Position of indices when and after being raised/lowered

625 Views Asked by At

I'm having some trouble with the rule of lowering and raising indices in my textbook: $$\begin{align} \tau_{j_1\ldots j_s j}^{i_1\ldots i_{k-1}\hspace{0.5em}i_{k+1}\hspace{0.5em}\ldots i_r} & \equiv d_{ji_k}\tau_{j_1\ldots j_s}^{i_1\ldots i_{k-1}\hspace{0.5em}i_k\hspace{0.2em}i_{k+1}\hspace{0.5em}\ldots i_r}\\ \tau_{j_1\ldots j_s}^{i_1\ldots i_{k-1}\hspace{0.5em}i_k\hspace{0.2em}i_{k+1}\ldots i_r} & \equiv d^{ji_k}\tau_{j_1\ldots j_sj}^{i_1\ldots i_{k-1}\hspace{0.5em}i_{k+1}\hspace{0.5em}\ldots i_r}\\ \end{align}$$ etc. What confuses me is the position of indices when and after being lowered/raised. These formulas seem to imply that one may raise any index to any place he likes. However, I found a contradiction when applying this rule to an antisymmetric second-order covariant tensor, for example: $$a^{ij}\equiv d^{ii'}d^{jj'}a_{i'j'}$$ While, if the raised indices can be placed anywhere: $$a^{ji}\equiv d^{jj'}d^{ii'}a_{i'j'}=a^{ij}$$ Contradicting the antisymmetry of $a^{ij}$ (it can be easily proved that $a^{ij}$ is antisymmetric due to the antisymmetry of $a_{ij}$.
I searched the Wikipedia, but the definition there seems to allow only raising the last index below to the first index above, which, I think, is not satisfying since one should be able to raise/lower any index. Did I take anything wrong? Or was my textbook making a mistake? More precisely, what's the restrictions for the position of indices being raised/lowered and after that?

1

There are 1 best solutions below

2
On BEST ANSWER

$T^{abc}_{def}$ is terrible notation. You should order all of the indices to avoid making mistakes when raising and lowering indices. E.g. you should distinguish between tensors of type ${T^{abc}}_{def}$ from those of type ${{{T^a}_{de}}^{bc}}_f$. Remember that a ${A^a}_b$ is really just a component of the tensor ${A^a}_be_a\otimes e^b$ while ${A_a}^b$ is a component of the tensor ${A_a}^be^a\otimes e_b$ which is clearly different from the first.

Here's the way to lower the indices of $a^{ij}$. Note that I'll explicitly add in the steps where I'm using the metric tensor $\mathbf g$ (or really the inverse metric tensor $\mathbf g^{-1}$):

$$\begin{align}a^{ij} &= \mathbf g^{-1}(e^i,{a_k}^je^k) \\ &= g^{ik}{a_k}^j \\ &= g^{ik}\mathbf g^{-1}(e^j,a_{kl}e^l) \\ &= g^{ik}g^{jl}a_{kl}\end{align}$$

vs

$$\begin{align}a^{ij} &= \mathbf g^{-1}(e^j,{a^i}_ke^k) \\ &= g^{jk}{a^i}_k \\ &= g^{jk}\mathbf g^{-1}(e^i,a_{lk}e^l) \\ &= g^{jk}g^{il}a_{lk}\end{align}$$

Notice that both of these are the same because $a$ contracts in its first index with $g^{i\square}$ where $\square$ is the summing index. And similarly for the second index on $a$.

For more information, I highly recommend Introduction to Tensor Calculus for General Relativity by Edmund Bertschinger as a reference for beginners.