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My Doubt:-

The author had pointed out the notations used by him. I have put that in a box. In the Proposition 3.29, the author used the notation for proving linear independence using $\sum c_{I}\alpha^{I}$. According to the paragraph in the box $c_I=(c_{i_1},...,c_{i_k})$ and $\alpha^I=\alpha_{i_1}\wedge ... \wedge \alpha_{i_k}$. On contradictory to the notation mentioned in the box auther had written $C_I \in \mathbb R$. How does the product $c_I \alpha^I$ defined? From Diego's answer. I have one more doubt. How can I expand the summation? For example, if the summation is like $\sum_{k=1}^n a_ic^i$. Then, without any confusion, I can expand it as $a_1c^1+...+a_nc^n$. Please help me.


Here $c_I$ isn't a tuple, but a number associated with the multi-index $I$. He specifically says so at "$c_I \in \mathbb{R}$".

  • $\begingroup$ How does the product $c_Iα^I$ defined? $\endgroup$ – Unknown x Jul 6 '18 at 16:38
  • $\begingroup$ Now I got one more doubt, How would the sum look like? $\endgroup$ – Unknown x Jul 6 '18 at 16:50
  • $\begingroup$ Is it like $\sum c_I\alpha^I=\sum_{i_1}\sum_{i_2}\sum_{i_3}...\sum_{i_k}c_{i_1,i_2,...,i_k}\alpha^{i_1}\wedge \alpha^{i_2}...\wedge \alpha^{i_k}$? $\endgroup$ – Unknown x Jul 7 '18 at 2:00
  • $\begingroup$ ....yes........ $\endgroup$ – janmarqz Jul 9 '18 at 14:13

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