I'm taking general relativity at the moment, and today in class the instructor gave us a definition of tangent vector as:

$v$ is a tangent vector based at $p\in M$ if $v_{p}$ is a linear combination of the directional derivative $\frac{\partial}{\partial x^\mu}$ $$ v=v^\mu \frac{\partial}{\partial x^\mu} \bigg|_{p} $$

He also mention that vectors and directional derivatives are in one-to-one correspondence.
I don't see why this definition is equivalent to the old Calculus definition of tangent vector: That given a curve $\gamma(\tau): (-\epsilon,\epsilon) \rightarrow M$ with parameter $\tau \in (-\epsilon,\epsilon)$ and $\gamma(0)=p$, the components of the tangent vector to the curve at $p$ is $$ v^{\mu}=\frac{\partial{x^\mu}}{\partial \tau}\bigg|_{\tau=0} $$

His new definition of tangent vector is a sum, while the old definition is a tuple. I have not yet understood manifolds, so can you explain this in a simple way with examples such that a student of Physics can understand?


For $v\in T_pM$, we want to view $v$ as a directional derivative operator. That is, if $c$ is a curve with $c'(0)=v$, then $$ v(f) : =\frac{d}{dt} f\circ c(t) $$

Here we must show that $v(f)$ is independent of curve $c$.

(1) Let $w:=(d\phi)^{-1}v $ Then if $c=\phi\circ \alpha,\ \alpha'(0)=w,$ $$ v(f)= \frac{d}{dt} f\circ c(t)=\frac{d}{dt} f\circ \phi\circ \alpha (t)=d(f\circ \phi)\ w $$ That is it is independent of curve. But we must consider $\phi,\ w$

(2) $\psi$ is another chart. Then $$ v(f)= d(f\circ \psi)\ x,\ x:=(d\psi)^{-1}(v) $$

Here $ d(\psi^{-1}\circ \phi)\ w=x $ so that $d(f\circ \psi)\ x=d(f\circ \phi)\ w $. That is it is independent of charts.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.