/sci/ - Science & Math » Thread #11311735

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Anonymous Sat 18 Jan 19:38:31 2020 No.11311735 View Reply Original Report

Quoted By: >>11311818

We just got into multilinear forms and I'm having quite a hard time trying to understand them.

Could someone please show me how this exercise can be done?

Thanks in advance!

Anonymous

Anonymous Sat 18 Jan 2020 19:54:31 No.11311776 Report

Quoted By: >>11311806

>Could someone please show me how this exercise can be done?
What have you tried?

Anonymous

Anonymous Sat 18 Jan 2020 20:10:18 No.11311806 Report

Quoted By:

>>11311776
I failed pretty early in. One idea would be to represent every $b_{i_k}$ using the $b^*_{i_k}$ s to then use fitting properties, but I do not know how to accomplish that.

Anonymous

Anonymous Sat 18 Jan 2020 20:17:18 No.11311818 Report

Quoted By: >>11311835 >>11311840

>>11311735
any vector $v$ can be written using the basis ${b_i}$ . do you see that the coefficients are precisely $b^*_i(v)$ ? i.e. it holds $v = \sum b^*_i(v)b_i$ .

Anonymous

Anonymous Sat 18 Jan 2020 20:22:52 No.11311835 Report

Quoted By: >>11311843

>>11311818
Yes, that makes sense to me, thanks. How can I use this property, as every vector in $\sum_{i_1\lt\cdots\lt i_k} D(b_{i_1},b_{i_2},\dots,b_{i_k}) \cdot b_{i_1}^* \wedge b_{i_2}^* \wedge \cdots \wedge b_{i_k}^* = D$ already is a basis vector?

Anonymous

Anonymous Sat 18 Jan 2020 20:25:47 No.11311840 Report

Quoted By:

>>11311818
Or would it make more sense to actually compute/represent $D(b_{i_1},b_{i_2},\dots,b_{i_k}) \cdot b_{i_1}^* \wedge b_{i_2}^* \wedge \cdots \wedge b_{i_k}^*$ , as it would maybe yield a analogous form?

Anonymous

Anonymous Sat 18 Jan 2020 20:28:54 No.11311843 Report

Quoted By: >>11311846

>>11311835
do you also see that (completely dually) any 1-form $\omega$ can be written using the $b^*_i$ and the coefficients are precisely $\alpha(b_i)$ ? i.e. $\alpha = \sum \alpha(b_i)b^*_i$

Anonymous

Anonymous Sat 18 Jan 2020 20:29:56 No.11311846 Report

Quoted By: >>11311868

>>11311843
$\omega$ should be $\alpha$ of course

Anonymous

Anonymous Sat 18 Jan 2020 20:43:19 No.11311868 Report

Quoted By: >>11311877

>>11311846
So the $\alpha$ you mentioned is $\in \Lambda^1 V*$ , right? If so, then yes, I see that $\alpha = \sum \alpha(b_i)b^*_i$ holds because of the very definition.

Anonymous

Anonymous Sat 18 Jan 2020 20:45:46 No.11311871 Report

Quoted By:

It's just so hard for me to connect everything together.

Anonymous

Anonymous Sat 18 Jan 2020 20:48:44 No.11311877 Report

Quoted By:

>>11311868
consider now a bilinear form, say $g \colon V \times V \to \mathbb{K}$ . not necessarily alternating, just bilinear. this can be written using $b^*_i \otimes b^*_j$ , i.e. $g = \sum K_{ij}b^*_i \otimes b^*_j$ . do you see what the coefficients $K_{ij}$ are ?

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