Mixing
Understanding the proof that the quaternionic projective space is diffeomorphic to the 4-sphere
$begingroup$
Let us consider the quaternionic non-commutative field $mathbf{H} = mathbf{R}^4$ and its projective space $mathbf{P}_1(mathbf{H})$ of dimension 1, which is, by definition, the set of lines in $mathbf{H}$ passing through the origin endowed with the unique structure of differential manifold for which the canonical left-action of the linear group $mathbf{GL}(1; mathbf{H})$ (where $mathbf{H}^d$ is considered as a left-module over $mathbf{H}$ for every $d in mathbf{N}$) on $mathbf{H}$ (the canonical action is, of course, $(s, L) mapsto s(L)$) is of class $mathscr{C}^infty.$
I want to understand the proof of the diffeomorphism between $mathbf{P}_1(mathbf{H})$ with $mathbf{S}_4$ (the unit sphere in $mathbf{R}^5$).
I sketch the proof now. Consider a point $(x, y) in mathbf{S}_7 subset mathbf{R}^8 = mathbf{R}^4 times mathbf{R}^4.$ Write $x = x_0 + ix_1 + jx_2 + kx_3$ and similarly for $y;$ define $$f(x, y) = (2x bar{y}, |x|^2 - |y|^2) = z = (z_0, z_1, z_2, z_3, z_4) in mathbf{H} times mathbf{R} = mathbf{R}^5.$$
Since $|(x, y)| = 1,$ it turns out $z_4 = 2|x|^2 - 1$ and for such $z_4 in (-1, 1)$ given, the point $2x bar{y}$ can be any in the sphere of radius $|2x bar{y}|^2 = 4 |x|^2 |y|^2 = (z_4+1)(1-z_4)=1-z_4^2,$ hence $f(x, y) = z in mathbf{S}_4$ and $f:mathbf{S}_7 to mathbf{S}_4$ is a surjective $mathscr{C}^infty$-function.
I already know that there exists a canonical submersion $mathbf{S}_7 xrightarrow{pi} mathbf{P}_1(mathbf{H})$ such that two points in $(x, y)$ and $(x', y')$ in $mathbf{S}_7$ are identified if there exists a quaternion $q$ such that $(x, y) = (x'q, y'q) = (x', y') q$ (where we are identifying some vector with matrices in the go, all using canonical bases). Now, observe that $f(x, y) = f(x', y')$ signifies $z = z'$ which is the same as the two relations $$xbar{y} = x' bar{y}' quad text{ and } quad 2|x|^2-1=2|x'|^2-1$$ and these imply that $|x| = |x'|$ so that $x' = x q$ for some quaternion $q$ of norm $1$ and then $xbar{y} = x' bar{y}' = x q bar{y}'$ implying $bar{y} = qbar{y}'$ or $y' = y bar{q}^{-1} = y q.$ That is to say, $f$ can be factorised:
$$f:mathbf{S}_7 xrightarrow{pi} mathbf{P}_1(mathbf{H}) xrightarrow{g} mathbf{S}_4,$$
where $g$ is a bijection, with no differentiability properties a priori.
Since $f$ is of class $mathscr{C}^infty$ as well as $pi$ and since $pi$ is a submersion, we know that $g$ is of class $mathscr{C}^infty.$ Hence, if we show $f$ is a submersion, then $g$ would have to be a submersion. If we had this ($g$ being a bijective submersion), it would turn out that $g$ is the desired diffeomorphism!
Here is the crux I cannot solve myself. How to show $f$ is a submersion $mathbf{S}_7 to mathbf{S}_4$?
Any help appreciated.
real-analysis manifolds differential-topology smooth-manifolds
asked Jan 13 at 2:06
Will M.Will M.
2,645315
$endgroup$
|
show 1 more comment
$begingroup$
Let us consider the quaternionic non-commutative field $mathbf{H} = mathbf{R}^4$ and its projective space $mathbf{P}_1(mathbf{H})$ of dimension 1, which is, by definition, the set of lines in $mathbf{H}$ passing through the origin endowed with the unique structure of differential manifold for which the canonical left-action of the linear group $mathbf{GL}(1; mathbf{H})$ (where $mathbf{H}^d$ is considered as a left-module over $mathbf{H}$ for every $d in mathbf{N}$) on $mathbf{H}$ (the canonical action is, of course, $(s, L) mapsto s(L)$) is of class $mathscr{C}^infty.$
I want to understand the proof of the diffeomorphism between $mathbf{P}_1(mathbf{H})$ with $mathbf{S}_4$ (the unit sphere in $mathbf{R}^5$).
