Analogy between vectors and covectors

[atyy]

P1: A curve defines a vector at point.
P2: A function defines a covector at a point.

F1a: A congruence of curves defines a vector field.
F1b:Every vector field corresponds to a congruence of curves.

Statements F1 are analogous to P1.

F2a: A "congruence of functions" defines a covector field?
F2b: Every covector field correspond to a "congruence of functions"?

What, if any, are the correct statements F2 that would be analogous to P2, just as F1 is analogous to P1?

[HallsofIvy]

First you will have to explain what you mean by a "congurence of functions".

[atyy]

First you will have to explain what you mean by a "congurence of functions".

The question is whether there is anything that when put in place "congruence of functions" will yield a true statement.

If there is, great, then there is a statement F2 analogous to P2, like F1 is analogous to P1.

Otherwise, is there any reason why such an analogy doesn't exist?

[quasar987]

I don't know if this counts as an analogy to you but for one, both vectors and covectors are vectors. And given a basis {v_1,...,v_n} of vectors, there is a natural basis {f_1,...,f_n} of covectors given by f_i(v_j)=1 if i=j and =0 otherwise.

http://en.wikipedia.org/wiki/Covector

[mathwonk]

what is a congruence of curves? and how does it define a vector field? i.e. before finding a statement analogous to yours i need to understand your statement.

[atyy]

I don't know if this counts as an analogy to you but for one, both vectors and covectors are vectors. And given a basis {v_1,...,v_n} of vectors, there is a natural basis {f_1,...,f_n} of covectors given by f_i(v_j)=1 if i=j and =0 otherwise.

Yes, it's because of this interplay of vectors and covectors - ie. a vector is a covector to a covector, and a covector is a vector when the original vector is considered a covector - that I wanted to know if it extended beyond vector spaces.

If you consider a trajectory, then you can define a velocity at every point along the trajectory. However, a velocity is actually meaningless on its own, and you have to specify a reference, such as a velocity with respect to some scalar field(s). So in some sense, it is better to consider the velocity vector as an operator, which is what a vector is with respect to a covector anyway. The scalar field defines a reference frame, which defines a covector at each point of the trajectory. The covector defined by the reference scalar field acting on the velocity vector gives the speed.

So anyway - trajectories or curves define vectors, and scalar fields define covectors.

what is a congruence of curves? and how does it define a vector field? i.e. before finding a statement analogous to yours i need to understand your statement.

Instead of just having a vector at a point, we can have vectors at every point in space. Since a trajectory gives rise to vectors, we can think of a vector field as being formed by many trajectories, all laid side by side so that they cover the entire space. That is what I mean by a congruence of curves, and why a congruence of curves gives rise to a vector field.

From the point of view of vector spaces, covectors and vectors are completely analogous. However, going to trajectories and scalar fields there is no analogy, because a trajectory is a map from the real numbers into the space, and a function is a map from the space into the real numbers (though I guess coordinates are sometimes specified as maps from the real numbers into the space, but you need N of them). I just wanted to know which wins out - the vector space analogy, or the lack of analogy between functions into and functions from the real numbers.

[mathwonk]

(in a sense you have things sort of backwards, as a vector field is more elementary than a congruence of curves, which in fact is a solution of a diff.eq. given by the vector field.)

but anyway, the correct analogy with your definition of a congruence of curves is just a single function, which gives a covector field.

[atyy]

but anyway, the correct analogy with your definition of a congruence of curves is just a single function, which gives a covector field.

Thanks, yes, a scalar function defines a covector field - so F2a works with "congruence of functions" being just "scalar function".

But it seems that not every covector field can derived from a scalar function. So in F2b "congruence of functions" cannot be "scalar function". Is there anything that would work there?

[mathwonk]

not all differential equations have solutions. this is true for vector and covector fields.

[atyy]

not all differential equations have solutions. this is true for vector and covector fields.

Thanks mathwonk!