Coordinate transformation of basis vectors

The question Let $$e_a$$ be the coordinate basis vectors in a manifold described by coordinate system $$x^a$$. The vector displacement between two nearby points is given by $$\begin{equation} ds=dx^ae_a=dx'^ae'_a \end{equation}$$ where a prime denotes the same quantity measured in a different reference frame.

Using the relationship $$dx^a=\frac{\partial x^a}{\partial x'^b}dx'^b$$, it is easy to see that $$\begin{equation} e'_a=\frac{\partial x^b}{\partial x'^a}e_b \end{equation}$$ However, my textbook (General Relativity An Introduction for Physicists by Hobson Lasenby and Efsthathiou, pg 60-61) says that the following must obviously be true as well: $$\begin{equation} e'^a=\frac{\partial x'^a}{\partial x^b}e^b \end{equation}$$ My question is how do I prove this?

Attempt at a solution

I can see that this would follow if we assume that $$dx_a$$ transforms as $$\begin{equation} dx'_a=\frac{\partial x^a}{\partial x'^b}dx_b \end{equation}$$ And the proof goes as: $$\begin{eqnarray} dx_ae^a&=&dx'_ae'^a\\ &=&\frac{\partial x^a}{\partial x'^b}dx_b e'^a \end{eqnarray}$$ Switching the dummy index to b on the left side we get: $$\begin{equation} dx_be^b=\frac{\partial x^a}{\partial x'^b}dx_b e'^a \end{equation}$$ which gives: $$\begin{equation} e^b=\frac{\partial x^a}{\partial x'^b} e'^a \end{equation}$$ Interchanging primed and unprimed variables as well as a and b we get $$\begin{equation} e'^a=\frac{\partial x'^b}{\partial x^a} e^b \end{equation}$$

This proof seems a bit clumsy to me, AND I am unable to prove my starting equation (my text uses the transformation rule for $$e'^a$$ to prove it later). Is there a better way of deriving this result WITHOUT using the equation I started with?

• Which text? Which page? – Qmechanic May 14 at 8:32
• The text is General Relativity An Introduction for Physicists by Hobson, Efsthathiou and Lasenby, page 60-61 – samgon May 14 at 8:57

I do not know what you mean by the "coordinate basis vectors" with their upstairs indices. The basis vectors of the tangent space $${\rm TM}_p$$ at a point $$p$$ are usually written with downstairs indices as $${\bf e}_\mu= \frac{\partial}{\partial x^\mu}.$$
The basis vectors with upstairs indices are $${\bf e}^{*\mu}= dx^\mu$$ and they are the basis vectors of the dual space $$({\rm TM}_p)^*$$.
If we insert a displacement vector $$\delta {\bf x}= \delta x^\mu \frac{\partial}{\partial x^\mu}$$ into the dual basis element $$dx^\nu$$ we use the definition of the dual basis $$dx^\nu(\partial_\mu)= \delta^\nu_\mu$$ to get $$dx^\nu( \delta {\bf x})= \delta x^\nu$$ Thus $$dx^\nu$$ is a machine into which we drop a vector a get back the numerical components of the vector.
If we change coordinates to $$x'^\mu$$ we have $${\bf e}'^{*\nu}= dx'^\mu= \frac{\partial x'^\mu}{\partial x^\nu} dx^\nu= \frac{\partial x'^\mu}{\partial x^\nu}{\bf e}^{*\nu}$$ and $${\bf e}'_\mu= \frac{\partial}{\partial x'^\mu}= \frac{\partial x^\nu }{\partial x'^\mu }\frac{\partial}{\partial x^\nu}= \frac{\partial x^\nu }{\partial x'^\mu }{\bf e}_\nu$$
• This is really helpful. In the book I am following, the $e_{\mu}$ are defined as $\partial s/\partial x^{\mu}$ where $\delta s$ is the infinitesimal vector separation between two nearby points. Thinking of them as operators $\partial /\partial x^{\mu}$instead makes this a lot easier. Thanks so much! – samgon May 15 at 11:29