- #1
Silviu
- 624
- 11
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!