What are all the functions that preserve the cross ratio?
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1
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Suppose a function $f:mathbb {RP}^1to mathbb {RP}^1$ satisfy:
$$
left[f(a),f(b);f(c),f(d)right]=left[a,b;c,dright]
$$
for all $a,b,c,d in mathbb {RP}^1$.
What can the function be in general? Möbius transformations are certainly one type, but are there any other? Suppose the function is linear, or differentiable, I can prove that there are none. But can we do this without these assumptions?
projective-geometry mobius-transformation cross-ratio
asked Oct 1 at 2:51
Trebor
54912
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up vote
1
down vote
favorite
Suppose a function $f:mathbb {RP}^1to mathbb {RP}^1$ satisfy:
$$
left[f(a),f(b);f(c),f(d)right]=left[a,b;c,dright]
$$
for all $a,b,c,d in mathbb {RP}^1$.
What can the function be in general? Möbius transformations are certainly one type, but are there any other? Suppose the function is linear, or differentiable, I can prove that there are none. But can we do this without these assumptions?
projective-geometry mobius-transformation cross-ratio
asked Oct 1 at 2:51
Trebor
54912
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Suppose a function $f:mathbb {RP}^1to mathbb {RP}^1$ satisfy:
$$
left[f(a),f(b);f(c),f(d)right]=left[a,b;c,dright]
$$
for all $a,b,c,d in mathbb {RP}^1$.
What can the function be in general? Möbius transformations are certainly one type, but are there any other? Suppose the function is linear, or differentiable, I can prove that there are none. But can we do this without these assumptions?
projective-geometry mobius-transformation cross-ratio
asked Oct 1 at 2:51
Trebor
54912
Suppose a function $f:mathbb {RP}^1to mathbb {RP}^1$ satisfy:
$$
left[f(a),f(b);f(c),f(d)right]=left[a,b;c,dright]
$$
for all $a,b,c,d in mathbb {RP}^1$.
What can the function be in general? Möbius transformations are certainly one type, but are there any other? Suppose the function is linear, or differentiable, I can prove that there are none. But can we do this without these assumptions?
projective-geometry mobius-transformation cross-ratio
projective-geometry mobius-transformation cross-ratio
asked Oct 1 at 2:51
Trebor
54912
asked Oct 1 at 2:51
Trebor
54912
asked Oct 1 at 2:51
Trebor
54912
asked Oct 1 at 2:51
Trebor
54912
asked Oct 1 at 2:51
Trebor
54912
54912
add a comment |
add a comment |
1 Answer
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1
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Möbius transformations are the only such functions. Indeed, just note that $[0,infty;x,1]=x$ for all $xinmathbb{RP}^1$, and so we must have $$x=[0,infty;x,1]=[f(0),f(infty);f(x),f(1)].$$ This implies $a=f(0)$, $b=f(infty)$, and $c=f(1)$ are all distinct (otherwise the cross-ratio on the right would be the same for all $x$). But given three distinct points $a,b,cinmathbb{RP}^1$, the function $g(x)=[a,b;x,c]$ is a Möbius transformation $mathbb{RP}^1tomathbb{RP}^1$. The equation above then says that $g(f(x))=x$ for all $x$ so $f$ must be the inverse of $g$, which is also a Möbius transformation.
answered Nov 13 at 6:19
Eric Wofsey
175k12202326
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Möbius transformations are the only such functions. Indeed, just note that $[0,infty;x,1]=x$ for all $xinmathbb{RP}^1$, and so we must have $$x=[0,infty;x,1]=[f(0),f(infty);f(x),f(1)].$$ This implies $a=f(0)$, $b=f(infty)$, and $c=f(1)$ are all distinct (otherwise the cross-ratio on the right would be the same for all $x$). But given three distinct points $a,b,cinmathbb{RP}^1$, the function $g(x)=[a,b;x,c]$ is a Möbius transformation $mathbb{RP}^1tomathbb{RP}^1$. The equation above then says that $g(f(x))=x$ for all $x$ so $f$ must be the inverse of $g$, which is also a Möbius transformation.
answered Nov 13 at 6:19
Eric Wofsey
175k12202326
add a comment |
up vote
1
down vote
accepted
Möbius transformations are the only such functions. Indeed, just note that $[0,infty;x,1]=x$ for all $xinmathbb{RP}^1$, and so we must have $$x=[0,infty;x,1]=[f(0),f(infty);f(x),f(1)].$$ This implies $a=f(0)$, $b=f(infty)$, and $c=f(1)$ are all distinct (otherwise the cross-ratio on the right would be the same for all $x$). But given three distinct points $a,b,cinmathbb{RP}^1$, the function $g(x)=[a,b;x,c]$ is a Möbius transformation $mathbb{RP}^1tomathbb{RP}^1$. The equation above then says that $g(f(x))=x$ for all $x$ so $f$ must be the inverse of $g$, which is also a Möbius transformation.
answered Nov 13 at 6:19
Eric Wofsey
175k12202326
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Möbius transformations are the only such functions. Indeed, just note that $[0,infty;x,1]=x$ for all $xinmathbb{RP}^1$, and so we must have $$x=[0,infty;x,1]=[f(0),f(infty);f(x),f(1)].$$ This implies $a=f(0)$, $b=f(infty)$, and $c=f(1)$ are all distinct (otherwise the cross-ratio on the right would be the same for all $x$). But given three distinct points $a,b,cinmathbb{RP}^1$, the function $g(x)=[a,b;x,c]$ is a Möbius transformation $mathbb{RP}^1tomathbb{RP}^1$. The equation above then says that $g(f(x))=x$ for all $x$ so $f$ must be the inverse of $g$, which is also a Möbius transformation.
answered Nov 13 at 6:19
Eric Wofsey
175k12202326
Möbius transformations are the only such functions. Indeed, just note that $[0,infty;x,1]=x$ for all $xinmathbb{RP}^1$, and so we must have $$x=[0,infty;x,1]=[f(0),f(infty);f(x),f(1)].$$ This implies $a=f(0)$, $b=f(infty)$, and $c=f(1)$ are all distinct (otherwise the cross-ratio on the right would be the same for all $x$). But given three distinct points $a,b,cinmathbb{RP}^1$, the function $g(x)=[a,b;x,c]$ is a Möbius transformation $mathbb{RP}^1tomathbb{RP}^1$. The equation above then says that $g(f(x))=x$ for all $x$ so $f$ must be the inverse of $g$, which is also a Möbius transformation.
answered Nov 13 at 6:19
Eric Wofsey
175k12202326
answered Nov 13 at 6:19
Eric Wofsey
175k12202326
answered Nov 13 at 6:19
Eric Wofsey
175k12202326
answered Nov 13 at 6:19
Eric Wofsey
175k12202326
175k12202326
add a comment |
add a comment |
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