# Basis for the set of all covariant $k$-tensors on V

Here's a proposition from Lee's Smooth Manifolds:

Let $V$ be a real vector space of dimension $n$, let $(E^i)$ be any basis for $V$, and let $(\epsilon^i)$ be the dual basis. The set of all $k$-tensors of the form $\epsilon^{i_1}\otimes...\otimes\epsilon^{i_k}$ for $1 \leq i_1,...,i_k \leq n$ is a basis for the set of all covariant $k$-tensors on $V$, denoted by $T^k(V)$, which therefore has dimension $n^k$.

I want to construct an example based on this proposition. Let $V=\mathbb{R}^3$, and let $e_1, e_2, e_3$ denote the usual basis for $\mathbb{R}^3$. Then the dual basis $(\epsilon^i)$ is $dx, dy$ and $dz$. Then $dx\otimes dy, dx\otimes dz, dy\otimes dz, dy\otimes dx, dz\otimes dy, dz\otimes dx, dx\otimes dx, dy\otimes dy$ and $dz\otimes dz$ form a basis for $T^2(\mathbb{R}^3)$.

Is the example correct until this point? If so, which of them are the symmetric 2-tensors?

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You have $dy\otimes dz$ listed twice but missed $dz\otimes dy$, otherwise the list is complete. – Alex Becker Dec 24 '11 at 23:58
@Alex I corrected it. – user20353 Dec 25 '11 at 0:02
@Alex I am not so sure that these form a dual basis for $\mathbb{R}^3$. – user20353 Dec 25 '11 at 0:07
I'm not sure what you're asking. I just assumed that you were using $dx,dy,dz$ to denote the three elements of the dual basis. If you are trying to attach a differential meaning to these, then you're incorrect. The dual basis is simply the set of covectors $e_1^T,e_2^T,e_3^T$. – Alex Becker Dec 25 '11 at 0:10
@user20353: both your dual basis $(dx,dy,dz)$ and the basis for $T^2(\mathbb{R}^3)$ are correct. Finding the symmetric ones should be a minor task. – M.B. Dec 25 '11 at 0:38

Yes, you are correct. As you have quoted, $T^k(V)$ has dimension $n^k$. Now $n=3$ and $k=2$. Therefore $3^2=9$ is the number of bases and there are $9$ elements in $$\{dx\otimes dy, dx\otimes dz, dy\otimes dz, dy\otimes dx, dz\otimes dy, dz\otimes dx, dx\otimes dx, dy\otimes dy, dz\otimes dz\}.$$

None of them is sysmmetric 2-tensor. By definition, symmemetric tensor is a tensor $T$ that is invariant under a permutation of its vector arguments. Therefore, permutate $x$ and $y$, we have $$dx\otimes dy\leftrightarrow dy\otimes dx,\hspace{2mm}dx\otimes dz\leftrightarrow dy\otimes dz, \hspace{2mm} dz\otimes dy\leftrightarrow dz\otimes dx,\hspace{2mm} dx\otimes dx\leftrightarrow dy\otimes dy.$$ On the other hand, permutate $x$ and $z$, we have $$dx\otimes dx\leftrightarrow dz\otimes dz.$$ Therefore, none of them is symmetric. However, you can find symmetric 2-tensor by forming the linear combination of them (since they form a basis), e.g. $$dx\otimes dy+dy\otimes dx+dx\otimes dz+dz\otimes dx+dy\otimes dz+dz\otimes dy$$ or $$dx\otimes dx+dy\otimes dy+dz\otimes dz$$ are symmetric 2-tensors.

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$dx \otimes dx$, $dy \otimes dy$, and $dz \otimes dz$ are indeed symmetric. Being invariant "under a permutation of its vector arguments" does not mean invariant under exchanging x,y,z---it means when viewing a tensor as a bilinear form, it is invariant under a permutation of the arguments of that form. – Santiago Canez Dec 25 '11 at 2:08

Just to elaborate on the comment of Santiago Canez:

We say that a covariant $k$-tensor $T$ is symmetric if it is unchanged under permutation of arguments, i.e: $$T(X_1, \ldots, X_k) = T(X_{\sigma(1)}, \ldots, X_{\sigma(k)})$$ where $\sigma$ is any permuation of $k$ elements.

So it is immediate that $dx\otimes dx$, $dy\otimes dy$ and $dz\otimes dz$ are symmetric. Furthermore, there is a natural projection from $T^k(V)$ to the set of symmetric tensors called symmetrization. Let $S_k$ be the group of permutations of $\{1, \ldots, k\}$ and for any $\sigma\in S_k$ define $$T^\sigma(X_1, \ldots, X_k) = T(X_{\sigma(1)}, \ldots, X_{\sigma(k)}).$$ Then we define $\text{Sym}: T^k(V) \to\text{symmetric tensors}$, by $$\text{Sym}\, T = \frac{1}{k!}\displaystyle\sum_{\sigma\in S_k}T^\sigma.$$ Then $\text{Sym}\, T$ is a symmetric tensor.

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