5
$\begingroup$

EDIT: I have boiled my question down to

How many independent components does a rank three totally symmetric tensor have in $n$ dimensions?

A derivation would be nice too.

OP: I know that I can represent $3\otimes 3$ by a rank two tensor. Then the symmetric part is $3(3+1)/2=6$ and the antisymmetric part is $3(3-1)/2=3$. Thus $$3\otimes 3=6\oplus\bar 3$$ I'm not quite sure about the bar though...It is not clear to me why there is a bar when I do this diagrammatically, because the $\bar 3$ has the same isospin and hypercharge states as a quark triplet. So that's my first question.

So then $$3\otimes 3\otimes 3=3\otimes(6\oplus \bar{3})=(3\otimes 6)\oplus(3\otimes \bar 3)$$ I can represent $3\otimes \bar 3$ by a mixed tensor and separate out the trace. Thus $$3\otimes \bar 3=8\oplus 1$$ I get this just fine.

But I'm not sure how to go about doing $3\otimes 6$. Since $6$ is a symmetric tensor, $3\otimes 6$ should be a rank three tensor, symmetric over two indices. Then I could form a totally symmetric tensor and a tensor antisymmetric over two indices. But I have no clue which one is $10$ and which one is $8$ and obviously no clue how to prove it.

Any help would be greatly appreciated.

EDIT I: I once heard that in three dimensions an antisymmetric tensor is a vector. Is this the reason for writing the second rank antisymmetric tensor as $\bar{3}$? But how does one identify an antisymmetric tensor with the conjugate representation?

EDIT II: Strike the first question. If $\varphi^{kl}$ is the antisymmetric tensor, then I can use the Levi-Civita symbol to write $\varphi_i=\varepsilon_{ikl}\varphi^{kl}$, which is the $\bar{3}$.

EDIT III: I know how to get the $8$. Represent $6$ by the symmetric $S^{ij}$ and $3$ by the vector $\varphi^k$. Then define $\varphi_{lm}=\varepsilon_{lmk}\varphi^k$ and $T^{ij}_{lm}=S^{ij}\varphi_{lm}$. The tensor $T^{ij}_{lj}$ is traceless and is thus a $3^2-1=8$. From this I see that this $8$ has mixed symmetry properties, as is should, because $$T^{ij}_{lj}=T^{ji}_{lj}=-T^{ji}_{jl}=-T^{ij}_{jl}$$ Now I know that $6\otimes 3=10\oplus 8$, but how do I figure out that the $10$ is an irrep and totally symmetric?

EDIT IV: Now I tried writing $T^{ijk}=S^{(ik}\varphi^{k)}$. This should be the $10$. I wrote down the 27 components and verified there are $10$ degrees of freedom, as expected. But this is kinda laborious. So my question now becomes: How many independent components does a rank three totally symmetric tensor have in $n$ dimensions? If the answer is 10 for $n=3$, then my question will be answered.

$\endgroup$

closed as off-topic by Danu, Jim, bobie, Qmechanic Jan 6 '15 at 13:21

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Qmechanic
If this question can be reworded to fit the rules in the help center, please edit the question.