nLab
Oberwolfach Workshop, June 2009 -- Friday, June 12

Talk notes for the session of

  • Friday June 12

of

Corbett Redden: String structures and 3-forms

  • notational conventions

    • M nM^n a smooth closed nn-manifold

    • gg a Riem. metric

    • SpinPMSpin \to P \to M a principal Spin(n)Spin(n)-bundle

    • AA a connection on PP

    • SS a string class

Outline

  • I {StringStructures}/ homotopy\{String Structures\}/_{homotopy}

  • II harmonic 4-forms on PP

  • III geometry and tmf?

I. String structures

definition

  • BStringB String is defined to be the homotopy fiber BString(n)BSpin(n)p 1/2K(,4)B String(n) \to B Spin(n) \stackrel{p_1/2}{\to} K(\mathbb{Z},4)

  • a String structure on π:PM\pi : P \to M is a lift of the classifying map to BSpinB Spin through BStringBSpinB String \to B Spin

    P BString M BSpin \array{ P && B String \\ \downarrow &\nearrow& \downarrow \\ M &\to& B Spin }

Prop

  • String structure exists iff 12p 1(P)=0H 4(X,)\frac{1}{2}p_1(P) = 0 \in H^4(X,\mathbb{Z}) (essentially by definition through homotopy fiber)

  • let i P *i^*_P the pullback to the fiber of PP, then we have
    $${String structures}/_{homotopy}
    \simeq_{canonical} {String classes}
    = {\rho \in H^3(P,\mathbb{Z}) s. t. i_P^* \rho = 1 \in H^3(Spin, \mathbb{Z})}
  • {stringclass}\{string class\} is a torsor for H 3(M,)H^3(M, \mathbb{Z}) under ρρ+π H3(M)\rho \mapsto \rho + \pi^ H^3(M)

Proof. universal example

K(,3) π *ESpin ESpin BString BSpin K(,4) \array{ K(\mathbb{Z},3) &\simeq& \pi^* E Spin &\to & E Spin \\ && B String &\to& B Spin &\to& K(\mathbb{Z}, 4) }

hm, here is my (Urs) description of the situation:

consider the following pasting diagram of homtopy pullbacks (PP is the given SpinSpin-bundle, P^\hat P its String-lift)

String P^ * Spin P B 2U(1) * * X BString(n) BSpin(n) \array{ String &\to& \hat P &\to& {*} \\ \downarrow && \downarrow && \downarrow \\ Spin &\to& P &\to& B^2 U(1) &\to& {*} \\ \downarrow && \downarrow && \downarrow && \downarrow \\ {*} &\to& X &\to& B String(n) &\to& B Spin(n) }

Why String structures?

String structure on PP transgresses\stackrel{transgresses}{\mapsto} Spin structure on loop space LSpinLPLML Spin \to L P \to L M

but all the reps of this loop group are projective, so there is actually a central S 1S^1 extension of LSpinL Spin in the game

LSpin^LSpin \hat {L Spin} \to L Spin

Need:

LSpin^ LP^ LSpin LP \array{ \hat {L Spin} &\to & \hat {L P} \\ \downarrow && \downarrow \\ L Spin &\to& L P }
ρH 3(P,) \rho \in H^3(P, \mathbb{Z}) \mapsto
(π !ev *)ρ H 2(LP,) iniv.ext. H 2(LG,) \array{ (\pi_! ev^*) \rho &\in& H^2(L P, \mathbb{Z}) \\ \downarrow && \downarrow \\ iniv. ext. && H^2(L G, \mathbb{Z}) }

(on the right: S 1S^1-bundles)

** String orientation ** of tmf = topological modular forms

M O\langle 8\rangle^{-n} = M String &\stackrel{\sigma}{\to}& tmf^{-n}(pt)

so give a String manifold with a String class on Spin(TM) M O\langle 8\rangle^{-n} = M String &Spin(T M)

M,ρ[M,ρ]σ(M,ρ) M , \rho \mapsto [M, \rho] \mapsto \sigma(M,\rho)
tmf σ MString Wittengenus ModForms \array{ && tmf \\ & {}^{\sigma}\nearrow & \downarrow \\ M String &\stackrel{Witten genus}{\to}& Mod Forms }

Wittengenus(M)=Witten genus(M) =index S 1D LMindex^{S^1} D_{L M}

(by the way, σ\sigma is surjective on homotopy classes)

warning: I think above my ρ\rho should really be an SS

II Harmonic representative of SS

reminder

(M,g)Δ=dd *+d *d(M,g) \mapsto \Delta = d d^* + d^* d

H k(M,) HodgeΔ g *Ω k(M) H^k(M,\mathbb{R}) \simeq_{Hodge} \Delta^*_g \subset \Omega^k(M)

construction

start with (π:PM,g m,A)(\pi : P \to M, g_m, A)

choose a bininvariant metric

g ρ:=π *g mg spin g_\rho := \pi^* g_m \oplus g_{spin}

where the direct sum comes from the splitting of tangent spaces T pPT_p P using the connection that we have

Introduce scaling factor δ>0\delta \gt 0

g ρ:=π *g mδ 2g spin g_\rho := \pi^* g_m \oplus \delta^2 g_{spin}

take the “adiabatic limit” lim δ0\lim_{\delta \to 0}

so now there is a 1-parameter family of metric on the bundle, and for each one can look at its Laplacian, so as δ\delta tends to 0 something becomes singular and one has to be careful, but fortunately others already did that for us…

theorem (Mazzeo-Melrose, Dai, Forman)

the kernel KerΔ g ρ Ker \Delta_{g_\rho}

  • extends smoothly to δ=0\delta = 0 (there is a smooth path in some Grassmannian space)

