斯托克斯定理，de Rham上同调习题

(1) For an orientable manifold MMM with boundary ∂M\partial M∂M, show that:
∙\bullet∙ there is a non-vanishing outer normal vector field NNN on ∂M\partial M∂M;
∙\bullet∙ given an orientation n-form Ω∈Λn(M)\Omega\in\Lambda^n(M)Ω∈Λn(M), the form:
Ω′=(ιNΩ)∣∂M,\Omega'=(\iota_N\Omega)|_{\partial M},Ω′=(ιNΩ)∣∂M,
is a non-vanishing (n−1)(n-1)(n−1) form on ∂M\partial M∂M, which gives an induced orientation on ∂M\partial M∂M.

Proof:
Firstly, we need to show a proposition:
For SSS is an immersed hypersurface in MMM, and NNN is a vector field along SSS that is nowhere tangent to SSS. Then SSS has a unique orientation. Also, if ω\omegaω is an orientation form for MMM, then (ιNω)∣S(\iota_N\omega)|_S(ιNω)∣S is an orientation form for SSS.

proof:
Suppose ω\omegaω is an orientation form for MMM, then σ=(ιNω)∣S\sigma=(\iota_N\omega)|_Sσ=(ιNω)∣S is an (n−1)(n-1)(n−1)-form on SSS. It follows that it’s an orientation form for SSS is we can show that it never vanishes. Given any basis (E1,…,En−1)(E_1,\dots,E_{n-1})(E1,…,En−1) for TpST_pSTpS, since NNN is nowhere tangent to SSS implies that (Np,E1,…,En−1)(N_p,E_1,\dots,E_{n-1})(Np,E1,…,En−1) is a basis for TpMT_pMTpM. The fact that ω\omegaω is nonvanishing implies that σp(E1,…,En−1)=ωp(Np,E1,…,En−1)≠0.\sigma_p(E_1,\dots,E_{n-1})=\omega_p(N_p,E_1,\dots,E_{n-1})\neq0.σp(E1,…,En−1)=ωp(Np,E1,…,En−1)=0.
Thus the orientation determined by σ\sigmaσ is the one defined in the statement of the proposition.
Suppose dim(M)=ndim (M)=ndim(M)=n, and ω\omegaω is an orientation form for MMM, let NNN be a smooth outward-pointing vector field along ∂M\partial M∂M. The (n−1)(n-1)(n−1)-form (ιNω)∣∂M(\iota_N\omega)|_{\partial M}(ιNω)∣∂M is an orientation form for ∂M\partial M∂M, so ∂M\partial M∂M is orientable.

(2) Show that the Stokes’ theorem is a generalization of the classical Green formula, Gauss formula and Stokes formula in multi-variable calculus.

Proof:
Apply Stokes’ theorem to 111-form ω=Pdx+Qdy\omega=Pdx+Qdyω=Pdx+Qdy, suppose DDD is a compact regular domain in R2\mathbb{R}^2R2, P,QP,QP,Q are smooth real-valued functions on DDD, thus we have Green formula.

Similarly, apply it to 222-form, ω=Pdy∧dz+Qdz∧dx+Rdx∧dy\omega=Pdy\wedge dz+Qdz\wedge dx+Rdx\wedge dyω=Pdy∧dz+Qdz∧dx+Rdx∧dy, we have Gauss formula.

And apply it to ω=Pdx+Qdy+Rdz\omega=Pdx+Qdy+Rdzω=Pdx+Qdy+Rdz, we have Stokes formula.

(3) Prove that the pull-back of a smooth map F:M→NF:M\rightarrow NF:M→N defines a linear map:
F∗:Hp(N)→Hp(M),F^*:H^p(N)\rightarrow H^p(M),F∗:Hp(N)→Hp(M),
which commutes with the wedge product.

Proof:
We set
Zp(M)=Ker(d:Ωp(M)→Ωp+1(M)))={closed p-forms on M},\mathcal{Z}^p(M)=Ker(d:\Omega^p(M)\rightarrow \Omega^{p+1}(M)))=\{\text{closed p-forms on }M\},Zp(M)=Ker(d:Ωp(M)→Ωp+1(M)))={closed p-forms on M},
Bp(M)=Im(d:Ωp−1(M)→Ωp(M)))={exact p-forms on M}.\mathcal{B}^p(M)=Im(d:\Omega^{p-1}(M)\rightarrow \Omega^{p}(M)))=\{\text{exact p-forms on }M\}.Bp(M)=Im(d:Ωp−1(M)→Ωp(M)))={exact p-forms on M}.

If ω\omegaω is closed, then d(F∗ω)=F∗(dω)=0d(F^*\omega)=F^*(d\omega)=0d(F∗ω)=F∗(dω)=0, so F∗ωF^*\omegaF∗ω is also closed. If ω=dη\omega=d\etaω=dη is exact, then F∗ω=F∗(dη)=d(F∗η)F^*\omega=F^*(d\eta)=d(F^*\eta)F∗ω=F∗(dη)=d(F∗η), which is also exact. Therefore, F∗F^*F∗ maps Zp(N)\mathcal{Z}^p(N)Zp(N) into Zp(M)\mathcal{Z}^p(M)Zp(M) and Bp(N)\mathcal{B}^p(N)Bp(N) into Bp(M)\mathcal{B}^p(M)Bp(M). The induced cohomology map F∗:Hp(N)→Hp(M)F^*:H^p(N)\rightarrow H^p(M)F∗:Hp(N)→Hp(M) is defines as follow:
F∗[ω]=[F∗ω].F^*[\omega]=[F^*\omega].F∗[ω]=[F∗ω].
If ω′=ω+dη\omega'=\omega+d\etaω′=ω+dη, then [F∗ω′]=[F∗ω+d(F∗η)]=[F∗ω][F^*\omega']=[F^*\omega+d(F^*\eta)]=[F^*\omega][F∗ω′]=[F∗ω+d(F∗η)]=[F∗ω], so this map is well defined.

(4) Show that the 111-st de Rham cohomology group of S1S^1S1 is H1(S1)=RH^1(S^1)=\mathbb{R}H1(S1)=R.

Proof:

For any orientation form on S1S^1S1 has nonzero integral, thus dimH1(S1)≥1dim H^1(S^1)\geq 1dimH1(S1)≥1. On the other hand, there is an injective linear map from H1(S1)H^1(S^1)H1(S1) into Hom(π1(S1,1),R)Hom(\pi_1(S^1,1),\mathbb{R})Hom(π1(S1,1),R), which is 111-dimensional, thus H1(S1)H^1(S^1)H1(S1) is dimension 111, and is spanned by the cohomology class of any orientation form, thus H1(S1)=RH^1(S^1)=\mathbb{R}H1(S1)=R.

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斯托克斯定理，de Rham上同调习题

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