Why does the Kronecker delta interchange with itself when j equals l?

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Discussion Overview

The discussion revolves around the properties and implications of the Kronecker delta, particularly when the indices are equal (j = l). Participants explore the interchangeability of delta functions, the implications of summation conventions, and the interpretation of expressions involving these deltas in mathematical contexts.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants question whether the expression \(\delta_{jm}\delta_{kl}\) simplifies to \(\delta_{km}\) when \(j = l\), with differing opinions on the validity of this interchange.
  • Others argue that the Kronecker delta can be viewed as a symmetric unit matrix, allowing for certain algebraic manipulations.
  • A participant suggests that the notation implies summation, leading to confusion about the treatment of indices and whether they can be equated when summed over.
  • Concerns are raised about the clarity of expressions and the implications of treating indices as dummy variables versus fixed values.
  • Some participants emphasize the importance of understanding the notation and the consequences of misinterpreting the relationships between indices in expressions.

Areas of Agreement / Disagreement

Participants express differing views on the implications of the Kronecker delta's properties, particularly regarding summation and interchangeability. There is no consensus on whether certain expressions are equivalent or how indices should be treated in various contexts.

Contextual Notes

Participants highlight limitations in understanding the notation and the implications of summation conventions. There are unresolved questions about the treatment of indices and whether certain expressions can be simplified or equated.

Who May Find This Useful

This discussion may be useful for those studying linear algebra, tensor calculus, or related fields where the Kronecker delta and index notation are frequently applied.

  • #31
quietrain said:
yes that's all the question wants. use quotient law to show that matrix is a 2nd order tensor.
I don't see how the quotient law can have anything to do with this.

quietrain said:
so how do i proceed? is diagonalizing it the right way?
I have no idea. What I do know is that it's pointless to do that unless you have a way to proceed once you have found the diagonal matrix.

quietrain said:
i think my prof also said that by the quotient law, if the matrix

(A) (scalar) = invariant, it shows that A is tensor. or something along the lines of these.
If A times a scalar is an invariant, then A is obviously invariant too. So I'm pretty sure that's not what your professor told you.

When you have a statement that you think might be useful, that may or may not be what you have been told by a professor or seen in a book, you should at least think about what it would mean if the statement is true. If you had done this here, it would have saved us both some time.

quietrain said:
anyway, i still don't really have any idea how to solve this quesiton
I don't either, and to be honest, I still doubt that you have stated the problem correctly.

Do you have any solved examples from your book or your lecture notes that you can show me? This could at least give us some idea about how the quotient law is relevant.
 
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  • #32
hmm its ok then , that's all the question says .

thanks for everything though, i finally understood what those indexes meant
 

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