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When an index of the Kronecker-delta tensor $\delta_a^b$ is lowered or raised with the metric tensor $g_{ab}$, i.e. $g_{ab}\delta^b_c$ or $g^{ab}\delta_b^c$, is the result another Kronecker-delta tensor?

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    $\begingroup$ No. It is the metric itself. $\endgroup$ – Prahar Oct 8 '15 at 21:10
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If you are dealing with spacetime indices (i.e. tensors over the spacetime), then symbols like $\delta^{ab}$ or $\delta_{ab}$ don't make sense. If you lower an index of $\delta^a_b$ you will end up with the metric $g_{ab}$, same for raising an index. This is clear from the definition of $\delta$:

$$g_{ab}\delta^b_c=g_{ac}$$

and

$$g^{ab}\delta_b^c=g^{ac}$$

since $\delta^b_c=1$ for $b=c$ and $\delta^b_c=0$ for $b\neq c$.

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  • $\begingroup$ But doesn't the metric tensor always lower/raises indices. By this rule the the Kronecker-delta tensor should raise and lower as $g_{ab}\delta_c^a=\delta_{bc}$ or $g^{ab}\delta_c^a=\delta^{bc}$. Is the result a Kronecker-delta function? $\endgroup$ – user37222 Apr 13 '14 at 13:13
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    $\begingroup$ @user37222 $\delta_{bc} = g_{bc}$, $\delta^{bc} = g^{bc}$, $g^b{}_c = \delta^b_c$. $\endgroup$ – Hunter Apr 13 '14 at 13:19
  • $\begingroup$ You can see it as using the definition $\delta^{ab}\equiv g^{ab}$ $\endgroup$ – Photon Apr 13 '14 at 13:19
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    $\begingroup$ The notation $\delta_{ab}$ doesn't really make sense. This is because you want $\delta$ to be the identity matrix, but $\delta_{ab}$ acts like $\delta_{ab} x^b = y_a$. You can't directly compare $x^b$ and $y_b$ because they don't belong to the same space. You could say that $y_b$ is the same as $x^b$ if $x^b = y^b$, the index raised with the metric. But this amounts to taking $\delta_{ab} = g_{ab}$. $\endgroup$ – Robin Ekman Apr 13 '14 at 13:36
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One answer is that the delta operates on the metric tensor, changing its (the metric tensor's) index (one of its indices).

Another answer is that the metric tensor operates on the delta, lowering one index.

Both answers must be correct and thus they are equal.

So you get delta with two bottom indices = g with two bottom indices.

The insight is that the delta tensor and the metric tensor are one and the same entity. When both indices are covariant or both are contravariant, we usually (always, in practice) use the symbol g; when one is covariant and the other is contravariant, we use the symbol delta.

When it's g, we think of it as the metric (or an index lowerer or raiser). When its delta we think of it as the identity (or as an index changer). But they are different aspects of the same thing.

For practical purposes, we think of them as different, but really, strictly, they're actually, ultimately the same tensor.

So it's not quite right to say there's no such thing as delta with two lower (or two upper) indices. It's just that we always write it as g.

Similarly, what is g with one lower and one upper index? Answer: it is delta with one lower and one upper index.

g is delta in disguise and delta is g in disguise.

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  • $\begingroup$ By the way, my understanding of this comes from watching this lecture: youtube.com/watch?v=jmAEo2Y8LOs&spfreload=10 $\endgroup$ – Hugh Jones Oct 8 '15 at 20:42
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    $\begingroup$ Hi. May I ask something: if the two tensors are the same then 1) why do we use both of them, 2) more importantly I have read that the metric tensor is equal with the δ in Euclidean space. How can it be that in general they are the same? Thank you. $\endgroup$ – Constantine Black Mar 27 '16 at 10:50
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You're confusing two things. In tensor calculus, the Kronecker delta should be visualized as basically the identity. What it does is relabel an index. Example:

$$g_{ab}\delta^b_c=g_{ac}$$

This has nothing whatsoever to do with the Dirac delta function (it's actually a distribution) in this context. In nonrelativistic quantum theory Dirac delta distributions might be used e.g. for orthogonality relations between eigenstates, just like the Kronecker delta, but this, in turn, has nothing to do with tensor calculus.

EDIT: This is an answer to your original question, which has since then been edited by Qmechanic, quite drastically changing the second part. I'm not sure this edit accurately reflects your confusion. If it does, please ignore the second part to my answer.

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  • $\begingroup$ Sorry I meant so say "is the result another Kronecker-Delta tensor". The metric tensor lowers and raises indices. By this rule the the Kronecker-delta tensor should raise and lower as $g_{ab}\delta_c^a=\delta_{bc}$ or $g^{ab}\delta_c^a=\delta^{bc}$. Is the result a Kronecker-delta function? $\endgroup$ – user37222 Apr 13 '14 at 13:11
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The result is the metric: the effect of the Kronecker delta in your examples is to set $b = c$. The Kronecker delta is really just the identity matrix.

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  • $\begingroup$ Hello. So an answer like the one by Hugh Jones doesn't make sense. I'm currently reading from the notes of a professor from my University and he writes the $δ_ab$. I can' t understand if he made a typo error or if he 's thinking is the same as of Hugh Jones. A reply would be most helpful. Thanks. $\endgroup$ – Constantine Black Mar 27 '16 at 10:49

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