Clarification question about definition Lorentz covariance for an arbitrary higher rank tensor

Question

What does it mean for a tensor of rank N to be Lorentz covariant. For a scalar (rank 0) it simply means that it is Lorentz invariant, in other words it remains unchanged under Lorentz transformations. For a (four-)vector is means that its inner product ($v^\mu v_\mu$) is Lorentz invariant scalar. Does this apply to all higher rank tensors? In other words, is it both necessary and sufficient for any tensor's inner product to Lorentz invariant scalar in order for the tensor to be considered a Lorentz covariant?

Answer 1

Tensor as invariant mathematical objects. Tensors are invariant by their very nature/definition: their the mathematical objects meant to write invariant equations. They can be written as the linear combination of base vectors, and the coefficients are called components of the tensor w.r.t. that basis. For a tensor to be invariant, components and vectors of the basis follow inverse transformations.

Some details. If you write a tensor $\mathbb{A}$ of order $N$ using a basis $\mathscr{B} = \{ \mathbf{b}_n\}_{n=1:N}$ of a space $V$ as

$$\mathbb{A} = A^{n_1 \dots n_N} \mathbf{b}_{n_1} \otimes \dots \otimes \mathbf{b}_{n_N} \ ,$$

being $A^{n_1 \dots n_N} $ the components of tensor $\mathbb{A}$ w.r.t. the tensor base build with vectors of $\mathscr{B}$.

The vector of another basis $\mathscr{\tilde{B}} = \{ \tilde{\mathbf{b}}_m \}_{m=1:N}$ can be written as a linear combination of the vector of base $\mathscr{B}$ as

$$\tilde{\mathbf{b}}_m = T^{n}_m \mathbf{b}_n \ ,$$

being $T^{n}_{m}$ the matrix for the change of basis, collecting the coefficients of the linear combinations. In this new basis, the tensor reads

$$\mathbb{A} = \tilde{A}^{m_1 \dots m_N} \tilde{\mathbf{b}}_{m_1} \otimes \dots \otimes \tilde{\mathbf{b}}_{m_N} \ ,$$

and using the relation between the vectors of the bases,

$$\begin{aligned} \mathbb{A} & = \tilde{A}^{m_1 \dots m_N} \left( T^{n_1}_{m_1} \mathbf{b}_{n_1} \right) \otimes \dots \otimes \left( T^{n_N}_{m_N}\mathbf{b}_{n_N} \right) = \\ & = \underbrace{\tilde{A}^{m_1 \dots m_N} T^{n_1}_{m_1} \dots T^{n_N}_{m_N} }_{=A^{n_1 \dots n_N}} \mathbf{b}_{n_1} \otimes \dots \otimes \mathbf{b}_{n_N} \ ,\end{aligned}$$

it's possible to find the rule of transformation of the components, resulting from the comparison of the expressions of the same tensor in 2 different basis,

$$A^{n_1 \dots n_N} = \tilde{A}^{m_1 \dots m_N} T^{n_1}_{m_1} \dots T^{n_N}_{m_N} \ .$$

or

$$\tilde{A}^{m_1 \dots m_N} = A^{n_1 \dots n_N} \tilde{T}^{m_1}_{n_1} \dots \tilde{T}^{m_N}_{n_N} \ ,$$

being $\tilde{T}_i^k T_k^j = \delta_{i}^j$.