Non-orientable Poincare Duality

This post is based on Differential Forms on Algebraic Topology written by Raoul Bott and Loring W. Tu. Without specifically emphasized, the coefficient ring of the cohomology is $\mathbb{R}$, and the cohomology ring is defined to be the ring of differential forms and wedge product. The nondegenerate pair, as appears in Algebraic Topology by Allen Hatcher called cap product $\cap$, is now the integration operation along (sub)manifolds.

The purpose of this article is to understand the Poincare Duality in the non-orientable manifolds. The mathematical objects that we are using is the twisted de Rham complex.

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Twisted de Rham complex

First we consider a simple case, when $V$ is a vector space.

Def. (Twisted differential forms) Let $M$ be a manifold and $V$ be a vector space. The space of twisted differential forms with values in $V$, denoted by $\Omega^*(M,V)$ is spanned by

$$\omega\otimes v$$

where $w\in\Omega^*(M)$ and $v\in V$. The differential operator $d$ is defined to be the linear expansion of

$$d(\omega\otimes v)=(d\omega)\otimes v.$$

Remark. With the above differential, the cohomology ring of $\Omega^*(M,V)$ can be defined. If $V$ is the vector space of dimension $n$, then

$$H^*(M,V)=\left(H^*(M)\right)^n.$$

To define the notion of twisted in a more general case, i.e. when $E$ is a vector bundle, we shall know how to differentiate the “vectors” in $E$. Think about “vectors in $E$” being “sections”, and the differentials can be defined through the process of local trivialization of $E$.

Def. ($E$-valued $q$-forms) Let $E$ be a vector bundle. Define the space of $E$-valued $q$-forms $\Omega^q(M,E)$ being expanded by global sections of $(\Lambda^qT^*M)\otimes E$, i.e.

$$s:M\rightarrow \left(\wedge^qT^*M\right)\otimes E.$$

Suppose that $\{(U_\alpha,\phi_\alpha)\}$ is a local trivialization of $E$, where

$$\phi_\alpha:E_{\alpha}\rightarrow U_\alpha\times\mathbb{R}^k.$$

Let $e_\alpha^1,\ldots,e_\alpha^k$ be the sections of $E_\alpha$ corresponding to the standard basis of $U_\alpha\times\mathbb{R}^k$, called a standard locally constant sections. $\forall \omega\in\Omega^q(M,E)$, it has a local expression on $U_\alpha$

$$\omega=\sum_i\omega_i\otimes e_\alpha^i,$$

where $\omega_i\in\Omega^q(M)$. With the local expression, define the differential on $U_\alpha$ to be

$$d\omega=\sum_i(d\omega_i)\otimes e_\alpha^i.$$

Compatibility: Suppose on the overlapping set $U_{\alpha\beta}$,

$$\omega=\sum\omega_i\otimes e_\alpha^i=\sum\tau_j\otimes e_\beta^j.$$

where $e_\alpha^i=\sum_jc_{ij}e_\beta^j$. Then

$$\begin{aligned} \sum_i (d\omega_i)\otimes e_\alpha^i &=\sum_{i,j}c_{ij}d\omega_i\otimes e_\beta^j\\ &=\sum_{j}d(\sum_i c_{ij}\omega_{i})\otimes e_\beta^j\\ &=\sum_j (d\tau_j)\otimes e_\beta^j. \end{aligned}$$

Hence $d\omega$ is globally defined on $M$. Therefore, the above procedure defines a cohomology group of $E$-valued forms $H_\phi^*(M,E)$, where the subscript cannot be ignored as the cohomology might not be independent of the choice of local trivialization $\{(U_\alpha,\phi_\alpha)\}$.

e.g. Suppose that $M=S^1$ and $E=S^1\times\mathbb{R}$. Consider two local trivialization of $E$ being

1. $U=S^1$, $\phi(x,h)\rightarrow h$;
2. $U_1,U_2$ where $\phi_1:U_1\times\mathbb{R}\rightarrow\mathbb{R}$ is

$$\phi_1(x,h)=h$$

and $\phi_2:U_2\times\mathbb{R}\rightarrow\mathbb{R}$ is

$$\phi_2(x,h)=\rho(x)h$$

for some (nowhere zero) function $\rho$. In the first local trivialization, $H_\phi^0(M,E)=\ker(d:\Omega^0\otimes E\rightarrow\Omega^1\otimes E)=\ker(d:\Omega^0\rightarrow\Omega^1)=\mathbb{R}$.

