# Can you write a Lorentz transformation as the tensor product of a 4-vector with a 4-covector?

I was wondering if there is a way to re-write a general Lorentz transformation

in a Lorentz covariant way. Since one index is up while the other is down, I assumed that it could be written as the tensor product of a 4-vector with a 4-covector. But the only covariant quantities I can think of using are the identity tensor, the 4-velocity and perhaps the levi-civita thensor?

Maybe Lorentz transformations are not tensors. The trace of the matrix is equal to $$2(1+\gamma)$$ which does not look invariant.

• hyperphysics.phy-astr.gsu.edu/hbase/Relativ/vec4.html#c2 Commented Sep 27, 2023 at 13:36
• Lorentz transformations transform coordinates from a frame into a primed frame and are often denoted as ${\Lambda_{\mu'}}^\nu\,.$ In contrast, components of a tensor are meaningful only in the same frame: $T_{\mu\nu}$ or $T_{\mu'\nu'}\,.$ Therefore I am not quite sure why that matrix above should be a tensor, and even if it were one, why this is physically meaningful. Commented Sep 27, 2023 at 13:56
• Note that a Lorentz Transformation has to be invertible. Commented Sep 27, 2023 at 14:11
• Quantities with indices are not always tensors. Commented Sep 27, 2023 at 16:51
• Since one index is up while the other is down You didn’t show any indices. Commented Sep 27, 2023 at 16:52

Some details. Lorentz transformation allows you to find the rule of transformation of the coordinates of a vector $$\mathbf{v}$$ as seen by two inertial observers $$O$$ and $$O'$$ in their own bases, i.e. $$$$\mathbf{v} = v^i \mathbf{b}_i = v'^j \mathbf{b}'_j \ ,$$$$ providing the rule of transformation $$$$v'^j = T^j_i v^i \ ,$$$$ so that $$$$\mathbf{v} = v^i \mathbf{b}_i = v^i T^j_i \mathbf{b}'_j \qquad \rightarrow \qquad \mathbf{b}_i = T^j_i \mathbf{b}'_j \ .$$$$
If you want to find the relation between the covariant coordinates as seen by the two observer, you first need to use the metric tensor $$g \hspace{-4pt}g$$ to transform from contravariant to covariant components $$$$v^i = g^{ik} v_k \qquad , \qquad v'^j = g'^{il} v'_l \ ,$$$$ and insert in the relation above $$$$g'^{jl}v'_l = T^j_i g^{ik} v_k \qquad \rightarrow \qquad v'_l = \underbrace{g'_{jl} T^j_i g^{ik}}_{={T^{cov \ }}^k_l} v_k \ ,$$$$ being the components of the metric tensor (special relativity, Minkowski metric) equal to $$diag(-1,1,1,1)$$ or $$diag(1,-1,-1,-1)$$ depending on the metric signature you're using.