Quaternion conjugation map is orthogonal linear transformation Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Rotation by quaternion conjugation and quaternion matrixAny material on complexification?Showing that the universal enveloping algebra of a Lie algebra $mathfrakg$ is isomorphic to its opposite ringWeyl group, bilinear form, and character/cocharacter pairing. Many questions!Expression transformation using quaternionShow that $overline xcdot y = bar y cdot bar x,$ where $x,y in mathbbH.$Why does the real part of quaternion conjugation with a pure quaternion stay 0?About the set of pure QuaternionAbout isomorphism between quaternion algebrasConstructing quaternions - proof that square of each imaginary unit is -1
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Quaternion conjugation map is orthogonal linear transformation
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Rotation by quaternion conjugation and quaternion matrixAny material on complexification?Showing that the universal enveloping algebra of a Lie algebra $mathfrakg$ is isomorphic to its opposite ringWeyl group, bilinear form, and character/cocharacter pairing. Many questions!Expression transformation using quaternionShow that $overline xcdot y = bar y cdot bar x,$ where $x,y in mathbbH.$Why does the real part of quaternion conjugation with a pure quaternion stay 0?About the set of pure QuaternionAbout isomorphism between quaternion algebrasConstructing quaternions - proof that square of each imaginary unit is -1
$begingroup$
The following concerns an exercise from an undergraduate level textbook on Lie groups, which provides an elementary proof of $SU(2)/mathbbZ_2 cong SO(3)$.
Think of $Sp(1)$ as the group of unit-length quaternions; that is, $Sp(1) = leftq.$ For every $q in Sp(1)$, show that the conjugation map $C_q: mathbb H to mathbb H$, defined as $C_q(V) = q V overlineq$, is an orthogonal linear transformation. Thus, with respect to the natural basis $left 1, i, j, k right$ of $mathbb H$, $C_q$ can be regarded as an element of $O(4)$.
My attempt so far has been as follows: To show it is an orthogonal linear transformation, we must show $langle C_q(V), C_q(W) rangle = langle V, W rangle$ for $V, W in mathbb H$. So we have
$LHS = (qV overline q) cdot (overlineqW overline q) = qV overline q cdot q overline W overline q = q V overline W overline q,$
and I just can't seem to figure out what to do from here. We know $q$, $overline q$ are unit-length, but that does not mean they commute with $V, overline W.$ I have a feeling that perhaps $V$ and $W$ must be pure imaginary quaternions, but even making that assumption doesn't seem to help. If anyone has any insight, I'd be very grateful. It's possible I'm using a bad/incorrect definition of orthogonal linear transformation, but it definitely seems like the error in question is something simple/obvious. Thanks for your time.
lie-groups lie-algebras quaternions
asked Apr 1 at 6:47
Elliot HerringtonElliot Herrington
1388
$endgroup$
add a comment |
$begingroup$
The following concerns an exercise from an undergraduate level textbook on Lie groups, which provides an elementary proof of $SU(2)/mathbbZ_2 cong SO(3)$.
Think of $Sp(1)$ as the group of unit-length quaternions; that is, $Sp(1) = leftq.$ For every $q in Sp(1)$, show that the conjugation map $C_q: mathbb H to mathbb H$, defined as $C_q(V) = q V overlineq$, is an orthogonal linear transformation. Thus, with respect to the natural basis $left 1, i, j, k right$ of $mathbb H$, $C_q$ can be regarded as an element of $O(4)$.
