2

If $M$ is a n-manifold, I've the definition that: $M$ is orientable if $\forall i, j$ the composition of the charts $\phi_{i}^{-1}\circ \phi_{j}$ has a positive Jacobian determinant (here $\phi_{\alpha}:U_{\alpha} \subset \mathbb{R}^{n} \rightarrow M$).

First of all, if the composition are always smooth (so, diffeomorphism and have Jacobian determinante non zero) and the determinant is continuous, why we don't have an orientation for all manifolds?

And second one, what's a conformal structure on a manifold and why an orientation is induced by this structure?

Reinaldo R.
  • 338
  • 2
    For your first question, the thing is the Jacobian for one composition $\phi_i^{-1} \circ \phi_j$ doesn't have to agree with the Jacobian of some other composition $\phi^{-1}_j \circ \phi_k$. In other words, "Jacobian of the composition" is not defined as a single continuous function over a whole manifold. – xyzzyz Oct 10 '16 at 23:29

1 Answers1

5

Consider for simplicity the case where $M$ can be covered by two coordinate charts $\phi_i \colon U_i \rightarrow M$. Then we only have a single non-trivial transition map $\phi_2^{-1} \circ \phi_1$ defined on $V = \phi_1^{-1}(\phi_1(U_1) \cap \phi_2(U_2)) \subseteq M_1$. The other transition maps are the identity maps $\phi_i^{-1} \circ \phi_i$ which have a positive Jacobian determinant and $\phi_1^{-1} \circ \phi_2$ which is the inverse of $\phi_2^{-1} \circ \phi_1$ and so the Jacobian determinants of $\phi_2^{-1} \circ \phi_1, \phi_2^{-1} \circ \phi_1$ have the same sign.

Assume first that $V$ is connected. Then we will have either $\det D(\phi_2^{-1} \circ \phi_1) > 0$ on the whole of $V$ and so the charts induce an orientation on $M$ or $\det D(\phi_2^{-1} \circ \phi_1) < 0$ on the whole of $V$. In the second case, by modifying one of the charts (say, by switching two coordinates or negating a coordinate) we can get an equivalent collection of charts for which $\det D(\phi_2^{-1} \circ \phi_1) > 0$ and so again an orientation.

However, if $V$ is not connected, then $\det D(\phi_2^{-1} \circ \phi_1)$ will have constant sign on each connected component of $V$ but in general, there isn't any modification we can do to one of the charts that will guarantee that $\det D(\phi_2^{-1} \circ \phi_1)$ will be positive. If $V$ has two connected components on which $\det D(\phi_2^{-1} \circ \phi_1)$ has different sign, then by modifying $\phi_i$ in the way described above will only switch the sign of the components.

In general, a conformal structure on a manifold is an equivalence class of Riemannian metrics and one can talk about conformal structures on non-orientable manifolds so you need to give more context so that your question makes sense.

levap
  • 65,634
  • 5
  • 79
  • 122
  • 1
    I get it, thank you. About the conformal structure, I was reading a book (osserman, minimal surfaces) that said: a conformal structure is when each composition of charts is a conformal map, then that induce a orientation in the manifold and he just talks about Riemannian metrics a little later than this. – Reinaldo R. Oct 11 '16 at 13:02
  • 2
    @ReinaldoR. When you talk about two-dimensional manifolds then indeed one common definition is that a conformal map is described by (an equivalence) of a conformal atlas but this doesn't work well in higher dimensions. Such an atlas determines an orientation and an equivalence class [g] of Riemannian metrics. Conversely, if $M$ is an oriented surface with an equivalence class of metrics $[g]$, one can find a collection of orientation preserving conformal charts on $M$ that induce the class $[g]$ but this is something special to dimension two! – levap Oct 11 '16 at 18:25