Is there a more precise definition of $dS^2$?

Question

In order to compute the interval between two events we perform the following calculation: $$dS^2=g_{\mu\nu}dx^{\mu}dx^{\nu}$$ I am A little confused by this notation. Are these $dx^{\mu}$'s (a basis of) one forms? What binary operation is going on here? Should it be a tensor product, a wedge product or something else entirely? I think I may have read somewhere that these $dx^{\mu}$'s are infinitesimals of the coordinates, but could someone please make this a little more precise in the terminology of differential geometry?

So, should I be interpreting $dS^2$ as something to which you feed two vectors and get out a real number? I.e. the distance between two events V and U is $$g(U,V)=dS^2(U,V)\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=g_{\mu\nu}(dx^{\mu}\otimes dx^{\nu})(V,U)\\~~~~~~~~~~~~~~~~~~~~~~~~~~~=g_{\mu\nu}dx^{\mu}(V)\cdot dx^{\nu}(U)\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=g_{\mu\nu}dx^{\mu}(v^i\partial_i) dx^{\nu}(u^j\partial_j)\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=g_{\mu\nu}v^iu^jdx^{\mu}(\partial_i) dx^{\nu}(\partial_j)\\~~~~~~~~~~~~~~~~=g_{\mu\nu}v^iu^j\delta^{\mu}_{~~i} \delta^{\nu}_{~~j}\\~~~~~~~~=g_{\mu\nu}v^{\mu}u^{\nu}$$

Where all components of the above tensor's are with respect to the coordinate basis $\{\partial_{i}\}$ and its dual basis $\{dx^{i}\}$. But then I often see expressions which involve the integral of dS and the integral of the $dx^{\mu}$, which leads me to believe that somewhere we need wedge products and all that jazz. So my question is how can we make $dS^2=g_{\mu\nu}dx^{\mu}dx^{\nu}$ more precise in the terminology of differential geometry, or is it perfectly fine how it is and I am just over thinking things?

Answer 1

Informally, $ds^2$ is the square of the distance taken when one travels from $x$ to $x+dx$.

Formally, the $dx^\mu$-s are basis 1-forms and we can define the symmetric product as $dx^\mu dx^\nu=\frac{1}{2}(dx^\mu\otimes dx^\nu+dx^\nu\otimes dx^\mu)$, then we have $$ ds^2\equiv g=g_{\mu\nu}dx^\mu dx^\nu. $$ This expression would be valid even if we used tensor products instead of symmetric products, but if a metric is written explicitly as, say $ds^2=-dt^2+2A(r)drd\vartheta+r^2(d\vartheta^2+\sin^2\vartheta d\varphi^2)$, then the commutativity of the differentials is usually assumed in the (informal) literature, which is only true for symmetric products. In particular, the above expression would contain the $A(r)dr\otimes d\vartheta +A(r)d\vartheta\otimes dr$ terms instead of the simplified $2A(r)dr d\vartheta$ term.

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So, basically, the symbol $ds^2$ is a relic from old-school differential geometry, when mathematicians weren't so anal about rigour as they are now in the post-Bourbaki world.

In order to keep up with traditions, we still use $ds^2$, but if you want to employ modern DG, you should interpret $ds^2$ to be the same as $g$ (the metric tensor field).

Note: About integration, integrating $ds$ is also informal. There does not exist an 1-form that assigns a curve's length to every curve (at least not to my knowledge - EDIT: There does not exist one.).

What really happens is that given a smooth curve $\gamma$, you either define its length as $$L[\gamma]=\int_{t_0}^{t_1}\sqrt{g(\dot{\gamma}(t),\dot{\gamma}(t))}dt,$$ which you can show to be reparametrization-invariant (and when written out in components, looks like $\int ds$, if you manipulate "differentials" heuristically), OR, if $\gamma:I\rightarrow M$, where $I$ is an (open) interval, you take the induced metric tensor $\gamma^*g$ (induced in the one-dimensional manifold $I$), and you integrate the volume form induced by $\gamma^*g$ on $I$ (which will be an 1-form, since $I$ is one dimensional). This 1-form will have the local component appearance of $đs=\sqrt{g_{\mu\nu}\frac{dx^\mu}{dt}\frac{dx^\nu}{dt}}dt,$ but it should be noted that this is an 1-form on $I$ and not on $M$!