# The Dimension of the Symmetric $k$-tensors

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.

• @Thomas They are equal indeed. Dec 25, 2011 at 13:57
• 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'.
– user20266
Dec 25, 2011 at 13:59

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 $$n!$$.)

• 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 $\{i_1,...,i_k+1-1:1\le i \le n\}$? Apr 24, 2013 at 19:15
• Thanks for such great hint. @Squirtle just modify your comment in the end a little bit Jan 4, 2016 at 1:55
• why is this set a basis? Mar 11, 2017 at 9:21