In the previous section, we discussed the theory of even dimensional characteristic forms and classes. i.e.
Let M be a smooth closed manifold and E be a vector bundle with a connection ∇E. We constructed a serial closed form
tr[f(√−12πRE)]∈Ωeven(M),where f is a power series in one variable. Then we obtain some even dimensional characteristic classes
[tr[f(√−12πRE)]]∈HevendR(M,C).In this section, we will discuss an odd dimensional analogue of this result.
Let M be a smooth closed manifold. Let g be a smooth map from M to the general linear group GL(N,C) with N>0 a positive integer:
g: M→GL(M,C),p→g(p)∈GL(N,C),i.e.g=(gij)N×N where gij∈C∞(M,C).
Let CN|M denote the trivial complex vector bundle of rank N over M. Then the above g can be viewed as a section of Aut(CN|M). Let d denote a trivial connection on CN|M. Then we consider n-form tr[(g−1dg)n].
- Assume n is a positive even integer,
tr[(g−1dg)n]=12tr[(g−1dg)n−1(g−1dg)+(g−1dg)(g−1dg)n−1]=12tr[(g−1dg)n−1(g−1dg)−(−1)(n−1)⋅1(g−1dg)(g−1dg)n−1]=tr[(g−1dg)n−1,(g−1dg)]=0.
- Assume n is a positive odd integer. Notice that I=gg−1, then dg−1=−g−1(dg)g−1. Hence
d[(g−1dg)n]=n∑i=1(−1)i−1(g−1dg)i−1d(g−1dg)(g−1dg)n−i=n∑i=1(−1)i−1(g−1dg)i−1(dg−1)(dg)(g−1dg)n−i=n∑i=1(−1)i−1(g−1dg)i−1(−g−1(dg)g−1)(dg)(g−1dg)n−i=n∑i=1(−1)i(g−1dg)n+1=−(g−1dg)n+1.[by n is a odd positive integer]
By the above case,
dtr[(g−1dg)n]=tr[d(g−1dg)n]=−tr[(g−1dg)n+1]=0.
Lemma 1. If gt: M→GL(N,C) depends smoothly on t∈[0,1], then for any positive odd integer n, the following identity holds,
∂∂ttr[(g−1tdgt)n]=nd[g−1t∂gt∂t(g−1tdgt)n−1].
∂∂ttr[(g−1tdgt)n]=nd[g−1t∂gt∂t(g−1tdgt)n−1].
If this lemma have been proved, one can integrate at both sides of above identity
tr[(g−11dg1)n]−tr[(g−10dg0)n]=∫10∂∂ttr[(g−1tdgt)n]dt=n∫10d[g−1t∂gt∂t(g−1tdgt)n−1]dt=d[n∫10[g−1t∂gt∂t(g−1tdgt)n−1]dt].
Let η=n∫10[g−1t∂gt∂t(g−1tdgt)n−1]dt∈Ωn−1(M), then tr[(g−11dg1)n]−tr[(g−10dg0)n]=dη, i.e.
[tr[(g−11dg1)n]]=[tr[(g−10dg0)n]].Now, we give the proof of lemma 1.
