How Does the Pushforward dφ Transform Vectors in Differential Geometry?

[demonelite123]

I am reading a derivation of the Euler Lagrange equations. They defined coordinate charts φ: U -> R^n where x -> φ(x) = (x¹, x², ... xⁿ). then they said for the one form dφ: TU -> TR^n ~= R^n X R^n where (x,y) -> dφ(x,y) = (x¹, ..., xⁿ, y¹, ..., yⁿ).

what I am confused about is that when i learned differential forms, i learned that they are always a map from the tangent bundle to R. they return real numbers. i don't understand how they got dφ: TU -> TR^n ~= R^n X R^n. can someone help to clarify this for me? thanks.

 

[gtoguy97]

What they are doing is defining a map from the tangent bundle of U to the tangent bundle of R^n. This is known as a "pushforward" of the differential form. To understand this, it is helpful to think of dφ as a linear transformation from the tangent space of U to the tangent space of R^n. It takes each vector in the tangent space of U and maps it to a corresponding vector in the tangent space of R^n. In this way, it can be thought of as an operator that transforms vectors from one coordinate system to another. When we say dφ: TU -> TR^n ~= R^n X R^n, what we mean is that the pushforward of the differential form dφ can be represented by a matrix in which each row corresponds to a vector in the tangent space of U and each column corresponds to a vector in the tangent space of R^n.