Recently, I had found some algebraic intuition by generalizing the construction.
We can define similar isomorphism $E^1_a$ of $X\oplus H$ for any $a\in X^{*}(=\text{Hom}_{\mathbb{Z}}(X,\mathbb{Z}))$$a\in X^{*}(=\text{Hom}_{R}(X,R)$, where $R$ is the coefficient ring of lattices).
To begin with, we define the isomorphism $E^1_a$ of $X\oplus\langle x\rangle$ as follows:
- For $\xi\in X$, $E^1_a(\xi):=\xi+a(\xi)x$,
- $E^1_a(x)=x$.
It seems that the construction of $E^1_a$ is natural and easy to come up with. And more, the following lemma holds if we assume the coefficient ring of lattices $R$ has 2 as an unit:
The isomorphism $E^1_a$ can be uniquely extended as isomorphism of $X\oplus H$.
Proof : Let's assume that there exists such an extension and write $E^1_a(y)=\omega+sx+ty$ by using some $\omega\in X$. Because $E^1_a$ preserves the metric, we have
- For $\xi\in X$, $0=E^1_a(\xi)\cdot E^1_a(y)=\xi\cdot\omega+ta(\xi)$,
- $1=E^1_a(x)\cdot E^1_a(y)=t$,
- $0=E^1_a(y)\cdot E^1_a(y)=\omega\cdot\omega +2st$.
Hence, we have $t=1$, for any $\xi\in X$, $\underline{\xi\cdot\omega+a(\xi)=0}_{(*)}$, and $s=-2^{-1}(\omega\cdot\omega)$. Note that there always exists the unique element $\omega\in X$ which satisfies $(*)$ (because $X$ has non-degenerate metric). This suggests the method of construction and completes the proof.
Finally, forin the case of $R=\mathbb{Z}$ and $\omega\in X$ with $\omega\cdot\omega\in 2\mathbb{Z}$, if we take $a\in X^*$ as $(*)$, $E^1_a$ agrees with the Siegel-Eichler transformation $E^1_\omega$.