# Inconsistency in the normal ordered Virasoro algebra

OP is right that one can superficially demonstrate (via heuristic manipulations of infinite series) that the Virasoro generators (2) satisfy the Witt algebra (3). The best would be to regularize the infinite sum (2), but physics textbooks rarely bother to do so.

Since the Virasoro generators $L_n$ should have a well-defined action on the vacuum state, we should as a minimum allow for an

*infinite*constant $C$ in OP's eqs. (2) & (3): $$L_n ~=~ \frac{1}{2} \sum_{k \in\mathbb{Z}} \alpha_{n-k} \alpha_k ~+~ C\delta_n^0{\bf 1}, \tag{2'}$$ $$[L_n, L_m] ~=~ (n-m)(L_{n+m}-C\delta_{n+m}^0{\bf 1}).\tag{3'}$$It is not useful/well-defined to work with an

*infinite*constant $C$. In particular, we cannot reliably determine the central charge in the Virasoro algebra (3'). Therefore we need to order the annihilation and creation operators $\alpha_k$ in the Virasoro generator $L_n$ in a manner so that only*finitely*many of its terms act non-trivially on the vacuum state. Whichever such operator-ordering-prescription we choose, we can always rewrite it using the standard normal-order as $$L_n ~=~ \frac{1}{2} \sum_{k \in\mathbb{Z}} :\alpha_{n-k} \alpha_k: ~+~ C\delta_n^0{\bf 1} \tag{2"}$$ by including a*finite*constant $C$. With eq. (2") the genie is out of the bottle: A non-zero central charge is now inevitable.In practice, authors now redefine the Virasoro generator (2") so that the constant $C=0$ is zero, cf. Refs. 1-2.

References:

B. Zwiebach,

*A first course in String Theory,*2nd edition, 2009; section 12.4.M.B. Green, J.H. Schwarz and E. Witten,

*Superstring theory,*Vol. 1, 1986; eq. (2.2.14).

The resolution is that the expression $$ L_n=\tfrac{1}{2}\sum_{m=-\infty}^{+\infty} \alpha_{n-m}\alpha_{m} $$ actually suffers from a divergence! It's easy to see it fails to be defined on the usual string Fock space $\mathcal H$ (built atop a vacuum state that is annihilated by non-negative oscillators). Therefore, while you can indeed calculate $$ [L_m,L_n]=(m-n)L_{m+n} $$ (no central charge!) just using canonical commutation relations, you have not exhibited a Hilbert space where $\alpha_n$ and thus $L_n$ correspond to honest operators.