Abstract

In our paper in Ann. I. H. Poincare – PR 41 (2005) 1101–1123, there is a mistake in the proof of the key Proposition 4.10: the use of dominated convergence on page 1112, line 5 from the bottom, is not justified since the dominating terms also vary when passing to the limit. Here is a correct proof that the error term R(d1,d2) of (4.11) tends to 0 when d1 is arbitrary (fixed) and d2 →∞. (See Fig. 5 on page 1111 for a quick understanding of the involved quantities.) Proof of Proposition 4.10. Applying (3.1) to the projection π1 gives G1(x1, y1)= ∑w2∈H(y2) G(x, y1w2). Let w2 ∈H(y2), where H(y2) is the horocycle of y2 in Tr . We write v2 = v(w2) for the unique element in H(x2) that satisfies v2 w2. By Lemma 4.4, the random walk has to pass through some point of the form in {u1v2: u1 ∈ H(x1)} on the way from x to y1w2, and it also has to pass through some point in {c1u2: u2 ∈ Tr , h(u2)=−h(c1)}. Therefore, the stopping time t = min{t1(c1), t2(v(w2))} is a.s. finite, and the random walk passes through Zt before reaching y1w2. We obtain the decomposition (modified with respect to the old one) G(x,y1w2)= Ex ( G(Zt, y1w2) ) = Ex ( 1[t2(v2)<t1(c1)]G(Zt2(v2), y1w2) )+ Ex(1[t1(c1)<t2(v2)]G(Zt1(c1), y1w2)). Now, if starting at x, we have t2(v2) < t1(c1), then Zt2(v2) = u1v2 for some random u1 ∈ H(x1) that must satisfy u(u1, y1) = u1 and d(u1, y1) = d1, since c1 cannot lie on x1 u1. But we also have u(v2,w2) = u2 = 0 and d(v2,w2)= d2. That is, the points u1v2 and y1w2 have the same relative position as the points x and y, and therefore G(u1v2, y1w2)=G(x,y) by Lemma 4.3. We get Ex ( 1[t2(v2)<t1(c1)]G(Zt2(v2), y1w2) ) = Prx[t2(v2) < t1(c1)]G(x,y). Now, given v2 ∈H(x2), there are precisely rd2 elements w2 ∈H(y2) with v(w2)= v2. Combining all these observations,

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