Exterior Derivative over Quaternions
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I was wondering if it is possible to define the exterior derivative of a quaternionic valued function. I am doing the quaternionic analogue of a previously complex valued computation, namely something like $omega = frac{h}{sqrt{1+|h|^2}}$ where $h$ is a quaternionic variable and I would like to compute $domega$.
Is this possible to do over $H$? In the complex case I treated $z$ and $bar{z}$ as independent variables and computed using $omega = frac{z}{sqrt{1+|z|^2}}$, with $domega$ being in terms of $dz$ and $dbar{z}$.
differential-geometry differential-forms quaternions exterior-algebra exterior-derivative
asked Nov 14 at 15:57
Bunneh
536
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show 2 more comments
up vote
2
down vote
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I was wondering if it is possible to define the exterior derivative of a quaternionic valued function. I am doing the quaternionic analogue of a previously complex valued computation, namely something like $omega = frac{h}{sqrt{1+|h|^2}}$ where $h$ is a quaternionic variable and I would like to compute $domega$.
Is this possible to do over $H$? In the complex case I treated $z$ and $bar{z}$ as independent variables and computed using $omega = frac{z}{sqrt{1+|z|^2}}$, with $domega$ being in terms of $dz$ and $dbar{z}$.
differential-geometry differential-forms quaternions exterior-algebra exterior-derivative
asked Nov 14 at 15:57
Bunneh
536
Maybe, you can do something similar with more variables. I mean, if $z=a+bi+cj+dk$, then you can define $overline z=a-bi+cj+dk$, $overline{overline z}=a+bi-cj+dk$ and $overline{overline{overline z}}=a+bi+cj-dk$. Then try to write $domega$ in terms of $dz$ and so on. I guess it won't work but I think it is worth trying.
– Dog_69
Nov 14 at 17:27
Thanks, I'll try that! I hadn't thought of doing it like that, my first assumption for the conjugate in this case was $bar{z} = a -bi -cj-dk$ but that could work. Though how I could relate that to $dh$ and $dbar{h}$ because I know that given $h = a+bi+cj+dk$ and $bar{h} = a-bi-cj-dk$, $|h|^2 = bar{h}h = hbar{h} = a^2 + b^2 + c^2 +d^2$.
– Bunneh
Nov 14 at 18:16
Your $overline z$ would be the composition of all my $overline z$. Do you see what I mean? I don't know how to type with symbols. Something like $a-bi-cj-dk =overline{overline{overline{(overline{overline{(overline z)}})}}}$.
– Dog_69
Nov 14 at 19:56
I'm quite sure that my method will fail at some point. But my intention posting that comment was just to shake your mind and to help you finding some different approach. Maybe the answer to your question is 'no, is not possible' but... until that moment...
– Dog_69
Nov 14 at 20:17
@Dog_69 Thank you for giving me a new angle to look at the problem^^ If you think about something too much you can get fixed on a certain method - a shake up is always appreciated!
– Bunneh
Nov 14 at 20:18
|
show 2 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I was wondering if it is possible to define the exterior derivative of a quaternionic valued function. I am doing the quaternionic analogue of a previously complex valued computation, namely something like $omega = frac{h}{sqrt{1+|h|^2}}$ where $h$ is a quaternionic variable and I would like to compute $domega$.
Is this possible to do over $H$? In the complex case I treated $z$ and $bar{z}$ as independent variables and computed using $omega = frac{z}{sqrt{1+|z|^2}}$, with $domega$ being in terms of $dz$ and $dbar{z}$.
differential-geometry differential-forms quaternions exterior-algebra exterior-derivative
asked Nov 14 at 15:57
Bunneh
536
I was wondering if it is possible to define the exterior derivative of a quaternionic valued function. I am doing the quaternionic analogue of a previously complex valued computation, namely something like $omega = frac{h}{sqrt{1+|h|^2}}$ where $h$ is a quaternionic variable and I would like to compute $domega$.