In the second local trivialization, the locally constant section $e_i:U_i\rightarrow\left(\wedge^0 T^*M\right)\otimes E=M\otimes E$ satisfies

• $e_0(x)=(x,1)$,
• $e_1(x)=(x,1/\rho(x))$: the second coordinate is because $\phi_2\circ e_1(x)=(x,\rho(x)*)$ should be locally constant near $x$, where $e_1(x)=(x,*)$.

Observe that a global $E$-valued $q$-form $\omega$ cannot be written as the “gluing” of the locally constant form on $U_1$ and $U_2$. Therefore $H_\psi^0(M,E)=0$.

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Cohomology with respect to local trivialization

Question: when will two local trivializations have the same cohomology group $H^*_\phi(M,E)$?

Prop. The twisted cohomology is invariant under the refinement of open covers.

Proof. The definition of differential operator is local. Thus if $U_\alpha$ is a refinement of $V_\beta$, then the two local homeomorphism $\phi$ and $\psi$ agrees on $U_\alpha$. Thus the differential operators coincide, and $\Omega_\phi^*(M,E)=\Omega^*_\psi(M,E)$. $\square$

Say that two local trivialization $\{(U_\alpha,\phi_\alpha)\}$ and $\{(U_\alpha,\psi_\alpha)\}$ differ by a locally constant comparison $0$-cochain, if $\exists a_\alpha^{ij}:U_\alpha\rightarrow \mathrm{GL}(k,\mathbb{R})$, s.t.

$$e_\alpha^i=\sum_j a_{\alpha}^{ij}f_\alpha^j.$$

Then there is an isomorphism

$$F:\Omega_\phi^q(M,E)\rightarrow\Omega_\psi^q(M,E)\\ e_\alpha^i\mapsto\sum_j a_\alpha^{ij}f_\alpha^j$$

which induces an isomorphism in cohomology $F^*:H^q_\phi(M,E)\rightarrow H_\psi^q(M,E)$. Therefore, if two local trivialization differs by a locally constant comparison $0$-cochain, then they have the same cohomology. By taking a common refinement, we can always compare two local trivialization on the same open cover $\{U_\alpha\}$.

Prop. Let $E$ be a flat vector bundle of rank $k$ and $\{g_{\alpha\beta}\}$ and $\{h_{\alpha\beta}\}$ be transition functions for $E$ relative to two locally constant trivialization $\phi$ and $\psi$ for the same open cover. If there exists a locally constant functions

$$\lambda_\alpha:U_\alpha\rightarrow\mathrm{GL}(k,\mathbb{R})$$

such that $g_{\alpha\beta}=\lambda_\alpha h_{\alpha\beta}\lambda_{\beta}^{-1}$, then there are isomorphisms

$$\Omega_\phi^*(M,E)\simeq\Omega_\psi^*(M,E)$$

and

$$H^*_\phi(M,E)\simeq H_\psi^*(M,E).$$

Orientation line bundle

To understand whether an manifold is orientable, we can consider its orientation bundle.

Def. (Orientation bundle) Let $g_{\alpha\beta}:=\phi_\alpha\circ\phi_{\beta}^{-1}$ be the transition map. Define the orientation bundle of $M$ to be the line bundle $L$ determined by

$$\mathrm{sgn} J(g_{\alpha\beta})$$

Remark. $M$ is orientable iff the $L$ is trivial.