The Poincare Plane distance
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The Poincare plane, $mathbb{H}$, is:
$$mathbb{H} = {(x,y) in mathbb{R^2} | y>0}$$
$$_{a}L = {(x,y) in mathbb{H} | x=a}$$
$$_{c}L_r = {(x,y) in mathbb{H} | (x-c)^2+y^2=r^2}.$$
The Poincare distance is given by the following if $x_1=x_2$:
$$d_mathbb{H}(P,Q)=|ln(y_2/y_1)|$$
My question is, why isn't the distance just $$d_mathbb{H}(P,Q)=|y_2-y_1|$$ instead since $x_1=x_2$? Why is the natural logarithm introduced for the distance?
geometry
asked Nov 13 at 4:57
Robben
1295
add a comment |
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The Poincare plane, $mathbb{H}$, is:
$$mathbb{H} = {(x,y) in mathbb{R^2} | y>0}$$
$$_{a}L = {(x,y) in mathbb{H} | x=a}$$
$$_{c}L_r = {(x,y) in mathbb{H} | (x-c)^2+y^2=r^2}.$$
The Poincare distance is given by the following if $x_1=x_2$:
$$d_mathbb{H}(P,Q)=|ln(y_2/y_1)|$$
My question is, why isn't the distance just $$d_mathbb{H}(P,Q)=|y_2-y_1|$$ instead since $x_1=x_2$? Why is the natural logarithm introduced for the distance?
geometry
asked Nov 13 at 4:57
Robben
1295
The formula comes from the general definition of the line element, $ds^2 = (dx^2 + dy^2)/y^2$. The distance function isn't arbitrary; it's chosen to make the resulting space negatively curved, and it has a very nice isometry group. If you just take the Euclidean distance on $mathbb{H}$, the resulting space is flat, and it's not even complete: consider the case $y_i to 0$.
– anomaly
Nov 13 at 5:08
@anomaly hmm I have not yet covered isometry nor calculus in the book. Is there a different approach in understanding why without introducing isometry or calculus?
– Robben
Nov 13 at 5:59
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The Poincare plane, $mathbb{H}$, is:
$$mathbb{H} = {(x,y) in mathbb{R^2} | y>0}$$
$$_{a}L = {(x,y) in mathbb{H} | x=a}$$
$$_{c}L_r = {(x,y) in mathbb{H} | (x-c)^2+y^2=r^2}.$$
The Poincare distance is given by the following if $x_1=x_2$:
$$d_mathbb{H}(P,Q)=|ln(y_2/y_1)|$$
My question is, why isn't the distance just $$d_mathbb{H}(P,Q)=|y_2-y_1|$$ instead since $x_1=x_2$? Why is the natural logarithm introduced for the distance?
geometry
asked Nov 13 at 4:57
Robben
1295
The Poincare plane, $mathbb{H}$, is:
$$mathbb{H} = {(x,y) in mathbb{R^2} | y>0}$$
$$_{a}L = {(x,y) in mathbb{H} | x=a}$$
$$_{c}L_r = {(x,y) in mathbb{H} | (x-c)^2+y^2=r^2}.$$
The Poincare distance is given by the following if $x_1=x_2$:
$$d_mathbb{H}(P,Q)=|ln(y_2/y_1)|$$
My question is, why isn't the distance just $$d_mathbb{H}(P,Q)=|y_2-y_1|$$ instead since $x_1=x_2$? Why is the natural logarithm introduced for the distance?
geometry
geometry
asked Nov 13 at 4:57
Robben
1295
asked Nov 13 at 4:57
Robben
1295
edited Nov 13 at 17:05
asked Nov 13 at 4:57
Robben
1295
asked Nov 13 at 4:57
Robben
1295
asked Nov 13 at 4:57
Robben
1295
1295
The formula comes from the general definition of the line element, $ds^2 = (dx^2 + dy^2)/y^2$. The distance function isn't arbitrary; it's chosen to make the resulting space negatively curved, and it has a very nice isometry group. If you just take the Euclidean distance on $mathbb{H}$, the resulting space is flat, and it's not even complete: consider the case $y_i to 0$.
– anomaly
Nov 13 at 5:08
@anomaly hmm I have not yet covered isometry nor calculus in the book. Is there a different approach in understanding why without introducing isometry or calculus?
– Robben
Nov 13 at 5:59
add a comment |
The formula comes from the general definition of the line element, $ds^2 = (dx^2 + dy^2)/y^2$. The distance function isn't arbitrary; it's chosen to make the resulting space negatively curved, and it has a very nice isometry group. If you just take the Euclidean distance on $mathbb{H}$, the resulting space is flat, and it's not even complete: consider the case $y_i to 0$.
– anomaly
Nov 13 at 5:08
@anomaly hmm I have not yet covered isometry nor calculus in the book. Is there a different approach in understanding why without introducing isometry or calculus?
– Robben
Nov 13 at 5:59
The formula comes from the general definition of the line element, $ds^2 = (dx^2 + dy^2)/y^2$. The distance function isn't arbitrary; it's chosen to make the resulting space negatively curved, and it has a very nice isometry group. If you just take the Euclidean distance on $mathbb{H}$, the resulting space is flat, and it's not even complete: consider the case $y_i to 0$.
– anomaly
Nov 13 at 5:08
The formula comes from the general definition of the line element, $ds^2 = (dx^2 + dy^2)/y^2$. The distance function isn't arbitrary; it's chosen to make the resulting space negatively curved, and it has a very nice isometry group. If you just take the Euclidean distance on $mathbb{H}$, the resulting space is flat, and it's not even complete: consider the case $y_i to 0$.
– anomaly
Nov 13 at 5:08
@anomaly hmm I have not yet covered isometry nor calculus in the book. Is there a different approach in understanding why without introducing isometry or calculus?
– Robben
Nov 13 at 5:59
@anomaly hmm I have not yet covered isometry nor calculus in the book. Is there a different approach in understanding why without introducing isometry or calculus?
– Robben
Nov 13 at 5:59
add a comment |
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The formula comes from the general definition of the line element, $ds^2 = (dx^2 + dy^2)/y^2$. The distance function isn't arbitrary; it's chosen to make the resulting space negatively curved, and it has a very nice isometry group. If you just take the Euclidean distance on $mathbb{H}$, the resulting space is flat, and it's not even complete: consider the case $y_i to 0$.
– anomaly
Nov 13 at 5:08
@anomaly hmm I have not yet covered isometry nor calculus in the book. Is there a different approach in understanding why without introducing isometry or calculus?
– Robben
Nov 13 at 5:59