Coordinate Vectors Span Tangent Spaces
Given a point 
in an 
-dimensional manifold 
, we have the vector space 
of tangent vectors at . Given a coordinate patch 
around , we’ve constructed coordinate vectors at , and shown that they’re linearly independent in . I say that they also span the space, and thus constitute a basis.
To see this, we’ll need a couple lemmas. First off, if 
is constant in a neighborhood of , then 
for any tangent vector 
. Indeed, since all that matters is the germ of , we may as well assume that is the constant function with value 
. By linearity we know that 
. But now since 
we use the derivation property to find
and so we conclude that 
.
In a slightly more technical vein, let 
be a “star-shaped” neighborhood of 
. That is, not only does contain 
itself, but for every point 
it contains the whole segment of points 
for 
. An open ball, for example, is star-shaped, so you can just think of that to be a little simpler.
Anyway, given such a and a differentiable function on it we can find functions 
with 
, and such that we can write
where 
is the 
th component function.
If we pick a point we can parameterize the segment 
, and set 
to get a function on the unit interval ![\phi:[0,1]\to\mathbb{R}](https://s0.wp.com/latex.php?latex=%5Cphi%3A%5B0%2C1%5D%5Cto%5Cmathbb%7BR%7D&bg=e6e6e6&fg=333333&s=0 "\phi:[0,1]\to\mathbb{R}")
. This function is clearly differentiable, and we can calculate
using the multivariable chain rule. We find
We can thus find the desired functions by setting
Now if we have a differentiable function defined on a neighborhood of a point 
, we can find a coordinate patch — possibly by shrinking — with 
and 
star-shaped. Then we can apply the previous lemma to 
to get
with ](https://s0.wp.com/latex.php?latex=%5Cpsi_i%280%29%3D%5Cleft%5B%5Cfrac%7B%5Cpartial%7D%7B%5Cpartial+x%5Ei%7D%28p%29%5Cright%5D%28f%29&bg=e6e6e6&fg=333333&s=0 "\psi_i(0)=\left[\frac{\partial}{\partial x^i}(p)\right](f)")
. Moving the coordinate map to the other side we find
Now we can hit this with a tangent vector
+x^i(p)v\left(\psi_i\circ x\right)\right)\\&=\sum\limits_{i=1}^nv(x^i)\psi_i(0)\\&=\sum\limits_{i=1}^nv(x^i)\left[\frac{\partial}{\partial x^i}(p)\right](f)\end{aligned}](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle%5Cbegin%7Baligned%7Dv%28f%29%26%3Dv%5Cleft%28f%28p%29%2B%5Csum%5Climits_%7Bi%3D1%7D%5Enx%5Ei%5Cleft%28%5Cpsi_i%5Ccirc+x%5Cright%29%5Cright%29%5C%5C%26%3Dv%5Cleft%28f%28p%29%5Cright%29%2B%5Csum%5Climits_%7Bi%3D1%7D%5Env%5Cleft%28x%5Ei%5Cleft%28%5Cpsi_i%5Ccirc+x%5Cright%29%5Cright%29%5C%5C%26%3D%5Csum%5Climits_%7Bi%3D1%7D%5En%5Cleft%28v%28x%5Ei%29%5Cleft%5B%5Cpsi_i%5Ccirc+x%5Cright%5D%28p%29%2Bx%5Ei%28p%29v%5Cleft%28%5Cpsi_i%5Ccirc+x%5Cright%29%5Cright%29%5C%5C%26%3D%5Csum%5Climits_%7Bi%3D1%7D%5Env%28x%5Ei%29%5Cpsi_i%280%29%5C%5C%26%3D%5Csum%5Climits_%7Bi%3D1%7D%5Env%28x%5Ei%29%5Cleft%5B%5Cfrac%7B%5Cpartial%7D%7B%5Cpartial+x%5Ei%7D%28p%29%5Cright%5D%28f%29%5Cend%7Baligned%7D&bg=e6e6e6&fg=333333&s=0 "\displaystyle\begin{aligned}v(f)&=v\left(f(p)+\sum\limits_{i=1}^nx^i\left(\psi_i\circ x\right)\right)\\&=v\left(f(p)\right)+\sum\limits_{i=1}^nv\left(x^i\left(\psi_i\circ x\right)\right)\\&=\sum\limits_{i=1}^n\left(v(x^i)\left[\psi_i\circ x\right](p)+x^i(p)v\left(\psi_i\circ x\right)\right)\\&=\sum\limits_{i=1}^nv(x^i)\psi_i(0)\\&=\sum\limits_{i=1}^nv(x^i)\left[\frac{\partial}{\partial x^i}(p)\right](f)\end{aligned}")
where we have used linearity, the derivation property, and the first lemma above. Thus we can write
and the coordinate vectors span the space of tangent vectors at .
As a consequence, we conclude that always has dimension — exactly the same dimension as the manifold itself. And this is exactly what we should expect; if is -dimensional, then in some sense there are independent directions to move in near any point , and these “directions to move” are the core of our geometric notion of a tangent vector. Ironically, if we start from a more geometric definition of tangent vectors, it’s actually somewhat harder to establish this fact, which is partly why we’re starting with the more algebraic definition.
Like this:
Like Loading...
Related
March 31, 2011 -
Posted by John Armstrong |
Differential Topology, Topology
[…] problem, then, is to calculate the th component — the one corresponding to — of . But we know that this coefficient comes from sticking into this vector and seeing what pops […]
Pingback by Derivatives in Coordinates « The Unapologetic Mathematician | April 6, 2011 |
[…] basis on , and thus get components of in those coordinates. As usual, we calculate the th component […]
Pingback by Curves « The Unapologetic Mathematician | April 8, 2011 |
[…] if has dimension , then we know is an -manifold, and thus we know that is an -dimensional vector space for every point . Now, of course all -dimensional vector […]
Pingback by The Tangent Bundle of a Euclidean Space « The Unapologetic Mathematician | April 11, 2011 |
[…] we know at each point that the coordinate vectors span the tangent space. So let’s take a vector field and break up the vector . We can […]
Pingback by Coordinate Vector Fields « The Unapologetic Mathematician | May 24, 2011 |
[…] is basically just like a lemma we proved for functions on star-shaped neighborhoods. Indeed, it suffices to […]
Pingback by Calculating the Lie Derivative « The Unapologetic Mathematician | June 16, 2011 |
Just after “multivariable chain rule”, f(p) – f(0) should be f(x) – f(0).
p lives in the manifold rather than R^n.
Hi, I’m learning differential geometry and run into this blog, I must say you did a great job. But I have a question: “What does \circ means?”
It sounds like for some reason the TeX is not being rendered in your browser. \circ is the TeX code to generate the composition symbol
.