Question
Consider two symmetric tensors $A_{ab}=A_{ba},\, B_{ab}=B_{ba}$ which are generally of full rank (non-degenerate, all eigenvalues non-zero but of general sign) and their tensor product under a particular symmetrization $$(A_{a(b} B_{cd)} - B_{a(b} A_{cd)})\,,$$ where $(bcd)$ denotes the totally symmetric part of the tensor $$C_{(abc)} = \frac{1}{6}\sum_{\pi \in S_3} C_{\pi(a) \pi(b) \pi(c)}\,.$$ Now I would like to figure out how is the space of vectors $v^a$ restricted by the condition (Einstein summation assumed) $$v^a (A_{a(b} B_{cd)} - B_{a(b} A_{cd)}) = 0\,.$$ I.e., how large is the space of vectors $v^a$ fulfilling the above condition?
(If it is of any help, I am mainly interested in dimension 4.)
Attempts at solution/comments
A tensor with three totally symmetric indices $b c d$ has 20 independent components (in dimension 4). This means that a condition of the type $$v^a D_{a(bcd)}$$ corresponds generally to twenty different conditions on the space of possible vectors $v^a$ and we thus have necessarily $v^a=0$.
(This would actually apply in any dimension because a 3-index totally symmetric tensor in dimension $n$ has $n+2 \choose 3$ independent components which is always $\geq$ than the critical $n$ conditions.)
However, if we were to only consider the condition $$v^a A_{a(b}B_{cd)}=0$$ it is unclear to me whether the fact that before symmetrization in $bcd$, we had $A_{a b} = A_{ba}$ affects the number of the resulting independent conditions on $v^a$. This symmetry does not mean $A_{a(b}B_{cd)} = A_{b(a}B_{cd)}$ and I am not quite sure how to approach it.
Then there is the antisymmetry with respect to the swapping of indices $ab \leftrightarrow cd$ which I am also unsure on how to handle. Normally, if we have a tensor $K_{(ab)(cd)}$ which is symmetric in $a\leftrightarrow b, c\leftrightarrow d$ and antisymmetric with respect to $ab \leftrightarrow cd$, it would have $${{n+1 \choose 2} \choose 2}|_{n=4} = 45$$ independent components.
But then again, the tensor $A_{ab}B_{cd} - A_{cd} B_{ab}$ possesses these symmetries but is antisymmetrized direct product of elements of two spaces. As we all know, the direct products of spaces always involve elements which are not expressible in this way (as a direct product of two elements from the respective spaces). Hence, the number of independent elements of $A_{ab}B_{cd} - A_{cd} B_{ab}$ will be surely smaller than that of a general $K_{(ab)(cd)}$.