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.

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    $\begingroup$ 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.) $\endgroup$ – pglpm Dec 7 '20 at 12:48
  • $\begingroup$ @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 $\nu$. $\endgroup$ – Ma Joad Dec 7 '20 at 13:09
  • $\begingroup$ 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 $\mathbf{R}^4$. $\endgroup$ – pglpm Dec 7 '20 at 13:20
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    $\begingroup$ Does this answer your question? Index notation Lorentz transfromation matrix $\endgroup$ – Dvij D.C. Dec 7 '20 at 13:37
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    $\begingroup$ Also related: physics.stackexchange.com/q/255933 $\endgroup$ – Dvij D.C. Dec 7 '20 at 13:37

Here's a fuller picture. Step by step:

enter image description here

A coordinate system $x$ can be seen as a manifold map from spacetime $M$ to $\mathbf{R}^4$. That is, $$x \colon M \to \mathbf{R}^4\ ,$$ so that $\bigl(x^0(P), \dotsc, x^3(P)\bigr)$ are the coordinates of the manifold point (event) $P$.

When we have two different coordinate systems $x$ and $y$, we consider the map from one copy of $\mathbf{R}^4$ to the other, going $\mathbf{R}^4\xrightarrow{y^{-1}}M\xrightarrow{x}\mathbf{R}^4$: $$x\circ y^{-1} \colon \mathbf{R}^4 \to \mathbf{R}^4 \ ,$$ that's the change of coordinates.

A coordinate system $x$ also has an associated tangent map $$x_P' \colon \mathrm{T}_PM \to \mathrm{T}_{x(P)}\mathbf{R}^4 \equiv \mathbf{R}^4 \ ,$$ where the last equivalence is a canonical isomorphism. This is the map through which we represent a tangent vector of $M$ as a quadruple of real numbers.

Also the coordinate-change map has an associated tangent map: $$(x \circ y^{-1})_{y(P)}' \colon \mathrm{T}_{y(P)}\mathbf{R}^4 \to \mathrm{T}_{x(P)}\mathbf{R}^4 \ ,$$ which gives the quadruple of real numbers associated with $y_P'$ to that associated with $x_P'$. And this is what $\Lambda$ actually is: it takes the components of a tangent vector in one coordinate system and yields the components in the other: $\Lambda_{y(P)} := (x \circ y^{-1})_{y(P)}'$.

This map can also be considered a so-called "two-point tensor": an object that belongs to the tensor product of the tangent space at a point of a manifold with the tangent space at a point of a different manifold, or at a different point of the same manifold. (A curiosity: two-point tensors were for example considered by Einstein in his teleparallel formulation of general relativity.)

Since this tangent map maps a vector $\pmb{u}$ (in $\mathrm{T}_{y(P)}\mathbf{R}^4$) to another vector $\pmb{v}$ (in $\mathrm{T}_{x(P)}\mathbf{R}^4$), we can write its operation with the usual "action on the right" notation: $$\pmb{v} = \Lambda\pmb{u}$$ typical of linear algebra (and linear algebra is just what we're doing!). Interpreted as tensor contraction, we're contracting with $\Lambda$'s tensor slot on its right side.

This is the reason why traditionally the lower index (which contracts with vectors) is on the right.

This is just to give you the full picture and the reason why, but you don't need to worry too much about it. If you're curious about two-point tensors and more about this, check for example

And for tangent maps, coordinate systems, and so on, an excellent reference is always

Additional note on raising or lowering the indices of $\Lambda$

$\Lambda\colon \mathrm{T}_{y(P)}\mathbf{R}^4 \to \mathrm{T}_{x(P)}\mathbf{R}^4$ is just a non-singular linear map between two vector spaces. So it induces an inverse map $$\Lambda^{-1}\colon \mathrm{T}_{x(P)}\mathbf{R}^4 \to \mathrm{T}_{y(P)}\mathbf{R}^4$$ and also a dual map (transpose) $$\Lambda^{\intercal} \colon \mathrm{T}^*_{x(P)}\mathbf{R}^{4} \to \mathrm{T}^*_{y(P)}\mathbf{R}^{4}$$ from the dual of the initial target, to the dual of the initial domain. And so on.

