Ordering of Indices in $\Lambda^\mu_{\space\space\nu}$

Question

I am having some questions on the ordering of indices that are both upstairs and downstairs. Let's take an example: $\Lambda^\mu_{\space\space\nu}$ is a Lorentz transfom if the following equation is satisfied: $$ \Lambda^\mu_{\space\space\sigma} \eta^{\sigma \tau}\Lambda^\nu_{\space\space\tau}=\eta^{\mu \nu}. $$ In matrix notation that means $$ \Lambda \eta^{-1}\Lambda^T =\eta^{-1}. $$

My question is: why must we place $\mu$ before $\nu$ in the expression $\Lambda^\mu_{\space\space\nu}$? (rather than just vertically above it)

I have thought about this for a moment, and get the following ideas:

1. Putting $\mu$ before $\nu$ reminds us to write the notation is the usual order of matrix multiplication. Usually we write $\Lambda^\mu_{\space\space\nu} x^\nu$ rather than $ x^\nu\Lambda^\mu_{\space\space\nu}$, because we'd like $\nu$'s to be "closer together". This matches our ordering of writing a matrix multiplying a (contravariant) vector $\Lambda \mathbf x$.
2. There are exceptions to point 1, for example $\Lambda^\mu_{\space\space\sigma} \eta^{\sigma \tau}\Lambda^\nu_{\space\space\tau}$, because here we are transposing the second Lorentz matrix.
3. However, if we have more than two indices, the above ideas make little sense. If we have and expression like $A^{\mu_1\mu_2 \ldots \mu_k}_{\nu_1 \nu_2 \ldots \nu_l} x^{\nu_1}\ldots x^{\nu_n}y_{\mu_1}\ldots y_{\mu_n}$, who knows what is the "correct" order of indices of $a$ and $x,y$? Mathematically there doesn't seem to be a reason for a particular ordering, because a tensor product of vector spaces does not depend on the order (up to isomorphism) in which we take the product.

Are the above observations correct? Are there any other reasons for the ordering?

Finally, will we ever see something like $$ \Lambda^{\space\space\mu}_{\sigma}? $$ i.e. downstairs before upstairs.

This post imported from StackExchange Physics at 2025-01-23 14:38 (UTC), posted by SE-user Ma Joad

Comments

Besides what the answers say, keep also in mind that $\Lambda_{\mu}^{\nu}$ is not a tensor, so the order of its indices doesn't denote the order of a tensor product. (Strictly speaking it's a two-point tensor, a notion that appears mainly in older texts and isn't used very much today.)

This post imported from StackExchange Physics at 2025-01-23 14:38 (UTC), posted by SE-user pglpm

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@pglpm Yes. I know that it is not a tensor. But similar things happen to tensors as well. I am looking for a reason why we want $\mu$ to be after $\nu$.

This post imported from StackExchange Physics at 2025-01-23 14:38 (UTC), posted by SE-user Ma Joad

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The full story is that $\Lambda$ is a so-called two-point tensor, a tensor that "lives" in two different manifolds. Equivalently it's a tangent map from the tangent vectors of $\mathbf{R}^4$ to the tangent vectors of $\mathbf{R}^4$. So the order of its indices reflect the usual operator notation $\Lambda\pmb{v}=\pmb{w}$, where $\Lambda$ operates (on vectors) on the right. The two copies of $\mathbf{R}^4$ are the coordinate manifolds: when we choose a system of coordinates we're simply choosing a map $M\to \mathbf{R}^4$ between manifolds, from spacetime to $\mathbf{R}^4$.

This post imported from StackExchange Physics at 2025-01-23 14:38 (UTC), posted by SE-user pglpm

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Does this answer your question? Index notation Lorentz transfromation matrix

This post imported from StackExchange Physics at 2025-01-23 14:38 (UTC), posted by SE-user user87745

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Also related: physics.stackexchange.com/q/255933

This post imported from StackExchange Physics at 2025-01-23 14:38 (UTC), posted by SE-user user87745

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It can be argued that the index positioning of $\Lambda^\mu{}_\nu$ is a convention (one index is always up and the other down but it doesn't matter which way around it is). Once this convention is set, we have $\Lambda_\mu{}^\nu = (\Lambda^{-1})^\nu{}_\mu$. This follows from the defining equation $\Lambda^T \eta \Lambda = \eta$.

