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Suppose that $\Gamma^{\beta}_{i\alpha}$ are Christoffel symbols for a connection with respct to a (local) basis $\{E_1,...,E_n\}$. I tried to prove that the Christoffel symbols for a dual connection with respect to the dual basis $\{\theta_1,...,\theta_n\}$ are then $-\Gamma_{i\alpha}^{\beta}$. I checked that the two expressions: one which defines the dual connection, and the expression $-\Gamma_{i\alpha}^{\beta}dx^i \otimes \theta_{\alpha}$ coincide but only for evaluated on pairs $(\partial_i,E_{\alpha})$. My question is: Is it enough in order to claim that the Christoffel's symbols of the dual connection are $-\Gamma$'s?

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You seem to have switched notation in your question as there appears $\partial_i$ and $dx^i$, but you didn't start with a coordinate frame. If I understand your question correctly, then the answer is yes. What you are trying to find is $C_{ij}^k$ given by the formula $$\nabla_{E_i}\theta_j=C_{ij}^k\theta_k$$To find these functions evaluate both sides of this equation at $E_l$.

We have that $C_{ij}^k(\theta_k)E_l=C_{ij}^k=C_{ij}^l$, while by definition $$(\nabla_{E_i}\theta_j)(E_l)=E_i(\theta_j(E_l))-\theta_j(\nabla_{E_i}E_l)=-\theta_j(\nabla_{E_i}E_l)=-\theta_j(\Gamma_{il}^pE_p)=-\Gamma_{il}^j$$

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It's been quite awhile since I've looked at a differential geometry book which covers this stuff, so this might not be exactly what is wanted here, but as I recall the connection's action on the dual basis $\theta_\mu$ may be found from its action in terms of the basis $E_\nu$, i.e., from the formula

$\nabla_\gamma E_\nu = \sum_\kappa \Gamma_{\nu \gamma}^\kappa E_\kappa \tag{1}$

as follows: the dual basis $\theta_\mu$ to $E_\nu$ satisfies

$\theta_\mu(E_\nu) = \delta_{\mu \nu} \tag{2}$

for all pairs of indices $\mu, \nu$. Applying $\nabla_\gamma = \nabla_{E_\gamma}$ to this equation yields

$\nabla_\gamma(\theta_\mu(E_\nu)) = 0, \tag{3}$

and since we want the Leibniz rule for derivatives of products to apply to $\theta_\mu(E_\nu)$ we must have

$\nabla_\gamma(\theta_\mu(E_\nu)) = (\nabla_\gamma \theta_\mu)(E_\nu) +\theta_\mu(\nabla_\gamma E_\nu); \tag{4}$

when (4) is combined with (3) we obtain

$(\nabla_\gamma \theta_\mu)(E_\nu) + \theta_\mu(\nabla_\gamma E_\nu) = 0, \tag{5}$

or

$(\nabla_\gamma \theta_\mu)(E_\nu) = -\theta_\mu(\nabla_\gamma E_\nu); \tag{6}$

using the formula (1) in (6) we see that

$(\nabla_\gamma \theta_\mu)(E_\nu) = -\theta_\mu(\sum_\kappa \Gamma_{\nu \gamma}^\kappa E_\kappa) = -\Gamma_{\nu \gamma}^\mu. \tag{7}$

(7) expresses the components of $\nabla_\gamma \theta_\mu$ via evaluation on the vector basis $E_\nu$; from this it follows from a standard but basic linear algebraic argument that

$\nabla_\gamma \theta_\mu = -\sum_\sigma \Gamma_{\sigma \gamma}^\mu \theta_\sigma. \tag{8}$

(8) is the formula for the covariant derivative of $\theta_\mu$; we see that the Christoffel symbols $\Gamma_{\rho \tau}^\sigma$ occuring in the expressions for $\nabla_\gamma E_\mu$ and $\nabla_\gamma \theta_\mu$ are indeed the negatives of one another, and summed over different indices as well, a lower for dual vectors and an upper index for members of the original basis vectors themselves. From this point of view, the Christoffel symbols for the dual basis $\theta_\mu$ are indeed the $-\Gamma$, where the $\Gamma$ are the Christoffel symbols for the basis $E_\mu$.

That's how I'd address it, in any event.

It should be observed that in the above discussion, the dual connection is in fact defined by (4); we stipulate the $\nabla_\gamma$ must so act on $\theta_\mu(E_\nu)$. With such stipulation and definition, the connection coefficients for the dual basis are as shown. The mixed notation $-\Gamma_{i\alpha}^{\beta}dx^i \otimes \theta_{\alpha}$ is one I have occasionally seen but with which I am not overly familiar, and since our OP truebaran doesn't specify how the corresponding mixed dual connection is defined, I will leave further discussion of this particular point without further comment.

Hope this helps. Cheers,

and as always,

Fiat Lux!!!

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  • $\begingroup$ As far as I understood, first You have proved that the dual connection (to a given one) is constructed uniquely: but I'm not quite sure why You argue that You want to have Leibnitz identity (in order to obtain (4)): You don't have a product but the connection acting on the form evaluated on vector field (more general: on section) which is the smooth function. $\endgroup$ – truebaran Sep 24 '14 at 21:15

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