Kronecker delta in tensor component form

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Problem 2.3b4

A First Course in String Theory

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Continue the previous calculation:

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\displaystyle{ \begin{aligned} \nu \neq \mu:&~~~~~~\sum_{\mu = 0}^4 L^\mu_{~\nu} M^{\beta}_{~\mu} &= 0 \\ \nu = \mu:&~~~~~~\sum_{\mu = 0}^4 L^\mu_{~\nu} M^{\beta}_{~\mu} &= 1 \\ \end{aligned}}

The two cases can be grouped into one, by replacing the right hand sides with the Kronecker delta. However, there are 4 possible forms and I am not sure which one should be used.

\displaystyle{\delta^i_{~j}}
\displaystyle{\delta_i^{~j}}
\displaystyle{\delta^{ij}}
\displaystyle{\delta_{ij}}

So I do a little research on Kronecker delta in this post.

— Me@2020-10-21 03:40:36 PM

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The inverse Lorentz transformation should satisfy \displaystyle{\left( \Lambda^{-1} \right)^\beta_{~\mu} \Lambda^\mu_{~\nu} = \delta^\beta_{~\nu}}, where \displaystyle{\delta^\beta_{~\nu} \equiv \text{diag}(1,1,1,1)} is the Kronecker delta. Then, multiply by the inverse on both sides of Eq. 4 to find

\displaystyle{ \begin{aligned} \left( \Lambda^{-1} \right)^\beta_{~\mu} \left( \Delta x' \right)^\mu &= \delta^\beta_{~\nu} \Delta x^\nu \\ &= \Delta x^\beta \\ \end{aligned}}

(6)

The inverse \displaystyle{\left( \Lambda^{-1} \right)^\beta_{~\mu}} is also written as \displaystyle{\Lambda_\mu^{~\beta}}. The notation is as follows: the left index denotes a row while the right index denotes a column, while the top index denotes the frame we’re transforming to and the bottom index denotes the frame we’re transforming from. Then, the operation \displaystyle{\Lambda_\mu^{~\beta} \Lambda^\mu_{~\nu}} means sum over the index \displaystyle{\mu} which lives in the primed frame, leaving unprimed indices \displaystyle{\beta} and \displaystyle{\nu} (so that the RHS of Eq. 6 is unprimed as it should be), where the sum is over a row of \displaystyle{\Lambda_\mu^{~\beta}} and a column of \displaystyle{\Lambda_{~\nu}^\mu} which is precisely the operation of matrix multiplication.

— Lorentz tensor redux

— Emily Nardoni

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This one is WRONG:

\displaystyle{(\Lambda^T)^{\mu}{}_{\nu} = \Lambda_{\nu}{}^{\mu}}

This one is RIGHT:

\displaystyle{(\Lambda^T)_{\nu}{}^{\mu} ~:=~ \Lambda^{\mu}{}_{\nu}}

— Me@2020-10-23 06:30:57 PM

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1. \displaystyle{(\Lambda^T)_{\nu}{}^{\mu} ~:=~\Lambda^{\mu}{}_{\nu}}

2. [Kronecker delta] is invariant in all coordinate systems, and hence it is an isotropic tensor.

3. Covariant, contravariant and mixed type of this tensor are the same, that is

\displaystyle{\delta^i_{~j} = \delta_i^{~j} = \delta^{ij} = \delta_{ij}}

— Introduction to Tensor Calculus

— Taha Sochi

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Raising and then lowering the same index (or conversely) are inverse operations, which is reflected in the covariant and contravariant metric tensors being inverse to each other:

{\displaystyle g^{ij}g_{jk}=g_{kj}g^{ji}={\delta ^{i}}_{k}={\delta _{k}}^{i}}

where \displaystyle{\delta^i_{~k}} is the Kronecker delta or identity matrix. Since there are different choices of metric with different metric signatures (signs along the diagonal elements, i.e. tensor components with equal indices), the name and signature is usually indicated to prevent confusion.

