Friday, 15 June 2018

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.




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