Dual vector spaces - differential forms

Question

I have a basic differential form problem that I have not understood for many years. Resources of exterior derivative refer to Wikipedia Exterior Derivative.

My problem is

(1) Why it is said that the vector basis is $\displaystyle e_j=\frac{\partial}{\partial x^j}\, $ ?

(2) Why it is said that the dual vector basis is $e^j=dx^j\,$?

Thank you a lot for your help. I would be appreciated if you can prove them to me.

Answer 1

I think the statements (1) and (2) can only be motivated but not proved. The introduction of differential forms by Elie Cartan was in my opinion one of the most ingenious and fruitful ideas in mathematics of the twentieth century. The heart of the idea must be the well known chain rule (here in two dimensions): $$\tag{A} \frac{df(x(t),y(t))}{dt}=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}\,. $$ Recall that a vector in $\mathbb R^2$ can be written as $$\tag{B} \boldsymbol{v}=v_1\boldsymbol{e}_1+v_2\boldsymbol{e}_2 $$ where $\boldsymbol{e}_i$ are the basis vectors, and $(v_1,v_2)$ the components of $\boldsymbol{v}\,.$ Since $t\mapsto (x(t),y(t))$ is a curve whose tangent vector has components $(\frac{dx}{dt},\frac{dy}{dt})$ we can write the RHS of (A) as $$ \boldsymbol{v}(f) $$ when we view $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$ as the basis vectors $\boldsymbol{e}_1$ and $\boldsymbol{e}_2\,$.

If we "drop" $dt$ in (A) we obtain $$ df=\frac{\partial f}{\partial x}\,dx+\frac{\partial f}{\partial y}\,dy\,. $$ which can be seen as a dual relationship between $(\frac{\partial}{\partial x},\frac{\partial}{\partial y})$ and $(dx,dy)$. More precisely, $$\tag{C} dx(\textstyle\frac{\partial}{\partial x})=1\,,\quad dy(\textstyle\frac{\partial}{\partial y})=1\,,\quad dx(\textstyle\frac{\partial}{\partial y})=0\,,\quad dy(\textstyle\frac{\partial}{\partial x})=0\,. $$ With $\boldsymbol{v}=v_1\frac{\partial}{\partial x}+v_2\frac{\partial}{\partial y}$ this leads directly to $$ df(\boldsymbol{v})=\frac{\partial f}{\partial x}v_1+\frac{\partial f}{\partial y}v_2=\boldsymbol{v}\cdot\nabla f $$ which is the directional derivative of $f$ in the direction of $\boldsymbol{v}\,.$ The dual relationships (C) are now seen as nothing else than describing the simple fact that

• the directional derivative of $f(x,y)=x$ in the direction of $\frac{\partial}{\partial x}$ is one and zero in the direction of $\frac{\partial}{\partial y}\,,$ and so on.

I stop here because there are much better pedagogical introductions to differential forms. My favourite one is pp. 53 in the book Gravitation by Misner, Thorne and Wheeler.