Proof . By an analogue of dg−1=−g−1(dg)g−1, one can obtain ∂∂tg−1t=−g−1t(∂gt∂t)g−1t. And, if A∈Ωodd(M,End(CN)), B∈Ωeven(M,End(CN)), one easily verifies that AB=BA by tr[A,B]=0. Hence
∂∂ttr[(g−1tdgt)n]=tr[∂∂t(g−1tdgt)n]=tr[n∑i=1(g−1tdgt)i−1∂∂t(g−1tdgt)(g−1tdgt)n−i]=tr[n∑i=1∂∂t(g−1tdgt)(g−1tdgt)n−1]by n is an odd integer=ntr[∂g−1t∂t(dgt)(g−1tdgt)n−1+g−1t(d∂gt∂t)(g−1tdgt)n−1]=ntr[−g−1t∂gt∂tg−1t(dgt)(g−1tdgt)n−1+g−1t(d∂gt∂t)(g−1tdgt)n−1]=ntr[−g−1t(∂gt∂t)(g−1tdgt)n+d(g−1t∂gt∂t(g−1tdgt)n−1)−(dg−1t)∂gt∂t(g−1tdgt)n−1–g−1t∂gt∂td((g−1tdgt)n−1)]=ntr[−g−1t(∂gt∂t)(g−1tdgt)n+d(g−1t∂gt∂t(g−1tdgt)n−1)+g−1t(dgt)g−1t∂gt∂t(g−1tdgt)n−1]=ntr[−g−1t(∂gt∂t)(g−1tdgt)n+d(g−1t∂gt∂t(g−1tdgt)n−1)+g−1t(∂gt∂t)(g−1tdgt)n]=ndtr[g−1t∂gt∂t(g−1tdgt)n−1].
∂∂ttr[(g−1tdgt)n]=tr[∂∂t(g−1tdgt)n]=tr[n∑i=1(g−1tdgt)i−1∂∂t(g−1tdgt)(g−1tdgt)n−i]=tr[n∑i=1∂∂t(g−1tdgt)(g−1tdgt)n−1]by n is an odd integer=ntr[∂g−1t∂t(dgt)(g−1tdgt)n−1+g−1t(d∂gt∂t)(g−1tdgt)n−1]=ntr[−g−1t∂gt∂tg−1t(dgt)(g−1tdgt)n−1+g−1t(d∂gt∂t)(g−1tdgt)n−1]=ntr[−g−1t(∂gt∂t)(g−1tdgt)n+d(g−1t∂gt∂t(g−1tdgt)n−1)−(dg−1t)∂gt∂t(g−1tdgt)n−1–g−1t∂gt∂td((g−1tdgt)n−1)]=ntr[−g−1t(∂gt∂t)(g−1tdgt)n+d(g−1t∂gt∂t(g−1tdgt)n−1)+g−1t(dgt)g−1t∂gt∂t(g−1tdgt)n−1]=ntr[−g−1t(∂gt∂t)(g−1tdgt)n+d(g−1t∂gt∂t(g−1tdgt)n−1)+g−1t(∂gt∂t)(g−1tdgt)n]=ndtr[g−1t∂gt∂t(g−1tdgt)n−1].
Corollary 2. If f,g: M→L(N,C) are two smooth maps from M to GL(N,C), then for any positive odd integer n, there exists ωn∈Ωn−1(M) such that the following transgression formula holds,
tr[((fg)−1d(fg))n]=tr[(f−1df)n]+tr[(g−1dg)n]+dωn.
tr[((fg)−1d(fg))n]=tr[(f−1df)n]+tr[(g−1dg)n]+dωn.
Proof . We consider the trivial vector bundle
C2N|M=CN|M⊕CN|M.We equip C2N|M with the trivial connection ˜d induced from d on CN|M. For any u∈[0,π2], let hu: M→GL(2N,C) be defined by
hu=(f001)(cos(u)sin(u)−sin(u)cos(u))(100g)(cos(u)−sin(u)sin(u)cos(u)).Obviously,
h0=(f00g),(fg001) Thus, hu provides a smooth homotopy between two sections (fg,1) and (f,g) in Γ( Aut(C2N|M)). By the above lemma, for any positive odd inetger n, there exists ωn∈Ωn−1(M) such that
tr[(h−10˜dh0)n]=tr[(h−1π2˜dh−1π2)n]+dωn
i.e.,
tr[((fg)−1d(fg))n]=tr[(f−1df)n]+tr[(g−1dg)n]+dωn.
