Two kinds of orientability/orientation for a differentiable manifold - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T02:32:15Z http://mathoverflow.net/feeds/question/10966 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/10966/two-kinds-of-orientability-orientation-for-a-differentiable-manifold Two kinds of orientability/orientation for a differentiable manifold Anweshi 2010-01-06T21:07:41Z 2010-01-20T23:57:16Z <p>Let $M$ be a differentiable manifold of dimension $n$. First I give two definitions of Orientability.</p> <p>The first definition should coincide with what is given in most differential topology text books, for instance Warner's book. </p> <blockquote> <p>Orientability using differential forms: There exists a nowhere vanishing differential form $\omega$ of degree $n$ on $M$.</p> </blockquote> <p>The second one is from Greenberg and Harper, "Algebraic Topology". This is the "fundamental class" approach. Let $x$ be a point on $X$, and let $R$ be a commutative ring and in the following the homologies are with coefficients in $R$.</p> <blockquote> <p>Local orientability: A local $R$ orientation of $X$ at $x$ is a choice of a generator of the $R$-module $H_n(X, X-x)$.</p> </blockquote> <p>By a simple application of Excision, it is seen that the above homology module is indeed isomorphic to $R$. We can also so arrange a neighborhood around every point that this local orientation can be "continued to a neighborhood" and is "coherent". Forgive me for being imprecise here; the detailed lemmas are in the reference given above. With this background in mind, we define:</p> <blockquote> <p>A Global $R$-orientation of $X$ consists of: 1. A family $U_i$ of open sets covering of $X$, 2. For each $i$, a local orientation $\alpha_i \in H_n(X, X -U_i)$ of along $X$, such that a "compatibility condition" holds.</p> </blockquote> <p>Here again I am imprecise about the compatibility condition; please check in the reference given above for details. I mean this basically as a question for those who already know both the definitions, as fully writing down the second definition would take 2-3 pages with all the necessary lemmas.</p> <p>Also we define "orientation" to be a such a global choice.</p> <p>Now the question:</p> <blockquote> <p>How do the two definitions, the first one using differential forms, and the second one using homology, match?</p> </blockquote> <p>Of course, to match we have to take $\mathbb{Z}$ to be the base ring for homology. A related question is about the meaning of orientability and orientation when we take a base ring other than $\mathbb{Z}$. It is nice when the base ring is $\mathbb{Z}/2\mathbb{Z}$; every manifold is orientable. But what on earth does it mean to have $4$ possible orientations for the circle or real line for instance, when you take the base ring to be $\mathbb{Z}/5\mathbb{Z}$?</p> <p>Also I ask, are there any additional ways to define orientability/orientation for a differentiable manifold(not just for a vector space)?</p> http://mathoverflow.net/questions/10966/two-kinds-of-orientability-orientation-for-a-differentiable-manifold/10968#10968 Answer by Emerton for Two kinds of orientability/orientation for a differentiable manifold Emerton 2010-01-06T21:37:45Z 2010-01-06T21:47:46Z <p>If $X$ is a differentiable manifold, so that both notions are defined, then they coincide.</p> <p>The patching'' of local orientations that you describe can be expressed more formally as follows: there is a locally constant sheaf $\omega_R$ of $R$-modules on $X$ whose stalk at a point is $H^n(X,X\setminus\{x\}; R).$ Of course, $\omega_R = R\otimes_{\mathbb Z} \omega_{\mathbb Z}$.</p> <p>This sheaf is called the orientation sheaf, and appears in the formulation of Poincare duality for not-necessarily orientable manifolds. It is not the case that <I> any</I> section of this sheaf gives an orientation. (For example, we always have the zero section.) I think the usual definition would be something like a section which generates each stalk.