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

[kiuhnm]

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 $\{\sigma^1,\ldots,\sigma^p\}$

 

[kiuhnm]

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]

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

 

[kiuhnm]

Thus the textbook 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.

 

[kiuhnm]

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.