If you fix a point and express everything in a local chart, then the required path is just the unique solution of the initial value problem .
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
so i try to show that with is a bijection.
Of course is well defined and injective, but i couldn't show that
is surjective!
If D is some arbitrary derivative at the point p. How can i find a path c, s.t. ?
Regards
Hello Rebesques,
thank you for your help! But i don't see what you mean. Ok lets choose a chart (U, ) around the fixed point p.
I think by your notation you mean , 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
whereas are the coordinate functions of our selected chart
Perhaps this equation can help? But i don't know how.
Not quite. We are looking for a curve -\epsilon,\epsilon)\rightarrow M" alt="c-\epsilon,\epsilon)\rightarrow M" /> with . There are many curves with this property; Let's construct one.
Choose a chart , and w.l.o.g. assume , where and I is the identity matrix.
Then, by letting and identifying , the curve has the required properties.
Ps. And is unique in the sence it belongs to the equivalence class of curves passing through and having tangent .