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The interiors of any two rectangles are conformally equivalent, by the Riemann mapping theorem. Suppose with each rectangle, we glue opposite sides together, and the metric on the quotient space, which is a torus, is that the distance between two points is the length of the shortest arc connecting them. Are the two quotient spaces still conformally equivalent, when the shapes of the rectangles differ? For which pairs of shapes is the answer "yes"?

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In general the answer is no, even in a slightly more general case. The tori you describe can be represented by $\mathbb{C} / \Gamma_\tau$, where $\Gamma_\tau$ is the lattice generated by $1$ and $\tau$ with $\Im \tau > 0$. (Rectangles correspond to $\tau$ being purely imaginary.) If you have a conformal map between two such tori, you can lift it to their universal covers, and so you get an affine map $T(z)=az+b$ in the plane. Composing with a translation (which projects to a conformal self-map of any such torus), you can assume that $T(0)=0$, so $T(z) = az$. This map has to map the lattice $\Gamma_{\tau_1}$ to $\Gamma_{\tau_2}$, so the question is which of these lattices are equivalent under multiplication with some complex scalar $a$. Obviously $\tau \Gamma_{-1/\tau} = \Gamma_\tau$, and $\Gamma_\tau = \Gamma_{\tau+1}$. The maps $\tau \mapsto -1/\tau$ and $\tau\mapsto \tau+1$ generate $PSL(2,\mathbb{Z}) = \left\{ \tau \mapsto \frac{a\tau+b}{c\tau+d}: a,b,c,d \in \mathbb{Z}, \, ad-bc=1 \right\}$, and it can be shown that $\tau_1$ and $\tau_2$ generate equivalent lattices iff they are in the same orbit of $PSL(2,\mathbb{Z})$ acting on the upper halfplane $\Im \tau > 0$. For the special case of rectangles, we get that $\tau_1 = i t_1$ and $\tau_2 = i t_2$ generate equivalent lattices iff $t_1=t_2$ or $t_1 = 1/t_2$, i.e., iff the rectangles are similar.

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"If you have a conformal map between two such tori, you can lift it to their universal covers, and so you get an affine map $T(z)=az+b$." It would have to be an affine map when restricted to the points to which the corners of the rectangles are lifted, but why would it have to be an affine map when applied to other values of $z$? –  Michael Hardy Oct 20 '12 at 16:13
    
You get a conformal map from the plane to itself, and the only such maps are affine maps. –  Lukas Geyer Oct 20 '12 at 18:13
    
I see: bijective and conformal in both directions, and the domain is the plane, implies it's affine. –  Michael Hardy Oct 20 '12 at 21:25

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