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I have a representation whose basis is $$\{e_{i_1} \wedge e_{i_2} \wedge \cdots e_{i_k} |\,\,1 \le i_1 <i_2 <\cdots <i_k\le n \}$$ and I want to apply elementary matrices on $e_{i_1} \wedge e_{i_2} \wedge \cdots e_{i_k}$ such that if $i <j$ we have $$E_{ij}(e_{i_1} \wedge e_{i_2} \wedge \cdots e_{i_k}) =0$$ but I want the case in which $i >j$ so that $E_{ij}(e_{i_1} \wedge e_{i_2} \wedge \cdots e_{i_k})$ can reach all possible elements in the basis defined above. The number of elements in that basis is $n_{C_{k}}$. It is possible that one apply product of this elementary matrices to get some elements in the basis.

Note that if we get $e_{i_1} \wedge e_{i_2} \wedge e_{i_2} \cdots e_{i_k}$, it will be zero and we dont want( we dont want any ith entry occuring twice.

Okay, lets try this interesting case, the case in which $n=4$ and $k=2$. Therefore we have the following elements in the basis $$\{ e_1 \wedge e_2 , e_1 \wedge e_3, e_1 \wedge e_4,e_2 \wedge e_3 , e_2 \wedge e_4, e_3 \wedge e_4\}$$ We want to apply elementary matrices on $e_1 \wedge e_2$ to obtain every element in the basis i.e $E{ij}(e_1 \wedge e_2)$( i must be greater than j). Note that $$E_{32}(e_1 \wedge e_2) = e_1 \wedge e_3$$ $$E_{21}(E_{32}(e_1 \wedge e_2))= E_{21}(e_1 \wedge e_3) = e_2 \wedge e_3$$ $$E_{43}E_{32}(e_1 \wedge e_2)= E_{43}(e_1 \wedge e_3)=e_1 \wedge e_4$$ All this actually makes sense because we found the resulting answers as members of our basis.

Honestly, I tried simpler cases and it makes a lot of sense but I cant seem to generalize it to my desired goal.

Any help and explanation on how the general case that I introduced above will be gladly appreciated.

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I am not really clear what you are asking, but we have $$E_{1,i_1}E_{2,i_2}\dots E_{r,i_r} (e_1\wedge e_2\wedge\dots\wedge e_r)= e_{i_1}\wedge e_{i_2}\wedge \dots \wedge e_{i_r}$$ for the relevant indices.

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  • $\begingroup$ Works perfectly fine. Thanks. $\endgroup$ – Jaynot Mar 31 '17 at 10:21
  • $\begingroup$ Well I changed a bit to work for what I wanted but once again your idea was helpful. $\endgroup$ – Jaynot Mar 31 '17 at 11:04

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