# Why is $d(xdx) = 0$?

If the idea behind the exterior derivative $$d$$ is that it tells us how quickly a $$k$$-form changes along every possible direction, why is $$d(xdx)=0$$ even though $$xdx$$ varies with $$x$$?

I understand the mechanics of exterior calculus but I'm trying to understand the intuition behind why it's the right tool for "differentiating" vector fields, rather than say working with vector-differences like a standard derivative (e.g. $$\frac {\partial\overrightarrow v}{\partial x}=\lim\limits_{h \to 0} \frac{\overrightarrow v(x+h,y)-\overrightarrow v(x,y)}{h}$$).

• The exterior derivative does not tell us how quickly a $k$ form changes along every possible direction, that is only the interpretation assigned to the exteriror derivative of $0$ forms. The exterior derivative is measuring exactly that - how much of an "exterior" or boundary structure exists and how it changes. $x dx = d\left(\frac{x^2}{2}\right)$ - which means that it is, in some heuristic sense, a "complete picture of a boundary" of the $0$ form. Commented Jun 24, 2023 at 9:58
• Think of how a hemisphere has a circular boundary but does not enclose a volume A full sphere encloses a volume, but it has no line boundary. What this means for us is $d^2\left(\frac{x^2}{2}\right) = 0$ Commented Jun 24, 2023 at 9:59
• $x \,\mathrm dx$ is constant when you use the right coordinates. So if you want to use this for e.g. physical theories, the result shouldn't depend on your coordinates. Commented Jun 24, 2023 at 10:00
• A $1$-form has exterior derivative equal to $0$ if the integral of it along any closed loop is zero. For $x\,dx$, this is intuitively plausible because the increase in $x$ as you move away from the starting point on the loop cancels out with the decrease in $x$ when you return to the starting point. Commented Jun 24, 2023 at 15:37

Consider a change of variables, $$y = \frac12 x^2$$, then $$\frac{\mathrm dx}{\mathrm dy} = \frac 1x$$. This implies $$x\, \mathrm dx = \mathrm dy$$, which is "constant"! This means that when we use different coordinates, we will disagree about what is "constant" for differential forms. We don't have this problem with functions: if $$f$$ is constant then no matter how you change variables the result is constant.
Therefore we have to accept that constancy is not a "real" concept for differential forms (unless you install additional structure called "connections" — just like how velocity is relative unless you designate a fixed body which we compare velocity to), and $$\mathrm d(x\,\mathrm d x) = 0$$ shouldn't be rejected solely because it is not constant from your viewpoint. In fact your first sentence is almost true: The exterior derivative shows all the "observable" rate of change of a $$k$$-form, and $$x\,\mathrm dx$$ only changes in a non-observable way.