This could be extremely trivial but i need to be sure I am not wrong. I am encountering many times statements where the author says
"Tensors are examples of representations for the Lorentz group". (A modern introduction to quantum field theory, Michele Maggiore. Page 20).
It is clear to me that if I consider for example $SO(3)$, or $SO(3,1)$ I can think to act its transformations on scalars, vectors or event tensors and find the related representations. So the previous sentence isn't an abuse of notation? I think I should say a tensorial representation of the Lorentz group is a representation that acts on tensors, is this correct?
I found a similar problem in a more physical context. With the statement
"$d$ degenerate eigenstates furnish a d-dimensional irreducible representation for the group $G$". (Group theory in a nutshelli for physicists, A. Zee. Page 163).
Here is the context. Given a hamiltonian $H$ and its symmetry group $G$, then $H$ is invariant under the transoformations $T$ of $G$ $$H=T^{-1}HT\rightarrow HT=TH$$ considering the eigenvalue problem $$H\psi=E\psi$$ $$H(T\psi)=TH\psi=E(T\psi)$$ so we can see that $T$ doesn't mix states of differents eigenspaces, and the eigenspaces are like invariant subspaces of the Hilbert space from the point of view of $T$. Using a suitable basis for the Hilbert space I can show $T$ has a block diagonal form and hence is reducible. But considering $H$ just on a specific eigenspace $V_E$ then the hamiltonian has the form $H=EI$ and following from Schur's lemma I can say that the specific block associated to $V_E$ of the reducible representation of T is an irreducible representation. Is this what it is meant with the previous statment? That a correct choiche of the basis $\psi$ in the eigenspace showes me what is the irreducible representation?