Let $\Lambda \subset \mathbb{Z}^{d}$ be finite and suppose we're given, for each $x=(x_{1},x_{2})\in \Lambda$, elements $H_{x}$ on an algebra. Let $e_{1}=(1,0)$ and $e_{2}=(0,1)$ e consider the sum: \begin{eqnarray} \frac{1}{2}\bigg{(}\sum_{x\in \Lambda}E_{1}H_{x+e_{1}}H_{x}+\sum_{x\in \Lambda}E_{1}H_{x-e_{1}}H_{x}+\sum_{x\in \Lambda}E_{2}H_{x}H_{x+e_{2}}+\sum_{x\in \Lambda}E_{2}H_{x}H_{x-e_{2}}\bigg{)}\tag{1}\label{1} \end{eqnarray} where $E_{1},E_{2}$ are both constants. Here some boundary conditions are needed in order to $x\pm e_{1}$ and $e_{2}\pm e_{2}$ to make sense when these vectors lie outside $\Lambda$, but this is not important for the question.

Question: Can the above sum can always be written as: $$\sum_{x\in \Lambda}E_{1}H_{x+e_{1}}H_{x}+\sum_{x\in \Lambda}E_{2}H_{x}H_{x+e_{2}}?$$ In other words, do the sums $\sum_{x\in \Lambda}E_{1}H_{x-e_{1}}H_{x}$ and $\sum_{x\in \Lambda}E_{2}H_{x}H_{x-e_{2}}$ repeat $\sum_{x\in \Lambda}E_{1}H_{x}H_{x+e_{1}}$ and $\sum_{x\in \Lambda}E_{2}H_{x}H_{x+e_{2}}$, respectivelly, so that I can always consider only forward translations in (\ref{1}), with an additional $2$ factor?


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