Orthogonal complement to a lattice

Question

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.

Answer 1

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.)