The Unapologetic Mathematician

Mathematics for the interested outsider

Tangent Vectors and Coordinates

Let’s say we have a coordinate patch (U,x) around a point p in an n-dimensional manifold M. We can use the function f to give us some tangent vectors at p called the “coordinate vectors”.

We define the coordinate vector \frac{\partial}{\partial x^i}(p)\in\mathcal{T}_pM as follows: given a smooth function f\in\mathcal{O}_p, we define

\displaystyle\left[\frac{\partial}{\partial x^i}(p)\right](f)=\left[D_i(f\circ x^{-1})\right](x(p))

Okay, I know that that’s confusing. But all we mean is this: start with a function f:U\to\mathbb{R}. We compose it with the inverse of the coordinate map to get f\circ x^{-1}:x(U)\to\mathbb{R}, where x(U) is some open neighborhood of the point x(p)\in\mathbb{R}^n. Now we can take that ith partial derivative of this function and evaluate it at the point x(p).

The first thing we really should check is that it doesn’t matter which representative f we pick. That is, if f=g in some neighborhood of p, do we get the same answer? Indeed, in that case f\circ x^{-1}=g\circ x^{-1} in some neighborhood of x(p), and so their partial derivatives are identical. Thus this operation only depends on the germ f\in\mathcal{O}_p.

But is it a tangent vector? It’s easy to see that it’s a linear functional, so we just have to check that it’s a derivation at p:

\displaystyle\begin{aligned}\left[\frac{\partial}{\partial x^i}(p)\right](fg)=&\left[D_i((fg)\circ x^{-1})\right](x(p))\\=&\left[D_i((f\circ x^{-1})(g\circ x^{-1}))\right](x(p))\\=&\left[D_i(f\circ x^{-1})\right](x(p))\left[g\circ x^{-1}\right](x(p))\\&+\left[f\circ x^{-1}\right](x(p))\left[D_i(g\circ x^{-1})\right](x(p))\\=&\left[\frac{\partial}{\partial x^i}(p)\right](f)g(p)+f(p)\left[\frac{\partial}{\partial x^i}(p)\right](g)\end{aligned}

And so we have at least these n vectors at each point p\in M. We can even tell that they much be distinct — and even linearly independent — since we can calculate

\displaystyle\begin{aligned}\left[\frac{\partial}{\partial x^i}(p)\right](x^j)&=\left[\frac{\partial}{\partial x^i}(p)\right](\pi^j\circ x)\\&=\left[D_i(\pi^j\circ x\circ x^{-1})\right](x(p))\\&=\left[D_i(\pi_j)\right](x(p))\end{aligned}

where \pi^j is the jth coordinate projection \pi^j:\mathbb{R}^n\to\mathbb{R}. But we know that D_i(\pi^j) is always and everywhere \delta_i^j — it takes the value 1 if i=j and 0 otherwise.

Thus \frac{\partial}{\partial x^i}(p) takes a different value on x^i than on all the other x^j. Further, any linear combination of the \frac{\partial}{\partial x^j}(p) for j\neq i must take the value 0 on x^i, while \frac{\partial}{\partial x^i}(p) takes the value 1; we see that none of the coordinate vectors can be written as a linear combination of the rest, and conclude that the dimension of \mathcal{T}_pM is at least n.

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March 30, 2011 - Posted by | Differential Topology, Topology

8 Comments »

  1. I’ve always been bothered by the really poor notation here. It’s very tedious,cumbersome, and confusing to wade through all the different senses of parentheses (function/operator application, grouping), operators, functions, and the sets and function spaces in which they live. I know that people like to draw a picture of all the parts living in M versus R^n or R, and that helps. Maybe typographically reminding the reader what space each symbol lives in would help (by typesetting the spaces for each letter on the line above, or using color). Just thinking out loud. Hell, I might right it up with type annotations a la Haskell just to make it clearer to myself lol.

    Sorry, I’m a visual thinker, and it helps to see visually just how this definition carries over our knowledge of R^n to M by all the coordinate patches.

    Excellent as always, though, thank you.

    Comment by Robert | March 31, 2011 | Reply

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