Is this possible to do over $H$? In the complex case I treated $z$ and $bar{z}$ as independent variables and computed using $omega = frac{z}{sqrt{1+|z|^2}}$, with $domega$ being in terms of $dz$ and $dbar{z}$.
differential-geometry differential-forms quaternions exterior-algebra exterior-derivative
differential-geometry differential-forms quaternions exterior-algebra exterior-derivative
asked Nov 14 at 15:57
Bunneh
536
asked Nov 14 at 15:57
Bunneh
536
asked Nov 14 at 15:57
Bunneh
536
asked Nov 14 at 15:57
Bunneh
536
asked Nov 14 at 15:57
Bunneh
536
536
Maybe, you can do something similar with more variables. I mean, if $z=a+bi+cj+dk$, then you can define $overline z=a-bi+cj+dk$, $overline{overline z}=a+bi-cj+dk$ and $overline{overline{overline z}}=a+bi+cj-dk$. Then try to write $domega$ in terms of $dz$ and so on. I guess it won't work but I think it is worth trying.
– Dog_69
Nov 14 at 17:27
Thanks, I'll try that! I hadn't thought of doing it like that, my first assumption for the conjugate in this case was $bar{z} = a -bi -cj-dk$ but that could work. Though how I could relate that to $dh$ and $dbar{h}$ because I know that given $h = a+bi+cj+dk$ and $bar{h} = a-bi-cj-dk$, $|h|^2 = bar{h}h = hbar{h} = a^2 + b^2 + c^2 +d^2$.
– Bunneh
Nov 14 at 18:16
Your $overline z$ would be the composition of all my $overline z$. Do you see what I mean? I don't know how to type with symbols. Something like $a-bi-cj-dk =overline{overline{overline{(overline{overline{(overline z)}})}}}$.
– Dog_69
Nov 14 at 19:56
I'm quite sure that my method will fail at some point. But my intention posting that comment was just to shake your mind and to help you finding some different approach. Maybe the answer to your question is 'no, is not possible' but... until that moment...
– Dog_69
Nov 14 at 20:17
@Dog_69 Thank you for giving me a new angle to look at the problem^^ If you think about something too much you can get fixed on a certain method - a shake up is always appreciated!
– Bunneh
Nov 14 at 20:18
|
show 2 more comments
Maybe, you can do something similar with more variables. I mean, if $z=a+bi+cj+dk$, then you can define $overline z=a-bi+cj+dk$, $overline{overline z}=a+bi-cj+dk$ and $overline{overline{overline z}}=a+bi+cj-dk$. Then try to write $domega$ in terms of $dz$ and so on. I guess it won't work but I think it is worth trying.
– Dog_69
Nov 14 at 17:27
Thanks, I'll try that! I hadn't thought of doing it like that, my first assumption for the conjugate in this case was $bar{z} = a -bi -cj-dk$ but that could work. Though how I could relate that to $dh$ and $dbar{h}$ because I know that given $h = a+bi+cj+dk$ and $bar{h} = a-bi-cj-dk$, $|h|^2 = bar{h}h = hbar{h} = a^2 + b^2 + c^2 +d^2$.
– Bunneh
Nov 14 at 18:16
Your $overline z$ would be the composition of all my $overline z$. Do you see what I mean? I don't know how to type with symbols. Something like $a-bi-cj-dk =overline{overline{overline{(overline{overline{(overline z)}})}}}$.
– Dog_69
Nov 14 at 19:56
I'm quite sure that my method will fail at some point. But my intention posting that comment was just to shake your mind and to help you finding some different approach. Maybe the answer to your question is 'no, is not possible' but... until that moment...
– Dog_69
Nov 14 at 20:17
@Dog_69 Thank you for giving me a new angle to look at the problem^^ If you think about something too much you can get fixed on a certain method - a shake up is always appreciated!
– Bunneh
Nov 14 at 20:18
Maybe, you can do something similar with more variables. I mean, if $z=a+bi+cj+dk$, then you can define $overline z=a-bi+cj+dk$, $overline{overline z}=a+bi-cj+dk$ and $overline{overline{overline z}}=a+bi+cj-dk$. Then try to write $domega$ in terms of $dz$ and so on. I guess it won't work but I think it is worth trying.
– Dog_69
Nov 14 at 17:27
Maybe, you can do something similar with more variables. I mean, if $z=a+bi+cj+dk$, then you can define $overline z=a-bi+cj+dk$, $overline{overline z}=a+bi-cj+dk$ and $overline{overline{overline z}}=a+bi+cj-dk$. Then try to write $domega$ in terms of $dz$ and so on. I guess it won't work but I think it is worth trying.
