Is a Lorentz transformation a tensor?

Question

I am confused whether a Lorentz transformation is a tensor or not, since it is a linear transform. If yes how can I verify that?

Answer 1

When we say something is a tensor, we're saying that it's a linear operator whose components have certain transformation properties under a change of coordinates. The underlying assumption is that this is a quantity that is a function of the state of a system. You can walk up to the system, choose your coordinate system, and measure a particular component of the tensor in that coordinate system.

The Lorentz transformation isn't a function of the state of the system, so it's not meaningful to talk about measuring its components. The Lorentz transformation is independent of the state of the system but does depend on other data: the data defining the two local frames of reference that this Lorentz transformation operates between.

So the general idea is that we have (A) tensor observables, and (B) transformations that convert the components of a tensor from one coordinate system to another. The Lorentz transformation is in category B, not A.

Another way to see this is that it never makes sense to write the Lorentz transformation in abstract index form. When we apply a Lorentz transformation between coordinates $x^\mu$ and coordinates $x^{\mu'}$ in concrete index notation, we have $T^\mu=\Lambda^\mu{}_{\mu'}T^{\mu'}$. If you try to write that relationship in abstract index notation, you can't, because the $\mu$ and $\mu'$ are there to explicitly name the two coordinate systems.

Answer 2

I know what you're thinking: A Lorentz matrix $\Lambda^{\mu}{}_{\nu}$ looks pretty much like a (1,1) tensor $T^{\mu}{}_{\nu}$, right? No, wrong. Perhaps it is easiest to explain with an analogy: If you are familiar with category theory and know that a category consists of objects and arrows, then a tensor is like a category, a reference frame is like an object, while a Lorentz transformation is like an arrow, which transforms objects. In other words, you're comparing apples and oranges.

Answer 3

A tensor can be defined and written down in any given reference frame. The Lorentz transformation is a statement about how things change when you switch from one frame to another.

Answer 4

Coordinate transforms from first principles

A coordinate transformation maps one tensor-space $\mathcal T_1$ to another tensor-space $\mathcal T_2$. All $[m, n]$-tensors in the one space get mapped to $[m, n]$-tensors in the other one, by different (but related) linear transforms, such that all inner products that result in scalars produce the same scalar. For example, given a covector $u$ (a $[0,1]$-tensor, a linear function mapping vectors to scalars) and a vector $v$ (a $[1, 0]$-tensor) we have that the coordinate transformations on each of these spaces $C_{1,0}[\bullet]$ and $C_{0,1}[\bullet]$ are related by the expression that,$$C_{0, 1}[u](C_{1,0}[v]) = u(v),$$ since the last is a scalar.

If we insert a bunch of basis vectors $\hat e_i$ into this picture we can describe any $v$ as being a sum $v = \sum_i v^i \hat e_i$ and we can invent basis covectors $\underline e^j$ such that $$\underline e^j(\hat e_i) = \{1 \text{ if } i = j \text{ else } 0\}.$$We can then characterize a linear coordinate transform by what it does to these basis vectors and covectors, $C_{1,0}[v] = \sum_i v^i~C_{1,0}[\hat e_i]$. Note that the components do not really "transform" here, what transforms is the basis vectors.

Where we get a matrix is when we have some other basis for the second space—who says we have to use the “natural” basis induced by $C$ that $\hat a_i = C_{1,0}[\hat e_i]$? So if we have this other basis then we have that $$C_{1,0}[\hat e_i] = \sum_{k'} C^{~k'}_i~\hat a_{k'},$$ where we use primes to mentally keep the two different spaces separate in our heads. Now we find that we can interpret the coefficients as changing via $$C_{1,0}[v] = \sum_{k'} v^{k'}\hat a_{k'},\\~~~\text{ where }~~~v^{k'} = \sum_i C_i^{~k'}~v^i.$$

Where it’s easy to get tripped up

Now within the first space $\mathcal T_1$ itself, there is a $[1,1]$-tensor which can be formed by those coefficients, simply by saying "well those primed indices are just numbers and the transform matrix is just a bunch of numbers so I will write some tensor, $$c(v) = \sum_{i,k'} \hat e_{k'}~ C_i^{~k'}~\underline e^i(v).$$This is a very general point so I am going to make it very loudly: In general for anything we say is “not a tensor”—Lorentz transforms, Christoffel symbols—it is possible to take their components and assemble a tensor! But what we mean is that it is illusorily a tensor, it looks like a tensor but it does not interact the way you would expect a tensor to interact. For example when you are dealing with curved spaces and you start to talk about Christoffel symbols, you can assemble a tensor from the components that the Christoffel symbol gives you. But that tensor will not give the Christoffel symbol after a coordinate transformation!

There is something similar happening here. This tensor is a composition of part of your coordinate transform $C_{1,0}$ going from $\mathcal T_1 \to \mathcal T_2$ with some other coordinate transform $D: \mathcal T_2\to \mathcal T_1$ that maps $\hat a_{k'} \to \hat e_{k'}$. That is why it has this weird mixup of primed and unprimed indices that is $\hat e_{k'}$, because you have "collapsed" both spaces together with this second coordinate transform that transforms the $\mathcal T_2$ basis vectors into the $\mathcal T_1$ basis vectors.

The illusion is revealed by the following failure-to-interact-the-way-you-expect: in fact we said that $C$ is not just $C_{1,0}$ but is some family of functions $C_{m,n}$ mapping $[m,n]$-tensors in $\mathcal T_1$ to $[m, n]$-tensors in $\mathcal T_2$. By deriving that there is this one $c(v)$ tensor for $C_{1,0}$ you might have expected that this mapping of vectors is all that there is, but it is not: actually even under this second coordinate transform $D$ you have to model this coordinate transform $C$ as a family of tensors, a different one from each $[m, n]$-tensor space onto itself. Each one of these is its own $[m+n,m+n]$-tensor, and they are certainly related to this $[1, 1]$-tensor $c(v)$, but even for constant $m+n$ they are not the same. So there is a different $[1, 1]$-tensor mapping covectors to covectors, it is not $\tilde c(u) = v \mapsto u\big(c(v)\big)$ the way such a tensor would normally transform a covector.

In fact the resulting math is no more interesting than just keeping the two spaces separate in the first place and on that basis we just do what’s lazy and say “it’s not a tensor:” yes, you can imagine it as studying a family of tensors within the space, but that just makes your life much harder than saying “the un-primed indices belong to space one and the primed indices belong to space two.”