Diff. forms: M_a = {u /\ a=0 | u in L}

  • #1
66
1

Main Question or Discussion Point

Here's exercise 1 of chapter 2 in Flanders' book.
Let ##L## be an ##n##-dimensional space. For each ##p##-vector ##\alpha\neq0## we let ##M_\alpha## be the subspace of ##L## consisting of all vectors ##\sigma## satisfying ##\alpha\wedge\sigma=0##. Prove that ##\dim(M_\alpha)\leq p##. Prove also that ##\dim(M_\alpha)=p## if and only if ##\alpha=\sigma^1\wedge\cdots\wedge\sigma^p## where ##\sigma^1,\ldots,\sigma^p## are vectors in ##L##.
(I wrote ##\sigma^1,\ldots## above, but Flanders wrote ##\sigma_1,\ldots## in this exercise. Weird: he uses upper indices in the theory sections.)

My question is this: Isn't it false that if ##\alpha=\sigma^1\wedge\cdots\wedge\sigma^p## where ##\sigma^1,\ldots,\sigma^p## are vectors in ##L##, then ##\dim(M_\alpha)=p##?
 

Answers and Replies

  • #2
if α=σ1∧⋯∧σp\alpha=\sigma^1\wedge\cdots\wedge\sigma^p where σ1,…,σp\sigma^1,\ldots,\sigma^p are vectors in LL, then dim(Mα)=p\dim(M_\alpha)=p?


equality ##\alpha\wedge \sigma=0##implies that ##\sigma## belongs to the subspace spanned on ##\{\sigma^1,\ldots,\sigma^p\}##
 
  • #3
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equality ##\alpha\wedge \sigma=0##implies that ##\sigma## belongs to the subspace spanned on ##\{\sigma^1,\ldots,\sigma^p\}##
Let ##\{u^1,u^2,u^3,u^4\}## be a base of ##L##. If we take ##\sigma^1=u^1+u^2## and ##\sigma^2=u^3+u^4## then ##\sigma^1## and ##\sigma^2## are clearly vectors in ##L##, but if ##\alpha=\sigma^1\wedge\sigma^2## then ##\dim(M_\alpha)=1## and not ##2(=p)## as claimed by the text.
The ##\sigma##s also need to be linearly independent!
 
  • #4
Infrared
Science Advisor
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The [itex]\sigma^i[/itex] are necessarily independent since [itex]\alpha\neq 0[/itex].

In your example, both [itex]\sigma^1[/itex] and [itex]\sigma^2[/itex] are in [itex]M_\alpha[/itex]. They are independent, so the dimension of [itex]M_\alpha[/itex] is not [itex]1[/itex].
 
  • #5
66
1
Let ##\{u^1,u^2,u^3,u^4\}## be a base of ##L##. If we take ##\sigma^1=u^1+u^2## and ##\sigma^2=u^3+u^4## then ##\sigma^1## and ##\sigma^2## are clearly vectors in ##L##, but if ##\alpha=\sigma^1\wedge\sigma^2## then ##\dim(M_\alpha)=1## and not ##2(=p)## as claimed by the text.
Nope, the dimension is still 2.
 
  • #6
then dim(Mα)=1\dim(M_\alpha)=1 and not 2(=p)2(=p) as claimed by the text.
Thus the textbook is wrong so am i, you have comprehended everything and there is no need for you to learn any more
 
  • #7
66
1
Thus the text book is wrong so am i, you have comprehanded everything and there is no need for you to learn any more
I thought we were having a civil discussion. Guess I was wrong.
 
  • #8
66
1
The [itex]\sigma^i[/itex] are necessarily independent since [itex]\alpha\neq 0[/itex].

In your example, both [itex]\sigma^1[/itex] and [itex]\sigma^2[/itex] are in [itex]M_\alpha[/itex]. They are independent, so the dimension of [itex]M_\alpha[/itex] is not [itex]1[/itex].
Yes, you're right. Thank you.
 

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