Definition of the wedge product on the exterior algebra of a vector space

[phibonacci]

Hi,

I am currently reading about differential forms in "Introduction to Smooth Manifolds" by J. M. Lee, and I was wondering exactly how you define the wedge product on the exterior algebra $\Lambda^*(V) = \oplus_{k=0}^n\Lambda^k(V)$ of a vector space $V$. I understand how the wedge product is defined as a map
$\Lambda^k(V)\times \Lambda^l(V) \rightarrow \Lambda^{k+l}(V)$.
Is the idea simply to extend this definition bilinearly to $\Lambda^*(V)$? I.e. given elements $(T_0, ..., T_n)$ and $(S_0, ..., S_n)$ in $\Lambda^*(V)$, do we simply calculate all the combinations $T_j \wedge S_k$, add the obtained forms that have the same degree, and finally let these sums make up the new element of $\Lambda^*(V)$? This seems to me like an intuitive definition, but Lee wasn't very specific about this, so I thought I should make sure this is correct.

Please let me know if there is something in my post that seems unclear.

 

[dextercioby]

Actually, the wedge product is placed as an internal operation on the exterior algebra right from the definition. That's the whole purpose of defining the direct sum of vector spaces, to make an external operation (from the pov of the vector spaces) internal. The wedge product takes an element of V* in another (typically different) element of V*, hence it's an internal operation on the exterior algebra. The direct sum vector space together with the wedge product is a Z_2 graded algebra.

 

[phibonacci]

Thanks for your reply! I'm not sure if I see what you mean, though. I was primarily interested in whether my "rule" for calculating the wedge product between two elements of the exterior algebra was correct or not.

 

[quasar987]

Hi,

I am currently reading about differential forms in "Introduction to Smooth Manifolds" by J. M. Lee, and I was wondering exactly how you define the wedge product on the exterior algebra $\Lambda^*(V) = \oplus_{k=0}^n\Lambda^k(V)$ of a vector space $V$. I understand how the wedge product is defined as a map
$\Lambda^k(V)\times \Lambda^l(V) \rightarrow \Lambda^{k+l}(V)$.
Is the idea simply to extend this definition bilinearly to $\Lambda^*(V)$? I.e. given elements $(T_0, ..., T_n)$ and $(S_0, ..., S_n)$ in $\Lambda^*(V)$, do we simply calculate all the combinations $T_j \wedge S_k$, add the obtained forms that have the same degree, and finally let these sums make up the new element of $\Lambda^*(V)$?

Yes, that is it!

 

[blue_raver22]

Thanks!

Hi there,

Thank you for your question about the wedge product on the exterior algebra of a vector space. You are correct in your understanding of how the wedge product is defined on \Lambda^*(V). The idea is to extend the definition bilinearly, as you suggested. This means that for two elements T and S in \Lambda^*(V), we can write T = \sum_{k=0}^n T_k and S = \sum_{l=0}^n S_l, where T_k and S_l are elements of \Lambda^k(V) and \Lambda^l(V) respectively. Then, the wedge product of T and S is defined as the sum of all possible wedge products T_k \wedge S_l, where k and l range from 0 to n, and the resulting forms have degree k+l.

I hope this answers your question and clarifies any confusion. If you have any further questions, please don't hesitate to ask. Have a great day!