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What's the relation between $\delta_{ij}$ and $\delta_{ji}$?

What about their mathematical and physical meanings?

Thank you!

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    @joriki: I'm sorry, I have committed a mistake in typing. There isn't the additional $\delta$ in the denominator... Is always a Kronecker's delta $\delta_{ij}$ equal to a partial derivative of a function $F_i$ with respect to a function $F_j$? thank you!2012-11-26

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It appears from the comments that you are referring to the Kronecker delta, $\delta_{ij} = \begin{cases}1, & i=j \\ 0, & \textrm{else.}\end{cases}$ An immediate consequence of the definition above is that the Kronecker delta is symmetric, $\delta_{ij} = \delta_{ji}$.

One can think of $\delta_{ij}$ as the $ij$th component of the identity matrix, $I$. Since $I^T = I$, $\delta_{ij} = I_{ij} = (I^T)_{ij} = I_{ji} = \delta_{ji}.$ It is natural to interpret the Kronecker delta as the Euclidean metric.

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    @sunrise: Glad to help. There are many representations of the [Kronecker delta](http://en.wikipedia.org/wiki/Kronecker_delta). For example, there is a nice integral representation. It can be written in terms of the Iverson bracket. I give another representation in my answer above. Sometimes it is useful to think of it as a tensor of type $(1,1)$, in which case the interpretation you refer to is used, though it would usually be written as $\delta_i^j = \partial x^j/\partial x^i$. We choose the representation depending on the problem.2012-12-02