Levi-Civita Connection: Properties and Examples

[Silviu]

Homework Statement

Let V be a Levi-Civita connection.
a) Let $f \in F(M)$, (function defined on the manifold M). Show that: $\nabla_\mu \nabla_\nu f = \nabla_\nu \nabla_\mu f$
b) Let $\omega \in \Omega^1(M)$ (one form on M). Show that $d \omega = (\nabla_\mu \omega)_\nu dx^\mu \wedge dx^\nu$

Homework Equations

Levi-Civita is a symmetric connection, i.e. $\Gamma^\alpha_{\mu \nu} = \Gamma^\alpha_{\nu \mu}$

The Attempt at a Solution

a) For any connection, function f and vector X, we have $\nabla_X f = X[f]$. So in our case $\nabla_\mu \nabla_nu f = \frac{\partial}{\partial x_\mu}\frac{\partial}{\partial x_\nu} f = \frac{\partial}{\partial x_\nu}\frac{\partial}{\partial x_\mu} f = \nabla_\nu \nabla_\mu f$. Is this correct? And if so, why does the connection has to be Levi-Civita, it seems to work for any connection?

b) $\omega = a_\nu dx^\nu$. By definition, $d\omega = \frac{\partial a_\nu}{\partial x_\mu}dx^\mu \wedge dx^\nu$. But $(\nabla_\mu \omega)_\nu = \frac{\partial a_\nu}{\partial x_\mu} - \Gamma^\lambda_{\mu\nu}a_\lambda$. I am a bit confused of what I did wrong here, as the 2 results don't match and $\Gamma^\lambda_{\mu\nu}$ doesn't vanish, even in a Levi Civita connection. Can someone help me? Thank you!

 

[Orodruin]

Is this correct?

No, it is not correct. $\nabla_\nu f$ is not a scalar. It is the components of a covector.

In (b) you are not taking into account that the wedge product is anti-symmetric.

 

[Silviu]

No, it is not correct. $\nabla_\nu f$ is not a scalar. It is the components of a covector.

In (b) you are not taking into account that the wedge product is anti-symmetric.

So for part b), $\Gamma^{\lambda}_{\mu \nu}a_\lambda dx^\mu \wedge dx^\nu$ vanishes because we multiply a symmetric term with an antisymmetric term. Thank you! For part a) I am not sure I understand. Isn't $\nabla_\nu f = \frac{\partial}{\partial x^\nu}f$, which is just the derivative of a function, which is a number, so it creates a scalar field?

 

[Orodruin]

Isn't ∇νf=∂∂xνf∇νf=∂∂xνf\nabla_\nu f = \frac{\partial}{\partial x^\nu}f, which is just the derivative of a function, which is a number, so it creates a scalar field?

No.

 

[Silviu]

No.

Then, what is the right formula for $\nabla_\nu f$?

 

[Orodruin]

Then, what is the right formula for $\nabla_\nu f$?

For a scalar field $\nabla_\nu f = \partial_\nu f$, but $\partial_\nu f$ is not a scalar field so generally $\nabla_\mu \partial_\nu f \neq \partial_\mu \partial_\nu f$. Compare to the components of the gradient in regular Euclidean space.

 

[Silviu]

For a scalar field $\nabla_\nu f = \partial_\nu f$, but $\partial_\nu f$ is not a scalar field so generally $\nabla_\mu \partial_\nu f \neq \partial_\mu \partial_\nu f$. Compare to the components of the gradient in regular Euclidean space.

So do you mean to treat $\partial_\nu f$ as a vector field?

 

[Orodruin]

So do you mean to treat $\partial_\nu f$ as a vector field?

You should, because it is a (dual) vector field.

 

[Silviu]

You should, because it is a (dual) vector field.

Thanks a lot!