Unitarity of a representation in the physics literature (in particular in CFT) In many places in the physics literature, and specifically in conformal field theory, a representation of the Virasoro algebra is defined to be unitary when $L_n^\dagger = L_{-n}$.
For example in the case of string theory this name seems to be motivated by the fact that demanding the Hamiltonian to be hermitian implies $L_n^\dagger = L_{-n}$ (see for example Tong's lecture notes).
However, to say that the theory has a hermitian Hamiltonian and to say that the representations used are unitary are two different things. 
Is it somehow true that $L_n^\dagger = L_{-n}$ holds iff the representation of the Virasoro algebra is unitary in the usual sense? 
My guess:
a Lie algebra representation $\pi$ is unitary if $\pi(X)^\dagger = -\pi(X)$ for all $X$. In this case this means in particular that $L_n^\dagger = -L_n$, and I do not think that $-L_n = L_{-n}$ or anything like that! 
So when physicists talk about unitary representations of the Virasoro algebras (like in the famous GKO papers) they do not really mean unitary in the usual sense? 
 A: Lets remark on what it means for a representation of the Virasoro algebra to be unitary. There are mathematical propositions involved in this, but at the core its a definition coming from phyiscs.
The Virasoro algebra enters like this: You have a symmetry group being given by the group of conformal maps on a two-dimensional signature $(1,1)$ space-time like $S^{1,1}$ or $\Bbb R^{1,1}$. This group of symmetries then necessarily has an action on the physical space of states. This means that there is a projective representation on $\Bbb P(H)$, the projective space of some Hilbert space.
Now its a lot easier to deal with unitary representations onto $H$ than projective representations onto $\Bbb P(H)$, thankfully there is a mathematical theorem telling you that for any group $G$ you can lift a projective representation on $\Bbb P(H)$ to a unitary representation of a central extension of $G$ by $U(1)$ on $H$. Unitary means that the elements of the group act by unitary maps.
In our case the situation is like this:


*

*The group of conformal symmetries is a big infinite dimensional thing. You have $\mathrm{Conf}^0(S^{1,1})\cong \mathrm{Diffeo}_+(S^1)\times \mathrm{Diffeo}_+(S^1)$.

*The Lie algebra of $\mathrm{Diffeo}_+(S^1)$ is the space of vector fields on $S^1$, here given by the closure of the real span of the fields $\frac d{d\theta}, \cos(n\theta)\frac d{d\theta}, \sin(n\theta)\frac d{d\theta}$, $n\in\Bbb N$.

*Note that the complexification of the span of those vector fields is the Witt algebra ($L_n = (-i\cos(n\theta)-\sin(n\theta))\frac d{d\theta}$ and $L_{-n}=(-i\cos(n\theta)+\sin(n\theta)) \frac{d}{d\theta}$ ).

*The Lie algebra of a central extension of the group is a central extension of the original Lie algebra.

*Hence if you have a projective representation of $\mathrm{Conf}^0(S^{1,1})$ you have a representation of (two copies of) the above Lie algebra so that its elements exponentiate to something unitary, i.e. are anti-selfadjoint.

*Since the Hilbert space is complex, you may extend the Lie algebra and the representation to their complexifications.


Now one more ingredient enters: The complexification process and the process of central extension commute for Lie algebras. Hence we could complexify before doing the central extension and thus what we get is a complex representation of the central extension of the Witt algebra, which is also known as the Virasoro algebra.
Now we need to understand how the condition of unitary enters. We understood what it means in point 6. above. We had some Lie algebra $\mathrm{Vir}_R$ (whose complexification $\mathrm{Vir}_R^{\Bbb C}$ is the Virasoro algebra) so that $\mathrm{Vir}_R$ is mapped by the representation to anti-selfadjoint (or at least anti-symmetric) operators. Concretely we know that $L_n = -i\cos(n\theta) \frac{d}{d\theta} - \sin(n\theta)\frac{d}{d\theta}$, hence
$$\rho(L_n)^* = \rho(-i\cos(n\theta)\frac d{d\theta})^* -\rho(\sin(n\theta)\frac{d}{d\theta})^*= (\overline{-i}) \rho(\cos(n\theta)\frac d{d\theta})^* - \rho(\sin(n\theta)\frac d{d\theta})^* \\
= (i)(-\rho(\cos(n\theta)\frac d{d\theta}) + \rho(\sin(n\theta)\frac d{d\theta})= \rho(L_{-n}).$$
Additionally going through the process you should find that the central charge must be symmetric in the representation.

To summarise:
There is a real form $\mathrm{Vir}_R\subset \mathrm{Vir}$ so that $\rho(X)^*=-\rho(X)$ for all $X\in \mathrm{Vir}_R$, that is the Lie-algebra representation on $\mathrm{Vir}_R$ is unitary in the usual sense. Since $\mathrm{Vir}$ is the complexification of this algebra, the extension of the representation to $\mathrm{Vir}$ is not unitary in that sense, but still unitary in the sense that it is coming from a unitary representation of some group.
There are more interesting things to be said. For example while (the completion of) $\mathrm{Vir}_R$ is real Lie algebra of some real infinite dimensional group, there doesn't exist any complex group having $\mathrm{Vir}$ as its complex Lie algebra. Another interesting point I like a quite a bit is remark 3 on page 4 of these elementary notes on the subject.