I sketch the proof now. Consider a point $(x, y) in mathbf{S}_7 subset mathbf{R}^8 = mathbf{R}^4 times mathbf{R}^4.$ Write $x = x_0 + ix_1 + jx_2 + kx_3$ and similarly for $y;$ define $$f(x, y) = (2x bar{y}, |x|^2 - |y|^2) = z = (z_0, z_1, z_2, z_3, z_4) in mathbf{H} times mathbf{R} = mathbf{R}^5.$$
Since $|(x, y)| = 1,$ it turns out $z_4 = 2|x|^2 - 1$ and for such $z_4 in (-1, 1)$ given, the point $2x bar{y}$ can be any in the sphere of radius $|2x bar{y}|^2 = 4 |x|^2 |y|^2 = (z_4+1)(1-z_4)=1-z_4^2,$ hence $f(x, y) = z in mathbf{S}_4$ and $f:mathbf{S}_7 to mathbf{S}_4$ is a surjective $mathscr{C}^infty$-function.
I already know that there exists a canonical submersion $mathbf{S}_7 xrightarrow{pi} mathbf{P}_1(mathbf{H})$ such that two points in $(x, y)$ and $(x', y')$ in $mathbf{S}_7$ are identified if there exists a quaternion $q$ such that $(x, y) = (x'q, y'q) = (x', y') q$ (where we are identifying some vector with matrices in the go, all using canonical bases). Now, observe that $f(x, y) = f(x', y')$ signifies $z = z'$ which is the same as the two relations $$xbar{y} = x' bar{y}' quad text{ and } quad 2|x|^2-1=2|x'|^2-1$$ and these imply that $|x| = |x'|$ so that $x' = x q$ for some quaternion $q$ of norm $1$ and then $xbar{y} = x' bar{y}' = x q bar{y}'$ implying $bar{y} = qbar{y}'$ or $y' = y bar{q}^{-1} = y q.$ That is to say, $f$ can be factorised:
$$f:mathbf{S}_7 xrightarrow{pi} mathbf{P}_1(mathbf{H}) xrightarrow{g} mathbf{S}_4,$$
where $g$ is a bijection, with no differentiability properties a priori.
Since $f$ is of class $mathscr{C}^infty$ as well as $pi$ and since $pi$ is a submersion, we know that $g$ is of class $mathscr{C}^infty.$ Hence, if we show $f$ is a submersion, then $g$ would have to be a submersion. If we had this ($g$ being a bijective submersion), it would turn out that $g$ is the desired diffeomorphism!
Here is the crux I cannot solve myself. How to show $f$ is a submersion $mathbf{S}_7 to mathbf{S}_4$?
Any help appreciated.
real-analysis manifolds differential-topology smooth-manifolds
asked Jan 13 at 2:06
Will M.Will M.
2,645315
$endgroup$
$begingroup$
Won't $g$ the stereographic projection be adequate ? Topologically it is well an homeomorphism, it should probably be deduced that it is a diffeomorphism.
$endgroup$
– R. Alexandre
Jan 15 at 19:30
1
$begingroup$
If $g$ is the stereopgraphic projection, then $g:mathbf{R}^{d + 1} to mathbf{S}_d,$ so I am not sure how the quaternionic projective space enters here. I mean, my question is, given the function $f$ above, how to show it is a submersion.
$endgroup$
– Will M.
Jan 15 at 19:43
$begingroup$
I meant the stereographic projection gives a diffeomorphism between $mathbf R^k$ and $S^k$. Since a quaternionic line is just a copy of $mathbf R^4$ ...
$endgroup$
– R. Alexandre
Jan 15 at 20:03
$begingroup$
I believe the stereographic projection is a diffeomorphism from $mathbf{R}^{d+1}$ onto $mathbf{S}_d-{e},$ where $e$ is the "pole of projection."
$endgroup$
– Will M.
Jan 15 at 20:28
$begingroup$
It's still the case at the compactification. If needed, take a second chart by taking an n-other pole and projection
$endgroup$
– R. Alexandre
Jan 16 at 16:24
|
show 1 more comment
$begingroup$
Let us consider the quaternionic non-commutative field $mathbf{H} = mathbf{R}^4$ and its projective space $mathbf{P}_1(mathbf{H})$ of dimension 1, which is, by definition, the set of lines in $mathbf{H}$ passing through the origin endowed with the unique structure of differential manifold for which the canonical left-action of the linear group $mathbf{GL}(1; mathbf{H})$ (where $mathbf{H}^d$ is considered as a left-module over $mathbf{H}$ for every $d in mathbf{N}$) on $mathbf{H}$ (the canonical action is, of course, $(s, L) mapsto s(L)$) is of class $mathscr{C}^infty.$
I want to understand the proof of the diffeomorphism between $mathbf{P}_1(mathbf{H})$ with $mathbf{S}_4$ (the unit sphere in $mathbf{R}^5$).