My attempt so far has been as follows: To show it is an orthogonal linear transformation, we must show $langle C_q(V), C_q(W) rangle = langle V, W rangle$ for $V, W in mathbb H$. So we have
$LHS = (qV overline q) cdot (overlineqW overline q) = qV overline q cdot q overline W overline q = q V overline W overline q,$
and I just can't seem to figure out what to do from here. We know $q$, $overline q$ are unit-length, but that does not mean they commute with $V, overline W.$ I have a feeling that perhaps $V$ and $W$ must be pure imaginary quaternions, but even making that assumption doesn't seem to help. If anyone has any insight, I'd be very grateful. It's possible I'm using a bad/incorrect definition of orthogonal linear transformation, but it definitely seems like the error in question is something simple/obvious. Thanks for your time.
lie-groups lie-algebras quaternions
asked Apr 1 at 6:47
Elliot HerringtonElliot Herrington
1388
$endgroup$
$begingroup$
You can be interested by this document gm.univ-montp2.fr/PERSO/mainprice/W_data/MTEX_Formation/…
$endgroup$
– Jean Marie
Apr 1 at 9:09
add a comment |
$begingroup$
The following concerns an exercise from an undergraduate level textbook on Lie groups, which provides an elementary proof of $SU(2)/mathbbZ_2 cong SO(3)$.
Think of $Sp(1)$ as the group of unit-length quaternions; that is, $Sp(1) = leftq.$ For every $q in Sp(1)$, show that the conjugation map $C_q: mathbb H to mathbb H$, defined as $C_q(V) = q V overlineq$, is an orthogonal linear transformation. Thus, with respect to the natural basis $left 1, i, j, k right$ of $mathbb H$, $C_q$ can be regarded as an element of $O(4)$.
My attempt so far has been as follows: To show it is an orthogonal linear transformation, we must show $langle C_q(V), C_q(W) rangle = langle V, W rangle$ for $V, W in mathbb H$. So we have
$LHS = (qV overline q) cdot (overlineqW overline q) = qV overline q cdot q overline W overline q = q V overline W overline q,$
and I just can't seem to figure out what to do from here. We know $q$, $overline q$ are unit-length, but that does not mean they commute with $V, overline W.$ I have a feeling that perhaps $V$ and $W$ must be pure imaginary quaternions, but even making that assumption doesn't seem to help. If anyone has any insight, I'd be very grateful. It's possible I'm using a bad/incorrect definition of orthogonal linear transformation, but it definitely seems like the error in question is something simple/obvious. Thanks for your time.
lie-groups lie-algebras quaternions
asked Apr 1 at 6:47
Elliot HerringtonElliot Herrington
1388
$endgroup$
The following concerns an exercise from an undergraduate level textbook on Lie groups, which provides an elementary proof of $SU(2)/mathbbZ_2 cong SO(3)$.
Think of $Sp(1)$ as the group of unit-length quaternions; that is, $Sp(1) = leftq.$ For every $q in Sp(1)$, show that the conjugation map $C_q: mathbb H to mathbb H$, defined as $C_q(V) = q V overlineq$, is an orthogonal linear transformation. Thus, with respect to the natural basis $left 1, i, j, k right$ of $mathbb H$, $C_q$ can be regarded as an element of $O(4)$.
My attempt so far has been as follows: To show it is an orthogonal linear transformation, we must show $langle C_q(V), C_q(W) rangle = langle V, W rangle$ for $V, W in mathbb H$. So we have
$LHS = (qV overline q) cdot (overlineqW overline q) = qV overline q cdot q overline W overline q = q V overline W overline q,$
and I just can't seem to figure out what to do from here. We know $q$, $overline q$ are unit-length, but that does not mean they commute with $V, overline W.$ I have a feeling that perhaps $V$ and $W$ must be pure imaginary quaternions, but even making that assumption doesn't seem to help. If anyone has any insight, I'd be very grateful. It's possible I'm using a bad/incorrect definition of orthogonal linear transformation, but it definitely seems like the error in question is something simple/obvious. Thanks for your time.
