# Raising or Lowering indices through metric tensor

• I
• cianfa72
cianfa72
TL;DR Summary
Rules to raise or lower indices through metric tensor
I'm still confused about the notation used for operations involving tensors.
Consider the following simple example:
$$\eta^{\mu \sigma} A_{\mu \nu} = A_{\mu \nu} \eta^{\mu \sigma}$$
Using the rules for raising an index through the (inverse) metric tensor ##\eta^{\mu \sigma}## we get ##A^{\sigma}{}_{\nu}##. However if we work out explicitly the contraction employing the operator ##C_{\alpha}^{\mu} ()## we get:

$$C_{\alpha}^{\mu} (A_{\alpha \nu} \eta^{\mu \sigma} e^{\alpha} \otimes e^{\nu} \otimes e_{\mu} \otimes e_{\sigma}) = A_{\mu \nu} \eta^{\mu \sigma} e^{\mu} (e_{\mu}) e^{\nu} \otimes e_{\sigma} = A_{\mu \nu} \eta^{\mu \sigma} e^{\nu} \otimes e_{\sigma}$$
The latter is a tensor, say ##T = T_{\nu} {}^{\sigma} e^{\nu} \otimes e_{\sigma}##.

Is it the same as ##A^{\sigma}{}_{\nu} e_{\sigma} \otimes e^{\nu}## ?

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ergospherical
\begin{align*}C_{\alpha}^{\mu} (A_{\alpha \nu} \eta^{\mu \sigma} e^{\alpha} \otimes e^{\nu} \otimes e_{\mu} \otimes e_{\sigma}) &= A_{\alpha \nu} \eta^{\mu \sigma} e^{\alpha} (e_{\mu}) e^{\nu} \otimes e_{\sigma} \\ &= A_{\mu \nu} \eta^{\mu \sigma} e^{\nu} \otimes e_{\sigma} \\ &={A^{\sigma}}_{\nu} e^{\nu} \otimes e_{\sigma}\end{align*}

cianfa72
\begin{align*}C_{\alpha}^{\mu} (A_{\alpha \nu} \eta^{\mu \sigma} e^{\alpha} \otimes e^{\nu} \otimes e_{\mu} \otimes e_{\sigma}) &= A_{\alpha \nu} \eta^{\mu \sigma} e^{\alpha} (e_{\mu}) e^{\nu} \otimes e_{\sigma} \\ &= A_{\mu \nu} \eta^{\mu \sigma} e^{\nu} \otimes e_{\sigma} \\ &={A^{\sigma}}_{\nu} e^{\nu} \otimes e_{\sigma}\end{align*}
Oops yes, from the RHS on the first line summing over ##\alpha## we get the second line and then (summing over ##\mu##) the result.

Maybe I'm missing the point, in your result the ##\nu## index in ##A^{\sigma}{}_{\nu}## actually refers to the first element in tensor product ##e^{\nu} \otimes e_{\sigma}##, not to the second one. Instead in the expression ##T_{\nu}{}^{\sigma} e^{\nu} \otimes e_{\sigma}## is the first index ##\nu## that refers to the first element.

Is the following correct ?

$${A^{\sigma}}_{\nu} e^{\nu} \otimes e_{\sigma} = T_{\nu}{}^{\sigma} e^{\nu} \otimes e_{\sigma}$$

ergospherical
Is the following correct ?
$${A^{\sigma}}_{\nu} e^{\nu} \otimes e_{\sigma} = T_{\nu}{}^{\sigma} e^{\nu} \otimes e_{\sigma}$$
Well, yes, but only because that’s how you defined ##{T_{\nu}}^{\sigma}##…

You are worrying too much. It’s conventional to maintain the same horizontal ordering of the component indices and the tensor arguments (so that it’s easy to tell which slot is which), but you can do whatever you want.

• cianfa72
cianfa72
Sorry, it seems to me there are actually two different answers we get reversing the order of the 'index raising' operation, namely:
$$C_{\alpha}^{\mu} (A_{\alpha \nu} \eta^{\mu \sigma} e^{\alpha} \otimes e^{\nu} \otimes e_{\mu} \otimes e_{\sigma}) = {A^{\sigma}}_{\nu} e^{\nu} \otimes e_{\sigma}$$
then if we reverse the order we get:
$$C_{\alpha}^{\mu} (\eta^{\mu \sigma} A_{\alpha \nu} e_{\mu} \otimes e_{\sigma} \otimes e^{\alpha} \otimes e^{\nu}) = {A^{\sigma}}_{\nu} e_{\sigma} \otimes e^{\nu}$$

The two are really two different tensors, where is the mistake ?

ergospherical
It’s nothing more significant than the ordering of the vector and co-vector arguments.

cianfa72
Suppose ##n=2## we get in the two cases:
$$A^{1}{}_{1} e^{1} \otimes e_{1} + A^{1}{}_{2} e^{2} \otimes e_{1} + A^{2}{}_{1} e^{1} \otimes e_{2} + A^{2}{}_{2} e^{2} \otimes e_{2}$$ $$A^{1}{}_{1} e_{1} \otimes e^{1} + A^{1}{}_{2} e_{1} \otimes e^{2} + A^{2}{}_{1} e_{2} \otimes e^{1} + A^{2}{}_{2} e_{2} \otimes e^{2}$$
So the difference is really just the slots order in which plug in the vector and co-vector.

Does the same thing hold for cases like the following ?

##\eta_{\mu \alpha} A^{\alpha \nu} \eta_{\sigma \nu} \Rightarrow A_{\mu \sigma} e^{\mu} \otimes e^{\sigma}##

##\eta_{\sigma \nu} A^{\alpha \nu} \eta_{\mu \alpha} \Rightarrow A_{\mu \sigma} e^{\sigma} \otimes e^{\mu}##

ergospherical
Yeah, chill, it’s just like the difference between ##f(x,y) = x^2 y## and ##g(x,y) = y^2 x##, whereby ##f(x,y) = g(y,x)##.

cianfa72
Yeah, chill, it’s just like the difference between ##f(x,y) = x^2 y## and ##g(x,y) = y^2 x##, whereby ##f(x,y) = g(y,x)##.
yes, the point confusing me is that when formally we raise and/or lower tensor indices through the metric tensor we need to take in account it (in other words we need to take in account the orders of the slots -- in the above example the order of the slots 'waiting' for vectors to be plugged in).

Hence from the point of view of the tensor we get from the raising/lowering operations through the metric tensor, the order makes the difference.

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