# Is Lorentz transform a tensor?

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.

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.