# Covariant derivative of a $(1,1)$ tensor field

I am trying to understand covariant derivatives in GR.

So far, I understand that if $$Z$$ is a vector field, $$\nabla Z$$ is a $$(1,1)$$ tensor field, i.e. the map $$\eta,X \mapsto \eta(\nabla_XZ)$$ is linear in $$\eta$$ and $$X$$. The map $$Z\mapsto\nabla_XZ$$ is not a tensor field since, by the Leibniz rule, it is not linear in $$Z$$: $$\nabla_X(fZ)=(\nabla_Xf)Z+f(\nabla_XZ)$$ where $$\nabla_Xf:=X(f)$$. My lecture notes then state that the action of $$\nabla$$ is extended to general $$(r,s)$$ tensor fields by the Leibniz rule, giving an $$(r,s+1)$$ tensor field as a result. The example given is $$\nabla \eta$$, with $$\eta$$ a $$(0,1)$$ tensor field:$$(\nabla\eta)(X,Y)\equiv\nabla_X\eta(Y):=\nabla_X(\eta(Y))-\eta(\nabla_XY)$$ which is easily seen to be linear in $$X$$, while the first terms from the Leibniz rule cancel to give linearity in $$Y$$ as well, so this is a $$(0,2)$$ tensor.

They then suggest to try and define $$\nabla T$$ for $$T$$ a $$(1,1)$$ tensor. My attempt at this is: $$\nabla_XT(\eta,Y):=\nabla_X(T(\eta,Y))+T(\nabla_X\eta,Y)-T(\eta,\nabla_XY)$$ Again, linearity in $$X$$ is easy and the Leibniz rule gives linearity in $$Y$$. However, this doesn't seem to be linear in $$\eta$$. First, I find that: $$\begin{eqnarray} \nabla_X(f\eta+\omega) (Y)&=& \nabla_X((f\eta+\omega)(Y)) - (f\eta+\omega)(\nabla_XY)\\ &=& (\nabla_Xf)\eta(Y)+f\nabla_X(\eta(Y))+\nabla_X(\omega(Y))-f\eta(\nabla_XY)-\omega(\nabla_XY)\\ &=& (\nabla_Xf)\eta(Y) + f \nabla_X\eta(Y)+\nabla_X\omega(Y) \end{eqnarray}$$ hence $$\nabla_X(f\eta+\omega) = (\nabla_Xf)\eta + f \nabla_X\eta+\nabla_X\omega$$. Therefore: $$\begin{eqnarray} \nabla_XT(f\eta+\omega,Y)&=& \nabla_X(T(f\eta+\omega,Y))+T(\nabla_X(f\eta+\omega),Y)-T(f\eta+\omega,\nabla_XY)\\ &=& (\nabla_Xf)(T(\eta,Y))+f\nabla_X(T(\eta,Y))+\nabla_X(T(\omega,Y))+(\nabla_Xf)T(\eta,Y)\\&&+fT(\nabla_X\eta,Y)+T(\nabla_X\omega,Y)-fT(\eta,\nabla_XY)-T(\omega,\nabla_XY)\\ &=& 2(\nabla_Xf)(T(\eta,Y)) + f\nabla_XT(\eta,Y)+\nabla_XT(\omega,Y) \end{eqnarray}$$ where the term $$2(\nabla_Xf)(T(\eta,Y))$$ shows that $$\nabla_XT(\eta,Y)$$ is not linear in $$\eta$$. However, if I defined it as:$$\nabla_XT(\eta,Y):=\nabla_X(T(\eta,Y))-T(\nabla_X\eta,Y)+T(\eta,\nabla_XY)$$ instead, then the same calculation shows that this is linear in $$\eta$$, but no longer linear in $$Y$$. So I am not sure which, if either, of them is the correct definition.

UPDATE: Thanks to Michael Albanese's answer I now understand that the correct definition is $$\nabla_XT(\eta,Y):=\nabla_X(T(\eta,Y))-T(\nabla_X\eta,Y)-T(\eta,\nabla_XY)$$ which is indeed linear in $$\eta$$, $$X$$ and $$Y$$. For the benefit of others, I'll mention that my confusion arose because I was trying to reverse engineer the $$+$$ and $$-$$ signs from the coordinate basis formula:

