Can someone explain (Differential Forms)

[latentcorpse]

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

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

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

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

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 d: \Omega^k \rightarrow \Omega^{k+1}

i.e. why d(\alpha^k \wedge \beta^l)=d \alpha^k \wedge \beta^l + (-1)^k \alpha^k \wedge d \beta^l
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

 

[tiny-tim]

Hi latentcorpse!

\alpha \wedge \beta = \sum_{i&lt;j} (\alpha_i \beta_j - \beta_j \alpha_i) dx_i \wedge dx_j

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 …

\alpha \wedge \beta = \sum_{i,j} (\alpha_i \beta_j) dx_i \wedge dx_j = \sum_{i&lt;j} (\alpha_i \beta_j - \beta_j \alpha_i) dx_i \wedge dx_j

since wedge is anti-commutative.

why d(\alpha^k \wedge \beta^l)=d \alpha^k \wedge \beta^l + (-1)^k \alpha^k \wedge d \beta^l
…
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

 

[latentcorpse]

cool i get the 2nd part now - still not sure why it's i<j or why \alpha_i \beta_j suddenly becomes \alpha_i \beta_j-\beta_j \alpha_i

 

[tiny-tim]

(have an alpha: α and a beta: β and a wedge: ∧ )

… still not sure why it's i<j or why \alpha_i \beta_j suddenly becomes \alpha_i \beta_j-\beta_j \alpha_i

because α_i ∧ β_j = -β_j ∧ α_i …

so if you sum α_i ∧ β_j over all i and j,

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