# Sums indexed by $2D$ integers

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?