C2N|M=CN|M⊕CN|M.We equip C2N|M with the trivial connection ˜d induced from d on CN|M. For any u∈[0,π2], let hu: M→GL(2N,C) be defined by
hu=(f001)(cos(u)sin(u)−sin(u)cos(u))(100g)(cos(u)−sin(u)sin(u)cos(u)).Obviously,
h0=(f00g),(fg001) Thus, hu provides a smooth homotopy between two sections (fg,1) and (f,g) in Γ( Aut(C2N|M)). By the above lemma, for any positive odd inetger n, there exists ωn∈Ωn−1(M) such that
tr[(h−10˜dh0)n]=tr[(h−1π2˜dh−1π2)n]+dωn
i.e.,
tr[((fg)−1d(fg))n]=tr[(f−1df)n]+tr[(g−1dg)n]+dωn.
Now, we consider the change of tr[(g−1dg)n] under different trivial connection. Let E be a trivial complex vector bundle on M. As we take a global basis of E, a trivialization is determined, and a trivial connection is determined. Let d be the trivial connection associated basis {e1,⋯,en}. Assume {e1′,⋯,e′n} is another basis of E, and
(e1′,⋯,e′n)=(e1,⋯,en)A,A∈Γ(Aut(CN|M)).
If d′ is the trivial connection associated {e1′,⋯,e′n}, then one can verify
d′=A−1∘d∘A.
Exercise 1. Please verify d′=A−1∘d∘A.
Corollary 3. Let g∈Γ(Aut(CN|M)). If d′ is another trivial connection on CN|M), then for any positive odd integer n, there exists ωn∈Ωn−1(M) such that the following transgression formula holds,
tr[(g−1dg)n]=tr[(g−1d′g)n]+dωn.
tr[(g−1dg)n]=tr[(g−1d′g)n]+dωn.
Proof . There exists A∈Γ(Aut(CN|M)) such that
d′=A−1∘d∘A=d+A−1∘(dA).One deduces that
g−1d′g=g−1∘d′∘g−d′=g−1∘A−1∘d′∘A∘g−A−1∘d∘A=A−1(A∘g−1∘A−1∘d′∘A∘g∘A−1−d)A=A−1((AgA−1)d(AgA−1))A.
From above corollary, there exists ωn∈Ωn−1(M) for any positive odd integer n such that
tr[(g−1d′g)n]=tr[A−1((AgA−1)d(AgA−1))nA]=tr[((AgA−1)d(AgA−1))n]=tr[(A−1dA)n]+tr[(AdA−1)n]+tr[(g−1dg)n]–dωn=tr[(g−1dg)n]−dωn.(It is from dA=−A(dA−1)A)
d′=A−1∘d∘A=d+A−1∘(dA).One deduces that
g−1d′g=g−1∘d′∘g−d′=g−1∘A−1∘d′∘A∘g−A−1∘d∘A=A−1(A∘g−1∘A−1∘d′∘A∘g∘A−1−d)A=A−1((AgA−1)d(AgA−1))A.
From above corollary, there exists ωn∈Ωn−1(M) for any positive odd integer n such that
tr[(g−1d′g)n]=tr[A−1((AgA−1)d(AgA−1))nA]=tr[((AgA−1)d(AgA−1))n]=tr[(A−1dA)n]+tr[(AdA−1)n]+tr[(g−1dg)n]–dωn=tr[(g−1dg)n]−dωn.(It is from dA=−A(dA−1)A)
Remark 1. The cohomology class determine by tr[(g−1dg)n] depends only on the homotopy class of g: M→GL(N,C).
When n is a positive odd integer, we call the closed n-form
(12π√(−1))n+12tr[(g−1dg)n]
the n-th Chern form associated to g,d and denote it by cn(g,d). The associated cohomology class will be called the n-th Chern class associated to the homotopy class [g], denote it by cn([g]).
We define the odd Chern character form associated to g,d by
ch(g,d)=∞∑n=0n!(2n+1)!c2n+1(g,d). Let ch([g]) denote the associated cohomology class which we call the odd Chern character associated to [g].
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