</p> <p>I will now work just with $\mathbb Z$ coefficients, and write $\omega = \omega_{\mathbb Z}$.</p> <p>Since the stalks of $\omega$ are free of rank one over $\mathbb Z$, to patch them together you end up giving a 1-cocyle with values in $GL_1({\mathbb Z}) = \{\pm 1\}.$ Thus underlying $\omega$ there is a more elemental sheaf, a locally constant sheaf that is a principal bundle for $\{\pm 1\}$. Equivalently, such a thing is just a degree two (not necessarily connected) covering space of $X$, and it is precisely the orientation double cover of $X$.</p> <p>Now giving a section of $\omega$ that generates each stalk, i.e. giving an orientation of $X$, is precisely the same as giving a section of the orientation double cover (and so $X$ is orientable, i.e. admits an orientation, precisely when the orientation double cover is disconnected).</p> <p>Instead of cutting down from a locally constant rank 1 sheaf over $\mathbb Z$ to just a double cover, we could also build up to get some bigger sheaves.</p> <p>For example, there is the sheaf ${\mathcal C}^{\infty}_X$ of smooth functions on $X$. We can form the tensor product ${\mathcal C}^{\infty}_X \otimes_{\mathbb Z} \omega,$ to get a locally free sheaf of rank one over ${\mathcal C}^{\infty}$, or equivalently, the sheaf of sections of a line bundle on $X$. This is precisely the line bundle of top-dimensional forms on $X$.</p> <p>If we give a section of $\omega$ giving rise to an orientation of $X$, call it $\sigma$, then we certainly get a nowhere-zero section of ${\mathcal C}^{\infty}_X \otimes_{\mathbb Z} \omega$, namely $1\otimes\sigma$.</p> <p>On the other hand, if we have a nowhere zero section of ${\mathcal C}^{\infty}_X \otimes_{\mathbb Z} \omega$, then locally (say on the the members of some cover $\{U_i\}$ of $X$ by open balls) it has the form $f_i\otimes\sigma_i,$ where $f_i$ is a nowhere zero real-valued function on $U_i$ and $\sigma_i$ is a generator of $\omega_{| U_i}.$</p> <p>Since $f_i$ is nowhere zero, it is either always positive or always negative; write $\epsilon_i$ to denote its sign. It is then easy to see that sections $\epsilon_i\sigma_i$ of $\omega$ glue together to give a section $\sigma$ of $X$ that provides an orientation.</p> <p>One also sees that two different nowhere-zero volume forms will give rise to the same orientation if and only if their ratio is an everywhere positive function. </p> <p>This reconciles the two notions.</p> http://mathoverflow.net/questions/10966/two-kinds-of-orientability-orientation-for-a-differentiable-manifold/10976#10976 Answer by Justin Curry for Two kinds of orientability/orientation for a differentiable manifold Justin Curry 2010-01-06T22:26:46Z 2010-01-07T01:38:23Z <blockquote> <p>Also I ask, are there any additional ways to define orientability/orientation for a differentiable manifold(not just for a vector space)?</p> </blockquote> <p>Another notion of orientability is the existence of an atlas whose transition functions have derivatives with everywhere positive determinant. This gives a clear cut way, along with the Cauchy-Riemann equations, of showing that every complex manifold (say, for simplicity, a Riemann surface) is orientable.</p> http://mathoverflow.net/questions/10966/two-kinds-of-orientability-orientation-for-a-differentiable-manifold/10985#10985 Answer by Ryan Budney for Two kinds of orientability/orientation for a differentiable manifold Ryan Budney 2010-01-06T23:33:24Z 2010-01-20T16:23:00Z <p>Your main question was answered by Emerton. Regarding other notions of orientability, there's many. A popular one is the obstruction-theoretic approach:</p> <p>1) A manifold $M$ is orientable if the tangent bundle $TM$ admits a trivialization when restricted to a $1$-skeleton of a CW-decomposition of $M$. An orientation of $M$ is