# Raising/lowering using the metric tensor

## Homework Statement

Given a N-dimensional manifold, let gab, be a metric tensor.
Compute
(i) gabgbc
(ii)gabgab

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

## 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 gac.

Which is correct?

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Hurkyl
Staff Emeritus
Gold Member
Because g is symmetric,

By symmetry of g: gcdTcd = gdcTcd
By renaming summed indices twice: gdcTcd = gzpTpz = gcdTdc

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

thanks. Any thoughts on the other bits?

i.e. gabgbc
gabgab?

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

Checked notes...yes it is.

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?

Good point.

It should simply be

gcdTdc=Tdc

right.

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

Dick
Homework Helper
Good point.

It should simply be

gcdTdc=Tdc

right.

I actually needed these for another problem. And, it was much easier if gcdTdc=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??

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