lie-groups lie-algebras quaternions
lie-groups lie-algebras quaternions
asked Apr 1 at 6:47
Elliot HerringtonElliot Herrington
1388
asked Apr 1 at 6:47
Elliot HerringtonElliot Herrington
1388
edited Apr 1 at 8:19
Elliot Herrington
asked Apr 1 at 6:47
Elliot HerringtonElliot Herrington
1388
asked Apr 1 at 6:47
Elliot HerringtonElliot Herrington
1388
asked Apr 1 at 6:47
Elliot HerringtonElliot Herrington
1388
1388
$begingroup$
You can be interested by this document gm.univ-montp2.fr/PERSO/mainprice/W_data/MTEX_Formation/…
$endgroup$
– Jean Marie
Apr 1 at 9:09
add a comment |
$begingroup$
You can be interested by this document gm.univ-montp2.fr/PERSO/mainprice/W_data/MTEX_Formation/…
$endgroup$
– Jean Marie
Apr 1 at 9:09
$begingroup$
You can be interested by this document gm.univ-montp2.fr/PERSO/mainprice/W_data/MTEX_Formation/…
$endgroup$
– Jean Marie
Apr 1 at 9:09
$begingroup$
You can be interested by this document gm.univ-montp2.fr/PERSO/mainprice/W_data/MTEX_Formation/…
$endgroup$
– Jean Marie
Apr 1 at 9:09
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You have $langle V,W rangle = frac12(V overlineW + W overlineV)$. Therefore
$$2langle C_q(V),C_q(W) rangle = q Voverlineq q overlineW overlineq + q W overlineq q overlineV overlineq = q VoverlineW overlineq + q WoverlineV overlineq = q (V overlineW + W overlineV) overlineq \ = q 2 langle V,W rangle overlineq .$$
But $2langle V,W rangle in mathbbR$, hence
$$q 2 langle V,W rangle overlineq = 2 langle V,W rangle q overlineq = 2 langle V,W rangle .$$
Added on request:
Write $V = a + bi + cj + dk, W = a' + b'i + c'j + d'k$. Then by definition $langle V, W rangle = aa' +bb' + cc' + dd' in mathbbR$. Moreover $VoverlineW = (a + bi + cj + dk)(a' - b'i - c'j - d'k) = aa' +bb' + cc' + dd' + r$ with a suitable $r = ui + vj + wk$. Hence $langle V, W rangle = textRe(VoverlineW)$. But $textRe(VoverlineW) = frac12(VoverlineW + overlineVoverlineW) = frac12(VoverlineW + WoverlineV)$.
answered Apr 1 at 8:35
Paul FrostPaul Frost
12.9k31035
$endgroup$
$begingroup$
Looks good. I'm just wondering where you got the first equality from, i.e., the statement $langle V, Wrangle = frac12 (V overline W + W overline V).$ Also, why is $2 langle V, W rangle in mathbb R$?
$endgroup$
– Elliot Herrington
Apr 1 at 9:22
$begingroup$
This is well-known. I add a proof in my answer.
$endgroup$
– Paul Frost
Apr 1 at 9:38
$begingroup$
Very nice! Thank you.
$endgroup$
– Elliot Herrington
Apr 1 at 9:47
add a comment |
Your Answer
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$begingroup$
You have $langle V,W rangle = frac12(V overlineW + W overlineV)$. Therefore
$$2langle C_q(V),C_q(W) rangle = q Voverlineq q overlineW overlineq + q W overlineq q overlineV overlineq = q VoverlineW overlineq + q WoverlineV overlineq = q (V overlineW + W overlineV) overlineq \ = q 2 langle V,W rangle overlineq .$$
But $2langle V,W rangle in mathbbR$, hence
$$q 2 langle V,W rangle overlineq = 2 langle V,W rangle q overlineq = 2 langle V,W rangle .$$
Added on request:
Write $V = a + bi + cj + dk, W = a' + b'i + c'j + d'k$. Then by definition $langle V, W rangle = aa' +bb' + cc' + dd' in mathbbR$. Moreover $VoverlineW = (a + bi + cj + dk)(a' - b'i - c'j - d'k) = aa' +bb' + cc' + dd' + r$ with a suitable $r = ui + vj + wk$. Hence $langle V, W rangle = textRe(VoverlineW)$. But $textRe(VoverlineW) = frac12(VoverlineW + overlineVoverlineW) = frac12(VoverlineW + WoverlineV)$.
answered Apr 1 at 8:35
Paul FrostPaul Frost
12.9k31035
$endgroup$
$begingroup$
Looks good. I'm just wondering where you got the first equality from, i.e., the statement $langle V, Wrangle = frac12 (V overline W + W overline V).$ Also, why is $2 langle V, W rangle in mathbb R$?
