# Inconsistency in the normal ordered Virasoro algebra

I seem to have found a basic contradiction when it comes to the commutation relations of the Virasoro algebra with normal ordered operators and I am not sure what the resolution is.

If we have a standard 1+1 D massless field with the (scaled by $$\sqrt{n}$$) creation and annihilation operators $$[\alpha_n, \alpha_m] =n \delta_{n+m, 0}\tag{1}$$ then the Virasoro generators are given by $$L_n = \frac{1}{2} \sum_{m = - \infty}^{\infty} \alpha_{n-m} \alpha_m.\tag{2}$$ One can explicitly verify by straightforwardly calculating the commutator that $$[L_n, L_m] = (n-m)L_{n+m}\tag{3}$$ so the generators satisfy the Witt algebra.

Now, we often choose to "normal order" the Virasoro generators, by putting the "creation operators" (of the form $$\alpha_{-k}$$ for $$k>0$$) to the left and the "annihilation operators" (of the form $$\alpha_{k}$$ for $$k > 0$$) to the right. However, the only generator which is changed by normal ordering is $$L_0$$. After using the commutation relation $$[\alpha_{m}, \alpha_{-m}] = m$$, one finds that the normal ordered $$L_0$$ differs from the non-normal ordered one by $$\tfrac{1}{2} \sum_{m = 0}^\infty m$$ which you might say is $$- \tfrac{1}{24}$$. The point, though, is not necessarily what the constant is (you could call it infinity if you like), but that it is a constant. In other words, the normal ordered Virasoro algebra differs from the non-normal ordered one merely by shifting $$L_0 \mapsto L_0 + \mathrm{const}$$.

Now, there are two sorts of central extensions that can be added to a Lie algebra: trivial and nontrivial. Trivial ones can always be given by just shifting the generators by a constant, while nontrivial ones cannot. It is well known that the central extension of the Virasoro algebra, which takes the form $$[L_n, L_m] = (n-m) L_{m+n} + \frac{1}{12}\delta_{m+n, 0}(n^3 - n)\tag{4}$$ is nontrivial (and this is not particularly hard to show). To me, this suggests that normal ordering the generators, which simply results in shifting $$L_0 \mapsto L_0+ \mathrm{const}$$, can not induce the nontrivial central extension of $$\tfrac{1}{12}\delta_{m+n, 0}(n^3 - n)$$.

However, as any student of string theory knows, this is not the case! In fact, we are told that normal ordering (which is an inherently quantum notion) turns the Witt algebra into the Virasoro algebra. I have just went through the work to verify this for myself (one source is here) and agree with it. The proof involves splitting up each of the generators $$L_n$$ and $$L_{-n}$$ into a part which is expressly normal ordered (with an infinite sum over creation operators multiplying annihilation operators) with a second part (which is either a finite sum of creation operators multiplying other creation operators, or annihilation operators multiplying other annihilation operators, depending on whether you are looking at $$L_n$$ or $$L_{-n}$$) and then computing the commutator carefully between each part. The final result will of $$[L_n, L_{-n}]$$ will then indeed be the normal ordered $$2 L_0$$ plus the central term. Here, instead of having to switch an infinite number of creation and annihilation operators, due to the careful way of splitting up the sums, you only end up having to switch a finite number of them to get the $$(n^3 - n)$$.

However, this to me seems like a few paradoxes, which I will list:

1. Paradox one: if normal ordering only shifts $$L_0$$ by a constant, how can it result in the nontrivial central extension of the Virasoro algebra?
2. Paradox two: It seems as though the method of calculating the central term $$(n^3 - n)$$, which I outlined above, did not really have anything to do with normal ordering. The method basically just amounted to computing an infinite sum of commutators in a careful way, and resulted in $$2 L_0 + \tfrac{1}{12} (n^3 - n)$$. In other words, the normal ordering did not really seem particularly important to attaining this result. It seems as though one method of calculation gets the $$(n^3-n)$$ central term while another does not?

It is often said that without normal ordering from quantum mechanics, you do not get the Virasoro algebra. However from recently going over the proofs of these calculations I am more confused than ever. Does anyone spot what I am overlooking? I feel it must be something very basic but I can't figure out what it is.

1. 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.

2. 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): $$L_n ~=~ \frac{1}{2} \sum_{k \in\mathbb{Z}} \alpha_{n-k} \alpha_k ~+~ C\delta_n^0{\bf 1}, \tag{2'}$$ $$[L_n, L_m] ~=~ (n-m)(L_{n+m}-C\delta_{n+m}^0{\bf 1}).\tag{3'}$$

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.

4. In practice, authors now redefine the Virasoro generator (2") so that the constant $$C=0$$ is zero, cf. Refs. 1-2.

References:

1. B. Zwiebach, A first course in String Theory, 2nd edition, 2009; section 12.4.

2. 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.

• In some sense I agree with this answer, that divergences are just "not allowed, end of story," but I also wonder if there is more to be said. The simple act of avoiding the divergence seems to caused a contradiction, where shifting a Lie algebra element by an (albeit infinite) constant has created a nontrivial central extension. Dec 13 '20 at 20:19
• @user1379857: You can only have a contradiction if your objects are defined. Perhaps I should rephrase this as: the contradiction you found proves that the Weyl-ordered Virasoros do not exist as operators on the Hilbert space.
– user21299
Dec 13 '20 at 22:30