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I would like to know the proof/explanation for the following three properties of the representation of $SO(6)$,

  • What is the importance of symmetric traceless tensors of arbitrary rank w.r.t $SO(6)$ representations?

  • Why can the $6$ dimensional vector representation of $SO(6)$ be thought of as the rank-$2$ anti-symmetric representation of $SU(4)$?

  • Why can the tensor product two of $6$ dimensional vector representations of $SO(6)$ be thought of as a sum of a symmetric traceless, anti-symmetric and a one-dimensional representation?

I would also like to know if there are some general properties of $SO(n)$ representations from which the above follow. Then may be someone can may be kindly also write down the explanation for $SO(n)$ in general!

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I don't understand how the first statement constitutes a property. Can you be more precise? –  Qiaochu Yuan Aug 12 '12 at 18:04
    
I'm confused on your third question-- doesn't the vector representation usual refer to the defining rep of $SO(6)$ on $\mathbb R^6$? But this is irreducible. –  Eric O. Korman Aug 13 '12 at 1:06
    
@Qiachu Yuan I am not sure how to make it precise. From certain literature I get the feeling that symmetric traceless tensors of arbitrary rank form a (irreducible?) representation of $SO(6)$. It would be great if you can elaborate on that point. Are these somehow the "fundamental representations"? If so then how? –  user6818 Aug 13 '12 at 22:57
    
@Eric I think I had a typo in the question. I tried editing it. Does it now make sense? –  user6818 Aug 13 '12 at 22:58
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1 Answer

up vote 3 down vote accepted

I'm not sure how to answer question 1-- it is a little vague. The second question is unique to $SO(6)$ and is answered by understanding how $SU(4)$ is the double cover of $SO(6)$. This double cover is constructed as follows: first notice that $\Lambda^2 \mathbb C^4$ is six dimensional and has a Hermitian inner product induced by the one on $\mathbb C^4$. This can be defined by saying if $\{e_1,\ldots,e_4\}$ is an orthonormal basis for $\mathbb C^4$ then $\{e_i \wedge e_j\}$ is an orthonormal basis for $\Lambda^2 \mathbb C^4$ or, more invariantly, $$ \langle v_1 \wedge v_2, w_1 \wedge w_2 \rangle = \det\langle v_i, w_j\rangle $$ where $\langle v, w\rangle$ is the Hermitian inner product on $\mathbb C^4$. Then it is easy to see that the action of $SU(4)$ on $\Lambda^2 \mathbb C^4$ preserves this inner product, giving a homomorphism $SU(4) \to U(\Lambda^2 \mathbb C^4) = U(6)$. But there is also a symmetric inner product on $\Lambda^2 \mathbb C^4$ $$ \Lambda^2 \mathbb C^4 \otimes \Lambda^2 \mathbb C^4 \to \Lambda^4 \mathbb C^4 \simeq \mathbb C $$ which is also preserved by the action of $SU(4)$ since elements in $SU(4)$ have determinant 1. Thus the image of the above homomorphism lies in $U(6) \cap SO(6, \mathbb C) = SO(6)$. One checks that the kernel of this map is $\{\pm 1\}$ and therefore, by dimensionality reasons, must be a surjection. Thus this is the double cover $SU(4) \to SO(6)$ but now it is transparent that the $\Lambda^2 \mathbb C^4$ rep of $SU(4)$ corresponds to the vector representation of $SO(6)$.

Your third question holds for all of $SO(n)$ (indeed for any self-dual representation). Since there is an invariant symmetric inner product on $V = \mathbb C^n$ preserved by $SO(n)$, it follows that the fundamental representation is self-dual. Thus $V\otimes V \simeq V^* \otimes V \simeq End(V)$. Now for any rep $V$ of any group, $V \otimes V$ always reduces into a direct sum of symmetric plus antisymmetric (this is immediate from the definition of the tensor product representation). Similarly, for any rep $V$ of any group, $End(V)$ has a one-dimensional subrepresentation spanned by the identity transformation. Now under the correspondence of $End(V) \simeq V\otimes V$, the symmetric tensors will contain the identity endomorphism but a complementary subspace is given by traceless endomorphisms.

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Thanks for the reply. Can you kindly make it explicit as to what is the $SU(4)$ action on $\wedge ^2 \mathbb{C}^4$ that you have in mind? Also given the 6 dimensional real representation of $SO(6)$ being spanned by say $\{ \phi^a \}_{a=1}^6$ how does one write down the basis of antisymmetric rank 2 tensors which span the same when thought of as a representation of $SU(4)$? Also if you could explain as to how the symmetric traceless tensors become irreducible representations of the $SO$ groups...and how/if they are all that can be gotten inside tensor powers. –  user6818 Aug 24 '12 at 15:28
    
@user6818: $A \in SU(4)$ acts on $v\wedge w \in \Lambda^2 \mathbb C^4$ by $(Av) \wedge (Aw)$ where $Av$ is the normal action of $SU(4)$ on $\mathbb C^4$. For the second question I suggest you work out what's happening with a basis of $\Lambda^2 \C^4$ to make a specific identification of $\mathbb R^6$. For the last question, I think it may take a little work to see its irreducible (unless you compute the highest weight and compare with the Weyl dimension formula): any symmetric matrix is diagonalizable via a matrix in $SO(6)$ so given two diagonal tracelss matrices try to find a matrix... –  Eric O. Korman Aug 25 '12 at 2:02
    
...in $SO(6)$ conjugating one into the other. –  Eric O. Korman Aug 25 '12 at 2:03
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Dear Eric, A small comment/query on your nice answer: is it standard to write $S^2 V$ for $V\otimes V$? I would rather have thought that $S^2 V$ stood for $Sym^2 V$ (although the intended meaning of your usage was clear in context). Regards, –  Matt E Oct 21 '12 at 20:08
    
@MattE : No, that was a mistake-- thanks for pointing it out. I also wrote "symmetric" somewhere where I should have said "tensor." Not sure why I kept conflating the two. I have edited it accordingly. –  Eric O. Korman Oct 21 '12 at 21:40
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