I sketch the proof now. Consider a point $(x, y) in mathbf{S}_7 subset mathbf{R}^8 = mathbf{R}^4 times mathbf{R}^4.$ Write $x = x_0 + ix_1 + jx_2 + kx_3$ and similarly for $y;$ define $$f(x, y) = (2x bar{y}, |x|^2 - |y|^2) = z = (z_0, z_1, z_2, z_3, z_4) in mathbf{H} times mathbf{R} = mathbf{R}^5.$$
Since $|(x, y)| = 1,$ it turns out $z_4 = 2|x|^2 - 1$ and for such $z_4 in (-1, 1)$ given, the point $2x bar{y}$ can be any in the sphere of radius $|2x bar{y}|^2 = 4 |x|^2 |y|^2 = (z_4+1)(1-z_4)=1-z_4^2,$ hence $f(x, y) = z in mathbf{S}_4$ and $f:mathbf{S}_7 to mathbf{S}_4$ is a surjective $mathscr{C}^infty$-function.
I already know that there exists a canonical submersion $mathbf{S}_7 xrightarrow{pi} mathbf{P}_1(mathbf{H})$ such that two points in $(x, y)$ and $(x', y')$ in $mathbf{S}_7$ are identified if there exists a quaternion $q$ such that $(x, y) = (x'q, y'q) = (x', y') q$ (where we are identifying some vector with matrices in the go, all using canonical bases). Now, observe that $f(x, y) = f(x', y')$ signifies $z = z'$ which is the same as the two relations $$xbar{y} = x' bar{y}' quad text{ and } quad 2|x|^2-1=2|x'|^2-1$$ and these imply that $|x| = |x'|$ so that $x' = x q$ for some quaternion $q$ of norm $1$ and then $xbar{y} = x' bar{y}' = x q bar{y}'$ implying $bar{y} = qbar{y}'$ or $y' = y bar{q}^{-1} = y q.$ That is to say, $f$ can be factorised:
$$f:mathbf{S}_7 xrightarrow{pi} mathbf{P}_1(mathbf{H}) xrightarrow{g} mathbf{S}_4,$$
where $g$ is a bijection, with no differentiability properties a priori.
Since $f$ is of class $mathscr{C}^infty$ as well as $pi$ and since $pi$ is a submersion, we know that $g$ is of class $mathscr{C}^infty.$ Hence, if we show $f$ is a submersion, then $g$ would have to be a submersion. If we had this ($g$ being a bijective submersion), it would turn out that $g$ is the desired diffeomorphism!
Here is the crux I cannot solve myself. How to show $f$ is a submersion $mathbf{S}_7 to mathbf{S}_4$?
Any help appreciated.
real-analysis manifolds differential-topology smooth-manifolds
asked Jan 13 at 2:06
Will M.Will M.
2,645315
$endgroup$
Let us consider the quaternionic non-commutative field $mathbf{H} = mathbf{R}^4$ and its projective space $mathbf{P}_1(mathbf{H})$ of dimension 1, which is, by definition, the set of lines in $mathbf{H}$ passing through the origin endowed with the unique structure of differential manifold for which the canonical left-action of the linear group $mathbf{GL}(1; mathbf{H})$ (where $mathbf{H}^d$ is considered as a left-module over $mathbf{H}$ for every $d in mathbf{N}$) on $mathbf{H}$ (the canonical action is, of course, $(s, L) mapsto s(L)$) is of class $mathscr{C}^infty.$
I want to understand the proof of the diffeomorphism between $mathbf{P}_1(mathbf{H})$ with $mathbf{S}_4$ (the unit sphere in $mathbf{R}^5$).