$$\begin{eqnarray} T^{{\mu_1}\ldots{\mu_r}}_{\phantom{{\mu_1}\ldots{\mu_r}}{\nu_1}\ldots{\nu_s};\rho} =T^{{\mu_1}\ldots{\mu_r}}_{\phantom{{\mu_1}\ldots{\mu_r}}{\nu_1}\ldots{\nu_s},\rho}&+\Gamma^{\mu_1}_{\sigma\rho}T^{{\sigma}\ldots{\mu_r}}_{\phantom{{\mu_1}\ldots{\mu_r}}{\nu_1}\ldots{\nu_s}}+\ldots+\Gamma^{\mu_r}_{\sigma\rho}T^{{\mu_1}\ldots{\sigma}}_{\phantom{{\mu_1}\ldots{\mu_r}}{\nu_1}\ldots{\nu_s}}\\ &-\Gamma^{\sigma}_{\nu_1\rho}T^{{\mu_1}\ldots{\mu_r}}_{\phantom{{\mu_1}\ldots{\mu_r}}{\sigma}\ldots{\nu_s}}-\ldots-\Gamma^{\sigma}_{\nu_s\rho}T^{{\mu_1}\ldots{\mu_r}}_{\phantom{{\mu_1}\ldots{\mu_r}}{\nu_1}\ldots{\sigma}} \end{eqnarray}$$ But I now see how they follow from the correct definition. In the case of a $$(1,1)$$ tensor we have $$T^{\mu}_{\phantom{\mu}\nu\,;\rho}:=\nabla_{e_\rho}T(f^\mu,e_\nu)$$, where I am using $$f^\mu$$ and $$e_\nu$$ for $$\mathrm{d}x^\mu$$ and $$\frac{\partial}{\partial x^\nu}$$. Hence we have: $$\begin{eqnarray} \nabla_{e_\rho}T(f^\mu,e_\nu) :=& \nabla_{e_\rho}(T(f^\mu,e_\nu))-T(\nabla_{e_\rho}f^\mu,e_\nu)-T(f^\mu,\nabla_{e_\rho}e_\nu) \end{eqnarray}$$ The first and third terms are easily evaluated: $$\nabla_{e_\rho}(T(f^\mu,e_\nu))=e_\rho(T(f^\mu,e_\nu))=e_\rho(T^{\mu}_{\phantom{\mu}\nu})=T^{\mu}_{\phantom{\mu}\nu,\rho}$$ $$T(f^\mu,\nabla_{e_\rho}e_\nu)=T(f^\mu,\Gamma^{\sigma}_{\nu\rho}e_\sigma)=\Gamma^{\sigma}_{\nu\rho}T(f^\mu,e_\sigma)=\Gamma^{\sigma}_{\nu\rho}T^\mu_{\phantom{\mu}\sigma}$$ For the second term, since $$f^\mu$$ is a $$(0,1)$$ tensor field, $$\nabla f^\mu$$ is a $$(0,2)$$ tensor field and hence $$\nabla_{e_\rho}f^\mu$$, for fixed $$e_\rho$$, is a $$(0,1)$$ tensor field. We evaluate: $$\begin{eqnarray} (\nabla_{e_\rho} f^\mu)_\lambda=\nabla_{e_\rho}f^\mu(e_\lambda) &:= &\nabla_{e_\rho}(f^\mu(e_\lambda))-f^\mu(\nabla_{e_\rho}e_\lambda) \\ &= & e_\rho(\delta^\mu_\lambda)-f^\mu(\Gamma_{\lambda\rho}^\sigma e_\sigma)\\ &= & 0 -\Gamma_{\lambda\rho}^\sigma f^\mu( e_\sigma)\\ &= & - \Gamma_{\lambda\rho}^\sigma \delta^\mu_\sigma\\ &= & - \Gamma_{\lambda\rho}^\mu \\ &= & - \Gamma_{\sigma\rho}^\mu \delta^\sigma_\lambda\\ &= & -\Gamma_{\sigma\rho}^\mu f^\sigma( e_\lambda)\\ \end{eqnarray}$$ Therefore $$\nabla_{e_\rho}f^\mu = -\Gamma_{\sigma\rho}^\mu f^\sigma$$ and hence: $$T(\nabla_{e_\rho}f^\mu,e_\nu)=T(-\Gamma_{\sigma\rho}^\mu f^\sigma,e_\nu)=-\Gamma_{\sigma\rho}^\mu T(f^\sigma,e_\nu)=-\Gamma_{\sigma\rho}^\mu T^\sigma_{\phantom{\sigma}\nu}$$ Thus, we have: $$\nabla_\rho T^{\mu}_{\phantom{\mu}\nu}=T^{\mu}_{\phantom{\mu}\nu,\rho}+\Gamma_{\sigma\rho}^\mu T^\sigma_{\phantom{\sigma}\nu}-\Gamma^{\sigma}_{\nu\rho}T^\mu_{\phantom{\mu}\sigma}$$ as expected.

In general, if $T$ is an $(r, s)$ tensor field, then $\nabla T$ is an $(r, s+1)$-tensor field given by

$$(\nabla T)(\eta_1, \dots, \eta_r, X, Y_1, \dots, Y_s) = (\nabla_XT)(\eta_1, \dots, \eta_r, Y_1, \dots, Y_s)$$

where the latter expression is intrinsically defined by the equation

\begin{align*} \nabla_X(T(\eta_1, \dots, \eta_r, Y_1, \dots, Y_s)) =&\ (\nabla_XT)(\eta_1, \dots, \eta_r, Y_1, \dots, Y_s)\\ &+ \sum_{i=1}^rT(\eta_1, \dots, \eta_{i-1}, \nabla_X\eta_i, \eta_{i+1}, \dots, \eta_r, Y_1, \dots, Y_s)\\ &+ \sum_{j=1}^sT(\eta_1, \dots, \eta_r, Y_1, \dots, Y_{j-1}, \nabla_XY_j, Y_{j+1}, \dots, Y_s). \end{align*}

In particular, if $T$ is a $(1, 1)$ tensor, then $(\nabla_XT)(\eta, Y)$ is intrisically defined by the equation

$$\nabla_X(T(\eta, Y)) = (\nabla_XT)(\eta, Y) + T(\nabla_X\eta, Y) + T(\eta, \nabla_XY)$$

and therefore the correct expression for $(\nabla_XT)(\eta, Y)$ is

$$(\nabla_XT)(\eta, Y) = \nabla_X(T(\eta, Y)) - T(\nabla_X\eta, Y) - T(\eta, \nabla_XY).$$

• Thank you, I understand now. The term $\nabla_X(T(\ldots))$ is defined since $T(\ldots)$ is a function, and all the terms $T(\ldots,\nabla_X\eta,\ldots)$ and $T(\ldots,\nabla_XY,\ldots)$ are defined since $\nabla_XY$ is defined and $\nabla_X\eta$ was defined explicitly. So the expression can be used to define $\nabla_XT(\ldots)$ implicitly. Mar 12, 2017 at 13:46
• I have added some details to my question to explain why I got confused. Mar 12, 2017 at 13:47