# Covariant derivative of contracted tensor product

I've followed a number of derivations of the coordinate expression for the covariant derivative of a one-form: $\nabla_\delta \omega_\alpha$. To summarize, one strategy is

We know how to take covariant derivatives of scalar functions and vector fields. Since the "product" of a one-form and a vector field is scalar, and we know how the covariant derivative behaves under "products", we can derive the covariant derivative for one-forms too.

So we start by examining $\nabla_u(\omega \otimes v)$: $$\nabla_u(\omega \otimes v) = \nabla_u \omega \otimes v + \omega \nabla_u \otimes v$$ or in coordinates $$\nabla_\delta(\omega_\alpha b^\beta) = (\nabla_\delta \omega_\alpha) v^\beta + \omega_\alpha (\nabla_\delta v^\beta)$$ $$= (\nabla_\delta \omega_\alpha) v^\beta + \omega_\alpha (\partial_\delta v^\beta) + \omega_\alpha \Gamma^\beta_{\gamma\delta}v^\gamma$$

Contracting on $\alpha$ and $\beta$ on both sides we get $$\nabla_\delta(\omega_\alpha v^\alpha) = (\nabla_\delta \omega_\alpha) v^\alpha + \omega_\alpha (\partial_\delta v^\alpha) + \omega_\alpha \Gamma^\alpha_{\gamma\delta}v^\gamma.$$

The left hand side $\omega_\alpha v^\alpha$ is a scalar function, so $$\nabla_\delta(\omega_\alpha v^\alpha) = (\partial_\delta \omega_\alpha)v^\alpha + \omega_\alpha(\partial_\delta v^\alpha)$$ so with the appropriate cancelling I can get my answer. Why was I allowed to treat the product $\omega_\alpha v^\alpha$ as a scalar function when I differentiated? I understand that the action of a one-form on a vector field is a scalar function, but I expanded this as some kind of product. It's not a tensor product of $\omega$ and $v$, rather a contracted tensor product. Should I think of this as just a set of $n^2$ derivatives? What is the correct interpretation of the covariant derivative of $\omega_\alpha v^\alpha$?

• One attempted explanation: $\omega_\alpha$ and $v^\beta$ are each a collection of $n$ functions. $\omega_\alpha v^\alpha$ is shorthand for $\omega_1v^1 + \ldots + \omega_n v^n$. This is in $\mathcal{F}(M)$ so the covariant derivative reduces to partial differentiation. $\partial_\delta(\omega_1v^1 + \ldots + \omega_n v^n)$ is shorthand for the $n$ partial derivatives of this function. Then, of course, the partial derivative expands as usual: $(\partial_\delta\omega_1)v^1 + \omega_1(\partial_\delta v^1) + \ldots + (\partial_\delta\omega_1)v^1 + \omega_1(\partial_\delta v^1)$ – orlandpm Aug 4 '16 at 6:49
• Second attempted explanation: the one-form $\omega$ expands as $(\omega_1 \epsilon^1 + \ldots + \omega_n \epsilon^n)$ with respect to basis covector fields $(\epsilon^j)$. By the time we've collected terms in the action on $(v^1 e^1 + \ldots + v^n e_n)$ we have a scalar function answer $\omega_\beta v^\alpha \delta_\alpha^\beta = \omega_\beta v^\beta$. By this point, we're not talking about the action of a covector on a vector as a linear functional, we're just talking about a sum of products of scalar functions. – orlandpm Aug 4 '16 at 6:56
• Your explanation are correct. – user99914 Aug 4 '16 at 6:59