This post imported from StackExchange Physics at 2025-01-23 14:38 (UTC), posted by SE-user Prahar

Answer 1

If you have $A^{\mu_1 \mu_2 \mu_3}$ you can think of it as a 3 Dimensional matrix, so you add a dimension to the idea $A^{\mu_1 \mu_2}$ as a matrix. You can imagine a new set of rows that go "inside" the page. You can understand how the order is important because the first index $\mu_1$ is labelling the "standard" rows, the second the columns and the third $\mu_3$ is labelling the "inside the page" row. Then if you exchange one of the indices you are picking a different element of the 3D matrix. And this idea can be extended to higher dimensions.

This post imported from StackExchange Physics at 2025-01-23 14:38 (UTC), posted by SE-user TheoPhy

Comments

And $ \Lambda_{a}{}^{b} $ is the transpose of $\Lambda^{b}{}_{a}$, obtained exchanging row and column indeces.

This post imported from StackExchange Physics at 2025-01-23 14:38 (UTC), posted by SE-user TheoPhy

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Well, of course in the case all indices are upstairs, the order matters. What I am asking about is the order in $\Lambda$ in your comment. Could you explain it a little bit more? Thanks.

This post imported from StackExchange Physics at 2025-01-23 14:38 (UTC), posted by SE-user Ma Joad

Answer 2

The simple answer is that we don't need to assign an order to the indices in ${\Lambda^\mu}_\nu$ to do calculations but it is necessary if we want to view them as matrices. I think I speak for a lot of people when I say that matrix notation is slightly easier to read/write down. But it might not always be clear how to translate the two and sometimes it's just not possible. Take for example the inner product which you can write as $$u\cdot v=u_\mu v^\mu=\mathbf u^T\mathbf v=\begin{pmatrix}u_1&u_2&u_3\end{pmatrix}\begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix}.$$ From this example you might argue that upper indices are associated with column vectors and lower indices with row vectors. You might be familiar from this from quantum mechanics. You have kets which are vectors and bras which eat vectors and they are each represented by column vectors or row vectors respectively. Let's take another example that reinforces this idea. $$(A\mathbf v)^i={A^i}_jv^j=\begin{pmatrix}A_{11}&A_{12}\\A_{21}&A_{22}\end{pmatrix}\begin{pmatrix}v_1\\v_2\end{pmatrix}$$ Again upper indices are associated with 'column-ness' and lower indices are associated 'rowness'. The matrix $A$ eats a vector (lower index $j$) and outputs another vector (upper index $i$). Now a counter example. What about $x^\mu g_{\mu\nu}y^\nu$? In this case $g$ has two lower indices. It eats two vectors. But how do we represent something that eats two vectors? There is a hack you can do. You can represent it as $$x^\mu g_{\mu\nu}y^\nu=\begin{pmatrix}x_1&x_2\end{pmatrix}\begin{pmatrix}g_{11}&g_{12}\\g_{21}&g_{22}\end{pmatrix}\begin{pmatrix}y_1\\y_2\end{pmatrix}$$ Note that it doesn't do justice to the nature of $g$. It is fundamentally something which eats two vectors but it is represented as something which eats one vector and spits out another. This is possible because linear functionals (things which eat a vector and spit out a vector) are dual to vectors. They can be changed into one another in an intuitive way.

So this is where I invite you to let loose a little of the idea of expressions like $g_{\mu\nu}$ 'being' matrices. Sometimes expressions in index notation can be expressed as matrices and vectors which is nice. It makes it easier to see what you are doing. But generally they are not equal to those matrices. Whenever you convert between the two you only have to make sure they are consistent. You have to make sure you sum over the right indices and get the right answer. When you are able to write an expression in the form $$A_{ij}B_{jk}v_k$$ where each of these indices could be upper or lower then you can safely write it as matrix multiplication. Like you mentioned we only need the summed over indices to be close together.

So how do you represent something like ${A^{\mu_1,\dots\mu_m}}_{\nu_1\dots\nu_n}x^{\nu_1}\dots x^{\nu_n}y_{\mu_1}\dots y_{\mu_m}$ as matrix multiplication? I wouldn't know!

This post imported from StackExchange Physics at 2025-01-23 14:38 (UTC), posted by SE-user AccidentalTaylorExpansion