What is difference between an infinitesimal displacement $dx$ and a basis one-form given by the gradient of a coordinate function?

Question

In general relativity, we introduce the line element as $$ds^2=g_{\mu \nu}dx^{\mu}dx^{\nu}\tag{1}$$ which is used to get the length of a path and $dx$ is an infinitesimal displacement But for a manifold $M$, we define the metric tensor as $$g=g_{\mu \nu}dx^{\mu}dx^{\nu}\tag{2}$$ where $dx$ is the basis of $T^*M$. My question is that what are difference between $ds^2$ and $dx$ in both cases? Do they have the same meaning? How we can identify if we consider the spacetime as manifold?

Answer 1

That notation is unfortunate, seeing as those $dx$'s do not refer to covectors. Physicists use the formula $$ds^2=g_{mn}dx^mdx^n \tag{1} \label{1}$$ to define the metric tensor because they mean to convey the notion that the metric tensor gives a sense of arc length locally. So, they introduce the notation $dx^m$ to signify this supposed infinitesimal displacement. This is fine as far as intuition is concerned; that is the real physical content of the metric. However, formally, it is not quite correct.

The metric $g$ loosely speaking is a function on the tangent space (a bilinear one), that gives you a notion of inner product on your manifold. It takes two vectors in the tangent space, $T_pM$, and outputs a real number that is thir inner product. That means the better notation is $$g(V,W)=g_{mn}V^mW^n$$ where we denote the components of the vectors $V,W \in T_pM$ by $V^m$. Thus you see that your $dx$'s are meant to be components of vectors, not of covectors.

Now, $ds^2$ is also an unfortunate notation, although not quite as much. You see, if you have a curve on the manifold, $\gamma: I \subset \mathbb{R} \rightarrow M$, and a metric at every tangent space, you can define the arclength of that path as $$s=\int_I \sqrt{g(\gamma',\gamma')} dt \tag{2} \label{2}$$ This is completely analogous to the usual formula for arclength for curves in $\mathbb{R}^n$ $$s=\int_I \sqrt{\gamma' \cdot \gamma'}dt$$ but the dot product is promptly generalized by the metric. Thus it is tempting to note $$ds=\sqrt{g(\gamma',\gamma')} dt$$ and to call $$ds^2=g(\gamma',\gamma') dt^2$$ But this of course is merely simbolic. Expressions like $ds^2$ don't make much sense. The real meaning of $ds$ is given by equation $\ref{2}$. So when people write $ds^2$ they mean that, if you had a curve of which velocity had components $dx^m$, then the "square" of the arclength of that curve would be given by equation $\ref{1}$ (or rather, really, equation $\ref{2}$).

In short, the $dx^m$ are not basis covectors, rather, they're components of vectors that physicist write that way because they want to convey they are in some way local objects, and $ds^2$ is a symbolic representation of the arclength of a certain path on the manifold

Answer 2

It is the same principle from polar coordinates in the plane, except that in the general case the manifold has non-zero curvature. In a given point $(r,\theta)$, the basis vectors are $\mathbf e_r$ and $\mathbf e_{\theta}$. A infinitesimal displacement can be obtained by Pythagoras: $ds^2 = (dr\mathbf e_r)\cdot(dr\mathbf e_r) + (d\theta \mathbf e_{\theta})\cdot(d\theta \mathbf e_{\theta}) = g_{11}(dX^1)(dX^1) + g_{22}(dX^2)(dX^2)$

Where $\mathbf e_r\cdot \mathbf e_r = g_{11}$, $\mathbf e_{\theta}\cdot \mathbf e_{\theta} = g_{22}$, $dr = dX^1$ and $d\theta = dX^2$

The metric tensor is the function that takes 2 times a contravariant infinitesimal vector and delivers a scalar. The $dX^i$ are the contravariant components of this vector, and $ds^2$ is the scalar generated by the function.