Gary
GaryVasco
Posts: 3,352 
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Post by Gary on Nov 26, 2016 23:04:33 GMT
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Post by Admin on Nov 27, 2016 8:58:53 GMT
Gary
I haven't had a look at that suggested exercise yet, so I will postpone looking at your document until I have had a go at it myself.
Vasco
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Post by Admin on Jan 25, 2017 17:02:56 GMT
Gary
In your document, doesn't the plane $\Pi$ through $z$ and $\widetilde{z}$, orthogonal to $\mathbb{C}$ have to also pass through $q$ the centre of $K$, because $z$ and $\widetilde{z}$ must lie on a Euclidean line through $q$?
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Jan 25, 2017 18:10:49 GMT
Gary In your document, doesn't the plane $\Pi$ through $z$ and $\widetilde{z}$, orthogonal to $\mathbb{C}$ have to also pass through $q$ the centre of $K$, because $z$ and $\widetilde{z}$ must lie on a Euclidean line through $q$? Vasco Vasco, I see the problem. I think it might be OK. It is stated that the plane $\Pi$ passes through $z$ and $\tilde{z}$ and is orthogonal to $\mathbb{C}$. I drew the arc from $z$ to $\tilde{z}$ as a great circle of $J$ because it seemed the most natural thing to do. But suppose the position of $z$ requires that we rotate $\Pi$ about the Z axis of $K$. If we don't rotate it beyond the circumference of $J$, and $z$ is still on J, and $\mathcal{I}_K$ maps $J$ to itself, one can still draw an h-line through the two points (see p. 325), so the inversion is still a motion in h-space. Since $zm$ still equals $\tilde{z}m$, it is still a reflection of hyperbolic space in $K$. Gary |
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Post by Admin on Jan 25, 2017 19:03:03 GMT
Gary In your document, doesn't the plane $\Pi$ through $z$ and $\widetilde{z}$, orthogonal to $\mathbb{C}$ have to also pass through $q$ the centre of $K$, because $z$ and $\widetilde{z}$ must lie on a Euclidean line through $q$? Vasco Vasco, I see the problem. I think it might be OK. It is stated that the plane $\Pi$ passes through $z$ and $\tilde{z}$ and is orthogonal to $mathbb{C}$. I drew the arc from $z$ to $\tilde{z}$ as a great circle of $J$ because it seemed the most natural thing to do. But suppose the position of $z$ requires that we rotate $\Pi$ about the Z axis of $K$. If we don't rotate it beyond the circumference of $J$, and $z$ is still on J, and $\mathcal{I}_K$ maps $J$ to itself, one can still draw an h-line through the two points (see p. 325), so the inversion is still a motion in h-space. Since $zm$ still equals $\tilde{z}m$, it is still a reflection of hyperbolic space in $K$. Gary Gary I've got myself a bit confused by this. I think I agree with you that $z$ and $\tilde{z}$ must lie on the great circle of $J$ so that it cuts $K$ orthogonally. In other words this plane containing the great circle of $J$ must also pass through $q$, and $J$ must cut $K$ orthogonally, so that the arc cuts $K$ orthogonally. Have I got that right? Vasco |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Jan 25, 2017 19:49:06 GMT
Vasco, I see the problem. I think it might be OK. It is stated that the plane $\Pi$ passes through $z$ and $\tilde{z}$ and is orthogonal to $mathbb{C}$. I drew the arc from $z$ to $\tilde{z}$ as a great circle of $J$ because it seemed the most natural thing to do. But suppose the position of $z$ requires that we rotate $\Pi$ about the Z axis of $K$. If we don't rotate it beyond the circumference of $J$, and $z$ is still on J, and $\mathcal{I}_K$ maps $J$ to itself, one can still draw an h-line through the two points (see p. 325), so the inversion is still a motion in h-space. Since $zm$ still equals $\tilde{z}m$, it is still a reflection of hyperbolic space in $K$. Gary Gary I've got myself a bit confused by this. I think I agree with you that $z$ and $\tilde{z}$ must lie on the great circle of $J$ so that it cuts $K$ orthogonally. In other words this plane containing the great circle of $J$ must also pass through $q$, and $J$ must cut $K$ orthogonally, so that the arc cuts $K$ orthogonally. Have I got that right? Vasco Vasco, I am arguing the contrary. If $z$ is on $J$, so is $\tilde{z}$, because $\mathcal{I}_K$ maps $J$ to itself, because $J$ is orthogonal to $K$. It doesn't matter whether $z$ is aligned with the centers of the two hemispheres. A plane passing through the Z axis of $K$ and through $z$ and $\tilde{z}$ will intersect $J$ in an h-line orthogonal to $K$ at a point m. Then $zm$ = $\tilde{z}m$ by the metric of the hemisphere. So this is a reflection of hyperbolic space in the h-plane K. I agree that $J$ must cut $K$ orthogonally. I think the arc of $z$ to $\tilde{z}$ must cut $K$ orthogonally. It is not intuitive that it does, but it is a segment of an h-line in $J$ and the plane of $\Pi$, which is orthogonal to $K$ at any $\theta$ angle. Gary |
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Post by Admin on Jan 27, 2017 15:13:39 GMT
Gary
I need to think more about this, before I reply.
Vasco
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Post by Admin on Jan 28, 2017 14:45:08 GMT
Gary
If $J$ and $K$ are orthogonal, they intersect in an h-line. If we imagine a vertical plane through the centres of the hemispheres then the arc of intersection with $J$ cuts this h-line orthogonally through its highest point. If we now rotate the plane about the $Z$-axis of $K$ the plane will still be orthogonal to $K$, but will the arc still be orthogonal to $K$? I don't think it will, but I find it hard to envisage and so I'm not 100% sure.
Vasco
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