Exterior derivative of 1

Question

In one of the examples of Loring W. Tu's Book "Introduction to Manifolds" (section 19.7 of 2nd ed.) the author takes the exterior derivative on both sides of the equation $x^2 + y^2 = 1$ wich results in $2x\,dx + 2y\,dy = 0$. The left-hand side makes sense to me and, intuitively, the right-hand side too. But I don't understand what the exterior derivative of 1 is supposed to mean, since the exterior derivative acts on differential forms, then 1 what kind of differential form is supposed to be? What is its degree? And why should its exterior derivation be 0? Thank you.

Answer 1

Smooth functions are differential zero-forms. Therefore, a real number, i.e. a constant function, is a differential zero-form.

Answer 2

The exterior derivative here is the one associated with the standard differential structure on $\mathbb{R}^2$, and both sides of the equation are one-forms (linear functions from $T\mathbb{R}^2 \simeq \mathbb{R}^2$ to $\mathbb{R}$.) In the case of the right-hand side, the one-form maps any vector $v$ to the derivative of the constant function (zero-form) $f(x,y)=1$ in the $(v_x,v_y)$ direction: $$df(x,y)v = \lim_{t\to 0} \frac{d}{dt} f(x+v_x t, y + v_y t) = 0.$$

Therefore the right-hand side doesn't depend on either $x,y$ or $v_x,v_y$ and always evaluates to zero. It is the zero element in the vector space of one-forms which is why it is written $0$ (you can think of it as $0dx + 0dy$ in the basis ($dx,dy$) if you prefer.)