Any 2-tensor = Sum of a symmetric 2-tensor + an alternating 2-tensor The following problem was asked in my assignment of linear algebra and I was not able to solve this.
Problem: Show that every 2-tensor can be uniquely written as sum of a symmetric 2-tensor and an alternating 2-tensor. Show that the corresponding statement of 3-tensors does not hold.
While web searching I found a solution here:https://www.youtube.com/watch?v=5BPmai9mTsw , but unfortunately I am not able to clearly understand the notations and I found the proof not rigorious enough. Can you please guide me through a proof.
For 2nd part, I found the following solution on MSE: https://math.stackexchange.com/a/95077/775699
but I have question in the solution and the OP was seen more than a month ago. So, I am asking them here:
Why does sum of alternating and symmetric tensor should have a value that is fixed under the action of $A_3$. I understand why symmetric tensor must follow this property but I am confused about alternating tensor. Can you please help me understand that?
Thanks!
 A: Let's view tensors as mathematicians first. The tensor product $V\otimes W$ of two vector spaces $V$ and $W$ is comprised of linear combinations of 'pure tensors' $v\otimes w$ (where $v\in V$, $w\in W$), subject to the assumption the tensor symbol $\otimes$ is bilinear, i.e. $(\lambda v)\otimes w=\lambda(v\otimes w)=v\otimes(\lambda w)$ for scalars $\lambda$ and the distributive property $(v_1+v_2)\otimes(w_1+w_2)=v_1\otimes w_1+v_1\otimes w_2+v_2\otimes w_1+v_2\otimes w_2$ holds. If $\{e_1,\cdots,e_r\}$ is a basis for $V$ and $\{f_1,\cdots,f_r\}$ a basis for $W$, then $\{e_i\otimes f_j\mid\substack{1\le i\le s \\ 1\le j\le r}\}$ is a basis for $V\otimes W$.
We can then define tensor powers $V^{\otimes n}=\overbrace{V\otimes\cdots\otimes V}^n$. The symmetric group $S_n$ acts linearly on this by permuting factors, namely $\sigma(v_1\otimes\cdots\otimes v_n)\sigma=v_{\sigma^{-1}(1)}\otimes\cdots\otimes v_{\sigma^{-1}(n)}$. For example if $\sigma=(123)$ in cycle notation, then $(123)(v_1\otimes v_2\otimes v_3)=v_3\otimes v_1\otimes v_2$ cycles the factors just as it cycles the numbers $1,2,3$ for any three vectors $v_1,v_2,v_3\in V$. The $\sigma^{-1}$s are necessary to ensure this is a left group action (using $\sigma$ instead of $\sigma^{-1}$ in the subscripts yield a right group action, which is useful for describing Schur-Weyl duality).
A symmetric tensor is one which is invariant under the action permutations. An antisymmetric tensor is one which either remains fixed or changes sign under the action of even or odd permutations, i.e. $\sigma \eta=\mathrm{sgn}(\sigma)\eta$ if $\eta\in V^{\otimes n}$ is antisymmetric and $\sigma\in S_n$. The alternating subgroup $A_n\subset S_n$ is comprised of even permutations, which are products of evenly-many transpositions and make up half of $S_n$.
Let's look at the tensor square $V^{\otimes 2}=V\otimes V$. Pick a basis $\{e_1,\cdots,e_n\}$ for $V$. Every tensor $\eta$ may be written as a linear combination of $\eta=\sum_{i,j}\eta^{ij}e_i\otimes e_j$s for some array of coefficients $\eta^{ij}$. I leave it as an exercise to verify that if $\eta\in V^{\otimes2}$ is symmetric, then $\eta^{ij}=\eta^{ji}$ for all indices $i,j$ and thus a basis for $S^2V$ (the symmetric square, the subspace of symmetric tensors) is $\{\tfrac{1}{2}(e_i\otimes e_j+e_j\otimes e_i)\mid{\small1\le i,j\le n}\}$ which has $n+\binom{n}{2}$ elements (the $n$ "diagonals" $e_i\otimes e_i$ and the $\binom{n}{2}$ off-diagonals for distinct pairs $i,j$). Meanwhile $\{\frac{1}{2}(e_i\otimes e_j-e_j\otimes e_i)\mid{\small1\le i<j\le n}\}$ is a basis for the antisymmetric tensors, which has $\binom{n}{2}$ elements.
If we reinterpret $\eta=\sum \eta^{ij}e_i\otimes e_j$ as an $n\times n$ matrix $N=[\eta^{ij}]$, then applying the transposition $(12)$ to $\eta$ corresponds to transposing the matrix. Thus, symmetric and antisymmetric tensors correspond to symmetric and antisymmetric matrices. Every matrix is uniquely a sum of a symmetric and antisymmetric matrix via $A=\frac{1}{2}(A+A^T)+\frac{1}{2}(A-A^T)$.
The corresponding decomposition of a tensor $\eta$ into symmetric and antisymmetric tensors is $v\otimes w=\frac{1}{2}(v\otimes w+w\otimes v)+\frac{1}{2}(v\otimes w-w\otimes v)$ for pure tensors $v\otimes w$. The sum of (anti)-symmetric tensors is (anti)-symmetric, so we can write any tensor as a sum of pure tensors, split each pure tensor into symmetric and antisymmetric components, then collect the symmetric and antisymmetric terms respectively. To see why the sum is unique, suppose $\eta=\alpha_1+\beta_1=\alpha_2+\beta_2$ are two different decompositions, then notice $\alpha_1-\alpha_2=\beta_2-\beta_1$ is both symmetric and antisymmetric, forcing it to be zero (why?).
By definition, when applying a transposition to any antisymmetric tensor, it changes its sign, as in $(12)\eta=-\eta$. That means applying an even number of transpositions, i.e. an even permutation, keeps it the same, because $(-1)^2=1$! Therefore, antisymmetric tensors are $A_n$-invariant. (Moreover, every $A_n$-invariant tensor is uniquely a sum of a symmetric and antisymmetric tensor.) In particular, $(123)=(12)(23)$ is a product of two transpositions.
You should be able to find bases for the symmetric and antisymmetric tensors in $V^{\otimes 3}$ and see that it's not enough to span the whole tensor cube. Examining your bases, find a tensor which is not in the span of the symmetric and antisymmetric tensors. (Equivalently, a tensor that is not invariant under cycling the tensor factors.)
In general, $V^{\otimes n}$ as a representation of ${\rm GL}_n(V)$ decomposes as a sum of Schur functors applied to $V$, indexed by integer partitions $\lambda$ of $n$. The subspaces of symmetric and antisymmetric tensors are just two of these functors.
