# 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

#### tiny-tim

Homework Helper
Hi latentcorpse!
$\alpha \wedge \beta = \sum_{i<j} (\alpha_i \beta_j - \beta_j \alpha_i) dx_i \wedge dx_j$

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.
It's i < j …

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

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

Homework Helper
(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 αi ∧ βj = -βj ∧ αi

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

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

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