The Unapologetic Mathematician

Mathematics for the interested outsider

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 ith 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.

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March 31, 2011 - Posted by | Differential Topology, Topology

8 Comments »

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  6. 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

  7. 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

  8. 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


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