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If $T$ is a torus and $\mathbb Z_2$ acts on it by $(z_1,z_2)\rightarrow(z^{-1}_1,-z_2)$, then is $T/\mathbb Z_2$ orientable?

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This argument is morally the same as Matt E's, but interpreted in de Rham cohomology. We denote the quotient by $X$; observe that $X$ inherits the structure of a smooth manifold from $T$ via the quotient map $h : T \to X$. Let $f : T \to T$ be the above-mentioned involution. It is not hard to check that $f$ is an orientation-reversing isometry of $T$, and that $h \circ f = h$.

Now, a smooth manifold $X$ is orientable if and only if there exists a nowhere-vanishing differential 2-form on $X$, so let us consider an arbitrary 2-form $\omega$ on $X$. The pullback 2-form $h^* \omega$ on $T$ satisfies $f^* h^* \omega = h^* \omega$. Now take de Rham cohomology classes; noting that $f^* \mu = - \mu$ where $\mu$ is the volume form on $T$ and the fact that $\mu$ generates $H^2 (T, \mathbb{R})$, we conclude that $h^* \omega$ must be cohomologous to $0$. In particular, $h^* \omega$ must vanish somewhere. But that means $\omega$ also vanishes somewhere – so $X$ is not orientable.

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Your involution is orientation reversing in the first variable, and orientation preserving (just a translation) on the second variable. Since the fundamental class of the torus is the product of the fundamental classes of each of the circle factors, we see that your involution acts by $-1$ on the fundamental class of the torus, and so is orientation reversing. Thus the quotient is not orientable.

(If you draw a picture of the torus as a square with identified edges, then you can see that your involution is like a glide-reflection, so that the quotient by your involution is a Klein bottle.)

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  • $\begingroup$ I'm afraid I'm stuck on some of the details of this argument. I can convince myself that the quotient map must induce the zero map in $H^2$, but why does this imply that the quotient is non-orientable? I wanted to say that $H^2$ preserves coequalisers, but I think this is only true for homotopy coequalisers... Clarification would be appreciated! Thanks, $\endgroup$ – Zhen Lin Jul 31 '12 at 5:45
  • $\begingroup$ @ZhenLin: Dear Zhen, The quotient is a closed surface with vanishing $H^2$; it is thus non-orientable. (The $H^2$ of the quotient with char. zero coeffs. is just the invariants in the $H^2$ of $T$, which vanish.) This is an instance of the general principle that quotienting out by an orientation reversing involution gives a non-orientable quotient. (Thinks about the antipodal map the sphere as another example.) Regards, $\endgroup$ – Matt E Jul 31 '12 at 5:56

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