On the scalar product of a vector field and the exterior derivative of a smooth map on a surface

Question

I am reading Roger Penrose's wonderful book 'The Road to Reality: A Complete Guide to the Laws of the Universe'. I need some help in understanding the scalar product of a vector field and the exterior derivative of a smooth map on a surface. This is all in section 10.4, for anyone who has the book. I will be consistent with the author's notation.

Let $\Phi:M \to \mathbb{R}$ be a map from a surface $M$ to the real line and let $\xi$ be a vector field on $M$. The author shows that we can write $\xi$ as $\xi=a \frac{\partial}{\partial x} + b \frac{\partial}{\partial y}$. This makes sense to me, as, harkening back to the days of multivariate calculus I recall that the directional derivative of a function on $\mathbb{R}^n$ can be given by the dot product of the gradient with a (unit) vector. So then, 'multiplying through' by $\Phi$ we obtain $\xi(\Phi)=a \frac{\partial \Phi}{\partial x} + b \frac{\partial \Phi}{\partial y}$, and this looks like the same thing that we would obtain in the situation that I am familiar with.

However, the author says that $\xi(\Phi)=\mathrm{d}\Phi \cdot \xi$.

He notes that we can write $\mathrm{d}\Phi = u\, \mathrm{d}x + v\, \mathrm{d}y$. Then we have $\xi(\Phi)=\mathrm{d}\Phi \cdot \xi = au\frac{\partial}{\partial x}\mathrm{d}x + bv\frac{\partial}{\partial y}\mathrm{d}y$. The author says that we find the value of $\xi(\Phi)$ to be $au+bv$.

What I don't understand is: how do the partial differentiation operators 'cancel' with the exterior derivatives of $x$ and $y$? Later, in section 12.3, he says that the covector $\mathrm{d}\Phi$ actually has the defining property that $\mathrm{d}\Phi \cdot \xi = \xi(\Phi)$. This feels alright to me since we have '(co)vectorize' $\Phi$ in order for the scalar product with $\xi$ to make sense. Even so, I'd like an intuitive explanation of how the two operators seem to undo each other.

Thanks for the help.

Answer 1

I do not know how these notations are introduced in your book, below is how I understand the notations. Let $p\in M$. The tangent space $T_pM$ at $p$ is the vector space of all derivations $\delta$ acting on local function $f$. A derivation is a linear map sending local function to local function such that

$$\delta(fg) = (\delta f) g + f (\delta g).$$

If you have a coordinate $(x, y)$ at $p$, then the vector space $T_pM$ is spanned by $\frac{\partial }{\partial x}$ and $\frac{\partial }{\partial y}$, with the obvious action on local function by differentiation: If $\xi = a\frac{\partial }{\partial x} + b\frac{\partial }{\partial y}$, then

$$\xi (\Phi) := a\frac{\partial \Phi}{\partial x} + b\frac{\partial \Phi}{\partial y}\ .$$

On the other hand, $T_pM^*$ is the dual space of $T_pM$. Given a coordinate around $p$, $dx$ and $dy$ are just dual vector of $\frac{\partial }{\partial x}$ and $\frac{\partial }{\partial y}$ respectively. Thus by definition,

$$(*) \ \ \ dx(\frac{\partial }{\partial x})=1,\ \ \ dx(\frac{\partial }{\partial y})=0$$

and similar for $dy$.

Given a function $\Phi$, $d\Phi$ is an element in $T_pM^*$ defined by

$$d\Phi (\xi) = \xi(\Phi),\ \ \ \ \forall \xi \in T_pM\ .$$

As an element in $T_pM^*$, $d\Phi$ can be written as $d\Phi = udx + vdy$. To find out $u$ and $v$, we apply $\frac{\partial }{\partial x}$:

$$ d\Phi(\frac{\partial }{\partial x}) = \big( udx+ vdy)(\frac{\partial }{\partial x}) = u = \frac{\partial }{\partial x}(\Phi) \Rightarrow u =\frac{\partial \Phi}{\partial x}$$

Similar for $\frac{\partial }{\partial y}$ we obtain $v = \frac{\partial \Phi}{\partial y}$. Thus

$$d\Phi = \frac{\partial \Phi}{\partial x} dx + \frac{\partial \Phi}{\partial y} dy\ .$$

Answer 2

Offhand, I don't recall Penrose's definition of coordinate differentials, but they're usually defined to be the covector fields whose values at a point $p$ are the dual basis to the (values of the) coordinate vector fields at $p$. That is (using $x_1$ and $x_2$ instead of $x$ and $y$), $dx_i(\partial/\partial x_j) = \delta_{ij}$ by definition.

Intuitively, you're meant to regard $dx_i(\partial/\partial x_j) = D_j x_i$ as the derivative of the coordinate function $x_i$ in the direction of the coordinate $x_j$.

Answer 3

Penrose just abuses the scalar product concept, what is meant is $d\Phi(\xi)$, 1-form acting on a tangent vector. When written in the corresponding bases the result $au+bv$ follows. As for the intuitive explanation, the differential represents function's change along a tangent given as an argument. One can also measure this change by taking the corresponding directional derivative of the function.