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

[trv]

Homework Statement

Given a tensor M^a_b, Prove that its trace is a scalar.

Homework Equations

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'^c_d=(dx'^c/dx^a)(dx^b/dx'^d)M^a_b

For the trace M^a_a, the transformation is as follows. I've used the Kronecker delta for changing M^a_b to M^a_a. ( Am I along the right lines in doing so?).
M'^c_d=(dx'^c/dx^a)(dx^b/dx'^d)M^a_b(Kronecker delta^a_b)

Then I thought using the delta I change dx^b to dx^a and cancel the two dx^a. However that would then the delta away from M^a_b, leaving it as M^a_b as before rather than M^a_a.

That's the best I can do, please help.

 

[Dick]

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.

 

[trv]

Ok, so its,

To get the transformation equation for the trace, multiply both sides by delta ^c<sub>d[/SUB}

M'^cd(delta ^cd)=(dx^b/dx'^d)(dx'^c/dx^a)M^ab(delta ^cd)

Thus,
M'^cc=(dx^b/dx'^c)(dx'^c/dx^a)M^ab

By chain rule,

M'^cc=(dx^b/dx^a)M^ab

Now dx^b/dx^a=delta ^ab

Thus,

M'^cc=M^aa

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=M^aa
i.e. M^aa is invariant under coordinate transformations. Therefore M^aa the trace of M^ab is a scalar.

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

 

[Dick]

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.

 

[trv]

Thanks :)

 

[trv]

If I may quickly add another quick question.

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

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

 

[Dick]

If I understand your question, then you just write the same formula with y instead of x'. But I probably don't. How do you indicate M is a tensor in the y coordinate system?

 

[trv]

Hi again Dick. Yeh, you misunderstood my question, but I managed to find out what I wanted.

Apparently M^a_b can be written as Mx^a_b for the tensor in the x-coordinate system and My^a_b for the y-coordinate system.