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.
The Dimension of the Symmetric $k$-tensors
8
$\begingroup$
linear-algebra
differential-geometry
tensors
-
1While 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'. – 2011-12-25
1 Answers
12
a basis for symmetric tensors, say $\otimes_1^r V$ with $\{v_1,...,v_n\}$ a basis for $V$, is given by the symmetrizations of $\{v_{i_1}\otimes ... \otimes v_{i_r} \ | \ 1\leq i_1\leq...\leq i_r\leq n\}$. so 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, ie counting the number of multisets of size $r$ with entries from $[1,n]$, and the answer you get is ${n+r-1\choose r}$.
-
1why is this set a basis? – 2017-03-11