Does $\mathbb Z^n$ contain $A_n$?

Question

Are there any positive integer $n > 3$ such that the root lattice $A_n$ is contained in $\mathbb Z^n$?

Answer 1

No, for no $n \neq 3$ does $A_n$ embed in $\mathbb{Z}^n$ (which I assume has the standard diagonal intersection form).

Call $r_1, \dots, r_n$ the roots generating $A_n$ with $r_i r_{i+1} = -1$ and $r_ir_j = 0$ otherwise. Suppose that you have an embedding of $A_n$ into $\mathbb{Z}^m$ for some $m$. Then, up to automorphisms of $\mathbb{Z}^m$, you can suppose that $r_1$ is sent to $e_1-e_2$. In fact, any $r_i$ is sent to $\pm e_j \pm e_k$ for some $j \neq k$ (since two can only be expressed as a sum of squares in a unique way). Up to another automorphism, you can suppose that $r_2$ is sent to $e_2-e_3$. Now, for $r_3$ you have a choice: either $e_3-e_4$ or $-e_1-e_2$. If you choose the latter, than you have no choice for $r_4$ (look at the intersection mod 2 with $r_1$ and $r_3$). If you choose the former, then you can no longer use $e_1$, $e_2$, or $e_3$ for any other $r_j$. An easy induction shows that the only choice is $r_i \mapsto e_i - e_{i+1}$, which implies that you need $m \ge n+1$.

For the record, this kind of argument is heavily used in smooth 4-dimensional topology, when applying Donaldson's diagonalisation theorem. In particular, Lisca used it to classify lens spaces that bound rational homology balls. (This is how I learnt about this proof and these ideas.)

Answer 2

The even sublattice of $\mathbb Z^n$ is the root lattice $D_n$ which does not contain $A_n$ for $n\geq 4$. ($D_2=A_1\oplus A_1$ and $D_3=A_3$ are generally omitted in lists of root-systems) This implies that $A_n$ (for $n\geq 4$) is not contained in $\mathbb Z^n$: Indeed, roots of $D_n$ are all elements of (squared-length) $2$ of $\mathbb Z^n$. An inclusion of $A_n$ in $\mathbb Z^n$ would therefore imply an inclusion of $A_n$ in $D_n$.

I will try to fill the gap : An elementary argument that $D_n$ does not contain $A_n$ (for $n\geq 4$).