Are there any advantages of treating four-dimensional Euclidean space as $mathbbC^2$ instead of $mathbbR^4$?
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When we treat four-dimensional Euclidean space as $mathbbR^4$ we can easily define a line, plane, and 3-plane by linear equations, whereas if we wish to do the same with $mathbbC^2$ we have a harder time with lines and 3-planes. Are there any circumstances where using $mathbbC^2$ is easier?
vector-spaces soft-question
edited Aug 15 at 10:30
Armando j18eos
2,43611226
asked Aug 15 at 4:45
Perry Ainsworth
344
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When we treat four-dimensional Euclidean space as $mathbbR^4$ we can easily define a line, plane, and 3-plane by linear equations, whereas if we wish to do the same with $mathbbC^2$ we have a harder time with lines and 3-planes. Are there any circumstances where using $mathbbC^2$ is easier?
vector-spaces soft-question
edited Aug 15 at 10:30
Armando j18eos
2,43611226
asked Aug 15 at 4:45
Perry Ainsworth
344
What do you mean by easier?
– Chickenmancer
Aug 15 at 4:46
1
Well, $mathbbC^2$ is a complex vector space (among other things). Expressing the complex eigenvector/eigenvalue pairs is going to be messier and less natural in $mathbbR^4$.
– Theo Bendit
Aug 15 at 4:47
2
Yes, e.g. for the Hopf fibration, IIRC there is a bice formulation of the map $S^3to S^2$ in terms of complex multiplication (to see it, look at the fact that the action of $S^1$ on $S^3$ can neatly be defined coordinatewise by complex multiplication)
– Max
Aug 15 at 6:03
2
What @Max described, also known as the hopf bundle is: let's say we want to move from $S^3$ to $S^2$, so let's say $(a,b,c,d)iff (z_0,z_1)=(a+ib, c+id)$ and $(a,b,c)iff (z,x)=(a+ib,c)$(move from $Bbb R^4$ and $Bbb R^3$ to $Bbb C^2,Bbb CtimesBbb R$). Now let's define Hopf bundle as $$p:S^3to S^2\(z_0,z_1)mapsto (2z_0barz_1,|z_0|^2-|z_1|^2).$$This is just an example for such bundle.
– Holo
Aug 15 at 6:42
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
When we treat four-dimensional Euclidean space as $mathbbR^4$ we can easily define a line, plane, and 3-plane by linear equations, whereas if we wish to do the same with $mathbbC^2$ we have a harder time with lines and 3-planes. Are there any circumstances where using $mathbbC^2$ is easier?
vector-spaces soft-question
edited Aug 15 at 10:30
Armando j18eos
2,43611226
asked Aug 15 at 4:45
Perry Ainsworth
344
When we treat four-dimensional Euclidean space as $mathbbR^4$ we can easily define a line, plane, and 3-plane by linear equations, whereas if we wish to do the same with $mathbbC^2$ we have a harder time with lines and 3-planes. Are there any circumstances where using $mathbbC^2$ is easier?
vector-spaces soft-question
edited Aug 15 at 10:30
Armando j18eos
2,43611226
asked Aug 15 at 4:45
Perry Ainsworth
344
edited Aug 15 at 10:30
Armando j18eos
2,43611226
edited Aug 15 at 10:30
Armando j18eos
2,43611226
edited Aug 15 at 10:30
Armando j18eos
2,43611226
2,43611226
asked Aug 15 at 4:45
Perry Ainsworth
344
asked Aug 15 at 4:45
Perry Ainsworth
344
asked Aug 15 at 4:45
Perry Ainsworth
344
344
What do you mean by easier?
– Chickenmancer
Aug 15 at 4:46
1
Well, $mathbbC^2$ is a complex vector space (among other things). Expressing the complex eigenvector/eigenvalue pairs is going to be messier and less natural in $mathbbR^4$.
– Theo Bendit
Aug 15 at 4:47
2
Yes, e.g. for the Hopf fibration, IIRC there is a bice formulation of the map $S^3to S^2$ in terms of complex multiplication (to see it, look at the fact that the action of $S^1$ on $S^3$ can neatly be defined coordinatewise by complex multiplication)
– Max
Aug 15 at 6:03
2
What @Max described, also known as the hopf bundle is: let's say we want to move from $S^3$ to $S^2$, so let's say $(a,b,c,d)iff (z_0,z_1)=(a+ib, c+id)$ and $(a,b,c)iff (z,x)=(a+ib,c)$(move from $Bbb R^4$ and $Bbb R^3$ to $Bbb C^2,Bbb CtimesBbb R$). Now let's define Hopf bundle as $$p:S^3to S^2\(z_0,z_1)mapsto (2z_0barz_1,|z_0|^2-|z_1|^2).$$This is just an example for such bundle.
– Holo
Aug 15 at 6:42
add a comment |Â
What do you mean by easier?
