Orientation character

Latest revision as of 10:24, 14 April 2014

1 Definition

Let ${{Authors|Daniel Kasprowski|Christoph Winges}}{{MediaWiki:Being reviewed}} == Definition == <wikitex>; The following definition can be found in \cite{Davis&Kirk2001|Section 5}. Let $M$ be a topological manifold. The orientation character is a homomorphism $$ w \colon \pi_1(M) \to \Zz/2 $$ which may be defined as follows. Take the [[orientation covering]] $p:\hat M\to M$ and let $x\in M$ be the base point with lifts $\tilde x_1$ and $\tilde x_{-1}$ in $\hat M$. For a loop $\gamma \colon ([0, 1], \{0, 1\}) \to (M, x)$ based at $x$, let $\tilde \gamma \colon [0, 1] \to M$ be the lift of $\gamma$ with $\tilde{\gamma}(0) = \tilde x_1$ and define $w$ on the homotopy class of $\gamma$ by $$w([\gamma]) := \left\{ \begin{array}{rl} 1~~ & \text{if $\tilde{\gamma}(1) = \tilde x_1$,} \ -1~~ & \text{if $\tilde{\gamma}(1) = \tilde x_{-1}$.} \end{array} \right. $$ The orientation character $w\colon\pi_1(M)\to \Zz/2$ induces a map $M \to B\Zz/2$ from $M$ to the [[Wikipedia:Classifying space|classifying space]] of $\Zz/2$, which is unique up to homotopy. By the first definition of the orientation character this map classifies the orientation covering. So $w$ is trivial if and only if $\hat M\cong M\times\Zz/2$, and therefore $M$ is orientable. </wikitex> == Alternative descriptions == <wikitex>; Let $\tilde M$ be the universal covering of $M$. Then $\pi_1(M)$ acts on $\tilde M$ by [[Wikipedia:Covering_space#Deck_transformation_group.2C_regular_covers|deck transformations]]. {{beginthm|Proposition}} Let $w\colon \pi_1(M)\to \Zz/2$ be the orientation character, then $w([\gamma])=1$ if and only if $[\gamma]$ acts [[Orientation of manifolds#Orientation of topological manifolds|orientation preserving]] on $\tilde M$. {{endthm}} {{beginproof}} If $M$ is orientable, then an orientation on $M$ induces an orientation on $\tilde M$ and every deck transformation is orientation preserving. If $M$ is non-orientable, then the standard orientation of the orientation covering $\hat M$ of $M$ induces an orientation on $\tilde M$. By construction the non-trivial deck transformation of $\hat M$ is orientation-reversing. Therefore, a deck transformation of $\tilde M\to M$ is orientation preserving if and only if it acts trivial on $\hat M$. {{endproof}} </wikitex> === The orientation character for smooth manifolds === <wikitex>; If the manifold $M^n$ carries a smooth structure, the orientation character can be additionally characterised in terms of tangential data. Define a map $w^\prime \colon \pi_1(M) \to \Zz/2$ in the following way: Let $x \in M$ be the base point. Pick a chart $\varphi \colon U \to V \subset \mathbb{R}^n$ around $x$. Let $\gamma$ be a based loop in $M$. The standard orientation $[e_1, \dots, e_n]$ of $\mathbb{R}^n$ corresponds to an orientation $o$ of $T_xM$ under the differential of $\varphi^{-1}$ (cf. [[Orientation of manifolds#Reformulations of orientation for smooth manifolds|Orientation of manifolds]]). Fibre transport along $\gamma$ yields another orientation $o^\prime$ of $T_xM$. Now set $$w^\prime([\gamma]) := \left\{ \begin{array}{rl} 1~~ & \text{if $o^\prime=o$,} \ -1~~ & \text{if $o^\prime=-o$.