By using the tangent maps $x'$ and $y'$ (and their duals) we can also map more general tensorial objects on $\mathrm{T}_PM$ to objects on $\mathrm{T}_{x(p)}\mathbf{R}^4$ and $\mathrm{T}_{y(p)}\mathbf{R}^4$ – the latter will be the coordinate representatives of those on $\mathrm{T}_PM$. This is also true for the metric tensor or its inverse on $M$. We have one coordinate proxy of it on $\mathrm{T}_{x(p)}\mathbf{R}^4$ (more precisely on $\mathrm{T}^*_{x(p)}\mathbf{R}^{4}\otimes\mathrm{T}^*_{x(p)}\mathbf{R}^{4}$) and another one on $\mathrm{T}_{y(p)}\mathbf{R}^4$.

The two-point tensor $\Lambda$ has one covariant leg (that's really the technical term) on $\mathrm{T}_{y(p)}\mathbf{R}^4$, since it must contract contravariant vectors there, and a contravariant leg on $\mathrm{T}_{y(p)}\mathbf{R}^4$, since it must "deposit" a contravariant vector there.

We can change the variance type of each leg. For example we can make the leg on $y(P)$ contravariant, by contracting it with the metric proxy that we made on $\mathrm{T}_{y(p)}\mathbf{R}^4$. The result is a new two-point tensor or linear map, which maps covectors in $\mathrm{T}^*_{y(p)}\mathbf{R}^{4}$ to vectors in $\mathrm{T}_{x(p)}\mathbf{R}^{4}$. This is a sort of mixed operation: we're taking a covector in the coordinate system $y$, contracting it with the inverse metric tensor, and giving the resulting vector in the new coordinate system $x$ (I personally think it's best not to mix these two different kinds of operations).

If we make the leg on $y(P)$ contravariant and the leg on $x(P)$ covariant using the proxy inverse metric tensor on $y(P)$ and the metric tensor on $x(P)$, then the result is $\Lambda^{-\intercal}$, the inverse of the transpose of $\Lambda$. But we could have used any other non-singular bilinear form instead of the metric tensor to perform this operation. What it does, indeed, is to take a covector in the coordinate system $y$, transform it into a vector by means of some transformation, change its coordinate representation to the system $y$, and finally transform it back to a covector using the inverse of the initial transformation (whatever it was).

  • $\begingroup$ If we define $\Lambda := (x \circ y^{-1})'$, how are ${\Lambda_i}^j$, $\Lambda_{ij}$, ${\Lambda^i}_j$ and $\Lambda^{ij}$ defined? As you said, we can identify $T_a\mathbb R^4$ with $\mathbb R^4$ for all $a\in\mathbb R^4$ , so my guess would be $\Lambda^i{}_j:=e^i\Lambda e_j$ (where $(e_1,\ldots,e_4)$ is just the standard basis of $\mathbb R^4$), but I don't know about the rest. $\endgroup$ – Filippo Dec 7 '20 at 14:32
  • $\begingroup$ If $\Lambda$ was a $2$-Form, it would be clear, $\Lambda_{ij}=\Lambda(v_i,v_j)$, $\Lambda_i{}^j=\Lambda(v_i,g^{-1}v^j)$ etc. (where $g$ is the metric tensor), but in this case I don't know... $\endgroup$ – Filippo Dec 7 '20 at 14:37
  • $\begingroup$ @Filippo The right index of $\Lambda$ is raised using the inverse metric tensor on $\mathrm{T}\mathbf{R}^4$ mapped by the coordinate system $y$ from the metric tensor in $M$. The left index is lowered with the metric tensor on $\mathrm{T}\mathbf{R}^4$ mapped by the other coordinate system $x$. $\endgroup$ – pglpm Dec 7 '20 at 14:43
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    $\begingroup$ Thank you, I'm excited to read the explanation :) $\endgroup$ – Filippo Dec 7 '20 at 18:32
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    $\begingroup$ Is $\Lambda^{\intercal} \colon \mathrm{T}^*_{x(P)}\mathbf{R}^{4} \to \mathrm{T}^*_{y(P)}\mathbf{R}^{4}$ defined by $\Lambda^{\intercal}v:=v\circ\Lambda\in \mathrm{T}^*_{y(P)}\mathbf{R}^{4}$ for all $v\in\mathrm{T}^*_{x(P)}\mathbf{R}^{4}$? $\endgroup$ – Filippo Dec 7 '20 at 20:27

