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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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Please provide some context about what you mean by $\delta_{ij}$. – Paresh Nov 26 '12 at 13:16
@Paresh: I'm studying Poisson brackets and I met this delta. For example: I have that $\frac {\partial q_k}{\partial \delta q_i}= \delta_{ki}$... – sunrise Nov 26 '12 at 13:25
@sunrise: That doesn't make much sense -- are you sure there's an additional $\delta$ in the denominator? Without it, the equation $\partial q_k/\partial q_i=\delta_{ki}$ would make a lot of sense, and in this case $\delta_{ki}$ would indeed refer to the Kronecker delta. – joriki Nov 26 '12 at 17:24
@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! – sunrise Nov 26 '12 at 20:21
up vote 3 down vote accepted

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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Thank you. Can I say that a Kronecker's delta $\delta_{ij}$ is always equal to the partial derivative of a function $F_i$ with respect to a function $F_j$? – sunrise Nov 30 '12 at 20:05
@sunrise: Glad to help. There are many representations of the 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. – user26872 Dec 2 '12 at 20:01

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