Geometric reason why even unimodular positive definite lattices exist only in dimensions divisble by $8$ It is a well-known fact that even unimodular rank $n$ lattices $L\subseteq \mathbb{R}^n$ only exist if $8\vert n$. 
The only proof of this that I know (in the book "Elliptic functions and modular forms" by Koecher/Krieg) is rather ingenious and uses the modularity of the associated theta function $$\Theta(\tau,L)=\sum_{\gamma\in L}e^{i\pi\Vert \gamma\Vert^2}$$ to conclude that 
$$\Theta(i,L)=e^{\frac{i\pi n}{4}}\Theta(i,L)$$
and hence $8\vert n$.
While it is quite natural to associate a theta function to a lattice, it seems to me that there has to be a deeper, somehow "purely geometric reason" for this phenomenon (i.e. the condition on the dimension) which does not use the theory of modular forms.
So my question is the following: 
What is the "geometric" reason why even unimodular positive definite lattices exist only in dimensions divisble by $8$? 
(I am aware that the term "geometric" is not well-defined and can be interpreted broadly: feel free to do so)

Instead of relying on the theory of modular forms for the proof, one can also use the methods from "geometry of numbers". Such a proof can be found, for example, in Serre's book A course in Arithmetic, on page $53$, Theorem $2$. Of course, the geometry of numbers might not give you an "geometric" reason in the sense you are looking for. I am not sure, that a "pure geometric argument" suffices.
Perhaps it is helpful to view the more general picture.
Consider unimodular symmetric bilinear modules, which are free $\mathbb{Z}$-modules endowed with an integral symmetric bilinear form of discriminant $\pm 1$. As a real form it has signature $(r,s)$. Then one can show 
with "geometric" methods that the following is true:
Proposition: The signature $(r,s)$ of an even unimodular symmetric bilinear module satisfies the congruence $r\equiv s\bmod 8$.
For even unimodular lattices of rank $r$ we obtain $r\equiv 0\bmod 8$. For a proof, see Serre, as above.
