Star operator in the simplest form Let E together with g be a inner product space(over field R) , dimE=n<infinity and {e1,⋯,en} is orthonormal basis of E that {e1,⋯,en} is its dual basis(for E∗). Now we define w:=e1 wedge⋯wedge en as an element of volume of E. I prove that g^flag :E→E∗ with rule (g^flag(u))(v)=g(u,v), is an isomorphism (forall u,wedge in E). Convention: u↦g^flag u~ i.e. u~:=g^flag (u). I wish to prove that for any p-form theta in wedge p(E), There exist a unique element eta in wedge (n−p)(E) such that eta (u1,⋯,un−p)w=theta wedge u~1⋯wedge u~n−p florall 1,⋯,un−p in E . How can I do this? Of course I guess that should be defined an inner product g wedge k(E) on wedge k(E) for any 0<k<n and then use g^flag wedge k(E). what is your method ? and How?

cuuhorre76

cuuhorre76

Answered question

2022-09-05

Star operator in the simplest form
Let E together with g be a inner product space(over field R) , dim E = n < and { e 1 , , e n } is orthonormal basis of E that { e 1 , , e n } is its dual basis(for E∗). Now we define ω := e 1 e n as an element of volume of E.
I prove that g : E E with rule ( g ( u ) ) ( v ) = g ( u , v ), is an isomorphism u , v E
Convention: u ~ := g ( u )
I wish to prove that for any p-form θ Λ p ( E ), There exist a unique element η Λ ( n p ) ( E ) such that
η ( u 1 , , u n p ) ω = θ u ~ 1 u ~ n p u 1 , , u n p E
How can I do this?
Of course I guess that should be defined an inner product g Λ k ( E ) on Λ k ( E ) for any 0<k<n and then use g Λ k ( E ) . what is your method ? and How?

Answer & Explanation

illpnthr21vw

illpnthr21vw

Beginner2022-09-06Added 17 answers

Given a p-form θ p E , we can define an alternating multilinear map h : E n p n E by
h ( u 1 , , u n p ) = θ u ~ 1 u ~ n p .
Let b : R n E be the linear map
b ( t ) = t ω .
Because n E is one-dimensional and ω is nonzero, this map is an isomorphism. You can think of it as the coordinate system for n E given by the basis { ω }. Observe that, for any α n E, the scalar b 1 ( α ) is the unique scalar with the property that b 1 ( α ) ω = α
Composing h with b 1 gives a map H : E n p R . Looking back at our discussions of h and b 1 , you can see that H ( u 1 , , u n p ) is the unique scalar with the property that
H ( u 1 , , u n p ) ω = θ u ~ 1 u ~ n p .
Since h is multilinear and alternating, H is too. The map H can therefore be expressed as an ( n p ) -form η, using the standard correspondence between elements of E n p R and alternating multilinear maps E n p R

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Linear algebra

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?