The restriction of differential form

[1591238460]

1. Assume M=xdy -ydx+dz ∈ Ω1(R^3). What's the restriction of M to the plane {z=2}? I think it's xdy-ydx. Is that right?
   

 

[lavinia]

1. Assume M=xdy -ydx+dz ∈ Ω1(R^3). What's the restriction of M to the plane {z=2}? I think it's xdy-ydx. Is that right?

Do the two forms have the same values on tangent vectors to z = 2?

 

[1591238460]

Thank you, so what should I do?

 

[lavinia]

Thank you, so what should I do?

Just check the values.

 

[1591238460]

I think the tangent vectors of the plane z=2 are in the form of a$\frac{\partial}{\partial x}$+b$\frac{\partial}{\partial y}$, to a$\frac{\partial}{\partial x}$+b$\frac{\partial}{\partial y}$, both of the two forms of the value, am I right?

 

[WWGD]

I think the tangent vectors of the plane z=2 are in the form of a$\frac{\partial}{\partial x}$+b$\frac{\partial}{\partial y}$, to a$\frac{\partial}{\partial x}$+b$\frac{\partial}{\partial y}$, both of the two forms of the value, am I right?

Use ## ##'s at the beginning and end if you want to do Latex editing here.

 

[WWGD]

I think the tangent vectors of the plane z=2 are in the form of a$\frac{\partial}{\partial x}+b\frac{\partial}{\partial y}$, to a $\frac{\partial}{\partial x}+b\frac{\partial}{\partial y}$, both of the two forms of the value, am I right?

 

[WWGD]

Notice that, as a plane $z=2$ is 2-dimensional. Equivalently, points in the plane are of the form $(x,y,2)$

 

[1591238460]

@WWGD , thank you very much, so I think I am right, doesn't it?

 

[mathwonk]

what about just saying z is constant along z=2 so dz is zero on vectors tangent to z=2? is that what you were thinking?

 

[1591238460]

@mathwonk yes, and in the embedded submanifold {z=2} the the tangent vectors are just in the form a∂∂x+b∂∂y, so the two values of two forms are the same.