6
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How many independent components does a rank three totally symmetric tensor have in $n$ dimensions?

Needed for the irrep decompositon of $3\otimes 3\otimes 3$ in here.

No idea where to start to prove this.

I did come up with a more clever way of figuring it out for $n=3$, but I don't think it can be generalized. In $3$ dimensions, a totally antisymmetric (rank three) tensor has one component. From here I just counted the components that are nonzero for a totally symmetric one. We have the $3$ diagonal components, obviously. Then there are the components of the form $iij$ (no sum). The $i$s run over $3$ values and $j$ over $2$. Since $2\cdot 3=6$ we have $1+3+6=10$ components.

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  • $\begingroup$ You'll have to elaborate on some notation here. What is $3 \otimes 3 \otimes 3$? Is that $\Bbb C^3 \otimes \Bbb C^3 \otimes \Bbb C^3$? Keep in mind that mathematicians and physicists often use different notation. $\endgroup$ – Omnomnomnom Jan 4 '15 at 21:32
  • $\begingroup$ $3$ is the $SU(3)$ quark triplet. The relation I'm looking to prove is $3\otimes3\otimes3=10\oplus8\oplus8\oplus1$. I need to show that $10$ is the number of independent components of rank three completely symmetric tensor in 3 dimensions. I did that explicitly by writing down the $27$ components and matching identical ones up. I am wondering if there is an elegant formula for $n$ dimensions. But the particle physics stuff is not important to this question. $\endgroup$ – Ryan Unger Jan 4 '15 at 21:37
  • $\begingroup$ See here for an additional answer to the question. $\endgroup$ – Dilaton Jun 29 '16 at 7:43
10
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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.


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 sequences 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

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

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  • $\begingroup$ 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? $\endgroup$ – Santi Sep 30 '15 at 13:15
  • $\begingroup$ @Santi : Although so late I think the number is 18. Take a look in this 3D-visualization of a totally symmetric tensor : imgur.com/a/BCfbF. If you draw two diagonal planes you restrict the 10 independent components if totally symmetric. But if you draw one diagonal plane you restrict the 18 independent components if symmetric in just two two of its indices (9 elements on the diagonal plane + 9 elements in the one of the two halves of the cube). $\endgroup$ – Frobenius Aug 30 '17 at 7:32

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