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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?

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A representation of a group $G$ is a pair $(\varphi,V)$ where $V$ is a vector space and $\varphi:G\to\mathrm{GL}(V)$ is a group homomorphism between the group $G$ and the set of invertible linear transformations over $V$. Sometimes authors call $\varphi$ the representations and sometimes authors call $V$ the representation.

You can say that the first way is fine since $\varphi$ carries information about the space it acts on. On the other hand the second one is, as you point out, an abuse of terminology. However, often (not always!) once $V$ is fixed, there is only one choice for $\varphi$, so it's fine.

For tensors, $V$ is the tensor product of $\mathbb{R}^n$ $\mathrm{rank}$ times and $\varphi$ is the unique homomorphism that makes $G$ acts on tensors in the usual way that we know.

For the Hamiltonian example you presented, $V$ is the eigenspace of energy $E$ and $\varphi$ is some unitary operator restricted to $V$, $U|_V$, that acts on the states and commutes with the Hamiltonian. Precisely because it commutes with it, we never go out of the eigenspace, so the operator $U$ can be restricted to $\mathrm{GL}(V)$.

By the way, you are using Schur's lemma backwards. It says that if $[H,U|_V]=0$ and the representation is irreducible, then $H \propto \mathbb{1}$ (which is true). Not the other way around. That means that $(U|_V,V)$ can still be a reducible representation if there are invariant subspaces. For example, suppose your Hamiltonian has degenerate energies for states with the same total spin $J = S+L$. Then if we look at energy eigenspaces and only care about spin $S$, and not $J$, we may find a direct sum of spin representations there.

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  • $\begingroup$ So it is in both cases the same situation, I just idenitfy the representation homomorphism with the space on which the matrices of the general linear group acts. Thanks a lot for the answer and for the precisation on Schurl's lemma. $\endgroup$
    – Ratman
    Apr 24, 2020 at 13:25
  • $\begingroup$ I wouldn't say identify. It's more a synecdoche: you refer to something by referring to a part of it instead of the whole. $\endgroup$
    – MannyC
    Apr 24, 2020 at 13:29
  • $\begingroup$ Perfectly clear, thanks a lot $\endgroup$
    – Ratman
    Apr 24, 2020 at 13:32

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