How to calculate $3\otimes 3$ and $3\otimes 3\otimes 3$ in $SU(3)$?

Question

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.

This post imported from StackExchange Mathematics at 2016-06-28 15:13 (UTC), posted by SE-user 0celo7

Comments

Related $SU(3)$ posts: physics.stackexchange.com/q/89173/2451 , physics.stackexchange.com/q/147243/2451 , physics.stackexchange.com/q/10403/2451 and links therein; especially the answer physics.stackexchange.com/a/14586/2451 and links therein.

This post imported from StackExchange Physics at 2015-01-05 12:15 (UTC), posted by SE-user Qmechanic

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That's the one part of this calculation that I understand.

This post imported from StackExchange Physics at 2015-01-05 12:15 (UTC), posted by SE-user 0celo7

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Do you know Young diagrams [pdf]? (The linked pdf explains how to use them to get arbitrary decompositions of $\mathrm{SU}(N)$ representations, though it does not fully derive why this works) The thinking about "tensors" is not very helpful in the general case, but the diagrams always work.

This post imported from StackExchange Physics at 2015-01-05 12:15 (UTC), posted by SE-user ACuriousMind

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I'll take a look at the diagrammatic approach later, but first I want to figure it out algebraically.

This post imported from StackExchange Physics at 2015-01-05 12:15 (UTC), posted by SE-user 0celo7

Answer 1

By a 'rank three tensor' I think you mean an element of $\bigotimes^3V$. This is not what 'rank' means in mathematics. An element of $\bigotimes^kV$ is said to have order (or degree) $k$.

A rank one tensor (or simple tensor) is a tensor of the form $v_1\otimes\dots\otimes v_m$. The rank of a tensor $T$ is the minimal number of rank one tensors needed to express $T$ as a sum.

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Let $V$ be an $n$-dimensional vector space, then for any $k \in \mathbb{N}$, let $\operatorname{Sym}^kV$ denote the collection of symmetric order $k$ tensors on $V$; note that $\operatorname{Sym}^kV$ is a vector space. Let $v_1, \dots, v_n$ be a basis for $V$, then a basis for $\operatorname{Sym}^kV$ is given by

$$\left\{\frac{1}{k!}\sum_{\sigma \in S_n}v_{\sigma(i_1)}\otimes\dots\otimes v_{\sigma(i_k)} \mid 1 \leq i_1 \leq \dots \leq i_k \leq n\right\}.$$

Therefore, $\dim\operatorname{Sym}^kV$ is equal to the number of non-decreasing sequence of $k$ integers in $\{1, \dots, n\}$. Let $x_1 = i_1 - 1$, $x_j = i_j - i_{j-1}$ for $j = 2, \dots, k$, and $x_{k+1} = n - i_k$. Note that the number of non-decreasing sequences of $k$ integers in $\{1, \dots, n\}$ is in one-to-one correspondence with the number of of solutions to $x_1 + \dots + x_{k+1} = n-1$ in non-negative integers. The latter number can be found using the stars and bars method from combinatorics; doing so, we see that

$$\dim\operatorname{Sym}^kV = \binom{(n-1)+(k+1)-1}{n-1} = \binom{n + k -1}{n-1} = \binom{n+k-1}{k}.$$

So every symmetric order $k$ tensor can be written uniquely as a linear combination of $\binom{n+k-1}{k}$ basis tensors. The coefficients in this linear combination are what you refer to as "independent components". In the case you worked out yourself, you had $n = 3$ and $k = 3$ in which case the number of independent components is

$$\binom{3+3-1}{3} = \binom{5}{3} = 10.$$

This post imported from StackExchange Mathematics at 2016-06-28 15:13 (UTC), posted by SE-user Michael Albanese

Comments

Hi, what would be the number of independent components of a tensor order k=3, but which is symmetric in just two of its indices?

This post imported from StackExchange Mathematics at 2016-06-28 15:13 (UTC), posted by SE-user Santi

Answer 2

Essentially for $SU(N)$ you find that tenors can be written in terms of upper and lower indices (except for $SU(2)$ which can be written solely in upper or lower indices due to the group being pseudo-real). In order to pick out the irreps using the highest weight procedure one finds that a tensor, $V_H$ which we describe by $(n,m)$ $n$ being the number of upper indices and $m$ the number of lower, has two properties when acted one by the generators of the group

1. $V_H$ is symmetric in the upper indices and symmetric in the lower indices
2. $V_H$ is traceless

Thus, in order to count the dimensionality of the tensor, we find that we can use simple counting. For the upper indices, that go from 1 to $N$, the number of ways to arrange them is

\[\binom{n + N -1}{N-1} = \frac{(n + N -1)!}{n!(N-1)!} \]

Again for the lower indices

\[\binom{m + N -1}{N-1} = \frac{(m + N -1)!}{m!(N-1)!} \]

Thus the total number of ways to have symmetric upper and lower indices is

\[B(n,m) = \frac{(n + N -1)!}{n!(N-1)!} \frac{(m + N -1)!}{m!(N-1)!} \]

However, the tensor is also required to be traceless. This says that the object
we get by taking the trace vanishes, and it is symmetric in $n-1$ upper indices
and $m-1$ lower indices. Thus this imposes $B(n-1, m-1)$ constraints, so the
total is

\[D(n,m) = B(n,m) - B(n-1,m-1)\]

Using, this equation for the case of N=3, n=3, m=0, you find that D = 10, as you expected.

I would highly suggest learning the Young Diagrams method that was listed in the comments, for it is extremely quick to calculate tensor products of irreps in $SU(N)$. The method I described above was found in Howard Georgi's Book Lie Algebras in particle physics, chapter 10.

EDIT: I just wanted to add that $B(n,m) \equiv 0$ if one of the terms is negative. This is just to say that one does not have any traceless constraints if there is nothing to contract.