The one-form basis and contraction

1. Oct 10, 2008

Ben Niehoff

I'm sure this question has been asked before, but I haven't been able to find a good answer using the search function.

I have a differentiable manifold M and some coordinate chart. I have vector and covector fields on this manifold, but no metric. At some point P, I know I can write the tangent vectors as a linear combination of directional derivative operators

$$v = \sum v_i \frac{\partial}{\partial x^i}$$

and I can write the covectors as a linear combination of one-forms

$$\omega = \sum \omega_j \, dx^j$$

Furthermore, I ought to be able to take the contraction $\langle \omega, v \rangle$ algebraically, using linearity and the relations

$$\langle dx^i, \frac{\partial}{\partial x^j} \rangle = \delta^i_j$$

Now, by definition, the one-form basis lies in the dual space of linear functionals on V, so the contraction above really means that the linear functionals dx^i must "act upon", in some sense, the basis vectors d/dx^j of V; i.e.

$$dx^i \left( \frac{\partial}{\partial x^j} \right) = \delta^i_j$$

But my question is, what in the world does this mean? I thought that this "canonical" set of bases was supposed to lead to some natural way to compute the contraction, but I don't see it.

What does it mean for dx( ) to act upon some other object?

2. Oct 10, 2008

George Jones

Staff Emeritus
I have to catch my bus to work in a few minutes, and I have a lot on my plate today, so I might not be able to respond today in as much detail as I'd like.

The indices on the components of v should be superscrpts.

Use this and linearity to calculate $\omega(v)$.

Last edited: Oct 10, 2008
3. Oct 10, 2008

Ben Niehoff

I think you missed my question. I was asking why

$$dx(\frac{\partial}{\partial x}) = 1$$

should be true in the first place.

4. Oct 10, 2008

George Jones

Staff Emeritus
I couldn't decide what you were asking. I though that first you wanted to get at the component version of contraction.

By definition!

5. Oct 10, 2008

Ben Niehoff

So then what was the point of using dx and d/dx as basis vectors? I thought the contraction was supposed to somehow arise naturally out of that choice, via calculus intuitions.

If we're just going to define the contraction ad hoc, then doesn't it increase readability to use e^i and e_i?

I guess the real question is, does the symbol dx( ) have some real meaning from which the contraction with d/dx naturally follows? Ideally, whatever it is should also have a sensible meaning in one-dimensional calculus; i.e., there should be some sense in which dx and d/dx are natural "inverses". But I don't see it.

6. Oct 10, 2008

dx

The 1-forms $$dx^i$$ are covectors representing linear approximations to the coordinate functions $$\pi^i$$, i.e they are linear functions. As you probably know, these form a vector space, and any gradient can be written as a linear combination of these coordinate gradients.

For example, let $$f$$ be a function defined on $$\mathbb{R} \times \mathbb{R}$$ to $$\mathbb{R}$$. What is the linear approximation to f at the origin? We want a linear function $$df : \mathbb{R}^2 \longrightarrow \mathbb{R}$$ such that for small $$(x,y) \in \mathbb{R}^2$$, we have $$df(x,y) \approx f(x,y)$$. Again, as you probably know, this function can be written as a linear combination of functions $$dx$$ and $$dy$$:

$$df = \partial_x{f} dx + \partial_{y}{f} dy$$.

This of course looks very familiar, and this is the reason we use this notation instead of $$e^i$$.

That takes care of functions, but what about their dual objects, i.e. parametrized curves? consider a curve $$\gamma : \mathbb{R} \longrightarrow \mathbb{R}^2$$ which passes through the origin with $$\gamma(0) = (0,0)$$. The linear approximation to this curve (the tangent vector) is a linear function of the form $$\gamma(u) = (ua,ub)$$, which can always be written as a combination of curves $$\gamma(u) = a\gamma_1(u) + b\gamma_2(u)$$ where $$\gamma_1(u) = (u,0)$$ and $$\gamma_2(u) = (0,u)$$. You can convince yourself that these tangent vectors to parametrized curves also form a vector space, and can always be written as a linear combination of curves determined by the coordinate system. $$\gamma_1$$ and $$\gamma_2$$ are usually denoted by

$$\gamma_1 = \frac{\partial}{\partial{x}}$$

$$\gamma_2 = \frac{\partial}{\partial{y}}$$

Now we see why $$dx \circ \partial_x$$ is 1: $$(dx \circ \partial_{x})(u)$$ = $$dx(u,0)$$= u.