Tangent Vectors and Coordinates
Let’s say we have a coordinate patch 
around a point 
in an 
-dimensional manifold 
. We can use the function 
to give us some tangent vectors at called the “coordinate vectors”.
We define the coordinate vector 
as follows: given a smooth function 
, we define
=\left[D_i(f\circ x^{-1})\right](x(p))](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle%5Cleft%5B%5Cfrac%7B%5Cpartial%7D%7B%5Cpartial+x%5Ei%7D%28p%29%5Cright%5D%28f%29%3D%5Cleft%5BD_i%28f%5Ccirc+x%5E%7B-1%7D%29%5Cright%5D%28x%28p%29%29&bg=e6e6e6&fg=333333&s=0 "\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 
. We compose it with the inverse of the coordinate map to get 
, where 
is some open neighborhood of the point 
. Now we can take that 
th partial derivative of this function and evaluate it at the point 
.
The first thing we really should check is that it doesn’t matter which representative we pick. That is, if 
in some neighborhood of , do we get the same answer? Indeed, in that case 
in some neighborhood of , and so their partial derivatives are identical. Thus this operation only depends on the germ .
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 :
=&\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}](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle%5Cbegin%7Baligned%7D%5Cleft%5B%5Cfrac%7B%5Cpartial%7D%7B%5Cpartial+x%5Ei%7D%28p%29%5Cright%5D%28fg%29%3D%26%5Cleft%5BD_i%28%28fg%29%5Ccirc+x%5E%7B-1%7D%29%5Cright%5D%28x%28p%29%29%5C%5C%3D%26%5Cleft%5BD_i%28%28f%5Ccirc+x%5E%7B-1%7D%29%28g%5Ccirc+x%5E%7B-1%7D%29%29%5Cright%5D%28x%28p%29%29%5C%5C%3D%26%5Cleft%5BD_i%28f%5Ccirc+x%5E%7B-1%7D%29%5Cright%5D%28x%28p%29%29%5Cleft%5Bg%5Ccirc+x%5E%7B-1%7D%5Cright%5D%28x%28p%29%29%5C%5C%26%2B%5Cleft%5Bf%5Ccirc+x%5E%7B-1%7D%5Cright%5D%28x%28p%29%29%5Cleft%5BD_i%28g%5Ccirc+x%5E%7B-1%7D%29%5Cright%5D%28x%28p%29%29%5C%5C%3D%26%5Cleft%5B%5Cfrac%7B%5Cpartial%7D%7B%5Cpartial+x%5Ei%7D%28p%29%5Cright%5D%28f%29g%28p%29%2Bf%28p%29%5Cleft%5B%5Cfrac%7B%5Cpartial%7D%7B%5Cpartial+x%5Ei%7D%28p%29%5Cright%5D%28g%29%5Cend%7Baligned%7D&bg=e6e6e6&fg=333333&s=0 "\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 vectors at each point 
. We can even tell that they much be distinct — and even linearly independent — since we can calculate
&=\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}](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle%5Cbegin%7Baligned%7D%5Cleft%5B%5Cfrac%7B%5Cpartial%7D%7B%5Cpartial+x%5Ei%7D%28p%29%5Cright%5D%28x%5Ej%29%26%3D%5Cleft%5B%5Cfrac%7B%5Cpartial%7D%7B%5Cpartial+x%5Ei%7D%28p%29%5Cright%5D%28%5Cpi%5Ej%5Ccirc+x%29%5C%5C%26%3D%5Cleft%5BD_i%28%5Cpi%5Ej%5Ccirc+x%5Ccirc+x%5E%7B-1%7D%29%5Cright%5D%28x%28p%29%29%5C%5C%26%3D%5Cleft%5BD_i%28%5Cpi_j%29%5Cright%5D%28x%28p%29%29%5Cend%7Baligned%7D&bg=e6e6e6&fg=333333&s=0 "\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 
is the 
th coordinate projection 
. But we know that 
is always and everywhere 
— it takes the value 
if 
and 
otherwise.
Thus 
takes a different value on 
than on all the other 
. Further, any linear combination of the 
for 
must take the value on , while takes the value ; we see that none of the coordinate vectors can be written as a linear combination of the rest, and conclude that the dimension of 
is at least .
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March 30, 2011 -
Posted by John Armstrong |
Differential Topology, Topology
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.
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