# Proving trace of (1,1) Tensor is a scalar

## Homework Statement

Given a tensor Mab, Prove that its trace is a scalar.

## The Attempt at a Solution

To prove the trace is a scalar, I know I have to prove it doesn't transform under coordinate transformations.

Now, we can transform M^a_b as follows.
M'cd=(dx'c/dxa)(dxb/dx'd)Mab

For the trace Maa, the transformation is as follows. I've used the Kronecker delta for changing Mab to Maa. ( Am I along the right lines in doing so?).
M'cd=(dx'c/dxa)(dxb/dx'd)Mab(Kronecker deltaab)

Then I thought using the delta I change dxb to dxa and cancel the two dxa. However that would then the delta away from Mab, leaving it as Mab as before rather than Maa.

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Dick
Homework Helper
The way you are doing it leaves your indices way out of balance. You've got three a indices and three b indices on the right side. Multiply both sides by $$\delta_c^d$$ and use the chain rule for partial derivatives.

Ok, so its,

To get the transformation equation for the trace, multiply both sides by delta cd[/SUB}

M'cd(delta cd)=(dxb/dx'd)(dx'c/dxa)Mab(delta cd)

Thus,
M'cc=(dxb/dx'c)(dx'c/dxa)Mab

By chain rule,

M'cc=(dxb/dxa)Mab

Now dxb/dxa=delta ab

Thus,

M'cc=Maa

Now, we can change the dummy variable c---> a. (I've seen this being done. Is it allowed only for indices that are being summed over, and thus don't affect other terms?)

Thus,
M'aa=Maa
i.e. Maa is invariant under coordinate transformations. Therefore Maa the trace of Mab is a scalar.

OK I get it, thanks Dick. Can someone check the dummy variable bit however.

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Dick
Homework Helper
Yes, if a variable is summed, it doesn't matter what it's name is. I would put interchange the c and d on your delta though. You've got two c indices up. In the summation convention you usually want one up and one down.

Thanks :)

If I may quickly add another quick question.

Mab in primed x-coordinates is denoted simply by adding the prime, i.e. M'ab.

Instead if I was transforming Mab into the y-coordinates. How can I denote that?

Dick