Virasoro algebra
The Virasoro algebra
where
Shifted equivalent extensions
Letting
for πΏ β² π = πΏ π + π½ π π and π½ π β π , and π β β€ , we get an ^equivalent extension, with the bracket now given by π β² : πΏ β² π β¦ π π [ πΏ β² π , πΏ β² π ] = ( π β π ) πΏ β² π + π + π 1 2 ( π 3 β π ) πΏ π + π + π ( π β π ) π½ π + π In particular,
removes the linear term. πΏ β² π = πΏ π β 1 2 4 πΏ π π
Proof of uniqueness
Let
be a central extension of π³ β² such that π‘ π³ = π΅ πΎ πΌ π π π‘ β π π and
[ π π , π π ] = ( π β π ) π π + π + πΎ π , π π [ π , π³ β² ] = 0 It follows from the alternating property and the ^Jacobi that
πΎ π , π + πΎ π , π = 0 ( π β π ) πΎ π + π , π + ( π β π ) πΎ π + π , π + ( π β π ) πΎ π + π , π = 0 for
. Assume π , π , π β β€ and π = 0 , so π + π β 0 β ( π + π ) πΎ π , π + ( π β π ) πΎ π + π , 0 = 0 whence
πΎ π , π = π β π π + π πΎ π + π , 0 By considering the equivalent shifted extension
π β² π = π π + π ( 1 β πΏ π ) 1 π πΎ π , 0 we can take
for πΎ π , π = 0 without loss of generality. The general solution to the constraints on π + π β 0 given πΎ π , π is then π + π + π = 0 πΎ π , β π = πΌ π 3 + π½ π where
, since any solution is determined by πΌ , π½ β π and πΎ 1 , β 1 . By shifting we can change πΎ 2 , β 2 arbitrarily and by rescaling π½ we can multiply π by any nonzero scalar. Thus the extension of πΌ given by π‘ is either Β ^equivalent to π³ β² or the ^trivial. π³
Properties
- The extension is the ^trivial restricted to
, since the central term becomes zero.π = π π β 1 + π π 0 + π π 1
Footnotes
-
β©c h a r β‘ π = 0 -
1988. Vertex operator algebras and the Monster, Β§1.9 pp. 32ff. β©