# Bijection between tangent spaces

• December 12th 2010, 09:44 AM
Sogan
Bijection between tangent spaces
Hello,

We have defined the tangent space at a point p of a manifold as the set of all derivatives.
Now i want to show, that there exist a bijection between all derivatives and the "Tangentvectors" defined by (equivalent classes of) paths.
More precisely:
We call two paths equivalent, iff they define the same derivation at a point p. Now i want to show that there is a bijection between the set of all equivalent classes and the Tangentspace (defined as the set of derivatives).

My Approach was the following.
Every path c:I->M into the manifold M defines a derivation by
$D(f):=\frac{d}{dt}_{|t=0}(f\circ c)(t)$

so i try to show that $\phi([c])=D$ with $D(f)=\frac{d}{dt}_{|t=0}(f\circ c)(t)$ is a bijection.

Of course $\phi$ is well defined and injective, but i couldn't show that
$\phi$ is surjective!

If D is some arbitrary derivative at the point p. How can i find a path c, s.t. $\phi(c)=D$?

Regards
• December 12th 2010, 02:05 PM
Rebesques
If you fix a point $p$ and express everything in a local chart, then the required path $c(s), \ -\epsilon is just the unique solution of the initial value problem $c'(0)=D, \ c(0)=p$.
• December 13th 2010, 08:39 AM
Sogan
Hello Rebesques,

thank you for your help! But i don't see what you mean. Ok lets choose a chart (U, $\phi$) around the fixed point p.
I think by your notation you mean $c'(0)=\frac{d}{dt}_{|t=0}(f\circ c)(t)$, is it right?

But How can i solve this differential equation D=c'(t)?

But i know by a thm., that every derivation D is of the form
$D=\sum_{i=1}^d D(x_i) \frac{d}{dx_i}_{|p}$

whereas $x_i$ are the coordinate functions of our selected chart $\phi$
Perhaps this equation can help? But i don't know how.
• December 16th 2010, 11:48 AM
Rebesques
Not quite. We are looking for a curve $c:(-\epsilon,\epsilon)\rightarrow M$ with $c(0)=p, \ c'(0)=(\frac{\partial}{\partial t}\vert_{0})c=D\in T_pM$. There are many curves with this property; Let's construct one.

Choose a chart $(U,\phi), \ \phi=(x^1,\ldots,x^n)$, and w.l.o.g. assume $\phi(p)=0, \ D(\phi)\vert_{p}=I$, where $D(\phi)\vert_{p}=(\frac{\partial}{\partial x^i}\vert_p(x^j))$ and I is the identity matrix.

Then, by letting $D=a^i\frac{\partial}{\partial x^i}$ and identifying $D=(a^1,\ldots,a^n)\in R^n$, the curve $c(t)=\phi^{-1}(tD), \ -\epsilon has the required properties.

Ps. And $c$ is unique in the sence it belongs to the equivalence class of curves passing through $p$ and having tangent $D\in T_pM$.