Coordinate Vectors Span Tangent Spaces

Given a point $p$ in an $n$ -dimensional manifold $M$ , we have the vector space $\mathcal{T}_pM$ of tangent vectors at $p$ . Given a coordinate patch $(U,x)$ around $p$ , we’ve constructed $n$ coordinate vectors at $p$ , and shown that they’re linearly independent in $\mathcal{T}_pM$ . 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 $f$ is constant in a neighborhood of $p$ , then $v(f)=0$ for any tangent vector $v$ . Indeed, since all that matters is the germ of $f$ , we may as well assume that $f$ is the constant function with value $c$ . By linearity we know that $v(c)=cv(1)$ . But now since $1=1\cdot1$ we use the derivation property to find

$\displaystyle v(1)=v(1\cdot1)=v(1)1(p)+1(p)v(1)=2v(1)$

and so we conclude that $v(1)=v(c)=0$ .

In a slightly more technical vein, let $U$ be a “star-shaped” neighborhood of $0\in\mathbb{R}^n$ . That is, not only does $U$ contain $0$ itself, but for every point $x\in U$ it contains the whole segment of points $tx$ for $0\leq t\leq1$ . An open ball, for example, is star-shaped, so you can just think of that to be a little simpler.

Anyway, given such a $U$ and a differentiable function $f$ on it we can find $n$ functions $\psi_i$ with $\psi_i(0)=D_if(0)$ , and such that we can write

$\displaystyle f=f(0)+\sum\limits_{i=1}^nu^i\psi_i$

where $u^i$ is the $i$ th component function.

If we pick a point $x\in U$ we can parameterize the segment $c(t)=tx$ , and set $\phi=f\circ c$ to get a function on the unit interval $\phi:[0,1]\to\mathbb{R}$ . This function is clearly differentiable, and we can calculate

$\displaystyle\phi'(t)=\sum\limits_{i=1}^n\frac{\partial f}{\partial x^i}\frac{dc^i}{dt}=\sum\limits_{i=1}^nD_if(tx)x^i$

using the multivariable chain rule. We find

$\displaystyle\begin{aligned}f(x)-f(0)&=\phi(1)-\phi(0)\\&=\int\limits_0^1\phi'(t)\,dt\\&=\sum\limits_{i=1}^nx^i\int\limits_0^1D_if(tx)\,dt\end{aligned}$

We can thus find the desired functions $\psi_i$ by setting

$\displaystyle\psi_i(x)=\int\limits_0^1D_if(tx)\,dt$

Now if we have a differentiable function $f$ defined on a neighborhood $U$ of a point $p\in M$ , we can find a coordinate patch $(U,x)$ — possibly by shrinking $U$ — with $x(p)=0$ and $x(U)$ star-shaped. Then we can apply the previous lemma to $f\circ x^{-1}$ to get

$\displaystyle f\circ x^{-1}=f(p)+\sum\limits_{i=1}^nu^i\psi_i$

with $\psi_i(0)=\left[\frac{\partial}{\partial x^i}(p)\right](f)$ . Moving the coordinate map to the other side we find

$\displaystyle f=f(p)+\sum\limits_{i=1}^nx^i\left(\psi_i\circ x\right)$

Now we can hit this with a tangent vector $v$

$\displaystyle\begin{aligned}v(f)&=v\left(f(p)+\sum\limits_{i=1}^nx^i\left(\psi_i\circ x\right)\right)\\&=v\left(f(p)\right)+\sum\limits_{i=1}^nv\left(x^i\left(\psi_i\circ x\right)\right)\\&=\sum\limits_{i=1}^n\left(v(x^i)\left[\psi_i\circ x\right](p)+x^i(p)v\left(\psi_i\circ x\right)\right)\\&=\sum\limits_{i=1}^nv(x^i)\psi_i(0)\\&=\sum\limits_{i=1}^nv(x^i)\left[\frac{\partial}{\partial x^i}(p)\right](f)\end{aligned}$

where we have used linearity, the derivation property, and the first lemma above. Thus we can write

$\displaystyle v=\sum\limits_{i=1}^nv(x^i)\frac{\partial}{\partial x^i}(p)$

and the coordinate vectors span the space of tangent vectors at $p$ .

As a consequence, we conclude that $\mathcal{T}_pM$ always has dimension $n$ — exactly the same dimension as the manifold itself. And this is exactly what we should expect; if $M$ is $n$ -dimensional, then in some sense there are $n$ independent directions to move in near any point $p$ , 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.

8 Comments »

[…] 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 | Reply
[…] 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 | Reply
[…] 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 | Reply
[…] 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 | Reply
[…] 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 | Reply
Just after “multivariable chain rule”, f(p) – f(0) should be f(x) – f(0).

p lives in the manifold rather than R^n.

Comment by Rory Molinari | July 6, 2011 | Reply
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?”

Comment by guo wei | October 16, 2012 | Reply
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 $\circ$ .

Comment by John Armstrong | October 16, 2012 | Reply

About this weblog

This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).

I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.

RSS Feeds

RSS - Posts
RSS - Comments
Feedback

Got something to say? Anonymous questions, comments, and suggestions at Formspring.me!
Subjects
Subjects
Archives

March 2011

M T W T F S S

« Feb Apr »

1 2 3 4 5 6

7 8 9 10 11 12 13

14 15 16 17 18 19 20

21 22 23 24 25 26 27

28 29 30 31
Search for:

The Unapologetic Mathematician

Mathematics for the interested outsider