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It is possible to define the tensor ${C^c}_{ab}$ without using contractions using the Christoffel symbols for the connections. If you remember that $$\nabla_a\omega_b=\partial_a\omega_b-\Gamma^c_{ab}\omega_c$$ $$\tilde{\nabla}_a\omega_b=\partial_a\omega_b-\tilde{\Gamma}^c_{ab}\omega_c$$ then the difference gives $$\tilde\nabla_a\omega_b-\nabla_a\omega_c=(\Gamma^c_{ab}-\tilde\Gamma^c_{ab})\omega_c\tag{1}\label{key}$$ so that what Wald has called the tensor ${C^c}_{ab}$ is $${C^c}_{ab}=\Gamma^c_{ab}-\tilde\Gamma^c_{ab}$$

It is a general result that the difference between two $\Gamma$ is a tensor, even though the $\Gamma$'s themselves are non-tensorial. You can check reasoning as follows: if the left-hand side of \eqref{key} is tensorial, because it is the difference between to covariant derivatives; then the right-hand side must also be a tensor. This is only possible if $\Gamma^c_{ab}-\tilde\Gamma^c_{ab}$ is a tensor. This is not just valid for the covariant vector $\omega_c$, you can do this reasoning in general by applying the covariant derivatives $\nabla$ and $\tilde\nabla$ to an arbitrary rank tensor ${T^{a_1a_2...a_m}}_{b_1b_2...b_n}$ and you will get the same result.

It is possible to define the tensor ${C^c}_{ab}$ without using contractions using the Christoffel symbols for the connections. If you remember that $$\nabla_a\omega_b=\partial_a\omega_b-\Gamma^c_{ab}\omega_c$$ $$\tilde{\nabla}_a\omega_b=\partial_a\omega_b-\tilde{\Gamma}^c_{ab}\omega_c$$ then the difference gives $$\tilde\nabla_a\omega_b-\nabla_a\omega_c=(\Gamma^c_{ab}-\tilde\Gamma^c_{ab})\omega_c\tag{1}\label{key}$$ so that what Wald has called the tensor ${C^c}_{ab}$ is $${C^c}_{ab}=\Gamma^c_{ab}-\tilde\Gamma^c_{ab}$$

It is a general result that the difference between two $\Gamma$ is a tensor. You can check reasoning as follows: if the left-hand side of \eqref{key} is tensorial, because it is the difference between to covariant derivatives; then the right-hand side must also be a tensor. This is only possible if $\Gamma^c_{ab}-\tilde\Gamma^c_{ab}$ is a tensor. This is not just valid for the covariant vector $\omega_c$, you can do this reasoning in general by applying the covariant derivatives $\nabla$ and $\tilde\nabla$ to an arbitrary rank tensor ${T^{a_1a_2...a_m}}_{b_1b_2...b_n}$ and you will get the same result.

It is possible to define the tensor ${C^c}_{ab}$ without using contractions using the Christoffel symbols for the connections. If you remember that $$\nabla_a\omega_b=\partial_a\omega_b-\Gamma^c_{ab}\omega_c$$ $$\tilde{\nabla}_a\omega_b=\partial_a\omega_b-\tilde{\Gamma}^c_{ab}\omega_c$$ then the difference gives $$\tilde\nabla_a\omega_b-\nabla_a\omega_c=(\Gamma^c_{ab}-\tilde\Gamma^c_{ab})\omega_c\tag{1}\label{key}$$ so that what Wald has called the tensor ${C^c}_{ab}$ is $${C^c}_{ab}=\Gamma^c_{ab}-\tilde\Gamma^c_{ab}$$

It is a general result that the difference between two $\Gamma$ is a tensor, even though the $\Gamma$'s themselves are non-tensorial. You can check reasoning as follows: if the left-hand side of \eqref{key} is tensorial, because it is the difference between to covariant derivatives; then the right-hand side must also be a tensor. This is only possible if $\Gamma^c_{ab}-\tilde\Gamma^c_{ab}$ is a tensor. This is not just valid for the covariant vector $\omega_c$, you can do this reasoning in general by applying the covariant derivatives $\nabla$ and $\tilde\nabla$ to an arbitrary rank tensor ${T^{a_1a_2...a_m}}_{b_1b_2...b_n}$ and you will get the same result.

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Urb
  • 2.7k
  • 4
  • 14
  • 26

It is possible to define the tensor ${C^c}_{ab}$ without using contractions using the Christoffel symbols for the connections. If you remember that $$\nabla_a\omega_b=\partial_a\omega_b-\Gamma^c_{ab}\omega_c$$ $$\tilde{\nabla}_a\omega_b=\partial_a\omega_b-\tilde{\Gamma}^c_{ab}\omega_c$$ then the difference gives $$\tilde\nabla_a\omega_b-\nabla_a\omega_c=(\Gamma^c_{ab}-\tilde\Gamma^c_{ab})\omega_c\tag{1}\label{key}$$ so that what Wald has called the tensor ${C^c}_{ab}$ is $${C^c}_{ab}=\Gamma^c_{ab}-\tilde\Gamma^c_{ab}$$

It is a general result that the difference between two $\Gamma$ is a tensor. You can check reasoning as follows: if the left-hand side of \eqref{key} is tensorial, because it is the difference between to covariant derivatives; then the right-hand side must also be a tensor. This is only possible if $\Gamma^c_{ab}-\tilde\Gamma^c_{ab}$ is a tensor. This is not just valid for the covariant vector $\omega_c$, you can do this reasoning in general by applying the covariant derivatives $\nabla$ and $\tilde\nabla$ to an arbitrary rank tensor ${T^{a_1a_2...a_m}}_{b_1b_2...b_n}$ and you will get the same result.