## Differential forms-Algebraic point of view

If we start with an ${R^{n}}$ space with linear co-ordinates ${x_{1}..x_{n}}$, we can define ${\Omega^{*}}$ to be Algebra over ${R}$ generated by ${dx_{1}...dx_{n}}$ with the relations, $dx_{i} \wedge dx_{j} = -dx_{j} \wedge dx_{i}$ One can write ${dx_{i} \wedge dx_{j}}$ as ${dx_{i}dx_{j}}$

Since ${dx_{i}^{2}=dx_{i}dx_{i}=-dx_{i}dx_{i}}$, we can set ${dx_{i}^{2} =0}$

As a vector space over ${R}$, ${\Omega^{*}}$ has basis ${\{dx_{i}\}}$.

Now we define differential forms over ${R^{n}}$ as elements of the set, $\Omega^{*}(R^{n})= \{C^{\infty}\thinspace \text{functions on} \thinspace R^{n}\otimes\Omega^{*}\}$

Example:

A zero form is considered as a scalar function. In ${R^{3}}$, $\omega_{0}=f(x,y,z)$

Whereas one form is written as $\omega_{1}=f_{x}(x,y,x)dx+f_{y}(x,y,x)dy+f_{z}(x,y,x)dz$

Note for one form we simply append ${dx_{i}}$.

Similarly for two forms, we append ${dx_{i}\wedge dx_{j}}$.

$\omega_{2}=f_{xy}(x,y,x)dx\wedge dy+f_{yz}(x,y,x)dy\wedge dz+f_{zx}(x,y,x)dz \wedge dx$

Three form is given below

$\omega_{3}=f_{xyz}(x,y,z)dx \wedge dy \wedge dz$