How many lattices are there which contain both the $A_n$ and $D_n$ lattices of the same dimension as sublattices? So far, I’ve found examples in 1D, 3D, 8D, and 24D.

## 2 Answers

This happens precisely in the dimensions of the form $k^2-1$. (Note that the OP has clarified in comments that rescaling of the lattice is fine.) I don't see anyway that we could hope to answer the question of "how many" such lattices there are: One you have one such lattice, any finite index sublattice has the same property.

Because we are allowing rescaling, the question is equivalent to asking whether there is a quadratic form on $\mathbb{Q}^n$ into which $A_n$ and $D_n$ both embed. In other words, this is asking whether the quadratic forms $A_n$ and $D_n$ are equivalent over $\mathbb{Q}$. I'll write $\mathbb{Q} A_n$ and $\mathbb{Q} D_n$ for the vector spaces which the two lattices span, with their induced quadratic forms.

Then, as rational vector spaces with quadratic forms, we have $$\mathbb{Q}^{n+1} = \mathbb{Q} A_n \oplus \mathbb{Q} (1,1,1,\ldots,1) = \mathbb{Q} D_n \oplus \mathbb{Q} (1,0,0,\ldots,0).$$ So $\mathbb{Q} A_n \cong \mathbb{Q} D_n$ if and only if $\mathbb{Q} (1,1,1,\ldots,1) \cong \mathbb{Q} (1,0,0,\ldots,0)$. Since $(1,1,1,\ldots,1) \cdot (1,1,1,\ldots,1) = n+1$ and $(1,0,0,\ldots,0) \cdot (1,0,0,\ldots,0) = 1$, this happens if and only if $n+1$ is square. $\square$

To be concrete, if $n = k^2-1$, then we can embed both $A_n$ and $D_n$ into $\mathbb{Q}^n$ with the standard quadratic form. This is standard for $D_n$: $$D_n = \{ (x_1, x_2, \ldots, x_n) \in \mathbb{Z}^n : \sum x_i \equiv 0 \bmod 2 \}.$$ For $A_n$, for $1 \leq i \leq n-1$, put $\alpha_i = e_i - e_{i+1}$. Put $$\alpha_n = ( \tfrac{1}{k+1}, \tfrac{1}{k+1}, \ldots, \tfrac{1}{k+1}, \tfrac{k+2}{k+1} ).$$ Then I claim that the $\alpha$'s pair with each other by the type $A$ Cartan matrix: The only nonobvious computations are $$\alpha_{n-1} \cdot \alpha_n = \tfrac{1}{k+1} - \tfrac{k+2}{k+1} = -1$$ and $$\alpha_n \cdot \alpha_n = (n-1) \tfrac{1}{(k+1)^2} + \tfrac{(k+2)^2}{(k+1)^2} = \tfrac{(k^2-2)+(k^2+4k+4)}{(k+1)^2} = \tfrac{2k^2+4k+2}{(k+1)^2} = 2.$$ So the integer span of the $\alpha_i$ is isomorphic to $A_n$. I think that, concretely, this integer span is $$\{ (x_1, x_2, \ldots, x_n) \in \tfrac{1}{k+1} \mathbb{Z}^n : x_i - x_j \in \mathbb{Z} \ \text{and} \sum x_i \in k \mathbb{Z} \}.$$

The lattices $D_n$ and $A_n$ are integral. There is a result by Krüskemper (Algebraic construction of bilinear forms) stating that any integral lattice can be constructed as an ideal lattice in some real number field. The proof is constructive and based on the previous work of O. Taussky (On the similarity transformation between an integral matrix with irreducible characteristic polynomial and its transpose), so in principle these lattices can be explicitly obtained as sublattices of some 'algebraic' lattices in any dimension. Other methods were applied to obtain such lattices over number fields, see for instance Theorem 3.20 in Damir–Mantilla-Soler - Bases of minimal vectors in tame lattices where the lattice consisting of zero trace elements in a tame number field of either conductor or degree prime is a scaled $A_n$. For the $D_n$ lattice see for example de Araujo and Jorge - Construction of full diversity $D_n$-lattices for all $n$.

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