What are all the functions that preserve the cross ratio?
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
add a comment |
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
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
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
54912
add a comment |
add a comment |
1 Answer
1
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.
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.
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.
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.
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
175k12202326
add a comment |
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2937507%2fwhat-are-all-the-functions-that-preserve-the-cross-ratio%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown