Something about the relationship between Plucker’s embedding and diff. forms

I hope everything is OK since I just start trying posting by mail.

The reason of being categorized into A.G. is that the idea is inspired by Plucker embedding in Algebraic geometry. Therefore I’d like to introduce this notion first.

There are several ways to give grassmanian a topology, manifold structure. Plucker embedding is one that in addition treat it as a variety.

$G(r,n)$ denotes all the subspaces with dimension r in a linear space $V=k^n$ . One may embed it into ${\bf P}V^{\wedge r}$ where P means projectivization, i.e. quotient scalars.(who tells me how to insert wedge in tex). Choose a subspace $W$ in $G(r,n)$ , choose any basis $v_1,...,v_r$ of $W$ , one may readily verify $v_1\wedge v_2...\wedge v_n$ always refers to the same element when the subspace is fixed. This is the Plucker’s embedding.

When taking scalars into consideration, if V is endowed with a pos. definite quadratic form, one may view $v_1\wedge v_2...\wedge v_n$ as a “cube” with some sort of “volume”. Therefore diff. forms can be viewed as continuingly varying cubes. Probably this is why we generalize the original concept of “differential” dated back to Newton’s age by introducing the anti-commutative tensors.

Written by lethear

December 25, 2010 at 7:05 am

Posted in A.G., Old things

Tagged with OLD JUNKS.

3 Responses

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Wedge in tex is \wedge. The formulas aren’t parsing as you have them.

$V^{\wedge r}$ mod scalars is just $V^{\otimes r}$ , right? You have, say, $v_1\wedge v_2=-v_2\wedge v_1=v_2\wedge v_1$ so the wedge is commutative.

From the geometry perspective, I always got the impression that “continuously varying cubes” are how you’re supposed to view differential forms. That is, (working on a manifold with $k=\mathbb{R}$ ), $r$ -forms are dual to singular $r$ -chains, and you pair the two via integration. So a $2$ -form is an “element of area,” something you can integrate over an area, and so on. I have no idea how it works on other fields.

thetwomeatmeal

December 28, 2010 at 7:49 am

Reply
- Ahh… probably it will take much time to practise compiling tex in brain.
  
  Indeed, what I mentioned is just the “element of area”. I just recalled this right after I publish it. It shouldn’t come to me with surprise. Therefore I changed the title.
  Actually when studying differential forms, we were just supposed to memorize Cartan’s lemma, the rules of anticommunitivity, bilinearity, and so on. We were just confused with the mysterious signs when integrating the highest forms and having no intuitive idea about the lower forms.
  But $V^{\wedge r}$ is not just $V^{\otimes r}$ . I think I chose the wrong term. The mod should be quotient, since I’m just constructing a top. space. After quotient it merely becomes a projective space. Ahh, 0 has to be excluded at first.
  
  By Cartan’s lemma, one may even find the defining polynomials of the the embedded grassmanian.
  
  By the way, do you know how to prove that de Rham cohomology is naturally isomorphic with singular cohomology in an orientable manifold? There is an arrow from the latter one to the former one assuming the Poincare duality and that there is smooth n-cycle in every equivalent class. But how to go further?
  
  lethear
  
  December 28, 2010 at 10:57 am
  
  Reply
  - Oh, I see where I went wrong. Projectivization makes the wedge product commutative rather than anticommutative, but the tensor product isn’t commutative in the first place. Silly me.
    
    For cohomology, I’d assume de Rham cochains include into singular cochains if we view the $k$ -form $\alpha$ as the cochain sending the singular $k$ -simplex $\beta$ to $\int_{\beta}\alpha$ . i. e. differential forms are just the singular cochains which happen to be smooth. You could prove that this becomes a cohomology isomorphism together with the map you gave.
    
    Actually, it might just be easier to prove that the Eilenberg-Steenrod axioms hold for de Rham cohomology, since any functor that satisfies them is naturally isomorphic to singular cohomology.
    
    I haven’t actually done this myself, but I think it would be cool to send it through Čech cohomology, which is defined by taking a “good open cover” of the manifold (all sets and all nonempty intersections contractible) and then turning that, combinatorially, into a simplicial complex, and taking the singular cohomology of that simplicial complex. For manifolds, the simplicial complex you get should be homeomorphic to the manifold, so it’s easy to see that this is isomorphic to singular cohomology, and I imagine de Rham wouldn’t be too much harder. The reason this is interesting is because it’s nontrivial to show that a good cover even exists! You can do it, I think, with geodesically convex balls.
    
    thetwomeatmeal
    
    December 30, 2010 at 5:01 am