# Are fully raised/lowered versions of Kronecker delta tensors?

I am confused. I have two textbooks contradicting each other, at least, it seems to me so. The first one – "Field theory" by Landau & Lifshitz says that by lowering or raising one index of Kronecker delta one gets covariant/contravariant metric tensor(in Minkowski space). The second one "Introducing Einstein’s Relativity (1992)" by Ray d’Inverno shows the opposite, namely, by lowering or raising an index of Kronecker delta one gets an object which apparently is not a tensor( the last sentence says that in this link). Is there a contradiction or the problem is with my understanding?

• I am very sorry for misleading all, saying that Ray 'Inverno shows the result given above, it was not him. It turns out that it is a solution for a problem from Ray d'Inverno's book given by some author on that website. – Arislan Makhmudov Nov 1 '15 at 12:44

The metric tensor $g : TM\times TM \to \mathbb{R}$ is by definition a $(0,2)$-tensor and transforms like one.

You question is not about the metric tensor, but about the Kronecker delta, and for which index positions it defines a tensor.

To define a tensor by its components, we must fix a coordinate system $x$ on our manifold $M$ since the components of a $(0,2)$-tensor $T$, for instance, are defined as coefficients in the expansion of $T$ in the basic tensors $\mathrm{d}x^\mu$: $$T= T_{\mu\nu}\mathrm{d}x^\mu\otimes\mathrm{d}x^\nu$$ and we now want to examine for which index positions on the $\delta$ the tensors $$\delta^{ab}\partial_a\otimes\partial_b\quad\text{and}\quad\delta_{ab}\mathrm{d}x^a\otimes\mathrm{d}x^b\quad\text{and}\quad{\delta^a}_b\mathrm{d}x^a\otimes\partial_b$$ are defined independent of the chosen coordinate system.

To that end, we examine the transformation behaviour of $\delta^{ab},{\delta^a}_b,\delta_{ab}$. What we find is that under a coordinate transformation $x\mapsto y(x)$, \begin{align} \delta^{ab} & \mapsto \sum_i \frac{\partial y^a}{\partial x^i}\frac{\partial y^b}{\partial x^i}\\ \delta_{ab} & \mapsto \sum_i \frac{\partial x^i}{\partial y^a}\frac{\partial x^i}{\partial y^b}\\ {\delta^a}_b &\mapsto {\delta^a}_b \end{align}

Therefore, a tensor that has components $\delta^{ab}$ or $\delta_{ab}$ in one coordinate system does not have those components in another system, so just writing down $\delta^{ab}$ does not specify a tensor unless you also specify a coordinate system in which the tensor has these components.

On the other hand, a $(1,1)$-tensor that has components ${\delta^a}_b$ in one system has those in all, therefore, ${\delta^a}_b$ defines a tensor without need for a particular coordinate system.

Now, since it defines a $(1,1)$-tensor, we can raise and lower its indices. However, contrary to what one might expect, the fully raised and fully lowered versions are not $\delta^{ab}$ and $\delta_{ab}$, but instead $g^{ab}$ and $g_{ab}$.

So, lowering/raising the index on the $\delta$ with the metric tensor1 gives proper tensors, "raising/lowering" the index by just writing both indices rasied on the $\delta$ does not give a well-defined tensor.

1Here, "raising/lowering" with the metric tensor refers to the musical isomorphism that a vector $v = v^\mu\partial_\mu$ has an associated covector $v^\flat = g_{\mu\nu}v^\nu\mathrm{d}x^\mu$ with components $(v^\flat)_\mu = g_{\mu\nu}v^\nu$, and similarily lowering an index of a tensor means contracting it with $g_{\mu\nu}$, and raising it contracting it with $g^{\mu\nu}$. So, lowering the index of ${\delta^a}_b$ means $(\delta^\flat)_{ab} = g_{ac}{\delta^c}_b = g_{ab}$ by definition of the Kronecker delta.

• Totally agree, I misnamed topic. It is now corrected. You confused me even more by telling that in order to specify a tensor you sometimes need to specify systems of coordinates, and why fully raised/lowered deltas become metric tensors? Could you answer me the question? Are the laws of transformation that you gave for the fully raised/lowered deltas tell us if they are tensors? Cause the definition of tensor is about transformation law, you know it. – Arislan Makhmudov Oct 29 '15 at 20:15
• @ArislanMakhmudov: I expanded the answer, tell me if you'd like further elaboration. – ACuriousMind Oct 29 '15 at 20:47
• sorry I still don't understand, this is because I am a newbie in this topic. Could you elaborate on how this author makes conclusion that expression (9) in that link tells us that it is not a tensor? Thank you very much! I really appreciate your help. – Arislan Makhmudov Oct 30 '15 at 10:24
• @ArislanMakhmudov: Well, I'm afraid that the usual physicists way of speaking about tensors only in those components is likely the cause of your confusion. To say "$\delta^{ab}$ is not a tensor" is silly. Of course $\delta^{ab}\partial_a\otimes\partial_b$ can be a tensor in some coordinate system! It just won't have components $\delta^{ab}$ in any other. So the conclusion is not that it "isn't a tensor", it is that it is not a coordinate-independent definition of a tensor. And since we would want the Kronecker delta to behave like $T_{ab}\delta{bc} = {T_a}^c$ always, that's not what we want. – ACuriousMind Oct 30 '15 at 14:19