I sketch the proof now. Consider a point $(x, y) in mathbf{S}_7 subset mathbf{R}^8 = mathbf{R}^4 times mathbf{R}^4.$ Write $x = x_0 + ix_1 + jx_2 + kx_3$ and similarly for $y;$ define $$f(x, y) = (2x bar{y}, |x|^2 - |y|^2) = z = (z_0, z_1, z_2, z_3, z_4) in mathbf{H} times mathbf{R} = mathbf{R}^5.$$
Since $|(x, y)| = 1,$ it turns out $z_4 = 2|x|^2 - 1$ and for such $z_4 in (-1, 1)$ given, the point $2x bar{y}$ can be any in the sphere of radius $|2x bar{y}|^2 = 4 |x|^2 |y|^2 = (z_4+1)(1-z_4)=1-z_4^2,$ hence $f(x, y) = z in mathbf{S}_4$ and $f:mathbf{S}_7 to mathbf{S}_4$ is a surjective $mathscr{C}^infty$-function.
I already know that there exists a canonical submersion $mathbf{S}_7 xrightarrow{pi} mathbf{P}_1(mathbf{H})$ such that two points in $(x, y)$ and $(x', y')$ in $mathbf{S}_7$ are identified if there exists a quaternion $q$ such that $(x, y) = (x'q, y'q) = (x', y') q$ (where we are identifying some vector with matrices in the go, all using canonical bases). Now, observe that $f(x, y) = f(x', y')$ signifies $z = z'$ which is the same as the two relations $$xbar{y} = x' bar{y}' quad text{ and } quad 2|x|^2-1=2|x'|^2-1$$ and these imply that $|x| = |x'|$ so that $x' = x q$ for some quaternion $q$ of norm $1$ and then $xbar{y} = x' bar{y}' = x q bar{y}'$ implying $bar{y} = qbar{y}'$ or $y' = y bar{q}^{-1} = y q.$ That is to say, $f$ can be factorised:
$$f:mathbf{S}_7 xrightarrow{pi} mathbf{P}_1(mathbf{H}) xrightarrow{g} mathbf{S}_4,$$
where $g$ is a bijection, with no differentiability properties a priori.
Since $f$ is of class $mathscr{C}^infty$ as well as $pi$ and since $pi$ is a submersion, we know that $g$ is of class $mathscr{C}^infty.$ Hence, if we show $f$ is a submersion, then $g$ would have to be a submersion. If we had this ($g$ being a bijective submersion), it would turn out that $g$ is the desired diffeomorphism!
Here is the crux I cannot solve myself. How to show $f$ is a submersion $mathbf{S}_7 to mathbf{S}_4$?
Any help appreciated.
real-analysis manifolds differential-topology smooth-manifolds
real-analysis manifolds differential-topology smooth-manifolds
asked Jan 13 at 2:06
Will M.Will M.
2,645315
asked Jan 13 at 2:06
Will M.Will M.
2,645315
edited Jan 13 at 2:12
Will M.
asked Jan 13 at 2:06
Will M.Will M.
2,645315
asked Jan 13 at 2:06
Will M.Will M.
2,645315
asked Jan 13 at 2:06
Will M.Will M.
2,645315
2,645315
$begingroup$
Won't $g$ the stereographic projection be adequate ? Topologically it is well an homeomorphism, it should probably be deduced that it is a diffeomorphism.
$endgroup$
– R. Alexandre
Jan 15 at 19:30
1
$begingroup$
If $g$ is the stereopgraphic projection, then $g:mathbf{R}^{d + 1} to mathbf{S}_d,$ so I am not sure how the quaternionic projective space enters here. I mean, my question is, given the function $f$ above, how to show it is a submersion.
$endgroup$
– Will M.
Jan 15 at 19:43
$begingroup$
I meant the stereographic projection gives a diffeomorphism between $mathbf R^k$ and $S^k$. Since a quaternionic line is just a copy of $mathbf R^4$ ...
$endgroup$
– R. Alexandre
Jan 15 at 20:03
$begingroup$
I believe the stereographic projection is a diffeomorphism from $mathbf{R}^{d+1}$ onto $mathbf{S}_d-{e},$ where $e$ is the "pole of projection."
$endgroup$
– Will M.
Jan 15 at 20:28
$begingroup$
It's still the case at the compactification. If needed, take a second chart by taking an n-other pole and projection
$endgroup$
– R. Alexandre
Jan 16 at 16:24
|
show 1 more comment
$begingroup$
Won't $g$ the stereographic projection be adequate ? Topologically it is well an homeomorphism, it should probably be deduced that it is a diffeomorphism.
$endgroup$
– R. Alexandre
Jan 15 at 19:30
1
$begingroup$
If $g$ is the stereopgraphic projection, then $g:mathbf{R}^{d + 1} to mathbf{S}_d,$ so I am not sure how the quaternionic projective space enters here. I mean, my question is, given the function $f$ above, how to show it is a submersion.