— Wikipedia on Raising and lowering indices

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So

{\displaystyle g^{ij}g_{jk}={\delta ^{i}}_{k}}

and

{\displaystyle g_{kj}g^{ji}={\delta _{k}}^{i}}

— Me@2020-10-19 05:21:49 PM

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\displaystyle{ T_{i}^{\; j} = \boldsymbol{T}(\boldsymbol{e}_i,\boldsymbol{e}^j) } and \displaystyle{T_{j}^{\; i} = \boldsymbol{T}(\boldsymbol{e}_j,\boldsymbol{e}^i) } are both 1-covariant 2-contravariant coordinates of T. The only difference between them is the notation used for sub- and superscripts;

\displaystyle{ T^{i}_{\; j} = \boldsymbol{T}(\boldsymbol{e}^i,\boldsymbol{e}_j) } and \displaystyle{ T^{j}_{\; i} = \boldsymbol{T}(\boldsymbol{e}^j,\boldsymbol{e}_i) } are both 1-contravariant 2-covariant coordinates of T. The only difference between them is the notation used for sub- and superscripts.

— edited Oct 11 ’17 at 14:14

— answered Oct 11 ’17 at 10:58

— EditPiAf

— Tensor Notation Upper and Lower Indices

— Physics StackExchange

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Rather, the dual basis one-forms are defined by imposing the following
16 requirements at each spacetime point:

\displaystyle{\langle \tilde{e}^\mu \mathbf{x}, \vec e_\nu \mathbf{x} \rangle = \delta^{\mu}_{~\nu}}

is the Kronecker delta, \displaystyle{\delta^{\mu}_{~\nu} = 1} if \displaystyle{\mu = \nu} and \displaystyle{\delta^{\mu}_{~\nu} = 0} otherwise, with the same values for each spacetime point. (We must always distinguish subscripts from superscripts; the Kronecker delta always has one of each.)

— Introduction to Tensor Calculus for General Relativity

— Edmund Bertschinger

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However, since \displaystyle{\delta_{~b}^a} is a tensor, we can raise or lower its indices using the metric tensor in the usual way. That is, we can get a version of \displaystyle{\delta} with both indices raised or lowered, as follows:

[\displaystyle{\delta^{ab} = \delta^a_{~c} g^{cb} = g^{ab}}]

\displaystyle{\delta_{ab} = g_{ac} \delta^c_{~b} = g_{ab}}

In this sense, \displaystyle{\delta^{ab}} and \displaystyle{\delta_{ab}} are the upper and lower versions of the metric tensor. However, they can’t really be considered versions of the Kronecker delta any more, as they don’t necessarily satisfy [0 when i \ne j and 1 when i = j]. In other words, the only version of \delta that is both a Kronecker delta and a tensor is the version with one upper and one lower index: \delta^a_{~b} [or \delta^{~a}_{b}].

— Kronecker Delta as a tensor

— physicspages

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Continue the calculation for the Problem 2.3b:

Denoting \displaystyle{ \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\nu \sigma}} as \displaystyle{L^{~\nu}_{\mu}} is misleading, because that presupposes that \displaystyle{ \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\nu \sigma}} is directly related to the matrix \displaystyle{L}.

To avoid this bug, instead, we denote \displaystyle{ \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\nu \sigma}} as \displaystyle{M ^\nu_{~\mu}}. So

\displaystyle{ \begin{aligned} (x')^\mu &= L^\mu_{~\nu} x^\nu \\ (x')^\mu (x')_\mu &= \left( L^\mu_{~\nu} x^\nu \right) \left( \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\beta \sigma} x_\beta \right) \\ (x')^\mu (x')_\mu &= \left( L^\mu_{~\nu} x^\nu \right) \left( M^{\beta}_{~\mu} x_\beta \right) \\ x^\mu x_\mu &= \left( L^\mu_{~\nu} x^\nu \right) \left( M^{\beta}_{~\mu} x_\beta \right) \\ \end{aligned}}

…

\displaystyle{ \begin{aligned} \nu \neq \mu:&~~~~~~\sum_{\mu = 0}^4 L^\mu_{~\nu} M^{\beta}_{~\mu} &= 0 \\ \nu = \mu:&~~~~~~\sum_{\mu = 0}^4 L^\mu_{~\nu} M^{\beta}_{~\mu} &= 1 \\ \end{aligned}}

Using the Kronecker Delta and Einstein summation notation, we have

\displaystyle{ \begin{aligned} L^\mu_{~\nu} M^{\beta}_{~\mu} &= M^{\beta}_{~\mu} L^\mu_{~\nu} \\ &= \delta^{\beta}_{~\nu} \\ \end{aligned}}

Note: After tensor contraction, the remaining left index should be kept on the left and the remaining right on the right.

— Me@2020-10-20 03:49:09 PM

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