# Is time reversal operator not a representation of Lorentz group?

I'm puzzled why every book says that time reversal operator is a representation of full Lorentz group. Because of physical consideration, time reversal is an antilinear operator. While the definition of group representation requires linear representation, time reversal operator can't be a representation of full Lorentz group.

So my question includes :

1) Can every antilinear operator be written in form of matrix?

2) What's the group representation of full Lorentz group $O(1,3)$ ? I only see some textbooks including the group representation of $SO^\uparrow(1,3)$. Is time reversal operator on complexified Lorentz Lie algrebra in QFT not a representation of full Lorentz group?

Another, equivalent, viewpoint is the following. States are density matrices, i.e. positive (thus self-adjoint), trace class operator with unit trace on the Hilbert space of the system. Pure states are extremal points in the space of that states (they, in fact, coincide to the rays of the other picture). The only possible compositions of states, in this picture, are real convex combinations: $$a \rho + b \rho'\quad a,b>0 \quad a+b=1\:.\tag{1}$$ The action of a symmetry has alway the form, where $U$ is the operator unitary or antiunitary prescribed by Wigner, $$\rho \mapsto U\rho U^\dagger$$ so that there is no way to distinguish the unitary case from the antiunitary one using the convex linear structure (1), as $a$ and $b$ are real. $$U(a \rho + b \rho') U^\dagger = a U\rho U^\dagger + b U\rho' U^\dagger\:.\tag{2}$$ Notice that if complex numbers $a,b$ were admitted, $U$ antiunitary would produce, in place of (2), $$U(a \rho + b \rho') U^\dagger = \overline{a} U\rho U^\dagger + \overline{b} U\rho' U^\dagger\:.\tag{3}$$
As a last, important, comment I stress that continuous symmetries (continuous projective representations of a continuos or Lie topological group) are always represented (up to phases) by unitary operators at least when representing the largest connected subgroup containing the identity of the group. Antiunitary operators may enter the representation when we consider other parts of the group which cannot be continuously conntected to the identity. This is the case for the time reversal with respect to the full Lorentz group, or the parity reflection which, instead, is usually unitarily represented. The representation of $O(1,3)$ in the Hilbert space of the theory is a mixed representation, generated by the unitary rep of $SO\uparrow(1,3)$, the unitary operator $P$ representing spatial reflection, the antiunitary operator $T$ representing the time reversal operation, and the mixed antiunitary operation $TP$. Each such operators $P,T,TP$ when composed with $SO\uparrow(1,3)$ produces a representation of a corresponding connected component of $O(1,3)$ which does not contain the identity.
Regarding your technical question, if an antiunitary operator can be represented by a matrix, the answer is the following. Fix a Hilbertian basis $\{\psi_i\}$ in the Hilbert space $\cal H$ and define the standard complex conjugation of components, $$C : \sum_i c_i \psi_i \mapsto \sum_i \overline{c_i} \psi_i\:.$$ $C$ is clearly antiunitary and depends on the fixed basis. For every antiunitary operator $U: \cal H \to \cal H$, there exists a unique unitary operator $V_U :\cal H \to \cal H$ such that $U= CV_U$. The proof is trivial. Obviously, $V_U$ admits a matrix representation $(V_U)_{ij} = \langle \psi_i| V_U\psi_j\rangle$. But one has to keep in his/her mind that also the complex conjugation $C$ acts after the action of the matrix. The said decomposition depends on the chosen basis.