– Dog_69
Nov 14 at 17:27
Thanks, I'll try that! I hadn't thought of doing it like that, my first assumption for the conjugate in this case was $bar{z} = a -bi -cj-dk$ but that could work. Though how I could relate that to $dh$ and $dbar{h}$ because I know that given $h = a+bi+cj+dk$ and $bar{h} = a-bi-cj-dk$, $|h|^2 = bar{h}h = hbar{h} = a^2 + b^2 + c^2 +d^2$.
– Bunneh
Nov 14 at 18:16
Thanks, I'll try that! I hadn't thought of doing it like that, my first assumption for the conjugate in this case was $bar{z} = a -bi -cj-dk$ but that could work. Though how I could relate that to $dh$ and $dbar{h}$ because I know that given $h = a+bi+cj+dk$ and $bar{h} = a-bi-cj-dk$, $|h|^2 = bar{h}h = hbar{h} = a^2 + b^2 + c^2 +d^2$.
– Bunneh
Nov 14 at 18:16
Your $overline z$ would be the composition of all my $overline z$. Do you see what I mean? I don't know how to type with symbols. Something like $a-bi-cj-dk =overline{overline{overline{(overline{overline{(overline z)}})}}}$.
– Dog_69
Nov 14 at 19:56
Your $overline z$ would be the composition of all my $overline z$. Do you see what I mean? I don't know how to type with symbols. Something like $a-bi-cj-dk =overline{overline{overline{(overline{overline{(overline z)}})}}}$.
– Dog_69
Nov 14 at 19:56
I'm quite sure that my method will fail at some point. But my intention posting that comment was just to shake your mind and to help you finding some different approach. Maybe the answer to your question is 'no, is not possible' but... until that moment...
– Dog_69
Nov 14 at 20:17
I'm quite sure that my method will fail at some point. But my intention posting that comment was just to shake your mind and to help you finding some different approach. Maybe the answer to your question is 'no, is not possible' but... until that moment...
– Dog_69
Nov 14 at 20:17
@Dog_69 Thank you for giving me a new angle to look at the problem^^ If you think about something too much you can get fixed on a certain method - a shake up is always appreciated!
– Bunneh
Nov 14 at 20:18
@Dog_69 Thank you for giving me a new angle to look at the problem^^ If you think about something too much you can get fixed on a certain method - a shake up is always appreciated!
– Bunneh
Nov 14 at 20:18
|
show 2 more comments
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Maybe, you can do something similar with more variables. I mean, if $z=a+bi+cj+dk$, then you can define $overline z=a-bi+cj+dk$, $overline{overline z}=a+bi-cj+dk$ and $overline{overline{overline z}}=a+bi+cj-dk$. Then try to write $domega$ in terms of $dz$ and so on. I guess it won't work but I think it is worth trying.
– Dog_69
Nov 14 at 17:27
Thanks, I'll try that! I hadn't thought of doing it like that, my first assumption for the conjugate in this case was $bar{z} = a -bi -cj-dk$ but that could work. Though how I could relate that to $dh$ and $dbar{h}$ because I know that given $h = a+bi+cj+dk$ and $bar{h} = a-bi-cj-dk$, $|h|^2 = bar{h}h = hbar{h} = a^2 + b^2 + c^2 +d^2$.
– Bunneh
Nov 14 at 18:16
Your $overline z$ would be the composition of all my $overline z$. Do you see what I mean? I don't know how to type with symbols. Something like $a-bi-cj-dk =overline{overline{overline{(overline{overline{(overline z)}})}}}$.
– Dog_69
Nov 14 at 19:56
I'm quite sure that my method will fail at some point. But my intention posting that comment was just to shake your mind and to help you finding some different approach. Maybe the answer to your question is 'no, is not possible' but... until that moment...
– Dog_69
Nov 14 at 20:17
@Dog_69 Thank you for giving me a new angle to look at the problem^^ If you think about something too much you can get fixed on a certain method - a shake up is always appreciated!
– Bunneh
Nov 14 at 20:18