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^{*}\}


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


Leave a comment

Filed under Uncategorized

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s