Orientation covering

Theorem 1.1 c.f. [Dold, VIII 2.11]. Let ${{Stub}}{{Authors|Matthias Kreck}} == Construction == <wikitex include="TeXInclude:">; Let $M$ be a $n$-dimensional topological (or smooth) manifold. We construct an oriented manifold $\hat M$ and a $-fold covering $p : \hat M \to M$ called the orientation covering. As a set $\hat M$ is the set of pairs $(x, o_x)$, where $o_x$ is a local orientation of $M$ at $x$, either given by a generator of $H_n(M, M-x;\mathbb Z)$ or by an orientation of $T_xM$ in the smooth case (for the equivalence of these data see the atlas page on orientation of manifold). The map $p$ assigns $x$ to $(x,o_x)$. Since there are precisely two local orientations, the fibres of this map have cardinality $. Next we define a topology on this set. Let $\varphi : U \to V\subset \mathbb R^n$ be a chart of $M$ (smooth, if $M$ is smooth). We orient $\mathbb R^n$ by the standard orientation given by the standard basis $e_1$, $e_2$, ..., $e_n$, from which we define a continuous local orientation by identifying the tangent space with $\mathbb R^n$. Since for a smooth manifold a tangential orientation defines a homological orientation, this also gives a homological orientation (see atlas page on orientation of manifolds). We call the standard local orientation at $x \in \mathbb R^n$ by $sto_x$. Using the chart we transport this standard orientation to $U$ by the induced map on homology or the differential in the case of tangential orientations. The local orientations given by this orientation of $U$ is a subset of $\hat M$, which we require to be open. Doing the same starting with the non-standard orientation of $\mathbb R^n$ we obtain another subset, which we also call open. We give $\hat M$ the topology generated by these open subsets, where we vary about all charts (smooth charts, if $M$ is smooth). By construction these open subsets are homeomorphic to an open subset of $\mathbb R^n$, and so we obtain an atlas of $\hat M$. In the smooth case this is a smooth atlas making $\hat M$ a smooth manifold. The map $p$ is by construction a $-fold covering, smooth, if $M$ is smooth. By construction $\hat M$ is oriented in a tautological way. Thus we have constructed a $-fold covering of $M$ by an oriented manifold $\hat M$, which is smooth, if $M$ is smooth. This covering is called the '''orientation covering'''. For more information, see \cite{Dold1995|VIII § 2}. </wikitex> == Some remarks== <wikitex include="TeXInclude:">; If $M$ is orientable, we pick an orientation and see that $\hat M$ is the disjoint union of $\{(x,o_x)| \,\, o_x \,\, is \,\, the \,\, local \,\, orientation \,\, given \,\, by \,\, the \,\, orientation \,\, of \,\, M\}$ and its complement, so it is isomorphic to the trivial covering $ M \times \mathbb Z/2$. In turn if the orientation covering is trivial it decomposes $\hat M$ into to open (and thus oriented) subsets homeomorphic to $M$ and so $M$ is orientable. By construction of $\hat M$ the deck transformation of the covering is orientation reversing. If $N $ is an oriented manifold and $p: N \to M$ is a $-fold covering with orientation reversing deck transformation, then it is isomorphic to the orientation covering. Namely we have a map $N \to \hat M$ by mapping $y \in N$ to $(p(y), orientation \,\, induced \,\, by \,\, p)$. This is an isomorphism of these two coverings. By the considerations above, $M$ is orientable if and only if this covering is trivial, or $M$ is non-orientable if and only if $N$ is connected. </wikitex> == References == {{#RefList:}} == External links == * The Encylopedia of Mathematics article on [http://www.encyclopediaofmath.org/index.php/Orientation orientation]. * The Wikipedia page on the [[Wikipedia:Orientability#Orientable_double_cover|orientability]]. [[Category:Theory]]M$ be a $n$-dimensional topological manifold. There is an oriented manifold $\hat M$ and a $2$-fold covering $p \colon \hat M \to M$ called the orientation covering. If $M$ is a smooth, resp. piecewise linear, manifold then $\hat M$ and the covering map $p$ are smooth, resp. piecewise linear.

Remark 1.2. The covering $p \colon \hat M \to M$ is called the orientation covering. For more information, see [Dold1995, VIII § 2].

Proof. As a set $\hat M$ is the set of pairs $(x, o_x)$, where $o_x$ is a local orientation of $M$ at $x$, either given by a generator of $H_n(M, M-x;\mathbb Z)$ or by an orientation of $T_xM$ in the smooth case (for the equivalence of these data see the atlas page on orientation of manifold). The map $p$ assigns $x$ to $(x,o_x)$. Since there are precisely two local orientations, the fibers of this map have cardinality $2$.

Next we define a topology on this set. Let $\varphi : U \to V\subset \mathbb R^n$ be a chart of $M$ (smooth, if $M$ is smooth). We orient $\mathbb R^n$ by the standard orientation given by the standard basis $e_1$, $e_2$, ..., $e_n$, from which we define a continuous local orientation by identifying the tangent space with $\mathbb R^n$. Since for a smooth manifold a tangential orientation defines a homological orientation, this also gives a homological orientation (see atlas page on orientation of manifolds). We call the standard local orientation at $x \in \mathbb R^n$ by $sto_x$. Using the chart we transport this standard orientation to $U$ by the induced map on homology or the differential in the case of tangential orientations. The local orientations given by this orientation of $U$ is a subset of $\hat M$, which we require to be open. Doing the same starting with the non-standard orientation of $\mathbb R^n$ we obtain another subset, which we also call open. We give $\hat M$ the topology generated by these open subsets, where we vary over all charts of $M$ (smooth charts, if $M$ is smooth). By construction these open subsets are homeomorphic to an open subset of $\mathbb R^n$, and so we obtain an atlas of $\hat M$. In the smooth case this is a smooth atlas making $\hat M$ a smooth manifold. The map $p$ is by construction a $2$-fold covering, smooth, if $M$ is smooth. By construction $\hat M$ is oriented in a tautological way. Thus we have constructed a $2$-fold covering of $M$ by an oriented manifold $\hat M$, which is smooth, if $M$ is smooth.

$\square$