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  1. K

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

    Yes, you're right. Thank you.
  2. K

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

    I thought we were having a civil discussion. Guess I was wrong.
  3. K

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

    Nope, the dimension is still 2.
  4. K

    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. K

    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. K

    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. K

    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...
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