# Search results

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

Yes, you're right. Thank you.
2. ### A Diff. forms: M_a = {u /\ a=0 | u in L}

I thought we were having a civil discussion. Guess I was wrong.
3. ### A Diff. forms: M_a = {u /\ a=0 | u in L}

Nope, the dimension is still 2.
4. ### A Diff. forms: M_a = {u /\ a=0 | u in L}

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

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...
6. ### A The Hodge star operator

I wasn't sure whether my formula for $*\lambda$ was correct. I had doubts because I got a wrong sign in example 1 on page 16. That example says that if the base of $V$ is $(dx^1,dx^2,dx^3,dt)$ where $(dx^i,dx^i)=1$ and $(dt,dt)=-1$, then $$*(dx^i dt) = dx^j dx^k,$$ where...
7. ### A The Hodge star operator

I'm reading section 2.7 of Flanders' book about differential forms, but I have some doubts. Let $\lambda$ be a $p$-vector in $\bigwedge^p V$ and let $\sigma^1,\ldots,\sigma^n$ be a basis of $V$. There's a unique $*\lambda$ such that, for all $\mu\in \bigwedge^{n-p}$, \lambda...