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I want to compute the dimension of the symmetric $k$-tensors. I know that a covariant $k$-tensor $T$ is called symmetric if it is unchanged under permutation of arguments. Also, I know that the dimension of covariant $k$-tensors is $n^k$ but how can I eliminate non-symmetric the cases? I found this but I could not get the intution. Also, this blog post answers my question but I don't see why we put | between different indices. Any concrete example would also help such as the symmetric covariant 2-tensors in $\mathbb{R^3}$, as I asked in this thread.

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    $\begingroup$ @Thomas They are equal indeed. $\endgroup$ Dec 25, 2011 at 13:57
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    $\begingroup$ While tensor (fields) are used extensively in differential geometry, your question is not specific for differential geometry. You might want to consider (re)tagging it with 'tensors', 'tensor-products', 'multilinear-algebra'. $\endgroup$
    – user20266
    Dec 25, 2011 at 13:59

1 Answer 1

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A basis for symmetric tensors, say $\text{Sym}_r(V)$ with $\{v_1,...,v_n\}$ a basis for $V$, is given by the symmetrizations of $\{v_{i_1}\otimes ... \otimes v_{i_r} \ | \ 1\leq i_1\leq...\leq i_r\leq n\}$. You must count the number of non-decreasing sequences (repetitions allowed) of length $r$ with entries in $[1,n]$. I've always heard the method of counting these referred to as stars and bars, i.e. counting the number of multisets of size $r$ with entries from $[1,n]$, and the answer you get is ${n+r-1\choose r}$.


You line up $r$ stars and insert $n-1$ bars, the first bar separating indicies 1 and 2, the second bar separating indicies 2 and 3, ..., the $(n-1)$st bar separating indicies $n-1$ and $n$.

For example, say $r=5$ and $n=3$. Here are some example of non-decreasing sequences of length $r=5$ with entries from $\{1,2,3\}$: \begin{align*} 11223 \ &: \ **|**|*\\ 22333 \ &: \ |**|***\\ 11122 \ &: \ ***|**|\\ 22222 \ &: \ |*****|\\ \end{align*} So there are $r+n-1$ ``things'' (stars and bars) and you're choosing $r$ of them to be stars (or $n-1$ of them to be bars).


As for why this determines a basis for symmetric tensors: any pure tensor on the chosen basis determines a symmetric tensor via $$ S(v_{i_1}\otimes ... \otimes v_{i_r})=\sum_{\pi\in S_n}v_{\pi(i_1)}\otimes ... \otimes v_{\pi(i_r)} $$ and two pure tensors have the same symmetrization if their indices determine the same multiset (i.e. non-decreasing sequence as described above). I'll leave it to the reader to show that these are independent and that they span the space of symmetric tensors. (On a technical note, the symmetrization needs to be modified in non-zero characteristic and some sources might divide by $n!$.)

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    $\begingroup$ Hint.... take $\{i_1\le \cdots \le i_k\}$ then shift this set in the following way $\{i_1+0<i_2+1<i_2+2< \cdots < i_k+k-1\}$ Now how big is $\{i_1,...,i_k+1-1:1\le i \le n\}$? $\endgroup$
    – Squirtle
    Apr 24, 2013 at 19:15
  • $\begingroup$ Thanks for such great hint. @Squirtle just modify your comment in the end a little bit $\endgroup$
    – Sushil
    Jan 4, 2016 at 1:55
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    $\begingroup$ why is this set a basis? $\endgroup$ Mar 11, 2017 at 9:21

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