Orthogonal complement to a lattice Suppose we have an even lattice of rank 2, $\Lambda$, with the following intersection form,
\begin{eqnarray}
\left( \begin{array}{cc}
2 & 3 \\
3 & 0
\end{array} \right)
\end{eqnarray}
As far as I know, there is a primitive embedding of this lattice into $U \oplus U$, where $U$ is the hyperbolic lattice,
\begin{eqnarray}
\Lambda \hookrightarrow U \oplus U.
\end{eqnarray}
I want to compute the orthogonal complement $\Lambda^{\perp}$ of $\Lambda$. Since $|disc(\Lambda)| = 9$, then $\Lambda^{\perp}$ is also not unimodular, and therefore I expect it to be $U[-3]$. 
Is this true?! 
How can I find $\Lambda^{\perp}$ explicitly anyways? 
Thank you.
The same question was asked in https://mathoverflow.net/q/355085/153717 , but the question is closed there now.
 A: First off: there exists such an embedding; in the "obvious" coordinates in which $U\oplus U$ is represented by a block diagonal matrix with $\left({\begin{array}{c} 0 & 1 \\ 1 & 0\end{array}}\right)$, we can just take the span of the two vectors $v = (1,1,-1,0)$ and $w = (1,1,1,-1)$.
We can compute the orthogonal to this specific embedding, which is given by all vectors $(x,y,x',y')$ such that
$\left\{
\begin{array}{l}
y+x-y' = 0\\
y+x+y'-x' = 0
\end{array}
\right.$. This set is spanned by $(1,0,2,1)$ and $(0,1,2,1)$; in this basis, the intersection form of the subspace they span is $\left({\begin{array}{c} 4 & 5 \\ 5 & 4\end{array}}\right)$, which is isomorphic to $\left({\begin{array}{c} -2 & 3 \\ 3 & 0\end{array}}\right)$.
However, this is not the only embedding: in the same coordinates, and in the same notation as above, we can take $v=(1,1,0,0)$ and $(3,0,0,0)$, and embed $\Lambda$ in $U$; in this case, the orthogonal is $0\oplus U$, but I guess that this is not what you mean as a primitive embedding.
Abstractly, there are only four possibilities for the orthogonal: $\Lambda$, $-\Lambda$, $U$, and $3U$. $U$ only comes from an embedding of $\Lambda$ into $U$, and then embedding $U$ into $U\oplus U$. We exhibited one with $\Lambda^\perp = -\Lambda$, but I can't exclude right now that the other two cases also occur. (Both $\Lambda \oplus 3U$ and $\Lambda\oplus\Lambda$ embed in unimodular lattices of rank 4, and you can explicitly determine these lattices. I haven't tried, though.)
