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))](http://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}](http://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}](http://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 .
March 30, 2011
Posted by John Armstrong |
Differential Topology, Topology |
8 Comments
