A somewhat similar question is this one but it is not quite the same.

I am getting used to the abstract index notation used for tensor algebra. So far so good, but the is one issue that concerns me, In General Relativity by Wald, it is discussed how the difference between two connections $\tilde{\nabla}$ and $\nabla$ induces a tensor $C$ of rank (1,2) in $T_pM$, and hence (by the index notation) we denote this tensor as ${C^c}_{ab}$. The problem is that Wald defines this tensor by the equation $$\nabla_a\omega_b=\tilde{\nabla}_a \omega_b- {C^c}_{ab}\omega_c $$ Which seems funny, we can think of the last term in the RHS of the latter as the contraction of the (1,3) tensor $C\otimes \omega $ with respect to the first dual vector slot and the third vector slot. But rather than this odd way of deffining the tensor $C$ we could make use of the index-free notation. So that we could define the tensor $C:V^{*}\times V\times V\to \mathbb{R}$ as $$\omega,t,s\mapsto (\tilde{\nabla}\omega)(t,s)-(\nabla\omega )(t,s) $$ which is what would be natural way of defining a tensor "by it's action". So the question is: Is my way of understanding the abstract index notation in this case is the correct one, or rather does the definition of Wald refers to $$\sum_{k=1}^nC(e^{k*},t,s)\cdot \omega(e_{k})=(\tilde{\nabla}\omega)(t,s)-(\nabla\omega)(t,s) $$ where $\{e_1,\ldots, e_n \}$ is some basis of $T_pM$ and $\{e^{1*},\ldots, e^{n*} \}$ is its dual basis. Which is it? How could one know using the Abstract index notation?

Note that the first definition (the one I assume is the correct one) does indeed define a tensor, meanwhile a tensor is hardly ever characterized by a contraction. But the notation, as defined does indeed suggest the contraction interpretation.


The equation $$\nabla_a\omega_b=\tilde{\nabla}_a \omega_b- {C^c}_{ab}\omega_c$$ can be contracted with vector components $X^a$ and $Y^b$:

$$C^{c}_{ab}X^aY^b\omega_c = (\tilde{\nabla}_a\omega_b) X^aY^b - (\nabla_a\omega_b) X^a Y^b$$ which can be expressed in coordinate-free notation as

$$\mathbf C(\boldsymbol\omega,\mathbf X,\mathbf Y) = (\nabla_{\mathbf X}\boldsymbol \omega)(\mathbf Y) - (\tilde{\nabla}_{\mathbf X}\boldsymbol \omega)(\mathbf Y)$$

where $\nabla_\mathbf{X} =X^a \nabla_a$. Note in particular that $\nabla_{\mathbf X}$, which is the covariant directional derivative along $\mathbf X$, maps a covector field $\boldsymbol \omega$ to a covector field

$$\nabla_\mathbf{X} \boldsymbol \omega = (X^a\nabla_a\omega_b)\hat \epsilon^b = X^a(\partial_a \omega_b -\omega_c \Gamma^c_{ab})\hat\epsilon^b$$

You can use this to go backwards to verify that the coordinate-free expression above does indeed yield the expression you started with.

Lastly, note that the $(0,2)$-tensor with components $C^c_{ab}\omega_c$ could equivalently viewed as the $(1,2)$-tensor $\mathbf C$ with the covector $\boldsymbol \omega$ plugged into the first slot (i.e. $\mathbf C(\boldsymbol \omega,\bullet,\bullet)$ ) or as the trace over the first and fourth index of the $(1,3)$-tensor ($\mathbf C \otimes \boldsymbol \omega$). If I'm interpreting the question correctly, the answer is that your two alternate interpretations are equivalent.

| cite | improve this answer | |
  • $\begingroup$ Yes. I noted that! And that was actually the answer I expected. Thanks $\endgroup$ – Victor Gustavo May May 12 at 1:20

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.

| cite | improve this answer | |
  • $\begingroup$ Thank you for your answer, but my question was about what the translation of the definition of the tensor would be in non-index notation? I think I could have made that clearer. For that I apologize, I will up-vote your answer since it was helpful. I successfully wrote a translation in index free notation that the contraction definition really defines a unique tensor. The proof was almost immediate, but doing it made evident an important observation that I missed when getting into the Abstact Index Notation. $\endgroup$ – Victor Gustavo May May 11 at 21:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.