3
$\begingroup$

I would like to derive the formula

$$\partial_{c}\vec{e}^{\,a}=-\Gamma_{bc}^{a}\vec{e}^{\,b}$$

where $\vec{e}_{a}$ are the basis vectors on a manifold.

In the lecture, we did it in the following way:

$$0=\partial_{c}(\delta_{b}^{a})=\partial_{c}(\vec{e}^{\,a}\cdot\vec{e}_{\,b})=\vec{e}_{\,b}\cdot\partial_{c}\vec{e}^{\,a}+\underbrace{\vec{e}^{\,a}\cdot\partial_{c}\vec{e}_{\,b}}_{=\Gamma_{bc}^{a}}$$

and therefore

$$\vec{e}_{\,b}\cdot\partial_{c}\vec{e}^{\,a}=-\Gamma_{bc}^{a}$$

Up to here I can follow. Then it is stated: "Multipliying with $\vec{e}^{\,b}$ yields the result", but I can't unterstand how.

If I multiply by $\vec{e}^{\,b}$ and then sum over $b$, we get

$$\vec{e}^{\,b}\cdot (\vec{e}_{\,b}\cdot\partial_{c}\vec{e}^{\,a})=-\Gamma_{bc}^{a}\vec{e}^{\,b}$$

How can we simplify the LHS?

$\endgroup$

1 Answer 1

3
$\begingroup$

The usual definition of the Christoffel symbols is $$\nabla_\mu {\bf e}_\nu = {\bf e}_\sigma {\Gamma^\sigma}_{\nu\mu} $$ where $\nabla_\mu $ is a covaraint derivative and the ${\bf e}_\mu$ are basis vectors of the tangent space $T(M)$. Your formula differs in sign for some reason. Where does it come from? Are the ${\bf e}_a$ an orthonormal basis, and if so what are the ${\bf e}^a$? Are they a coframe? If so, the sign is right, but they are not the basis vetors of the tansgent space, but rather of its dual $T^*(M)$. Also if this is the case then for a vector $X= X^a {\bf e}_a$ we us the fact that evalauting a covector on a vector returns its components, i.e. ${\bf e}^a(X)=X^a$, to see that $$ {\bf e}_a {\bf e}^a(X)= {\bf e}_a X^\mu=X. $$ Thus $\sum_a {\bf e}_a {\bf e}^a$ is the identity map from $T(M)\to T(M)$. No "$\cdot$" is needed between the ${\bf e}_a$ and the ${\bf e}^a$.

$\endgroup$
3
  • 1
    $\begingroup$ You are doing a covariant basis while the OP used a contravariant, hence the sign diff. $\endgroup$
    – user196418
    Commented Feb 4, 2021 at 13:40
  • $\begingroup$ @ggcg Yes. That's why I said the sign was correct if he was using a basis for $T^*(M)$ rather than for $T(M)$. It was not clear what the ${\bf e}^a$ were. I usually use the notation ${\bf e}^{*a}$ for the dual basis so as to avoid the ambiguity of whether a number $e_1^3$ is the third component of the vctor ${\bf e}_1= e_a^\mu \partial_\mu$ or the first component of the covector ${\bf e}^{*3}= e^{*3}_\mu dx^\mu $. $\endgroup$
    – mike stone
    Commented Feb 4, 2021 at 14:00
  • 1
    $\begingroup$ I see. Your previous statement was "for some reason" which kind of made it look like you weren't sure. Thanks for the reply. $\endgroup$
    – user196418
    Commented Feb 4, 2021 at 14:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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