I am currently taking an introductory differential geometry course concentrating on curves and surfaces (we are using Pressley's Elementary Differential Geometry) and we recently covered the first fundamental form.

I am confused by how we are dealing with the symbols du and dv alone. Unfortunately, I didn't cover differential forms in real analysis so I am not sure how to rigorously treat infinitesimal without outside of the integral sign.

In the textbook they state du and dv are linear maps which project vectors in the tangent space onto surface patch derivatives ($\sigma_u $ and $\sigma_v $ in the book's notation). This seems related in a way to the derivative of a smooth map between surfaces but I am not entirely sure how.

If anyone could clear this up and/or suggest an accessible text to read up on the subject I would very much appreciate it! I am looking if possible for both an intuitive meaning behind this concept and a rigorous treatment of it.


1 Answer 1


First, let me say that I think its a serious deficit in our educational system, (oops, sorry, I mean your system, since I can no longer identify with it) that differential notation is being ignored, I see this oversite causing problem with many math students.

But rants aside you can think of $du$ as the derivative of $u$, that is, in the $u$ direction, but without specifying with respect to which variable this derivative is to be taken. Thus if $u=u(t)$ is thought of as being a function of the variable $t$ then $$du\left(\frac{d}{dt}\right)=\frac{du}{dt}$$ and of course $$du\left(\frac{d}{du}\right)=1$$

This is useful since in differential geometry we deal often with different parameterisations. For example arc length is a preferred parameterisation for a curve, but in practice we need formulas for valid for all parameters since arc length is hard to calculate.

So the meaning of $$ds^2=Edu^2+Fdudv+Gdv^2$$ is that if $t$ is any parameter representing a curve in the surface, and so $u=u(t)$ and $v=v(t)$, then $$\left(\frac{ds}{dt}\right)^2=E\left(\frac{du}{dt}\right)^2+F\left(\frac{du}{dt}\right)\left(\frac{ dv}{dt}\right)+G\left(\frac{dv}{dt}\right)^2 $$

  • $\begingroup$ Thanks! So I guess du is just a linear operator like d/dt is, makes sense. $\endgroup$
    – Louis
    Mar 19, 2017 at 19:24

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