} \end{array} \right. $$ {{beginthm|Proposition}}\label{x} The map $w^\prime$ is a well-defined group homomorphism and coincides with the orientation character $w$.{{endthm}} {{beginproof}} This follows from the comparison between homological and tangential orientation. See [[Orientation of manifolds]]. {{endproof}} In addition to this geometric characterisation, the orientation character also admits a description in terms of characteristic classes: Under the "evaluation" homomorphism of the Universal Coefficient Theorem $H^1(M;\mathbb{Z}/2) \to \hom(H_1(M;\mathbb{Z}), \mathbb{Z}/2)$, the first Stiefel-Whitney class $w_1(M)$ corresponds to a homomorphism $w^{\prime\prime} \colon H_1(M;\mathbb{Z}) \to \mathbb{Z}/2$. {{beginthm|Proposition}}\label{sw} Let $\rho \colon \pi_1(M) \to H_1(M;\mathbb{Z})$ be the [[Wikipedia:Hurewicz theorem|Hurewicz homomorphism]]. Then the orientation character of $M$ is given by the composotion $w = w^{\prime\prime} \circ \rho$.{{endthm}} <!--{{beginthm|Proposition}}\label{sw} Upon precomposition with the Hurewicz homomorphism $\pi_1(M) \to H_1(M;\mathbb{Z})$, the homomorphism $w^{\prime\prime}$ coincides with the orientation character $w$.{{endthm}}--> {{beginproof}} All [[1-manifolds]] are orientable, so suppose that $n = \textup{dim}(M) > 1$. Let $\tau\colon M \to BO(n)$ classify the tangent bundle of $M$ and consider the induced map $\tau_* \colon \pi_1(M) \to \pi_1(BO(n))$. Now $\pi_1(BO(n)) \cong \pi_0(O(n)) = \Z/2$ is generated by the classifying map of the [[Wikipedia:Mobius strip|Mobius bundle]], denote this class by $[\eta]$. Hence for a based loop $\gamma \in M$ with homotopy class $[\gamma] \in \pi_1(M)$, we see that the tangent bundle of $M$ restricted to $\gamma$ is non-trivial if and only if $\tau_*([\gamma]) = [\eta]$. By Proposition \ref{x} we see that $w([\gamma]) = 1$ if and only if $\tau_*([\gamma]) = [\eta]$. Applying the universal coefficient theorem in $BO(n)$, we see that universal first Stiefel-Whitney class $w_1$, which generates $H^1(BO(n); \Z/2)$, has the property that $w_1(\eta) = 1$. Hence we see that $w([\gamma]) = 1$ if and only if $\tau_*([\gamma]) = [\eta]$ if and only if $w_1(\tau_*[\gamma]) = 1$ if and only if $w_1(M)([\gamma]) = 1$. Since $w \colon \pi_1(M) \to \Zz/2$ and $w_1(M) \circ \rho \colon \pi_1(M) \to \Zz/2$ are both homomorphisms we see that they agree. {{endproof}} </wikitex> == Examples == <wikitex>; # Since $\mathbb RP^n$ is orientable if and only if $n$ is odd, the orientation character $w\colon\pi_1(\Rr P^n)\to \Zz/2$ is trivial if and only if $n$ is odd. If $n$ is even $w\colon\pi_1(\Rr P^n) \cong \Zz/2\to \Zz/2$ is the identity. # The open [[Wikipedia:Mobius_strip|Möbius strip]] has fundamental group $\Zz$ and is non-orientable. Therefore, the orientation character is given by the projection $\Zz\twoheadrightarrow\Zz/2$. # If $i\colon N\to M$ is an embedding of a manifold of the same dimension (possibly with boundary), then the orientation character of $N$ is given as the composition $\pi_1(N)\xrightarrow{i_*} \pi_1(M)\xrightarrow{w}\Zz/2$. This follows from the fact that the atlas of $M$ with local orientations induces by restriction an atlas of $N$ with local orientations. Therefore, the orientation covering of $N$ is the orientation covering of $M$ restricted to $N$. # The preimage of $D^n\subseteq