Can someone explain (Differential Forms)

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Homework Help Overview

The discussion revolves around differential forms, specifically the wedge product of two forms and the Leibniz rule for the exterior derivative. Participants are exploring the properties and definitions related to these mathematical concepts.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are questioning the notation and steps involved in the wedge product, particularly the transition from a summation over indices to the expression involving anti-commutativity. There is also inquiry into the origin of the (-1)^k factor in the Leibniz rule for the exterior derivative.

Discussion Status

Some participants have provided clarifications regarding the notation used in the wedge product and the reasoning behind the anti-commutative property. Others are still seeking further understanding of specific steps and definitions, indicating an ongoing exploration of the topic.

Contextual Notes

Participants are navigating through complex definitions and properties of differential forms, with some expressing uncertainty about the implications of the notation and the rules governing the operations involved.

latentcorpse
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(i) if [itex]\alpha=\sum_i \alpha_i(x) dx_i \in \Omega^1, \beta=\sum_j \beta_j(x) dx_j[/itex]

then[itex]\alpha \wedge \beta = \sum_{i,j} \alpha_i(x) \beta_j(x) dx_i \wdge dx_j \in \Omega^2[/itex]

NOW THE STEP I DON'T FOLLOW - he jumps to this in the lecture notes:

[itex]\alpha \wedge \beta = \sum_{i<j} (\alpha_i \beta_j - \beta_j \alpha_i) dx_i \wedge dx_j[/itex]

the subscript on the sum was either i<j or i,j - could someone tell me which as well as explaing where on Earth this step comes from.

(ii) could someone explain the Leibniz rule for exterior derivative [itex]d: \Omega^k \rightarrow \Omega^{k+1}[/itex]

i.e. why [itex]d(\alpha^k \wedge \beta^l)=d \alpha^k \wedge \beta^l + (-1)^k \alpha^k \wedge d \beta^l[/itex]
note that the superscript on the differential form indicates that it's a k form or and l form

my main problem here is where the (-1)^k comes from

cheers for your help
 
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Hi latentcorpse! :smile:
latentcorpse said:
[itex]\alpha \wedge \beta = \sum_{i<j} (\alpha_i \beta_j - \beta_j \alpha_i) dx_i \wedge dx_j[/itex]

the subscript on the sum was either i<j or i,j - could someone tell me which as well as explaing where on Earth this step comes from.

It's i < j …

[itex]\alpha \wedge \beta = \sum_{i,j} (\alpha_i \beta_j) dx_i \wedge dx_j = \sum_{i<j} (\alpha_i \beta_j - \beta_j \alpha_i) dx_i \wedge dx_j[/itex]

since wedge is anti-commutative.
why [itex]d(\alpha^k \wedge \beta^l)=d \alpha^k \wedge \beta^l + (-1)^k \alpha^k \wedge d \beta^l[/itex]

my main problem here is where the (-1)^k comes from

d() is like another wedge, so if you move d() through k elementary forms, you multiply it by (-1)k :wink:
 
cool i get the 2nd part now - still not sure why it's i<j or why [itex]\alpha_i \beta_j[/itex] suddenly becomes [itex]\alpha_i \beta_j-\beta_j \alpha_i[/itex]
 
(have an alpha: α and a beta: β and a wedge: ∧ :wink:)
latentcorpse said:
… still not sure why it's i<j or why [itex]\alpha_i \beta_j[/itex] suddenly becomes [itex]\alpha_i \beta_j-\beta_j \alpha_i[/itex]

because αi ∧ βj = -βj ∧ αi

so if you sum αi ∧ βj over all i and j,

then for i > j you just "turn it round" :wink:
 

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