What does it mean to evaluate a vector field on a 2-form?

**gummy_ratz** · December 13th 2012, 12:15 PM

For example, I've read the definition of the interior product a dozen times, Interior product - Wikipedia, the free encyclopedia , but my understanding is that a 2-form is something that eats two tangent vectors and spits out a number. A vector field on a manifold is a map v(x) = (x,u) for x in M and u in T_xM, so I don't understand how to plug a map into a form.

**Drexel28** · December 13th 2012, 02:24 PM

My guess would ge the following. So, as you've noted a vector field $X$ is just a section (smooth as you please) of the tangent bundle. So, now, if you instead think about the map $Y:X\to TM\oplus T$ $ defined by $Y(x)=(X(x),X(x))$ . You see then that if $\omega\in \bigwedge^2(T^\ast M)$ then you can think about $\omega(Y)$ defined by $\omega(Y)(x)=\omega(Y(x))=\omega(X(x),X(x))$ . That's my best guess without more context.

**xxp9** · December 13th 2012, 04:30 PM

Yes a 2-form eats two vectors and outputs a number, but if you only feed one input it outputs a map:
$\omega_X : V \rightarrow V$ defined by $\omega_X=\omega(X, . ) : Y \mapsto \omega(X,Y)$

Thread: What does it mean to evaluate a vector field on a 2-form?

LinkBack

Thread Tools

Display

What does it mean to evaluate a vector field on a 2-form?

Re: What does it mean to evaluate a vector field on a 2-form?

Re: What does it mean to evaluate a vector field on a 2-form?

Similar Math Help Forum Discussions

Ring, field, Galois-Field, Vector Space

set of symbols form a field

Smith Normal Form over a field

Vector equations in parametric form and in scalar product form

the component form of a vector + unit vector with same direction

Search Tags