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

• A
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##?

wrobel
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\}##

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!

Infrared
Gold Member
The $\sigma^i$ are necessarily independent since $\alpha\neq 0$.

In your example, both $\sigma^1$ and $\sigma^2$ are in $M_\alpha$. They are independent, so the dimension of $M_\alpha$ is not $1$.

kiuhnm
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.

wrobel
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

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.

The $\sigma^i$ are necessarily independent since $\alpha\neq 0$.

In your example, both $\sigma^1$ and $\sigma^2$ are in $M_\alpha$. They are independent, so the dimension of $M_\alpha$ is not $1$.

Yes, you're right. Thank you.