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!

  • $\begingroup$ So, the order of indices in $\Lambda^{\mu}_{\space\space \nu}$ does not matter (as long as $\mu$ is up and $\nu$ is below it), and they are in this way ($\mu$ left, $\nu$ right) just for the sake of "keeping indices close together"? $\endgroup$ – Ma Joad Dec 7 '20 at 13:12
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    $\begingroup$ @MaJoad I actually had to think about this because I have seen both $\Lambda^\mu_\nu$ and ${\Lambda^\mu}_\nu$. But the order does matter. If you transpose a rotation matrix (which is one example of a Lorentz transformation) it gets a minus sign for the off-diagonals. This answer physics.stackexchange.com/a/144475/93729 makes it slightly more clear. From what I've seen only the sloppy authors write it as $\Lambda^\mu_\nu$. $\endgroup$ – AccidentalTaylorExpansion Dec 7 '20 at 13:48
  • $\begingroup$ I'll edit my answer to remove this confusion. $\endgroup$ – AccidentalTaylorExpansion Dec 7 '20 at 13:50

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.

  • $\begingroup$ And $ \Lambda_{a}{}^{b} $ is the transpose of $\Lambda^{b}{}_{a}$, obtained exchanging row and column indeces. $\endgroup$ – TheoPhy Dec 7 '20 at 12:04
  • $\begingroup$ 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. $\endgroup$ – Ma Joad Dec 7 '20 at 13:08

$\Lambda$ is just a matrix, not a tensor. The index on the left denotes the row and the index on the right denotes the column. Positioning one index higher than the other is simply practical for using Einstein summation. There's not a deeper meaning like in the case of tensors.

To answer your last question: \begin{equation} {\Lambda_j}^i:={\left(\Lambda^{T}\right)^j}_i={\Lambda^i}_j \end{equation}

  • $\begingroup$ But why we need to order the indices in such a way? (Note: I am NOT asking about why the index should be higher or lower; I am asking about whether it should be on the left or right) $\endgroup$ – Ma Joad Dec 7 '20 at 13:00
  • $\begingroup$ @MaJoad It's just a convention. If everyone wanted to we could introduce the convention that the index denoting the row is written above the index denoting the column. $\endgroup$ – Filippo Dec 7 '20 at 13:04
  • $\begingroup$ @MaJoad As Phranceso pointed out, some people define ${\Lambda_j}^i:={\left(\Lambda^{T}\right)^j}_i={\Lambda^i}_j$. $\endgroup$ – Filippo Dec 7 '20 at 13:08
  • $\begingroup$ There actually is a deeper meaning, but it isn't usually stressed because it isn't so important for the usual calculations. $\Lambda$ is a so-called two-point tensor living in two different copies of $\mathbf{R}^4$, or equivalently a tangent map from $\mathrm{T}\mathbf{R}^4$ to $\mathrm{T}\mathbf{R}^4$. $\endgroup$ – pglpm Dec 7 '20 at 13:14
  • $\begingroup$ @Filippo But - that's nothing different from $\Lambda^{j}_{\space\space i}$ - just like the first equation I write in the question. That also means the transpose, isn't it? $\endgroup$ – Ma Joad Dec 7 '20 at 13:18

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