$endgroup$
– Elliot Herrington
Apr 1 at 9:22
$begingroup$
This is well-known. I add a proof in my answer.
$endgroup$
– Paul Frost
Apr 1 at 9:38
$begingroup$
Very nice! Thank you.
$endgroup$
– Elliot Herrington
Apr 1 at 9:47
add a comment |
$begingroup$
You have $langle V,W rangle = frac12(V overlineW + W overlineV)$. Therefore
$$2langle C_q(V),C_q(W) rangle = q Voverlineq q overlineW overlineq + q W overlineq q overlineV overlineq = q VoverlineW overlineq + q WoverlineV overlineq = q (V overlineW + W overlineV) overlineq \ = q 2 langle V,W rangle overlineq .$$
But $2langle V,W rangle in mathbbR$, hence
$$q 2 langle V,W rangle overlineq = 2 langle V,W rangle q overlineq = 2 langle V,W rangle .$$
Added on request:
Write $V = a + bi + cj + dk, W = a' + b'i + c'j + d'k$. Then by definition $langle V, W rangle = aa' +bb' + cc' + dd' in mathbbR$. Moreover $VoverlineW = (a + bi + cj + dk)(a' - b'i - c'j - d'k) = aa' +bb' + cc' + dd' + r$ with a suitable $r = ui + vj + wk$. Hence $langle V, W rangle = textRe(VoverlineW)$. But $textRe(VoverlineW) = frac12(VoverlineW + overlineVoverlineW) = frac12(VoverlineW + WoverlineV)$.
answered Apr 1 at 8:35
Paul FrostPaul Frost
12.9k31035
$endgroup$
$begingroup$
Looks good. I'm just wondering where you got the first equality from, i.e., the statement $langle V, Wrangle = frac12 (V overline W + W overline V).$ Also, why is $2 langle V, W rangle in mathbb R$?
$endgroup$
– Elliot Herrington
Apr 1 at 9:22
$begingroup$
This is well-known. I add a proof in my answer.
$endgroup$
– Paul Frost
Apr 1 at 9:38
$begingroup$
Very nice! Thank you.
$endgroup$
– Elliot Herrington
Apr 1 at 9:47
add a comment |
$begingroup$
You have $langle V,W rangle = frac12(V overlineW + W overlineV)$. Therefore
$$2langle C_q(V),C_q(W) rangle = q Voverlineq q overlineW overlineq + q W overlineq q overlineV overlineq = q VoverlineW overlineq + q WoverlineV overlineq = q (V overlineW + W overlineV) overlineq \ = q 2 langle V,W rangle overlineq .$$
But $2langle V,W rangle in mathbbR$, hence
$$q 2 langle V,W rangle overlineq = 2 langle V,W rangle q overlineq = 2 langle V,W rangle .$$
Added on request:
Write $V = a + bi + cj + dk, W = a' + b'i + c'j + d'k$. Then by definition $langle V, W rangle = aa' +bb' + cc' + dd' in mathbbR$. Moreover $VoverlineW = (a + bi + cj + dk)(a' - b'i - c'j - d'k) = aa' +bb' + cc' + dd' + r$ with a suitable $r = ui + vj + wk$. Hence $langle V, W rangle = textRe(VoverlineW)$. But $textRe(VoverlineW) = frac12(VoverlineW + overlineVoverlineW) = frac12(VoverlineW + WoverlineV)$.
