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Is there any nice example of a 1-dimensional non-hausdorff manifold that is not oriented? I have tried the line with two origins, but maybe something more exotic is needed?

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How about the quotient of the real line by the relation $x\sim y$ iff $x=-y$ and $|x|>1$? It looks sort of like this:

enter image description here

(The two black dots are the equivalence classes of $+1$ and $-1$.)

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  • $\begingroup$ why is this non-orientable?? $\endgroup$ – janmarqz Jan 12 '14 at 18:27
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    $\begingroup$ Intuitively, if your orient the left-hand loop in the clockwise direction, then approaching from above forces the right-hand ray to be oriented to the right, while approaching from below forces it to be oriented to the left. You can turn this into a rigorous argument by covering the space with two coordinate charts (images of $(-1,\infty)$ and $(-\infty,1)$ in the quotient space), and considering the various possibilities for choices of orientations on the two components of the overlap. $\endgroup$ – Jack Lee Jan 12 '14 at 18:30
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Here is a more technical construction:

Let $X = (0, 2]$ with a basis consisting of open intervals in $(0, 2)$ as well as subsets of the form $(1 - \epsilon, 1) \cup (2 - \epsilon, 2]$ for $\epsilon > 0$.

This space is clearly locally Euclidean for $x \in X - \{2\}$. For $2$, use the following homeomorphism from $V = (1 - \epsilon, 1) \cup (2 - \epsilon, 2]$ onto the open interval $(1 - \epsilon, 1 + \epsilon)$ in $\Bbb R$: $$ f(x) = \begin{cases} x &: x \in (1 - \epsilon, 1) \\ 3 - x &: x \in (2 - \epsilon, 2] \end{cases} $$

The points $1$ and $2$ have no disjoint neighborhoods, so $X$ is not Hausdorff. $X$ is second countable, since it has a basis consisting of open sets with rational end points.

Finally, $X$ is non-orientable. One possible generator for $H_1(V, V - \{2\})$ is the following singular $1$-simplex: $$ \tau(t) = \begin{cases} 1 - \dfrac{\epsilon}{2} + t\epsilon &: t \in \left[0, \dfrac{1}{2}\right) \\ 2 - \left(t - \dfrac{1}{2}\right)\epsilon &: t \in \left[\dfrac{1}{2}, 1\right] \end{cases} $$

No matter what orientation you choose for $(0, 2)$, $\tau$ cannot be compatible with it. Verify this! This is also true for the other generator of $H_1(V, V - \{2\})$.

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