General relativity: Why don't these two differentials commute?

I encountered some metric today defined by

$$ds^2 = - \left( 1 - \frac{2GM}{r} \right) dv^2 + (dv dr + dr dv) + r^2 d \Omega^2$$

In all education I've done until now

$$dv dr = dr dv.$$ Why is this no longer the case? I suspect this has something to do with tensors, but I am not sure why.

Answer

1. 
   

   More generally, a metric tensor

   $$\mathbb{g} ~\in~ \Gamma\left( {\rm Sym}^2(T^{\ast}M)\right)$$ is a section in the symmetric tensor product $${\rm Sym}^2(T^{\ast}M)~=~T^{\ast}M\odot T^{\ast}M$$ over the cotangent bundle $T^{\ast}M$. In other words, $\mathbb{g}$ is a symmetric $(0,2)$ covariant tensor field.
   
   


2. 
   

   In a coordinate chart $U\subseteq M$, it takes the form

   $$\mathbb{g}|_U ~=~ g_{\mu\nu}\mathrm{d}x^{\mu}\odot \mathrm{d}x^{\nu},$$ with the manifest rule $$\mathrm{d}x^{\mu}\odot \mathrm{d}x^{\nu}~=~\mathrm{d}x^{\nu}\odot \mathrm{d}x^{\mu},$$ cf. OP's question.