Raising/lowering using the metric tensor

[trv]

Homework Statement

Given a N-dimensional manifold, let g_ab, be a metric tensor.
Compute
(i) g_abg^bc
(ii)g_abg^ab

Also, just need a clarification on something similar.
g^cdT_cd=g^cdT_dc=tr T?
I'm pretty sure its yes. Probably even a stupid question but a clarification would be useful.

Homework Equations

The Attempt at a Solution

(i)tr g ?
(ii)Here I'm caught between tr g, since I'm thinking its still the tensor and its inverse, and the indices don't matter.

The other option is g_a^c.

Which is correct?

 

[Hurkyl]

Because g is symmetric,

By symmetry of g: g^cdT_cd = g^dcT_cd
By renaming summed indices twice: g^dcT_cd = g^zpT_pz = g^cdT_dc

(I only used p and z to make things overly clear. You could do the renaming one step, and would normally do so)

 

[trv]

thanks. Any thoughts on the other bits?

i.e. g_abg^bc
g_abg^ab?

 

[Dick]

g_ab*g^bc=delta_a^c, right? The g with upper indices is defined as the inverse of the one with lower indices.

 

[trv]

Checked notes...yes it is.

 

[eok20]

g_{ab}g^{ab} = g_{ab}g^{ba} = delta_a^a = tr delta = N.

also, I don't see why g^{cd}T_{cd} would be tr T-- what does the metric have to do with the trace of a tensor?

 

[trv]

Good point.

It should simply be

g^cdT_dc=T^dc

right.

I actually needed these for another problem. And, it was much easier if g^cdT_dc=tr T :p. Guess I'll have to have a look at it again.

 

[Dick]

Good point.

It should simply be

g^cdT_dc=T^dc

right.

I actually needed these for another problem. And, it was much easier if g^cdT_dc=tr T :p. Guess I'll have to have a look at it again.

The metric has a lot to do with the trace of a tensor. You are right the second time. g^cd*T_dc is tr(T). You are forgetting the index balance again. In the first form the left side has no free indices and the right side has two. How can that be??

 

[eok20]

The metric has a lot to do with the trace of a tensor. You are right the second time. g^cd*T_dc is tr(T). You are forgetting the index balance again. In the first form the left side has no free indices and the right side has two. How can that be??

Ah...ok, sorry for the wrong advice-- I think I see why I was wrong. You can't really get the trace of a (2,0) tensor directly so tr T is the trace of the corresponding (1,1) tensor, so the computation is like g^{cd} T_{dc} = g^{cd} g_{ac} T_d^a = delta_a^d T_d^a = T_d^d = tr T, right?

 

[Dick]

Well, I would say g_cd*T^cd=tr(T) is pretty direct, but yes, what you say is also correct.

 

[trv]

thanks for the clarification Dick. And interesting way of looking at what's happening eok.