- #1

latentcorpse

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The wedge product defines a bilinear multiplication in [itex]\Lambda(V^{*})[/itex] which is associateive i.e. [itex](\alpha \wedge \beta) \wedge \gamma = \alpha \wedge (\beta \wedge \gamma) \forall \alpha,\beta,\gamma \in \Lambda(V^{*})[/itex] and which is supercommutative i.e. if [itex]\alpha^{k} \in \Lambda^{k}(V^{*}),\beta^{l} \in \Lambda^{l}(V^{*})[/itex] then [itex]\alpha^{k} \wedge \beta^{l} = (-1)^{kl} \beta^{l} \wedge \alpha^{k}[/itex]

Question 1: What is a bilinear multiplication?

Question 2: Is my definition of supercommutative ok? I wasn't sure if it was [itex](-1)^{kl}[/itex] or [itex](-1)^{k+l}[/itex]? What does supercommutative mean?

Then as an example of this, we considered the following:

[itex]\alpha=e_1^* \wedge e_2^* - 3 e_2^* \wedge e_4^* = e_{12}^* -3 e_{24}^* \in \Lambda^{2}(V^{*}), \beta=3 e_1^* + 4e_2^*[/itex]

Then,

[itex]\alpha \wedge \beta=6 e_{12}^* \wedge e_1^* +8 e_{12}^* \wedge e_2^* - 9 e_{24}^* \wedge e_1^* - 12 e_{24}^* \wedge e_2^*[/itex]

but we have[itex]e_{ij}^* \wedge e_j^* = e_i^* \wedge (e_j^* \wedge e_j^*)=0[/itex] so 3 terms drop out and we're left with

[itex]\alpha \wedge \beta =- 9 e_{24}^* \wedge e_1^* = -9 e_2^* \wedge e_4^* \wedge e_1^8=\mathbf{- (-1)^{2} e_1^* \wedge e_2^* \wedge e_4^*=-e_{124}^*}[/itex]

Question 3: Is [itex]\beta \in \Lambda^{1}(V^{*})[/itex]?

Question 4: I don't understand how he got the part in bold. I have a feeling he just forgot to write in the 9 but the rearranging of that step uses the supercommutative algebra described above and I don't follow the logic there?

Any help would be greatly appreciated

cheers