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In the same spirit of my question here I want to decompose $\mathfrak{sl}_2 \otimes \mathfrak{sl}_2$ into a direct sum of irreducible representations of $\mathfrak{sl}_2$. There is a unique irreducible representation of dimension of $\mathfrak{sl}_2$ $n\in\mathbb N^*$ which is denoted $V(n)$.

I know that $W\otimes W=S^2W\oplus \bigwedge^{2}W$, here if we write the basis $\{e,f,h\}=\{v_0,v_1,v_2\}$ the exterior power would be generated by $\{e\otimes f,e\otimes h,f\otimes h\}$ and the symmetric part by the remaining tensor products between elements of the basis.

We see here that $\mathfrak{sl}_2\otimes \mathfrak{sl}_2\cong V(4)\oplus V(2)\oplus V(0)$. I tried to show that $\bigwedge^2 \mathfrak{sl}_2$ is irreducible by saying that if there is a sub representation of dimension 1, it corresponds to the trivial one so if $\sum_{i,j}a_{ij}v_i\otimes v_j$ spans this space we must have $$[e,u]=\sum_{i,j}a_{ij}([e,v_i]\otimes v_j+v_i\otimes [e,v_j])$$ and from this I get that $a_{ij}=0$ $\forall i,j$.

So I concluded that it's irreducible, so the 1 dimensional subrepresentation must be in $S^2\mathfrak{sl}_2$, but by doing the same process with an element of $S^2\mathfrak{sl}_2$ I get that there is no one dimensional subrepresentation in $S^2\mathfrak{sl}_2$.

So how do we get the decomposition here ?

raisinsec
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    Are you familiar with the Clebsch-Gordon formula? Whenever $m\ge n$ we have $$V(m)\otimes V(n)\simeq V(m+n)\oplus V(m+n-2)\oplus\cdots\oplus V(m-n).$$ The highest weights go down from $m+n$ to $m-2$ in decrements by two.

    Anyway, $\mathfrak{sl}_2\simeq V(2)$ as a rep, so all you need to do is to expand $V(2)\otimes V(2)$. There are even formulas for finding the explicit highest weight vectors of the summands (in terms of some standard basis of weight vectors).

     – Jyrki Lahtonen May 11 '23 at 18:43
  • The first link anyway has $V(2)\otimes V(2)\cong V(4)\oplus V(2)\oplus V(0)$. So you want to know $\Lambda^2(V(2))$ and $S^2(V(2))$, I suppose? – Dietrich Burde May 11 '23 at 18:44
  • Anyway, it seems to me that Clebsch-Gordon is a better tool here than thinking about in terms of even/odd tensors. – Jyrki Lahtonen May 11 '23 at 18:44
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    The one dimensional rep you mention is precisely the span of the Killing form (under appropriate identification) so is certainly in $S^2\mathfrak{sl}_2$ – Callum May 11 '23 at 18:46
  • @JyrkiLahtonen No I'm not familiar with it but I think this the result I'm using from the link to decompose the tensor product into sums. I just don't know how to describe the sums or decompose them further – raisinsec May 11 '23 at 20:50
  • @Callum Yes I guess you're true, but I don't know why. Taking a linear combination of ${e\otimes e, f\otimes f, h\otimes h, f\otimes e, h\otimes e, h\otimes f}$ and applying $[e,-]$ and $[f,-]$ separately I get that coefs are zero. Here $[e,v\otimes w]=[e,v]\otimes w+v\otimes [e,w]$, am I misunderstanding something ? – raisinsec May 11 '23 at 20:54
  • @DietrichBurde I guess, if I'm right $V(2)$ is the exterior power and the rest is the symmetric part. So it's enough to describe how $V(4),V(2),V(0)$ look like in the tensor product. I'm looking for explicit generators but what I tried to get $V(0)$ didn't work although it should be quite direct. – raisinsec May 11 '23 at 20:59
  • Writing it as $[e,v\otimes w]$ is a bit of an abuse of notation but that is the correct formula. However, you are listing only 6 elements there as a basis and there are 9. Note $f \otimes e \neq e\otimes f$. I get $h \otimes h + 2e \otimes f + 2f\otimes e$ as an element of the trivial rep. – Callum May 12 '23 at 10:14
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    Additionally the exterior product is not generated by $f\otimes e$, etc. Indeed that is not even an element of the exterior power which instead consists of things like $f\otimes e - e\otimes f$. I think perhaps you have misunderstood how tensor products work – Callum May 12 '23 at 10:16
  • @Callum Isn't the basis of the exterior algebra given by ${e_{i_1}\otimes \dots \otimes e_{i_n}|i_1<\dots < i_n}$ – raisinsec May 14 '23 at 19:44
  • @Callum I listed 6 elements which I consider to be the basis of the symmetric algebra, which is of dimension 6. But apparently there is a problem with my basis elements. – raisinsec May 14 '23 at 19:48
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    @raisinsec If you write wedge products instead of tensor products that would be a basis of $\bigwedge^* V$. But we are finding $\bigwedge^2 V$ as a subspace of $\bigotimes^2 V$. So antisymmetric tensors are of the form $v \otimes w - w\otimes v$. Indeed we could define that as $v \wedge w$. Meanwhile the symmetric square would be tensors of the form $v \odot w := v \otimes w + w\otimes v$ (which includes $v \otimes v$ of course). – Callum May 15 '23 at 15:54

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