Can one prove that orientability(of a manifold)is a topological property without using algebraic topology? That is, using a combination of general topology,linear algebra,and topological groups(such as Lie groups-notably the orthogonal group $O(n)$). The basic idea I have would be to show that the group actions of orientation reversing orthogonal linear transformations(elements of $O(n)$ with $Det=-1$) on a vector field defined on an orientable surface are not homotopic to the actions by an orientation preserving transformation(elements of $O(n)$ with $Det=+1$). An example I could think of are the actions of $I_3$ and $-I_3$(identity and antipodal map)on a normal vector field defined on the unit sphere $S^2$ embedded in $\mathbb R^3$. To reverse the direction of a normal vector on S2 involves rotating it about an orthogonal axis tangent to $S^2$ and passing it through the surface.

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    $\begingroup$ A topological property is something that is preserved by homeomorphisms. If you define orientability in terms of the tangent bundle, then there is no obvious sense in which this is true since homeomorphisms are not necessarily differentiable. Indeed some manifolds have multiple incompatible smooth structures. The only way I know how to define an orientation in a way that doesn't reference a smooth structure uses homology theory. $\endgroup$ Oct 25, 2013 at 20:32
  • $\begingroup$ How else can the orientability of a manifold M be defined? Also what do you mean by "incompatible" in terms of smooth structure(s) on a Manifold that has more than 1 such structure.... $\endgroup$
    – Mr X
    Oct 25, 2013 at 21:34
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    $\begingroup$ An orientation of a topological $n$-manifold can be defined as a consistent choice of generator for $H_n(M,M\setminus\{p\})\cong\mathbb Z$ for each $p\in M$. $\endgroup$ Oct 25, 2013 at 22:55
  • $\begingroup$ Also by "incompatible" I mean that the different smooth structures are not diffeomorphic. $\endgroup$ Oct 25, 2013 at 22:56


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