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David E Speyer
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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 $$\{ (x_1, x_2, \ldots, x_n) \in \tfrac{1}{k+1} \mathbb{Z}^n : x_i - x_j \in \mathbb{Z} \ \text{and} \sum x_i \in k \mathbb{Z} \}.$$

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. $\square$

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 $$\{ (x_1, x_2, \ldots, x_n) \in \tfrac{1}{k+1} \mathbb{Z}^n : x_i - x_j \in \mathbb{Z} \ \text{and} \sum x_i \in k \mathbb{Z} \}.$$

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David E Speyer
  • 156.3k
  • 14
  • 421
  • 763

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. $\square$