# Non-orientable Poincare Duality

This post is based on

Differential Forms on Algebraic Topologywritten 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 ofdifferential formsandwedge product. The nondegenerate pair, as appears inAlgebraic Topologyby 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*.

### 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

- $U=S^1$, $\phi(x,h)\rightarrow h$;
- $U_1,U_2$ 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 $H_\psi^0(M,E)=0$.

### 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.