  • and comes from a filtration isomorphic to Serre SS for (SpinPM)(Spin \to P \to M)

this means that

H k(P,)lim δ0KerΔ g ρ=:H k(P) \Rightarrow H^k(P, \mathbb{Z}) \stackrel{\simeq}{\to} \lim_{\delta \to 0} Ker \Delta_{g_\rho} =: H^k(P)

Theorem (Redden)

Given Pπg,AP \stackrel{\pi}{\to} g, A and 12p 1(P)=0\frac{1}{2}p_1(P) = 0

then

H 3(P,) H 3(P,) H 3(P) S CS 3(A) CS 3(A)π *H \array{ H^3(P, \mathbb{Z}) &\to& H^3(P,\mathbb{R}) &\to& H^3(P) \\ S &&\mapsto& CS_3(A)&& CS_3(A) - \pi^* H }

here CS 3(A)CS_3(A) is the Chern-Simons 3-form of the spin-connection

and recall H 3(P)H^3(P) here denotes harmonic forms on PP (should really be script font

remark

in genral

[ρ] g δ=CS 3(A)π *H+O(δ)¬π *Ω 3(M) [\rho]_{g_\delta} = CS_3(A) - \pi^* H + O(\delta) \not\in \pi^* \Omega^3(M)

if we have a product of two groups we accordingly would get CS of one connection minus CS of the other.

What is HH?

(first digression)

theorem (Chern-Simons,…)

given (PM,A)frac12p 1^(A)H^ 4(M)(P \to M , A) \mapsto \hat {frac{1}{2} p_1}(A) \hat H^4(M)

Ω 3 (M)Ω 3(M) toa H^ 4(M) H 4(M,) 0 H 12p 1(A)^ 12p 1(P)=0 \array{ \Omega^3_{\mathbb{Z}} &\to& (M)\Omega^3(M) &\stackrel{a}{to}& \hat H^4(M) &\to& H^4(M,\mathbb{Z}) &\to& 0 \\ && H &\mapsto& \hat{\frac{1}{2}p_1(A)} &\mapsto& \frac{1}{2}p_1(P) = 0 && }

in particular

H 3(M) 12p 1(A)^ / \array{ && \mathbb{R} \\ & {}^{\int H} \nearrow & \downarrow \\ \mathbb{Z}_3(M) &\stackrel{\hat {\frac{1}{2}p_1(A)}}{\to}& \mathbb{R}/\mathbb{Z} }

and secondly d *H=0d^* H = 0

these two properties determine HH uniquely up to harmonic forms 3(M)=kerΔ g\mathcal{H}^3_{\mathbb{Z}}(M) = ker \Delta_g

Equivariance

H ρ+π *ψ=H ρ+π *H4 H_{\rho + \pi^* \psi} = H_\rho + \pi^* H4

where ψH 3(M,)\psi \in H^3(M, \mathbb{Z}) H 4 3(M)\mapsto H_4 \in \mathcal{H}^3(M)

over all what this says is that if we go from

Metr(M)×A(P)×{StringClass} (g,A,S) Ω 3(P) CS 3(A)π *H q,A,S Ω 3(M) H g,Aρ \array{ Metr(M)\times A(P) \times \{String Class\} & (g,A,S) \\ \downarrow & \downarrow \\ \Omega^3(P) & CS_3(A) - \pi^* H_{q,A,S} \\ \downarrow & \downarrow \\ \Omega^3(M) & H_{g, A \rho} }
tmf n MString MF \array{ && tmf^{-n} \\ & \nearrow & \downarrow \\ M String &\stackrel{}{\to}& MF }

conjecture (S. Stolz) if MM is String and admits a positive Ricci curvature metric, then Witten(M)=0Witten(M) = 0

question: also σ(M,ρ)=0\sigma(M,\rho) = 0? no, no way!

hypothesis

If MM is a Spin manifold that admits a metric and String structure (g,S)(g, S) and AA is the Levi-Civita connection

such that

  • Ric(g)>0Ric(g) \gt 0

  • H g,s=0Ω 3(M)H_{g,s} = 0 \in \Omega^3(M)

σ(M,S)=0tmf n(pt)\Rightarrow \sigma(M,S) = 0 \in tmf^{-n}(pt)

example M=S 3=SU(2)M = S^3 = SU(2)

p 1H 4(S 3)=0 p_1 \in H^4(S^3) = 0
H 3(S 3,)==numberofstringclasses H^3(S^3, \mathbb{Z}) = \mathbb{Z} = number of string classes
dH=d *H=0HH 3(S 3,) d H = d^* H = 0 \Rightarrow H \in H^3(S^3, \mathbb{R}) \simeq \mathbb{R}
StringClasses MString 3=π 3 S=tmf 3=/24 \array{ String Classes \\ \downarrow \\ M String^{-3} = \pi_3^S = tmf^{-3} = \mathbb{Z}/{24} }

(can’t type the full diagram…)

consider a 1-parameter family of “berger metrics” on S 3S^3

rescaling the fiber in the Hopf fibration S 1S 3S 2S^1 \to S^3 \to S^2

Konrad Waldorf

Urs: I had to miss that and the following two talks, hopefully somebody else has notes. Konrad’s talk is based on his new article

one more

Mike Hopkins : Kervaire invariant one


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Revised on June 15, 2009 06:10:33 by Urs Schreiber (134.100.222.156)