Hi. Is there a result in the spirit of Poincaré duality but for non-oriented manifolds? Thanks.
$\begingroup$
$\endgroup$
3
-
8$\begingroup$ Yes, it gives an isomorphism between ordinary homology and cohomology of the local system of orientations (locally isomorphic to $\mathbb Z$) of closed manifolds. This local system is trivial precisely when the manifold is orientable. $\endgroup$– Torsten EkedahlCommented Apr 10, 2011 at 8:59
-
10$\begingroup$ Also, any closed manifold satisfies Poincaré duality when homology and cohomology are taken with coefficients in the integers mod $2$. See Hatcher (math.cornell.edu/~hatcher/AT/ATpage.html) Chapter 3.3 (the twisted version is also treated in the additional topics for Chapter 3). $\endgroup$– Mark GrantCommented Apr 10, 2011 at 10:34
-
1$\begingroup$ For smooth manifolds the twisted version is also treated at the end of the first chapter of Bott & Tu. $\endgroup$– FaisalCommented Apr 10, 2011 at 20:07
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
2
A high level purely homotopical answer that does not require explicit consideration of homology or cohomology is given in Theorem 19.1.5 of Parametrized Homotopy Theory, by Johann Sigurdsson and myself. That result works equivariantly, where orientation is not as well understood as one would like. Nonequivariantly, the result gives a description of the spectrum $k\wedge M_+$ for any spectrum $k$ and smooth closed manifold $M$ as a function spectrum defined in terms of parametrized spectra. That may sound daunting, but it is really very natural. Here $k$ doesn't even have to be a ring spectrum.
answered Apr 10, 2011 at 23:14
-
4$\begingroup$ It does sound daunting :) (There are three users with your name; if you'd like them to be merged into one, you could ask on tea.mathoverflow.net, by the way) $\endgroup$ Commented Apr 10, 2011 at 23:27
-
$\begingroup$ And I suspect that Peter has more than 21 rep :) $\endgroup$– David Roberts ♦Commented Apr 11, 2011 at 1:17
$\begingroup$
$\endgroup$
1
Another kind of answer is Atiyah duality: if we write $D(X)$ for the Spanier-Whitehead dual of $X$ then $D(M/\partial M)$ is the Thom spectrum $M^{-TM}$. For any finite spectrum $X$ we have $H_k(DX)=H^{-k}(X)$ and $H^k(DX)=H_{-k}(X)$. For any finite complex $X$ and any orientable virtual vector bundle $V$ of virtual dimension $d$ we also have $H^k(X^V)=H^{k-d}(X)$ and $H_k(X^V)=H_{k-d}(X)$. If $M$ is an orientable manifold of dimension $n$ this gives $H^{k-n}(M^{-TM})=H^k(M)$. By combining these facts we can recover Poincaré duality from Atiyah duality.
answered Apr 10, 2011 at 12:32
-
2$\begingroup$ @Neil: just a minor comment: both answers are really the same. If we put $-TM$ in dimension $-d$ then we get $H^*(M^{-TM})$ coincides with $H^{*+d}(M;\mathcal{L})$ where $\mathcal{L}$ is the orientation sheaf. $\endgroup$ Commented Apr 10, 2011 at 22:20
$\begingroup$
$\endgroup$
1
Yes. For simplicity, assume $M^n$ closed and connected. For each $x \in M$, let $\omega_{M,x}:= H_n (M; M \setminus x)$. This is a coefficient system of groups isomorphic to $\mathbb{Z}$ (or any other ring, if you start with that). Now you can observe that $H_n(M,M\setminus x; \omega_M)$ is \emph{canonically} isomorphic to $\mathbb{Z}$ (the isomorphism depends on the choice of an isomorphism $H_n (R^n, R^n \setminus 0 ) \cong Z$. The customary proofs of Poincare duality now show: -there exists a unique element $[M] \in H_n (M; \omega_M)$ which restricts to the given generator in $H_n (M, M \setminus 0;\Omega_M)$ for each $x \in M$. -cap product with the fundamental class induces a isomorphisms $H^{k}(M; \omega_M) \to H_{n-k}(M; \mathbb{Z})$ and $H^{k}(M; \mathbb{Z}) \to H_{n-k}(M;\omega_M)$.
An orientation is by definition a trivialization of $\omega_M$; from that you recover the oriented Poincare duality theorem, with the familiar dependence on the orientation.
answered Apr 10, 2011 at 11:05
-
3$\begingroup$ The isomorphism $H_n(M,M-x;\omega_M)$ and $\mathbb Z$ does not depend on anything; the former group is canonically isomorphic to $\omega_{M,x}\otimes \omega_{M,x}$. $\endgroup$ Commented Apr 10, 2011 at 23:59