# Do generators of translations transform as *covariant* vectors under a homogeneous Lorentz transformation?

Using the composition law of Poincaré transformations, it is easy to see (cf. e.g. Ref. 1 this answer) that under a Lorentz transformation $$\underbrace{U(\Lambda,0)P^\mu U(\Lambda,0)^{-1}}_{P'^{\mu}}=\Lambda_\nu{}^{\mu}P^\nu\tag{1}\label{1}.$$ Evidently, \eqref{1} is behaviour of a covariant vector$$^1$$ as it's transforming with the inverse Lorentz matrix $$\Lambda_\nu{}^{\mu}=(\Lambda^{-1})^\mu{}_{\nu}$$. On the other hand, the upper notation would at least suggest to expect a contravariant behaviour. Although is not to expect anything from notation alone, can we conclude this is a misleading notation? Is there anything wrong with what I said?

References:

1. The Quantum Theory Of Fields. Vol. I, Foundations, S. Weinberg. Chapter 2, sec. 2.3-2.4.

$$^1$$ I disagree with the comments below the answer linked above, which state that we obviously find the inverse matrix considering the inverse Lorentz transformation. \eqref{1} is the transformation we should consider.

• The index placement in the RHS of equation (1) is wrong. It should read $\Lambda^{\mu}_{~~\nu}$. Feb 28 at 17:54
• @DanielC That's exactly the problem of the question. That is equation $(2.4.9)$ in Weinberg. Feb 28 at 17:56
• Mr. Feynman I do not have the book handy right now. If this is the case, it must be a mistake. Feb 28 at 17:59

There's a bit of confusion here around what it means for the generators of translation to "transform as a vector". Looking at a particular unitary representation - as the $$U(\Lambda,0)$$ notation suggests - isn't really the right thing to look at.

Abstractly, we have the Poincaré group $$P$$ and its Lie algebra $$\mathfrak{p}$$, and there is the natural adjoint representation of $$P$$ on $$\mathfrak{p}$$. Since the Lorentz group $$\mathrm{SO}(1,3)$$ is a subgroup of $$P$$, this induces a representation of the Lorentz group on $$\mathfrak{p}$$. As representations of the Lorentz group, this decomposes as $$\mathfrak{so}(1,3)\oplus\mathfrak{t}$$, where $$\mathfrak{t}\cong \mathbb{R}^4$$ is the algebra of the generators of translation.

Now, from the definition of the Poincaré group as the semi-direct product $$\mathrm{SO}(1,3)\rtimes \mathbb{R}^4$$ with the multiplication $$(\Lambda_1,a)\cdot (\Lambda_2,b) = (\Lambda_1\Lambda_2, a + \Lambda_1 b)$$ it follows that the Lie bracket on $$\mathfrak{so}(1,3)\oplus\mathfrak{t}$$ is $$[(A,a), (B,b)] = ([A,B], Ab - Ba).$$ where $$Ba$$ is the fundamental action of a matrix $$\mathfrak{so}(1,3)$$ on a vector in $$\mathbb{R}^4$$. One way to derive this is to explicitly model the Poincaré group as the subgroup of $$\mathrm{GL}(\mathbb{R}^5)$$ of the form $$\begin{pmatrix} \Lambda & a \\ 0 & 1 \end{pmatrix}, \Lambda\in\mathrm{SO}(1,3), a\in\mathbb{R}^4$$ whose Lie algebra is the matrices of the form $$\begin{pmatrix} A & a \\ 0 & 0 \end{pmatrix}, A\in\mathfrak{so}(1,3), a \in\mathbb{R}^4.$$

Anyway, the adjoint representation of $$\mathfrak{so}(1,3)$$ on $$\mathfrak{t}$$ is therefore given by $$\mathrm{ad}_A(b) = [(A,0),(0,b)] = Ab,$$ for any $$b\in\mathfrak{t}$$ and so, indeed, the algebra of translation transforms as a vector under $$\mathfrak{so}(1,3)$$ in the sense that it transforms in the fundamental representation.

This is the unambiguous and representation-independent way of saying what we mean by "the generators of translation transform as a 4-vector under Lorentz transformations". So far we haven't used an index anywhere.

When you have a representation $$V_\rho$$, then $$V_\rho\otimes V_\rho$$ defined by the representation map $$g\mapsto \rho(g)\otimes \rho(g)$$ can be a different representation from $$V_\rho \otimes V^\ast_\rho$$ where $$V^\ast_\rho$$ is the dual representation and the representation is $$g\mapsto \rho(g)\otimes \rho(g^{-1})^T$$ even if $$V^\ast_\rho \cong V_\rho$$ and so the distinction matters. When you only have $$V_\rho$$ alone, it is isomorphic to $$V^\ast_\rho$$ for all representations of the Lorentz group and there isn't really any meaning to asking whether the representation alone is "covariant" or "contravariant".
And unfortunately the meaning of the index position in non-isolated contexts is ambiguous because we tend to use indices for both "objects" and components: Consider expanding a tangent vector as $$v = v^\mu \partial_\mu$$. We say the components $$v^\mu$$ "transform like a vector" but mathematically it's the $$\partial_\mu$$ who are the vectors (they're a basis of our vector space after all) and the $$v^\mu$$ are just numbers! So does the $$\partial_\mu$$ "transform like a vector" or "transform like a co-vector"? How do you distinguish this from the components $$A_\mu$$ of a covector $$A = A_\mu\mathrm{d}x^\mu$$?