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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)

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I don't understand. If I am reading your question correctly, you are asking how to prove that an ${\rm SU}(2)$-equivariant map $W_1\otimes W_2 \to W_3$ is a scalar, where the $W_i$ are irreps. But it will not be: indeed, unless $W_1$ is the trivial irrep, then $W_1\otimes W_2$ is not an irrep, and so certainly not isomorphic to $W_3$, and so a map between them cannot be a multiple of the identity, because there is no identity between them. –  Theo Johnson-Freyd May 16 '10 at 18:27
    
What if $W_1\bigotimes W_2$ decomposes to a sum that contains $W_3$ ? –  Pedro May 16 '10 at 18:49
    
So what Baez means in gr-qc/9905087 in pp 18? >"each basis of intertwiners $\iota:\; j_1 \bigotimes j_2 \rightarrow j_3$ contains at most one element." –  Pedro May 16 '10 at 18:55
    
Why do you use $\bigotimes$ instead of just $\otimes$? –  Kevin H. Lin May 17 '10 at 1:01
    
Sorry I thought that the bug is general, in my browser (chrome) $\otimes$ does not appear. –  Pedro May 17 '10 at 1:05

2 Answers 2

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 $\mathrm{SU}(2)$.

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it's precisely what bothers me: * why $\mathrm{Hom}_G(V_1\bigotimes V_2,V_3)$ is nonzero only if $j_3$has same parity as $j_1+j_2$ and... * why $\mathrm{Hom}_G(V_1\bigotimes V_2,V_3)$ is one dimensional if the previous condition is fullfiled –  Pedro May 16 '10 at 12:06
1  
This is a special fact about $\mathrm{SU}(2)$. One looks at the action of diagonal matrices $\mathrm{diag}(u,1/u)$ on the representation. An irreducible representaion breaks up into "weight spaces" where this matrix multiplies the vector by $u^j$. The representation $V_j$ splits into one-dimensional weight spaces of weights $-j$, $-(j-2)$, $-(j-4),\dots,j$. To count the number of irreps of a given type in a finite-dimensional representation one does a census of the weight spaces. If you do that for $V_i\otimes V_j$ you find that the stated irreps turn up just once each. –  Robin Chapman May 16 '10 at 12:26
    
I know that the a given irrep space shows up just one time in the tensorial sum decomposition of a tensorial product, my question is perhaps silly but why if I have $\mathrm{Hom}_G(V_1\bigotimes V_2,V_3) \equiv \mathrm{Hom}_G(...\oplus V_3 \oplus ...,V_3)$ then $\mathrm{Hom}_G$ is 1-dimensional, and zero otherwise. –  Pedro May 16 '10 at 12:55
    
A question that may shed light on my confusion: If I have two irreps $\rho_{j_1}$ and $\rho_{j_2}$ of SU(2), does the condition $\iota \circ \rho_{j_1} = \rho_{j_2} \circ \iota$ implies that $\iota$ is identically zero? –  Pedro May 16 '10 at 13:17
2  
I'm not sure what your $\rho$s are, but by Schur's lemma if $V$ and $W$ are finite-dimensional complex irreducible representations of a group $G$ then either $\mathrm{Hom}_G(V,W)$ is one-dimensional or zero-dimensional according to whether or not $V$ and $W$ are isomorphic. Also can I recommend Fulton and Harris's book Representation Theory as an excellent introduction to the representation theory of classical Lie groups. –  Robin Chapman May 16 '10 at 14:02

By definition, an intertwiner between a representation $V$ to a representation $W$ of a group $G$, say, is a $G$-equivariant linear map from $V$ to $W$.

In your example, assuming that $j_3 \in \lbrace |j_1-j_2|,|j_1-j_2|+1,\dots,j_1+j_2\rbrace$, there is precisely one $V_{j_3}$ factor in the decomposition of $V_{j_1} \otimes V_{j_2}$ into irreducibles: $$ V_{j_1} \otimes V_{j_2} \cong V_{|j_1-j_2|} \oplus V_{|j_1-j_2|+1} \oplus \cdots \oplus V_{j_1+j_2}.$$ (This is the Clebsch-Gordan series.)

The map $\iota$ is then simply the projection onto $V_{j_3}$ along its complement.

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Is it the fact that $\iota$ is an intertwinet that renders it the projection onto $V_{j_3}$? –  Pedro May 16 '10 at 13:04
1  
Yes, in the sense that there is no nonzero $G$-equivariant linear map from $V_j$ to $V_k$ unless $j=k$. –  José Figueroa-O'Farrill May 16 '10 at 13:44
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Above I should have simply put $G=\mathrm{SU}(2)$. –  José Figueroa-O'Farrill May 16 '10 at 13:45
    
Can you give a simple proof of that, or at least a justification? –  Pedro May 16 '10 at 14:04
1  
@Pedro: $V_j$ and $V_k$ are irreducible. Try to work out what the kernel and cokernel of such an equivariant map could be. –  S. Carnahan May 16 '10 at 16:05

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