# Covariant and Contravariant Kronecker Delta operating on Tensor

1. Aug 22, 2011

### lewis198

I am aware that the following operation:

$mathbf{M}_{ij} \delta_{ij}$

produces

$mathbf{M}_{ii} or mathbf{M}_jj$

However, if we have the following operation:

$mathbf{M}_{ij} \delta^i{}_j$

will the tensor M be transformed at all?

2. Aug 22, 2011

### WannabeNewton

I can't read your latex well but $M_{ij}\delta _{ij} \neq M_{ii}, M_{ij}\delta _{ij} \neq M_{jj}$ as there is no implied summation. In order to contract indeces you have to employ the summation convention i.e. $M_{ij}\delta ^{ij} = M^{i}_{i} = M^{j}_{j}$ which is just the trace of the tensor M and $M_{ij}\delta ^{i}_{j} = M_{jj}$.

3. Aug 22, 2011

### lewis198

is M_{ij}\delta ^{i}_{j} = M_{ii} also valid?

4. Aug 22, 2011

### WannabeNewton

No, $M_{ij}\delta ^{i}_{j}\neq M_{ii}$ because you are contracting with the i index which is what the summation is implied over; contract the i's and move in the j where the i is. However, $M_{ij}\delta ^{j}_{i}= M_{ii}$.

5. Aug 23, 2011

### lewis198

Ok thanks, got it. By the way, how come your Latex is showing and mine isn't?

6. Aug 23, 2011

### Fredrik

Staff Emeritus
You typed the tags wrong in the first post and used no tags at all in the second. You need to type itex or tex, not latex. Hit the quote button next to a post with math, and you'll see how it's done.

That first claim is wrong. $M_{ij}\delta ^{i}_{j}$ is definitely $=M_{ii}$. Edit: Uh, wait, if the convention is that there's no implied summation when both indices are downstairs, you're right and I was wrong. We have $M_{ij}\delta^i_j=M_{jj}$.

7. Aug 24, 2011

### lewis198

Thanks - I have another question if you don't mind- Is $$\delta_{i}^{i}$$ summed over? i.e Is the above equal to $$\displaystyle\sum\limits_{i=0}^n \delta_{i}^{i}$$ If so what determines n?

8. Aug 24, 2011

### dextercioby

Usually yes, the summation is assumed. The dimension of the vector space (or the upper limit of summation) should also be specified.