# Meaning of a notation regarding Gradient and the line element

In the book 'Spacetime and Geometry' by Sean Caroll, in chapter $$1$$, he writes that the notation for the gradient is d$$\phi$$ for some scalar function $$\phi$$. Then in section 2.4, he uses the expression $$ds^2=g_{\mu\nu}\text{d}x^\mu \text{d}x^\nu$$ and goes on to explain that here $$\text{d}x$$ here is not a differential element but the gradient(please note the different $$d$$ and $$\text{d}$$). My question is if that is the gradient in the equation of line element, then how does the line element equation justify as being correct?

I don't know how familiar you are with differential geometry, so I'll try to give a self contained answer. First, as you might know, the metric tensor $$g$$ is a bilinear map taking two (tangent) vectors, e.g. velocity vectors. This object can be defined without choosing coordinates of our space(time). Now take coordinate functions $$x^0,...,x^d$$, then we can introduce a basis $$e_\mu$$ in the spaces of tangent vectors in a natural way (this basis is often denoted in Terms of differentials $$\partial_\mu$$). Thus, a velocity vector $$v$$ for example can be represented as $$v=v^\mu e_\mu$$.
However, to get to your question, the gradients of the coordinate functions d$$x^\mu$$ give you the coefficients of a Vector with respect to the chosen coordinates, i.e. d$$x^\mu(v)=v^\mu$$. Note that here the gradient d$$f$$ is not a vector(field), but a linear function taking tangent vectors. A map taking tangent vectors can now be represented in terms of this gradients (i.e. they define a basis). The metric can be expressed as $$g=g_{\mu\nu}\text{d}x^\mu\text{d}x^\nu.$$ So pluging in two tangent vectors $$v,w$$ gives $$g(v,w)=g_{\mu\nu}\text{d}x^\mu(v)\text{d}x^\nu(w)=g_{\mu\nu}v^\mu w^\nu.$$ Now, the line element $$ds^2$$ is defined to be the map $$v\mapsto g(v,v),$$ so in coordinates we get essentially the same representation as for the metric tensor, just that we plug in the same vector in both slots. So as a last note, the term "$$ds^2$$" should be taken as a single term, as you take $$\frac{df}{dx}$$ as a single term and not as a fraction.