– Chickenmancer
Aug 15 at 4:46
1
Well, $mathbbC^2$ is a complex vector space (among other things). Expressing the complex eigenvector/eigenvalue pairs is going to be messier and less natural in $mathbbR^4$.
– Theo Bendit
Aug 15 at 4:47
2
Yes, e.g. for the Hopf fibration, IIRC there is a bice formulation of the map $S^3to S^2$ in terms of complex multiplication (to see it, look at the fact that the action of $S^1$ on $S^3$ can neatly be defined coordinatewise by complex multiplication)
– Max
Aug 15 at 6:03
2
What @Max described, also known as the hopf bundle is: let's say we want to move from $S^3$ to $S^2$, so let's say $(a,b,c,d)iff (z_0,z_1)=(a+ib, c+id)$ and $(a,b,c)iff (z,x)=(a+ib,c)$(move from $Bbb R^4$ and $Bbb R^3$ to $Bbb C^2,Bbb CtimesBbb R$). Now let's define Hopf bundle as $$p:S^3to S^2\(z_0,z_1)mapsto (2z_0barz_1,|z_0|^2-|z_1|^2).$$This is just an example for such bundle.
– Holo
Aug 15 at 6:42
What do you mean by easier?
– Chickenmancer
Aug 15 at 4:46
What do you mean by easier?
– Chickenmancer
Aug 15 at 4:46
1
1
Well, $mathbbC^2$ is a complex vector space (among other things). Expressing the complex eigenvector/eigenvalue pairs is going to be messier and less natural in $mathbbR^4$.
– Theo Bendit
Aug 15 at 4:47
Well, $mathbbC^2$ is a complex vector space (among other things). Expressing the complex eigenvector/eigenvalue pairs is going to be messier and less natural in $mathbbR^4$.
– Theo Bendit
Aug 15 at 4:47
2
2
Yes, e.g. for the Hopf fibration, IIRC there is a bice formulation of the map $S^3to S^2$ in terms of complex multiplication (to see it, look at the fact that the action of $S^1$ on $S^3$ can neatly be defined coordinatewise by complex multiplication)
– Max
Aug 15 at 6:03
Yes, e.g. for the Hopf fibration, IIRC there is a bice formulation of the map $S^3to S^2$ in terms of complex multiplication (to see it, look at the fact that the action of $S^1$ on $S^3$ can neatly be defined coordinatewise by complex multiplication)
– Max
Aug 15 at 6:03
2
2
What @Max described, also known as the hopf bundle is: let's say we want to move from $S^3$ to $S^2$, so let's say $(a,b,c,d)iff (z_0,z_1)=(a+ib, c+id)$ and $(a,b,c)iff (z,x)=(a+ib,c)$(move from $Bbb R^4$ and $Bbb R^3$ to $Bbb C^2,Bbb CtimesBbb R$). Now let's define Hopf bundle as $$p:S^3to S^2\(z_0,z_1)mapsto (2z_0barz_1,|z_0|^2-|z_1|^2).$$This is just an example for such bundle.
– Holo
Aug 15 at 6:42
What @Max described, also known as the hopf bundle is: let's say we want to move from $S^3$ to $S^2$, so let's say $(a,b,c,d)iff (z_0,z_1)=(a+ib, c+id)$ and $(a,b,c)iff (z,x)=(a+ib,c)$(move from $Bbb R^4$ and $Bbb R^3$ to $Bbb C^2,Bbb CtimesBbb R$). Now let's define Hopf bundle as $$p:S^3to S^2\(z_0,z_1)mapsto (2z_0barz_1,|z_0|^2-|z_1|^2).$$This is just an example for such bundle.
– Holo
Aug 15 at 6:42
add a comment |Â
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What do you mean by easier?
– Chickenmancer
Aug 15 at 4:46
1
Well, $mathbbC^2$ is a complex vector space (among other things). Expressing the complex eigenvector/eigenvalue pairs is going to be messier and less natural in $mathbbR^4$.
– Theo Bendit
Aug 15 at 4:47
2
Yes, e.g. for the Hopf fibration, IIRC there is a bice formulation of the map $S^3to S^2$ in terms of complex multiplication (to see it, look at the fact that the action of $S^1$ on $S^3$ can neatly be defined coordinatewise by complex multiplication)
– Max
Aug 15 at 6:03
2
What @Max described, also known as the hopf bundle is: let's say we want to move from $S^3$ to $S^2$, so let's say $(a,b,c,d)iff (z_0,z_1)=(a+ib, c+id)$ and $(a,b,c)iff (z,x)=(a+ib,c)$(move from $Bbb R^4$ and $Bbb R^3$ to $Bbb C^2,Bbb CtimesBbb R$). Now let's define Hopf bundle as $$p:S^3to S^2\(z_0,z_1)mapsto (2z_0barz_1,|z_0|^2-|z_1|^2).$$This is just an example for such bundle.
– Holo
Aug 15 at 6:42