process

338 U. Bunke, T. Schick, M. Spitzweck, and A. Thomcorresponding to E univ ! R n as in 6.3.1. In [BRS] we have shown that H 3 .R n I Z/ Š 0.The diagram (65) now implies that there is a unique Picard stack P univ 2 Q Euniv withOc.P univ / DOc univ and underlying pair up.P univ / Š .E univ ;H univ / (see 6.3.5).6.4.2 Let us fix a T n -principal bundle E ! B, or the corresponding extension ofsheaves E. Let us furthermore fix a class h 2 F 2 H 3 .EI Z/. In [BRS] we haveintroduced the set Ext.E; h/ of extensions of .E; h/ to a T -duality triple. The maintheorem about this extension set is [BRS, Theorem 2.24].Analogously, in the present paper we can consider the set of extensions of .E; h/to a Picard stack P with underlying T n -bundle E ! B and f.P/ D h, wheref W Q E ! F 2 H 3 .EI Z/ is as in (65). In symbols we can write f1 .h/ for this set.The main goal of the present section is to compare the sets Ext.E; h/ and f1 .h/ Q E . In the following paragraphs we construct maps between these sets.6.4.3 We fix a T n -principal bundle E ! B and let E 2 Sh Ab S=B be the correspondingextensions of sheaves. Let Triple E .B/ denote the set of isomorphism classes of triplest such that c.t/ D c.E/. We first define a mapˆW Triple E .B/ ! Q E :Let t 2 Triple E .B/ be a triple which is classified by a map f t W B ! R n . Pulling backthe group stack P univ 2 PIC.S=R n / we get an element ˆ.t/ WD ft .Puniv/ 2 PIC.S=B/.If t 2 Triple E .B/, then we have ˆ.t/ 2 Q E . We further haveOc.ˆ.t// D ft Oc.P univ / D ft Oc univ DOc.t/: (67)6.4.4 In the next few paragraphs we describe a map‰ W Q E ! Triple E .B/;i.e. a construction of a T -duality triple ‰.P/ 2 Triple E .B/ starting from a Picard stackP 2 Q E .6.4.5 Consider P 2 PIC.S=B/. We have already constructed one pair .E; H /. Thedual D.P / WD HOM PIC.S=B/ .P; BT jB / is a Picard stack with (see Corollary 5.10)H 0 .D.P // Š D.H 1 .P // Š D.T jB / Š Z jB ;H 1 .D.P // Š D.H 0 .P // Š D.E/:In view of the structure (57) of E, the equalities D.T n jB / Š Zn jB , D.Z jB / Š T jB , andExt 1 Sh Ab S=B .T jB ; T jB / Š 0 we have an exact sequence0 ! T jB ! D.E/ ! Z n jB! 0: (68)