Covariant and Contravariant Kronecker Delta operating on Tensor

[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?


Thank you for your time.

 

[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 $M_{ij}\delta ^{i}_{j} = M_{jj}$.

 

[lewis198]

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

 

[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, $M_{ij}\delta ^{j}_{i}= M_{ii}$.

 

[lewis198]

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

 

[Fredrik]

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

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.

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, $M_{ij}\delta ^{j}_{i}= M_{ii}$.

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 $M_{ij}\delta^i_j=M_{jj}$.

 

[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?

 

[dextercioby]

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