I was reading Sean Carroll's Introduction to General Relativity

On Page 12 there is an equation given for defining the Lorentz group as a collection of $4\times 4$ matrices that satisfy

$$ \Lambda^T\eta_{\mu \nu} \Lambda = \eta_{\mu \nu} $$

Now he also presents the same defining condition in terms of tensors and index notation on page 13 as:

$$ \eta_{\rho \sigma} = \Lambda^{\mu'}_{\ \ \ \rho} \eta_{\mu' \nu'} \Lambda^{\nu'}_{\ \ \ \sigma} $$

Problem Statement:

I'm trying to figure out what this statement concretely means and prove that it is equivalent to the matrix definition. Unpacking the summation convention I have:

$$ \Lambda^{\mu'}_{\ \ \ \rho} \eta_{\mu' \nu'} \Lambda^{\nu'}_{\ \ \ \sigma} = \sum_{\mu' = 0}^{4} \left[ \Lambda^{\mu'}_{\ \ \ \rho} \eta_{\mu' \nu'} \Lambda^{\nu'}_{\ \ \ \sigma} \right] = \sum_{\nu'=0}^{4} \left[ \sum_{\mu' = 0}^{4} \left[ \Lambda^{\mu'}_{\ \ \ \rho} \eta_{\mu' \nu'} \Lambda^{\nu'}_{\ \ \ \sigma} \right] \right] = \\$$

$$ \left(\Lambda^{0}_{\ \ \ \rho}\eta_{0,0}\Lambda^{0}_{\ \ \ \sigma} + \Lambda^{1}_{\ \ \ \rho}\eta_{1,0}\Lambda^{0}_{\ \ \ \sigma} + ... \right) + \left(\Lambda^{0}_{\ \ \ \rho}\eta_{0,1}\Lambda^{1}_{\ \ \ \sigma} + \Lambda^{1}_{\ \ \ \rho}\eta_{1,1}\Lambda^{1}_{\ \ \ \sigma} + ... \right) + ... $$

Now I don't know how to evaluate $\Lambda^{\beta}_{\ \ \ \alpha} $ Does this refer to the element in row $\beta$ column $\alpha$ or in column $\alpha$ row $\beta$ assuming $\Lambda$ is a $n\times n$ matrix. And generally speaking if I encounter something like $$\Lambda^{i_{0,0} i_{0,1} ... }_{\ \ \ \ \ \ \ \ \ \ \ \ \ i_{1,0} i_{1,1} ...} $$

How do I figure out which element of $\Lambda$ I should be accessing/referring to?

Work attempted:

This person: Definition of the Lorentz group was reading the exact same page, but the OP and answeres did not clarify exactly what was meant by the tensor notation, if someone needs to convert back to matrix notation.

Generally if $A$ is a matrix, I usually take $A_{\alpha, \beta}$ to refer to row $\alpha$, column $\beta$.

  • $\begingroup$ $\Lambda^{\mu}_{\; \rho}$ is the Lorentz transformation - it's a linear mapping - it's not a tensor. The tensor is $\eta^{uv}$. There are no $2$ dimensional objects involved - they all live in a $4$ dimensional space. The $\eta^{uv}$ tensor has $2$ indices which makes it an order $2$ tensor - and has type $(0,2)$. $\endgroup$ Aug 28, 2019 at 7:10

1 Answer 1


You can’t (or at least shouldn’t) think of tensors with more than two indices as matrices. There is no need to think of tensors with two indices as matrices. But if you do, the usual convention is that the first index specifies the row and the second the column.

Tensors are more general and more important than matrices. They come in any integer rank, and they have transformation rules under transformations of the coordinate system. You can just think of the tensor component $T^2{}_{03}$ as the tensor component $T^2{}_{03}$, without trying to envision “where it is” in some $4\times 4\times 4$ “matrix”.


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