Given a manifold $\mathcal{M}$, the notion of a vector field $\xi$ on $\mathcal{M}$ can be interpreted as a collection of arrows on the manifold. In the book The Road to Reality by Roger Penrose, Chapter 12, Section 4, he mentions that the vector field $\xi$ can act on any (smooth) scalar field $\Phi$ to produce another scalar field $\xi(\Phi)$ in the manner of a differential operator. He uses the following definition:

The interpretation of $\xi(\Phi)$ is to be the 'rate of increase' of $\Phi$ in the direction indicated by the arrows that represent $\xi$ ...

The book would like the reader to show that the scalar product \begin{align*} \alpha\cdot\xi=\alpha_1\xi^1+\alpha_2\xi^2+\cdots+\alpha_n\xi^n \end{align*} is consistent with $d\Phi\cdot\xi=\xi(\Phi)$, in the particular case where $\alpha=d\Phi$ using the chain rule, but unfortunately without knowing the precise definition of $\xi(\Phi)$, I have no way to prove this. There seems to be a similarity between this definition and the directional derivative, and I was always taught that the directional derivitive in multivariable calculus was $\nabla_\mathbf{v}f=\nabla f\cdot\mathbf{v}$ by definition. But if that were true, there would be nothing to prove.

  • $\begingroup$ You can think of $\text{d}\Phi\cdot\xi=\xi(\Phi)$ as being a generalization of $\nabla_\mathbf{v}f=\nabla f\cdot\mathbf{v}$. To a first order approximation, the rate of increase of $\Phi$ in the direction of $\xi$ at a specific point $P$ is a linear function of $\xi$. We denote this linear function by the covector $\text{d}\Phi$. $\endgroup$ – gandalf61 Jan 15 at 10:51
  • $\begingroup$ @gandalf61 Okay, I thought that it was related to directional derivatives. How is this definition more general though? It seems to me that they're identical. $\endgroup$ – Alex S Jan 15 at 11:21
  • $\begingroup$ @Alex_S $\nabla$ can be applied to functions from $\mathbb{R}^n$ to $\mathbb{R}$ whereas $\text{d} \Phi$ can be applied to scalar functions on any differentiable manifold, which only needs to resemble $\mathbb{R}^n$ locally. $\endgroup$ – gandalf61 Jan 15 at 12:25

We begin by defining the scalar product of a covector and a vector as \begin{align*} \alpha\cdot\xi=\alpha_1\xi^1+\alpha_2\xi^2+\cdots+\alpha_n\xi^n \end{align*}

Since $\alpha=d\Phi$, we have \begin{align*} \alpha\cdot\xi&=d\Phi\cdot\xi \end{align*} The covector $d\Phi$ is equal to \begin{align*} d\Phi=\left(\frac{\partial\Phi}{\partial x^1},\frac{\partial\Phi}{\partial x^2},\ldots,\frac{\partial\Phi}{\partial x^n}\right) \end{align*} Using the expression for the scalar product defined earlier, \begin{align*} d\Phi\cdot\xi&=\frac{\partial\Phi}{\partial x_1}\xi^1+\frac{\partial\Phi}{\partial x_2}\xi^2+\cdots+\frac{\partial\Phi}{\partial x_n}\xi^n \end{align*} The vector field $\xi$ can be represented by its set of components in terms of partial differential operators, and we will of course choose the coordinates to be the same as the previously used coordinates for $\Phi$: \begin{align*} \xi=\xi^1\frac{\partial}{\partial x^1}+\xi^2\frac{\partial}{\partial x^2}+\cdots+\xi^n\frac{\partial}{\partial x^n} \end{align*} Therefore, \begin{align*} \xi(\Phi)&=\left(\xi^1\frac{\partial}{\partial x^1}+\xi^2\frac{\partial}{\partial x^2}+\cdots+\xi^n\frac{\partial}{\partial x^n}\right)\Phi\\ &=\xi^1\frac{\partial\Phi}{\partial x^1}+\xi^2\frac{\partial\Phi}{\partial x^2}+\cdots+\xi^n\frac{\partial\Phi}{\partial x^n} \end{align*}

Hence, the definition of the scalar product implies $d\Phi\cdot \xi=\xi(\Phi)$ as required.

  • $\begingroup$ Note that $\alpha=d\Phi$ is a 1-form, i.e. a (covector)-field, and the 'scalar product' above is a natural map which inputs a covector and a vector, and is not an inner product of two vectors. $\endgroup$ – Berci Jan 15 at 13:14
  • $\begingroup$ @Berci Yes, that was mentioned in the text. It should be clear by the use of indices though. Also I forgot to mention that $\mathcal{M}$ is a finite dimensional manifold so doesn't every one-form essentially act like an inner product with respect to vectors? I read somewhere it has to do with the Riesz representation theorem but I'm not sure. $\endgroup$ – Alex S Jan 15 at 13:50

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