If I want to covariantly differentiate a one form then I can write:

$\nabla_\beta \tilde p = \dfrac{\partial p_\alpha}{\partial x^\beta} \tilde \omega^\alpha + p_\alpha \dfrac{\partial \tilde \omega^\alpha}{\partial x^\beta} $

So I can write, $\nabla_\beta \tilde p = \dfrac{\partial p_\alpha}{\partial x^\beta} \tilde \omega^\alpha + p_\alpha \bigg(\dfrac{\partial \tilde \omega^\alpha}{\partial x^\beta} \bigg)_\gamma \tilde \omega^\gamma $

Manipulating with the dummy indices a little bit, I get,

$\nabla_\beta \tilde p = \dfrac{\partial p_\alpha}{\partial x^\beta} \tilde \omega^\alpha + p_\gamma \bigg(\dfrac{\partial \tilde \omega^\gamma}{\partial x^\beta} \bigg)_\alpha \tilde \omega^\alpha = \bigg[ \dfrac{\partial p_\alpha}{\partial x^\beta} + p_\gamma \bigg(\dfrac{\partial \tilde \omega^\gamma}{\partial x^\beta} \bigg)_\alpha \bigg] \tilde \omega^\alpha $

This is what I get using the definition of covariant differentiation. The way my book uses is a bit different. It uses the fact that the covariant differentiation of a scalar is just simple differentiation along with the fact that contraction of a vector on a one form is a scalar. i.e.

$\phi= <\tilde p,\vec{V}>=p_\alpha V^\alpha$

Therefore, $\nabla_\beta\phi=\nabla_\beta(p_\alpha V^\alpha)$.

Therefore, $\nabla_\beta(p_\alpha V^\alpha) = \dfrac{\partial(p_\alpha V^\alpha)}{\partial x^\beta}$

A little manipulation leads to the expression

$\nabla_\beta \phi = (\dfrac{\partial p_\alpha}{\partial x^\beta} - p_\gamma \Gamma^\gamma_{\alpha\beta})V^\alpha+p_\alpha V^\alpha_{;\beta}$.

Now considering the product rule for the covariant differentiation to have the usual properties ( i.e. $\nabla_\beta(p_\alpha V^\alpha) = p_{\alpha;\beta}V^\alpha + p_\alpha V^\alpha_{;\beta} $ ), they arrive at the conclusion that the term in the bracket must be $p_{\alpha;\beta}$. And thus,

$\nabla_\beta \tilde p= (\dfrac{\partial p_\alpha}{\partial x^\beta} - p_\gamma \Gamma^\gamma_{\alpha\beta})\tilde \omega^\alpha $.

Thus, $\bigg(\dfrac{\partial \tilde \omega^\gamma}{\partial x^\beta} \bigg)_\alpha = -\Gamma^\gamma_{\alpha\beta}$.

My question is whether or not there exist a more direct way to conclude the above equality without making the a priori assumption that $\nabla_\beta(p_\alpha V^\alpha) = p_{\alpha;\beta}V^\alpha + p_\alpha V^\alpha_{;\beta} ? $


1 Answer 1


The first line is incorrect. The action of the connection $\nabla$ on the 1-form is:$$ \nabla_{\beta} (p_{\alpha}w^{\alpha})=(\partial_{\beta}p_{\alpha})w^{\alpha} + p_{\alpha}\nabla_{\beta}w^{\alpha}=(\partial_{\beta}p_{\alpha})w^{\alpha} + p_{\alpha}\tilde\Gamma^{\alpha}_{\beta \gamma}w^{\gamma}$$ by definition of connection (which requires the product rule with scalar functions used in my first equality) and the fact that the $w^{\alpha}$ form a local basis of the cotangent bundle ( so, since $\nabla_{\beta}w^{\alpha}$ belongs in it, it must be a linear combination of the type $\tilde\Gamma^{\alpha}_{\beta \gamma}w^{\gamma}$). By then using the method you stated, one finds that $\tilde\Gamma^{\alpha}_{\beta \gamma}= -\Gamma^{\alpha}_{\beta \gamma}$ (I have been assuming from the beginning that your $w^{\alpha}$ are the dual of $\partial_{\alpha}$).

The fact that, if X,Y are vector fields and $w$ a 1-form then $\nabla_X(w(Y))=w(\nabla_X Y) + \nabla_X(w)Y$, is not an assumption and can be demonstrated. I won't write everything down (i believe you can find it in any differential geometry book) but the idea is the following: having a connection on the tangent bundle, one extends it uniquely to all tensor bundles, asking that it satisfies some properties ( like $\nabla_X (t_1 \otimes t_2 )=(\nabla_X t_1)\otimes t_2 + t_1\otimes\nabla_X(t_2)$ ). It can then be demonstrated that it commutes with contractions, and since $w(Y)$ is a contraction, one gets the result we wanted.

p.s.:I only noticed that the question was 1 month old when I had already written most of the stuff down. If you had already solved this, sorry for the notification.


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