Lattice with only isotropic vectors (does not exist)

Question

Consider a lattice $ L:=(X, \phi)$, i.e. $X$ is a free $\mathbb Z$ module of finite rank and $\phi:X\times X\to \mathbb Z$ is a non-degenerate, symmetric, bilinear form.

An element $x \in X$ is said to be isotropic if $\phi(x,x) =0$.

I would like to find an elementary proof that is impossible that $\forall x\in X$, $x$ is isotropic.

A way to prove this, but it's an overkill, is the following: if our lattice $L$ is definite then there are not isotropic vectors.
If instead $L$ is indefinite then we have a classification, $L\simeq m (+1)\oplus n(-1)$ or $L\simeq m E_8 \oplus n H$, and in all these cases there exists vectors which are not isotropic.

However I'm sure there is a more elementary way to see this.

Answer 1

$ \newcommand\Q{\mathbb Q} \newcommand\Z{\mathbb Z} $Extend scalars to $\Q$. Because $\phi$ is bilinear it extends to $\phi_\Q : (\Q\otimes_\Z X) \times (\Q\otimes_\Z X) \to \Q$. Since $\Q$ is a field with characteristic $\ne2$ the symmetric bilinear form $\phi_\Q$ is completely determined by the quadratic form $q : \Q\otimes_\Z X \to \Q$ with $q(x) = \phi_\Q(x,x)/2$ via polarization. But then $$ \phi_\Q(x, y) = q(x + y) - q(x) - q(y) = 0 $$ for all $x, y$, which is impossible because $\phi$ is non-degenerate.

---

Alternatively, stick to $X$ and define $q(x) = \phi(x,x)$. Then $$ 0 = q(x + y) - q(x) - q(y) = 2\phi(x,y). $$ Because multiplication by $2$ is injective it follows that $\phi(x,y) = 0 $ for all $x, y$, which is impossible for non-degenerate $\phi$.

Answer 2

Let $A\in GL_n(\mathbb Z)$ be a matrix representing $\phi$ with respect to some basis of $X$.

Since $A$ is unimodular and simmetric, is diagonalizable over $\mathbb R$. Thus exists a unit eigenvector $x\in \mathbb R^n$ such that $A x = \lambda x$, with $\lambda = \pm 1$.

Since $\mathbb Q^n$ is dense in $\mathbb R^n$, for every $\epsilon> 0$ exists $x_\epsilon \in \mathbb Q^n$, such that $||x - x_\epsilon||<\epsilon$, $||x_\epsilon ||- 1 < \epsilon$ and $|| (A-\lambda) x_\epsilon ||<\epsilon$.

Consequently $$ \langle x_\epsilon, A x_\epsilon\rangle = \lambda||x_\epsilon|| + \langle x_\epsilon, (A-\lambda) x_\epsilon\rangle, $$and $|\langle x_\epsilon, (A-\lambda) x_\epsilon\rangle|< (1-\epsilon)\epsilon < 2\epsilon$ hence for $\epsilon>0 $ small enough, $\langle x_\epsilon, A x_\epsilon\rangle\neq 0$. From now one we will consider such $\epsilon$ fixed.

Now since $x_\epsilon \in \mathbb Q^n$, exists $q_\epsilon \in \mathbb Z \setminus (0)$ such that $q_\epsilon x_\epsilon \in \mathbb Z^n$ and

$$ \langle q_\epsilon x_\epsilon, A q_\epsilon x_\epsilon\rangle = (q_\epsilon^2) \langle x_\epsilon, A q_\epsilon x_\epsilon\rangle \neq 0.$$

Consequently $q_\epsilon x_\epsilon $ is the seeked anisotropic vector.

Answer 3

If $x,y\in X$, then $$\phi(x+y,x+y)=\phi(x,x)+2\phi(x,y)+\phi(y,y)$$ so $2\phi(x,y)=0$, which means that $\phi(x,y)=0$.

So this is the zero bilinear form, which is obviously not non-degenerate.