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Given that $\mu(A) := \iint_{A}\frac{\mathrm dx\mathrm dy}{y^2}$ where $A \subset H$ and $H$ is the upper half-plane, I need to show that:

a. The measure $\mu$ is invariant under all $g \in SL_2(\mathbb R)$ i.e. $\mu(gA) = \mu(A)$.
b. Compute the hyperbolic area of the standard fundamental domain for the modular group $SL_2(\mathbb Z)$ ( $ A = \{\tau | -0.5 \leq \Re(\tau) \leq 0.5 , |\tau| \geq 1\} $ ).

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I don't know measure theory, but my guess is that the trick is in recognising that measure density here is equal to the square root of the determinant of Riemannian metric up to multiplication by a constant (see en.wikipedia.org/wiki/Poincar%C3%A9_half-plane_model) so it must be preserved by all isometries, and the group of isometries is exactly $PS^*L(2, \mathbb{R})$. $PSL(2, \mathbb{R}) < PS^*L(2, \mathbb{R})$, and $SL(2, \mathbb{R})$ is obviously a double covering of $PSL(2, \mathbb{R})$, so if in this case it acts on $H$ by this covering, you should be OK :) –  Alexei Averchenko Apr 2 '11 at 13:49
    
@Alexei: I find your comment rather misleading. Yes, everything works out fine because $\operatorname{SL}(2,\mathbb{R})$ acts isometrically on $\mathbb{H}$ and thus preserves the gramian matrix (rather than the trace) of the metric, but that's exactly what I outlined in my answer below. By the way, how do you see that $\operatorname{PS^{\ast}L}(2,\mathbb{R})$ is the isometry group of $\mathbb{H}$? In other words, there's no way around getting one's hands a bit dirty. –  t.b. Apr 2 '11 at 14:02
    
Yes, you are right :) Yet I think that proving isometry is slightly simpler, and for $PSL(2, \mathbb{R})$ the proof is very beautiful when you identify its elements with a subgroup of Möbius transformations. –  Alexei Averchenko Apr 2 '11 at 14:16
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1 Answer

up vote 6 down vote accepted

Hints:

  1. That the measure is (left) invariant is essentially shown here.

  2. You want to compute the integral $$\mu(A) = \int_{-\frac{1}{2}}^{\frac{1}{2}} \int_{\sqrt{1-x^2}}^{\infty} \frac{1}{y^2}\,dy\,dx$$ and that shouldn't be too hard. To check your computations, the answer is $\frac{\pi}{3}$.


On Pavel's request I elaborate on point 1.

Write $z = x + iy$ and $w = g\cdot z = \frac{az + b}{cz + d} = u + iv$ where $g = \begin{bmatrix}a & b \\\ c & d \end{bmatrix}\in \operatorname{SL}(2,\mathbb{R})$. Then the Jacobian determinant is $$\frac{\partial (u,v)}{\partial(x,y)} = \frac{\partial u}{\partial x} \frac{\partial v}{\partial y} - \frac{\partial u}{\partial y} \frac{\partial v}{\partial x}$$ and by appealing to the Cauchy-Riemann equations we see that this is equal to $$\frac{\partial (u,v)}{\partial(x,y)} = \left(\frac{\partial u}{\partial x}\right)^2 + \left(\frac{\partial v}{\partial x}\right)^2 = \cdots = \frac{1}{|cz + d|^4}$$ using the calculations in the Wikipedia link I gave above.

Recalling that $v = \frac{y}{|cz + d|^2}$ we can compute $$ \mu(gA) = \int\!\!\!\!\int_{gA} \frac{du\,dv}{v^2} = \int\!\!\!\!\int_{A} \frac{|cz + d|^4}{y^2}\,\frac{\partial (u,v)}{\partial(x,y)}\, dx\, dy = \int\!\!\!\!\int_{A} \frac{dx\, dy}{y^2} = \mu(A)$$ as we wanted.

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Thanks. About 1 - I didn't learn tensors, so I didn't understand why it is invariant and how the volume element derived from there. Can you explain some more? –  Pavel Apr 2 '11 at 12:37
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