M^n$ in the orientation covering $\hat M$ is a disjoint union of two disks. The orientation covering of a [[Connected sum|connected sum]] $M\#N$ along $D^n$ is the "double connected sum" of the orientation coverings along the preimages of $D^n$ in the orientation coverings. How to pair the disks is determined by the local orientations. From this it follows that the orientation character of $M\#N$ is given by the pushout $$\xymatrix{\pi_1(S^{n-1})\ar[r]\ar[d]&\pi_1(M\backslash D^n)\ar[d]\ar[ddr]&\ \pi_1(N\backslash D^n)\ar[r]\ar[drr]&\pi_1(M\#N)\ar@{-->}[dr]&\&&\Zz/2}$$ </wikitex> == The orientation character via Poincaré duality== <wikitex>; Even though the original definition is very geometric, the orientation character is already completely determined by the homotopy type of a given closed manifold. We already know this for differentiable manifolds by Proposition \ref{sw}. In general this is most easily seen in the more abstract setting of [[Poincaré complexes]], see \cite{Lück2001|Section 3.1}. {{beginthm|Lemma}} If $(X, [X], w)$ is a Poincaré complex, then $w$ is the only homomorphism $y\colon\pi_1(X) \to \Zz/2$ such that $X$ is a Poincaré complex with orientation character $y$. {{endthm}} {{beginproof}} Suppose $(X,[X],y)$ is a Poincaré complex. We first introduce some notation: # $\mathbb{Z}^y$ denotes $\mathbb{Z}$ as a right $\mathbb{Z}[\pi_1(X)]$-module, where the $\pi_1(X)$-action is given by $z \cdot g = y(g) \cdot z$. # $\mathbb{Z}[\pi_1(X)]^y := \mathbb{Z}^y \otimes_{\mathbb{Z}[\pi_1(X)]}\mathbb{Z}[\pi_1(X)]$, where the tensor product is equipped with the diagonal action. # For a left $\mathbb{Z}[\pi_1(X)]$-chain complex $C_*$, we define $C_*^y$ analogous to (2.). Applying $H_*$ to the $\mathbb{Z}[\pi_1(X)]$-chain equivalence $? \cap [X] \colon C^{n-*}(\tilde{X}) \to C_*(\tilde{X})^y$, we obtain an isomorphism of $\mathbb{Z}[\pi_1(X)]$-modules $$H_n(C^{n-*}(\tilde{X})) \xrightarrow{\cong} H_0(C_*(\tilde{X})^y) \cong \mathbb{Z}^y.$$ Note that the left hand side is independent of the orientation character $y$. Therefore, $\Zz^y$ is isomorphic to $\Zz^w$ as a $\Zz[\pi_1(X)]$-module and $y$ and $w$ have to agree. {{endproof}} {{beginthm|Corollary}} If $(X, [X], w)$ is a Poincaré complex, then $w$ depends only on the homotopy type of $X$. {{endthm}} {{beginproof}} Let $(X,[X],w)$ and $(X',[X'],w')$ be Poincaré complexes and $f:X\rightarrow X'$ be a homotopy equivalence. Then $(X,[X],w'\circ f_*\colon\pi_1(X)\to\Zz/2)$ is a Poincaré complex and $w$ and $w'\circ f_*$ have to agree. {{endproof}} {{beginthm|Remark}} The above statements are also true for Poincaré pairs in general. {{endthm}} Proposition \ref{sw} generalises to the following: {{beginthm|Proposition|\cite{Byun1999|Lemma 4.3}}} Using the Universal Coefficients Theorem and the Hurewicz map, the orientation character of a Poincaré complex coincides with the first Stiefel-Whitney class of its [[Spivak normal fibration]]. {{endthm}} </wikitex> ==References== {{#RefList:}} ==External links== * The Wikipedia page about the [[Wikipedia:Orientation character|orientation character]] [[Category:Definitions]]M$ be a connected topological manifold throughout. The following definition can be found in [Davis&Kirk2001, Section 5]. The orientation character is a homomorphism

