Let $\mathbb{F}$ be a field of characteristic $2$ and define $S$ to be the set of all triples $(i,j,k)\in\lbrace 1,\dotsc,n\rbrace^3$ with $\left|i-j\right|=1$, $\left|i-k\right|>1$, and $\left|j-k\right|>1$. Define $V=\bigoplus\limits_{(i,j)\in\lbrace 1,\dots,n\rbrace^2}\mathbb{F}\sigma_{i,j}$ to be the vector space with basis the quotient of $\lbrace \sigma_{i,j}\rbrace_{(i,j)\in\lbrace 1,\dots,n\rbrace^2}$ by the relation $\sigma_{i,j}=\sigma_{j,i}~\forall~(i,j)\in\lbrace 1,\dots,n\rbrace^2$. We can consider the vector subspace $V_S=\sum_{(i,j,k)\in S}\mathbb{F}(\sigma_{i,k}+\sigma_{j,k})$ (I don't know if this is an standard notation, but I want to represent the sum of subspaces). It is obvious that the above sum is not direct, since for every tuple $(i-1,i,j,j+1)$ with $i<j$ and $\left|j-i\right|>1$ we have $$(\sigma_{i-1,j}+\sigma_{i-1,j+1})+(\sigma_{i,j}+\sigma_{i,j+1})+(\sigma_{i-1,j}+\sigma_{i,j})+(\sigma_{i-1,j+1}+\sigma_{i,j+1})=0.$$ My question is if we can find the number of summands needed in $V_S$ such that the vector space is represented as a direct sum, which is equivalent to finding $\dim(V_S)$. It is easy to prove that $\left|S\right|=(n-2)(n-3)$ and with a bit of extra work one can see that the problem mentioned above can be solved by deleting $\frac{(n-3)^2}{4}$ summands in case $n$ is odd and $\frac{(n-2)(n-4)}{4}$ summands in case $n$ is even. However, this is not enough, for the case $n=6$ there is one extra summand to delete which does not correspond to the above situation. I guess there must be many more extra summands to delete for bigger $n$.
This problem appeared when trying to compute the homology of a certain family of groups, and I am not an expert on combinatorics, so I'm quite lost.
Any hints or help will be appreciated.