answered Apr 1 at 8:35
Paul FrostPaul Frost
12.9k31035
$endgroup$
You have $langle V,W rangle = frac12(V overlineW + W overlineV)$. Therefore
$$2langle C_q(V),C_q(W) rangle = q Voverlineq q overlineW overlineq + q W overlineq q overlineV overlineq = q VoverlineW overlineq + q WoverlineV overlineq = q (V overlineW + W overlineV) overlineq \ = q 2 langle V,W rangle overlineq .$$
But $2langle V,W rangle in mathbbR$, hence
$$q 2 langle V,W rangle overlineq = 2 langle V,W rangle q overlineq = 2 langle V,W rangle .$$
Added on request:
Write $V = a + bi + cj + dk, W = a' + b'i + c'j + d'k$. Then by definition $langle V, W rangle = aa' +bb' + cc' + dd' in mathbbR$. Moreover $VoverlineW = (a + bi + cj + dk)(a' - b'i - c'j - d'k) = aa' +bb' + cc' + dd' + r$ with a suitable $r = ui + vj + wk$. Hence $langle V, W rangle = textRe(VoverlineW)$. But $textRe(VoverlineW) = frac12(VoverlineW + overlineVoverlineW) = frac12(VoverlineW + WoverlineV)$.
answered Apr 1 at 8:35
Paul FrostPaul Frost
12.9k31035
edited Apr 7 at 17:18
answered Apr 1 at 8:35
Paul FrostPaul Frost
12.9k31035
answered Apr 1 at 8:35
Paul FrostPaul Frost
12.9k31035
answered Apr 1 at 8:35
Paul FrostPaul Frost
12.9k31035
12.9k31035
$begingroup$
Looks good. I'm just wondering where you got the first equality from, i.e., the statement $langle V, Wrangle = frac12 (V overline W + W overline V).$ Also, why is $2 langle V, W rangle in mathbb R$?
$endgroup$
– Elliot Herrington
Apr 1 at 9:22
$begingroup$
This is well-known. I add a proof in my answer.
$endgroup$
– Paul Frost
Apr 1 at 9:38
$begingroup$
Very nice! Thank you.
$endgroup$
– Elliot Herrington
Apr 1 at 9:47
add a comment |
$begingroup$
Looks good. I'm just wondering where you got the first equality from, i.e., the statement $langle V, Wrangle = frac12 (V overline W + W overline V).$ Also, why is $2 langle V, W rangle in mathbb R$?
$endgroup$
– Elliot Herrington
Apr 1 at 9:22
$begingroup$
This is well-known. I add a proof in my answer.
$endgroup$
– Paul Frost
Apr 1 at 9:38
$begingroup$
Very nice! Thank you.
$endgroup$
– Elliot Herrington
Apr 1 at 9:47
$begingroup$
Looks good. I'm just wondering where you got the first equality from, i.e., the statement $langle V, Wrangle = frac12 (V overline W + W overline V).$ Also, why is $2 langle V, W rangle in mathbb R$?
$endgroup$
– Elliot Herrington
Apr 1 at 9:22
$begingroup$
Looks good. I'm just wondering where you got the first equality from, i.e., the statement $langle V, Wrangle = frac12 (V overline W + W overline V).$ Also, why is $2 langle V, W rangle in mathbb R$?
$endgroup$
– Elliot Herrington
Apr 1 at 9:22
$begingroup$
This is well-known. I add a proof in my answer.
$endgroup$
– Paul Frost
Apr 1 at 9:38
$begingroup$
This is well-known. I add a proof in my answer.
$endgroup$
– Paul Frost
Apr 1 at 9:38
$begingroup$
Very nice! Thank you.
$endgroup$
– Elliot Herrington
Apr 1 at 9:47
$begingroup$
Very nice! Thank you.
$endgroup$
– Elliot Herrington
Apr 1 at 9:47
add a comment |
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$begingroup$
You can be interested by this document gm.univ-montp2.fr/PERSO/mainprice/W_data/MTEX_Formation/…
$endgroup$
– Jean Marie
Apr 1 at 9:09