$$\displaystyle w \colon \pi_1(M) \to \{ \pm 1 \} =: C_2$$

which may be defined as follows. Take the orientation covering $p:\widehat M\to M$ and let $x\in M$ be the base point with lifts $\tilde x_1$ and $\tilde x_{-1}$ in $\widehat M$. For a loop $\gamma \colon ([0, 1], \{0, 1\}) \to (M, x)$ based at $x$, let $\tilde \gamma \colon [0, 1] \to M$ be the lift of $\gamma$ with $\tilde{\gamma}(0) = \tilde x_1$ and define $w$ on the homotopy class of $\gamma$ by

$$\displaystyle w([\gamma]) := \left\{ \begin{array}{rl} 1~~ & \text{if $\tilde{\gamma}(1) = \tilde x_1$,} \\ -1~~ & \text{if $\tilde{\gamma}(1) = \tilde x_{-1}$.} \end{array} \right.$$

The orientation character $w\colon\pi_1(M)\to C_2$ induces a map $M \to BC_2$ from $M$ to the classifying space of $C_2$, which is unique up to homotopy. By the definition of the orientation character, this map classifies the orientation covering. So $w$ is trivial if and only if $\widehat M \cong M\times C_2$, and therefore $M$ is orientable.

2 Alternative descriptions

Let $\widetilde M$ be the universal covering of $M$. Then $\pi_1(M)$ acts on $\widetilde M$ by deck transformations.

Proof. If $M$ is orientable, then an orientation on $M$ induces an orientation on $\widetilde M$ and every deck transformation is orientation preserving. If $M$ is non-orientable, then the standard orientation of the orientation covering $\widehat M$ of $M$ induces an orientation on $\widetilde M$. By construction the non-trivial deck transformation of $\widehat M$ is orientation-reversing. Therefore, a deck transformation of $\widetilde M\to M$ is orientation preserving if and only if it acts trivially on $\widehat M$.

$\square$

If the manifold $M^n$ carries a smooth structure, the orientation character can be additionally characterised in terms of tangential data. Define a map $w^\prime \colon \pi_1(M) \to C_2$ in the following way: Let $x \in M$ be the base point. Pick a chart $\varphi \colon U \to V \subset \mathbb{R}^n$ around $x$. Let $\gamma$ be a based loop in $M$. The standard orientation $[e_1, \dots, e_n]$ of $\mathbb{R}^n$ corresponds to an orientation $o$ of $T_xM$ under the differential of $\varphi^{-1}$ (cf. Orientation of manifolds). Fibre transport along $\gamma$ yields another orientation $o^\prime$ of $T_xM$. Now set

$$\displaystyle w^\prime([\gamma]) := \left\{ \begin{array}{rl} 1~~ & \text{if $o^\prime=o$,} \\ -1~~ & \text{if $o^\prime=-o$.} \end{array} \right.$$

Using the comparison between homological and tangential orientation (see Orientation of manifolds), one can show the following:

In addition to this geometric characterisation, the orientation character also admits a description in terms of characteristic classes: Under the "evaluation" homomorphism of the Universal Coefficient Theorem $H^1(M;\Zz/2) \to \hom(H_1(M;\Zz), \Zz/2)$, the first Stiefel-Whitney class $w_1(M)$ corresponds to a homomorphism $H_1(M;\Zz/2) \to \Zz/2$ which we also call $w_1(M)$.

Proof. All 1-manifolds are orientable, so suppose that $n = \textup{dim}(M) > 1$. Let $\tau\colon M \to BO(n)$ classify the tangent bundle of $M$ and consider the induced map $\tau_* \colon \pi_1(M) \to \pi_1(BO(n))$. Now $\pi_1(BO(n)) \cong \pi_0(O(n)) \cong C_2$ is generated by the classifying map of the Möbius bundle; denote this class by $[\eta]$. Hence, for a based loop $\gamma$ in $M$ with homotopy class $[\gamma] \in \pi_1(M)$, we see the pullback $\gamma^*TM$ is non-trivial if and only if $\tau_*([\gamma]) = [\eta]$. By Proposition 2.1, we see that $w([\gamma]) = 1$ if and only if $\tau_*([\gamma]) = [\eta]$. Applying the Universal Coefficient Theorem for $BO(n)$, we see that the universal first Stiefel-Whitney class $w_1$, which generates $H^1(BO(n); \Zz/2)$, has the property that $w_1(\rho([\eta])) = 1$. Hence, we see that $w([\gamma]) = 1$ if and only if $\tau_*([\gamma]) = [\eta]$ if and only if $w_1(\rho(\tau_*[\gamma])) = 1$ if and only if $w_1(M)(\rho([\gamma])) = 1$.