taken to be a (homotopy class of) trivialization of $TM_{|M^0}$ that extends over $M^1$. </p> <p>2) [Corrected to take into account Chris's comment] You can restate definition 1 in a way that avoids skeleta. A popular one is to define the associated orthogonal (principal) bundle to $TM$, lets call it $O(TM)$. This is the bundle over $M$ whose fibers over points $p \in M$ is the linear isomorphisms between $\mathbb R^m$ and $T_pM$. Then $M$ is orientable if every loop $S^1 \to M$ lifts to a loop $S^1 \to O(TM)$. </p> <p>3) There's a small variant on these ideas called the "orientation cover", this is a 2-sheeted covering space of $M$, and it is connected if and only if $M$ is non-orientable. This has the additional assumption that $M$ is connected. </p> <p>4) Another variant on this comes from bundle classifying-space machinery. Every vector bundle has a classifying map $M \to B(GL_m)$, and $GL_m$ has a subgroup of positive-determinant matrices, call it $GL^+_m$. $M$ is orientable if and only if the classifying map $M \to BGL_m$ lifts to a map $M \to BGL^+_m$, and an orientation is a homotopy-class of such lifts (flexible enough to allow homotopy of the original classifying map). </p> <p>Anyhow, those are a few. There's of course more since all these ideas admit perturbations in various directions. For example, another small variant would be that the 1st Stiefel-Whitney class is trivial. One advantage to approaches (1), (2), (4) is that any of them are natural lead-in to other notions of orientation, like $spin$ or $spin^c$ structures. </p> http://mathoverflow.net/questions/10966/two-kinds-of-orientability-orientation-for-a-differentiable-manifold/12386#12386 Answer by SB for Two kinds of orientability/orientation for a differentiable manifold SB 2010-01-20T06:37:00Z 2010-01-20T06:37:00Z <p>Speaking personally, I was not really comfortable with the notion of orientation until I understood the notion for vector bundles, so I will tell you about that.</p> <p>Given a vector bundle $\pi: E \to B$, first select an orientation for each fiber $\pi^{-1}(b)$. The bundle will be oriented is you made these choices in a coherent manner and the following two are equivalent notions of 'coherent'.</p> <p>1) For every point $b$ in $B$ has a neighborhood $N$ such that there are sections $s_1, \ldots, s_r: N \to E$ such that for all $n \in N$: {$s_1(n), \ldots, s_r(n)$} is an oriented basis for the fiber $\pi^{-1}(n)$.</p> <p>2) Every point $b$ in $B$ is in a vector bundle chart $\phi:N \times \mathbb{R}^r \to \pi^{-1}(N)$ such that $\phi(n,\cdot): n \times \mathbb{R}^r \to \pi^{-1}(n)$ is orientation preserving.</p> <p>Forgetting about picking an orientation for each fiber ahead of time, being orientable is also equivalent to:</p> <p>3) You can cover $B$ with vector bundle charts $\phi:N \times \mathbb{R}^r \to \pi^{-1}(N)$ such that for any two $\phi$ and $\psi$ the linear isomorphism $n \times \mathbb{R}^r \stackrel{\phi(n,\cdot)}{\to} \pi^{-1}(n) \stackrel{\psi(n,\cdot)^{-1}}{\to} n \times \mathbb{R}^r$ is orientation preserving.</p> <p>4) There is a nonzero section of the line bundle $\wedge^rE \to B$.</p> <p>Now a manifold $M$ being orientable is equivalent to its tangent bundle being orientable. Given what has been said, the quickest way to see this is to note that a nonzero n-form on $M$ is by definition a nonzero section of the bundle $\wedge^n (T^*M)$. (Note: $T^*M$ is orientable iff $TM$ is orientable since they are isomorphic as bundles by picking a Riemannian metric.)</p> <p>The canonical example of a nonorientable bundle is the Mobius bundle which is the line bundle over the circle whose total space looks like a Mobius band. In terms of 1) this bundle is not orientable since if you pick a nonzero section (vector) at a point and try to extend to the whole circle, by the time you get back to where you started your vector is now pointing the other way.</p>