $j$-dimensional (or maybe $(j+1)$-dimensional) representation of $\mathrm{SU}(2)$. I wouldn't write something like $j_i\otimes j_2\to j_3$ since that notation confuses the represntation $V_j$ with the label $j$. I would write $\iota:V_{j_i}\otimes V_{j_2}\to V_{j_3}$. This notation is honest in the sense that it exhibits the spaces between which $\iota$ is a map.
To simplify notation further I would write $\iota : W_1\otimes W_2\to W_3$ and assume that the $W_i$ are repsentations of any group $G$. Then $\iota$ is an intertwiner (or module homomorphism) if $$\iota(g v_1\otimes g v_2)=g \iota(v_1\otimes v_2)$$ for any $g\in G$ and $v_i\in V_i$. The intertwiners form a vector space $\mathrm{Hom}_G(V_1\otimes V_2,V_3)$.
In your example, take $V_i$ to be the $(j_i+1)$-dimensional irreducible representation of $\mathrm{SU}(2)$ then $\mathrm{Hom}_G(V_1\otimes V_2,V_3)$ is nonzero if and only if $j_3$ has the same parity as $j_1+j_2$ and $ | j_1-j_2 |< j_3\le j_1+j_2 $. In this case $\mathrm{Hom}_G(V_1\otimes V_2,V_3)$ is one-dimensional, but note that this is a very special property of $\mathrm{SU}(2)$.