$endgroup$
– Will M.
Jan 15 at 19:43
$begingroup$
I meant the stereographic projection gives a diffeomorphism between $mathbf R^k$ and $S^k$. Since a quaternionic line is just a copy of $mathbf R^4$ ...
$endgroup$
– R. Alexandre
Jan 15 at 20:03
$begingroup$
I believe the stereographic projection is a diffeomorphism from $mathbf{R}^{d+1}$ onto $mathbf{S}_d-{e},$ where $e$ is the "pole of projection."
$endgroup$
– Will M.
Jan 15 at 20:28
$begingroup$
It's still the case at the compactification. If needed, take a second chart by taking an n-other pole and projection
$endgroup$
– R. Alexandre
Jan 16 at 16:24
$begingroup$
Won't $g$ the stereographic projection be adequate ? Topologically it is well an homeomorphism, it should probably be deduced that it is a diffeomorphism.
$endgroup$
– R. Alexandre
Jan 15 at 19:30
$begingroup$
Won't $g$ the stereographic projection be adequate ? Topologically it is well an homeomorphism, it should probably be deduced that it is a diffeomorphism.
$endgroup$
– R. Alexandre
Jan 15 at 19:30
1
1
$begingroup$
If $g$ is the stereopgraphic projection, then $g:mathbf{R}^{d + 1} to mathbf{S}_d,$ so I am not sure how the quaternionic projective space enters here. I mean, my question is, given the function $f$ above, how to show it is a submersion.
$endgroup$
– Will M.
Jan 15 at 19:43
$begingroup$
If $g$ is the stereopgraphic projection, then $g:mathbf{R}^{d + 1} to mathbf{S}_d,$ so I am not sure how the quaternionic projective space enters here. I mean, my question is, given the function $f$ above, how to show it is a submersion.
$endgroup$
– Will M.
Jan 15 at 19:43
$begingroup$
I meant the stereographic projection gives a diffeomorphism between $mathbf R^k$ and $S^k$. Since a quaternionic line is just a copy of $mathbf R^4$ ...
$endgroup$
– R. Alexandre
Jan 15 at 20:03
$begingroup$
I meant the stereographic projection gives a diffeomorphism between $mathbf R^k$ and $S^k$. Since a quaternionic line is just a copy of $mathbf R^4$ ...
$endgroup$
– R. Alexandre
Jan 15 at 20:03
$begingroup$
I believe the stereographic projection is a diffeomorphism from $mathbf{R}^{d+1}$ onto $mathbf{S}_d-{e},$ where $e$ is the "pole of projection."
$endgroup$
– Will M.
Jan 15 at 20:28
$begingroup$
I believe the stereographic projection is a diffeomorphism from $mathbf{R}^{d+1}$ onto $mathbf{S}_d-{e},$ where $e$ is the "pole of projection."
$endgroup$
– Will M.
Jan 15 at 20:28
$begingroup$
It's still the case at the compactification. If needed, take a second chart by taking an n-other pole and projection
$endgroup$
– R. Alexandre
Jan 16 at 16:24
$begingroup$
It's still the case at the compactification. If needed, take a second chart by taking an n-other pole and projection
$endgroup$
– R. Alexandre
Jan 16 at 16:24
|
show 1 more comment
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$begingroup$
Won't $g$ the stereographic projection be adequate ? Topologically it is well an homeomorphism, it should probably be deduced that it is a diffeomorphism.
$endgroup$
– R. Alexandre
Jan 15 at 19:30
1
$begingroup$
If $g$ is the stereopgraphic projection, then $g:mathbf{R}^{d + 1} to mathbf{S}_d,$ so I am not sure how the quaternionic projective space enters here. I mean, my question is, given the function $f$ above, how to show it is a submersion.
$endgroup$
– Will M.
Jan 15 at 19:43
$begingroup$
I meant the stereographic projection gives a diffeomorphism between $mathbf R^k$ and $S^k$. Since a quaternionic line is just a copy of $mathbf R^4$ ...
$endgroup$
– R. Alexandre
Jan 15 at 20:03
$begingroup$
I believe the stereographic projection is a diffeomorphism from $mathbf{R}^{d+1}$ onto $mathbf{S}_d-{e},$ where $e$ is the "pole of projection."
$endgroup$
– Will M.
Jan 15 at 20:28
$begingroup$
It's still the case at the compactification. If needed, take a second chart by taking an n-other pole and projection
$endgroup$
– R. Alexandre
Jan 16 at 16:24