$\square$

3 Examples

1. Since $\mathbb RP^n$ is orientable if and only if $n$ is odd, the orientation character $w\colon\pi_1(\Rr P^n)\to C_2$ is trivial if and only if $n$ is odd. If $n$ is even, $w\colon\pi_1(\Rr P^n) \to C_2$ is an isomorphism.
2. The open Möbius strip has fundamental group $\Zz$ and is non-orientable. Therefore, the orientation character is given by the surjection $\Zz\twoheadrightarrow\Zz/2 \cong C_2$.
3. If $i\colon N\to M$ is an embedding of a manifold of the same dimension (possibly with boundary), then the orientation character of $N$ is given as the composition $\pi_1(N)\xrightarrow{i_*} \pi_1(M)\xrightarrow{w}C_2$. This follows from the fact that the atlas of $M$ with local orientations induces by restriction an atlas of $N$ with local orientations. Therefore, the orientation covering of $N$ is the orientation covering of $M$ restricted to $N$.
4. Let $n \geq 2$. The preimage of an embedded disk $D^n\subseteq M^n$ in the orientation covering $\widehat M$ is a disjoint union of two disks. The orientation covering of a connected sum $M\#N$ along $D^n$ is the "double connected sum" of the orientation coverings along the preimages of $D^n$ in the orientation coverings. How to pair the disks is determined by the local orientations. From this it follows that the orientation character of $M\#N$ is given by the pushout
   

4 The orientation character via Poincaré duality

Even though the original definition is very geometric, the orientation character is already completely determined by the homotopy type of a given closed manifold. Both the Hurewicz homomorphism and the first Stiefel-Whitney class are homotopy invariants (for the second point, see the Manifold Atlas page Wu class), so we already know this for differentiable manifolds by Proposition 2.3. In general this is most easily seen in the more abstract setting of Poincaré complexes, see [Lück2001, Section 3.1].

Proof. Suppose $(X,[X],y)$ is a connected Poincaré complex. We first introduce some notation:

1. $\mathbb{Z}^y$ denotes $\mathbb{Z}$ as a right $\mathbb{Z}[\pi_1(X)]$-module, where the $\pi_1(X)$-action is given by $z \cdot g = y(g) \cdot z$.
2. $\mathbb{Z}[\pi_1(X)]^y := \mathbb{Z}^y \otimes_{\mathbb{Z}[\pi_1(X)]}\mathbb{Z}[\pi_1(X)]$, where the tensor product is equipped with the diagonal action.
3. For a left $\mathbb{Z}[\pi_1(X)]$-chain complex $C_*$, we define $C_*^y$ analogous to (2.).

Applying $H_*$ to the $\mathbb{Z}[\pi_1(X)]$-chain equivalence $? \cap [X] \colon C^{n-*}(\widetilde{X}) \to C_*(\widetilde{X})^y$, we obtain an isomorphism of $\mathbb{Z}[\pi_1(X)]$-modules

$$\displaystyle H_n(C^{n-*}(\widetilde{X})) \xrightarrow{\cong} H_0(C_*(\widetilde{X})^y) \cong \mathbb{Z}^y.$$

Note that the left hand side is independent of the orientation character $y$. Therefore, $\Zz^y$ is isomorphic to $\Zz^w$ as a $\Zz[\pi_1(X)]$-module and $y$ and $w$ have to agree.

$\square$

Proof. Let $(X,[X],w)$ and $(X',[X'],w')$ be connected Poincaré complexes and $f:X\rightarrow X'$ be a homotopy equivalence. Then $(X,[X],w'\circ f_*\colon\pi_1(X)\to C_2)$ is a Poincaré complex and $w$ and $w'\circ f_*$ have to agree.

$\square$

Proposition 2.3 generalises to the following:

5 References

• [Byun1999] Y. Byun, Tangent fibration of a Poincaré complex, J. London Math. Soc. (2) 59 (1999), no.3, 1101–1116. MR1709099 (2000f:57023) Zbl 0935.57032
• [Davis&Kirk2001] J. F. Davis and P. Kirk, Lecture notes in algebraic topology, American Mathematical Society, 2001. MR1841974 (2002f:55001) Zbl 1018.55001
• [Lück2001] W. Lück, A basic introduction to surgery theory, 9 (2001), 1–224. Available from the author's homepage. MR1937016 (2004a:57041) Zbl 1045.57020

6 External links

• The Wikipedia page about the orientation character