Mixing
Reflection in Geometric Algebra
Clash Royale CLAN TAG#URR8PPP
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When I reflect $v$ through the hyperplane orthogonal to $v-w$, where $w = f(v)$ and $f$ is an isometry on $mathbbR^n$ that preserves the origin. I believe I am supposed to get $w$, however whenever I do the calculation I get:
$$textRef_v-w(v) = -(v-w)v(v-w)^-1 = frac-(v-w)v(v-w)(v-w)^2 =
$$
$$= frac(w-v)v(v-w)(v-w)^2 = frac(wv-v^2)(v-w)(v-w)^2 = fracwv - v^2v-w = -v$$
No matter what I do I can't seem to get away from this conclusion. Can anybody help me understand what I'm doing wrong?
linear-algebra geometry reflection geometric-algebras
asked Jul 22 at 2:38
Emilio Minichiello
1917
 |Â
show 1 more comment
up vote
0
down vote
favorite
When I reflect $v$ through the hyperplane orthogonal to $v-w$, where $w = f(v)$ and $f$ is an isometry on $mathbbR^n$ that preserves the origin. I believe I am supposed to get $w$, however whenever I do the calculation I get:
$$textRef_v-w(v) = -(v-w)v(v-w)^-1 = frac-(v-w)v(v-w)(v-w)^2 =
$$
$$= frac(w-v)v(v-w)(v-w)^2 = frac(wv-v^2)(v-w)(v-w)^2 = fracwv - v^2v-w = -v$$
No matter what I do I can't seem to get away from this conclusion. Can anybody help me understand what I'm doing wrong?
linear-algebra geometry reflection geometric-algebras
asked Jul 22 at 2:38
Emilio Minichiello
1917
1
Take $w=0$: should the reflection of $v$ through the hyperplane orthogonal to $v$ be $0$? Your belief of what you are "supposed" to get is a source of error.
– KCd
Jul 22 at 3:35
But what about this result math.stackexchange.com/questions/465304/… ? This is where this is coming from. I'm trying to translate it into Geometric Algebra and don't understand what I'm doing wrong.
– Emilio Minichiello
Jul 22 at 3:48
1
There are many hyperplanes orthogonal to $v-w$. Only one of them is going to produce a reflection that maps $v$ onto $w$. You’ve chosen a different one.
– amd
Jul 22 at 4:52
How do I know which one ive chosen?
– Emilio Minichiello
Jul 22 at 5:26
1
To make the comment by @amd clear, we might not be dealing with hyperplanes through the origin. See Theorem A.3 on p. 15 of math.uconn.edu/~kconrad/blurbs/grouptheory/isometryRn.pdf, which uses notation from Lemma A.1.
– KCd
Jul 22 at 5:31
 |Â
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
When I reflect $v$ through the hyperplane orthogonal to $v-w$, where $w = f(v)$ and $f$ is an isometry on $mathbbR^n$ that preserves the origin. I believe I am supposed to get $w$, however whenever I do the calculation I get:
$$textRef_v-w(v) = -(v-w)v(v-w)^-1 = frac-(v-w)v(v-w)(v-w)^2 =
$$
$$= frac(w-v)v(v-w)(v-w)^2 = frac(wv-v^2)(v-w)(v-w)^2 = fracwv - v^2v-w = -v$$
No matter what I do I can't seem to get away from this conclusion. Can anybody help me understand what I'm doing wrong?
linear-algebra geometry reflection geometric-algebras
asked Jul 22 at 2:38
Emilio Minichiello
1917
When I reflect $v$ through the hyperplane orthogonal to $v-w$, where $w = f(v)$ and $f$ is an isometry on $mathbbR^n$ that preserves the origin. I believe I am supposed to get $w$, however whenever I do the calculation I get:
$$textRef_v-w(v) = -(v-w)v(v-w)^-1 = frac-(v-w)v(v-w)(v-w)^2 =
$$
$$= frac(w-v)v(v-w)(v-w)^2 = frac(wv-v^2)(v-w)(v-w)^2 = fracwv - v^2v-w = -v$$
No matter what I do I can't seem to get away from this conclusion. Can anybody help me understand what I'm doing wrong?
linear-algebra geometry reflection geometric-algebras
asked Jul 22 at 2:38
Emilio Minichiello
1917
edited Jul 22 at 12:22
asked Jul 22 at 2:38
Emilio Minichiello
1917
asked Jul 22 at 2:38
Emilio Minichiello
1917
asked Jul 22 at 2:38
Emilio Minichiello
1917
1917
1
Take $w=0$: should the reflection of $v$ through the hyperplane orthogonal to $v$ be $0$? Your belief of what you are "supposed" to get is a source of error.
– KCd
Jul 22 at 3:35
But what about this result math.stackexchange.com/questions/465304/… ? This is where this is coming from. I'm trying to translate it into Geometric Algebra and don't understand what I'm doing wrong.
– Emilio Minichiello
Jul 22 at 3:48
1
There are many hyperplanes orthogonal to $v-w$. Only one of them is going to produce a reflection that maps $v$ onto $w$. You’ve chosen a different one.
– amd
Jul 22 at 4:52
How do I know which one ive chosen?
– Emilio Minichiello
Jul 22 at 5:26
1
To make the comment by @amd clear, we might not be dealing with hyperplanes through the origin. See Theorem A.3 on p. 15 of math.uconn.edu/~kconrad/blurbs/grouptheory/isometryRn.pdf, which uses notation from Lemma A.1.
– KCd
Jul 22 at 5:31
 |Â
show 1 more comment
1
Take $w=0$: should the reflection of $v$ through the hyperplane orthogonal to $v$ be $0$? Your belief of what you are "supposed" to get is a source of error.
– KCd
Jul 22 at 3:35
But what about this result math.stackexchange.com/questions/465304/… ? This is where this is coming from. I'm trying to translate it into Geometric Algebra and don't understand what I'm doing wrong.
– Emilio Minichiello
Jul 22 at 3:48
1
There are many hyperplanes orthogonal to $v-w$. Only one of them is going to produce a reflection that maps $v$ onto $w$. You’ve chosen a different one.
– amd
Jul 22 at 4:52
How do I know which one ive chosen?
– Emilio Minichiello
Jul 22 at 5:26
1
To make the comment by @amd clear, we might not be dealing with hyperplanes through the origin. See Theorem A.3 on p. 15 of math.uconn.edu/~kconrad/blurbs/grouptheory/isometryRn.pdf, which uses notation from Lemma A.1.
– KCd
Jul 22 at 5:31
1
1
Take $w=0$: should the reflection of $v$ through the hyperplane orthogonal to $v$ be $0$? Your belief of what you are "supposed" to get is a source of error.
– KCd
Jul 22 at 3:35
Take $w=0$: should the reflection of $v$ through the hyperplane orthogonal to $v$ be $0$? Your belief of what you are "supposed" to get is a source of error.
– KCd
Jul 22 at 3:35
But what about this result math.stackexchange.com/questions/465304/… ? This is where this is coming from. I'm trying to translate it into Geometric Algebra and don't understand what I'm doing wrong.
– Emilio Minichiello
Jul 22 at 3:48
But what about this result math.stackexchange.com/questions/465304/… ? This is where this is coming from. I'm trying to translate it into Geometric Algebra and don't understand what I'm doing wrong.
– Emilio Minichiello
Jul 22 at 3:48
1
1
There are many hyperplanes orthogonal to $v-w$. Only one of them is going to produce a reflection that maps $v$ onto $w$. You’ve chosen a different one.
– amd
Jul 22 at 4:52
There are many hyperplanes orthogonal to $v-w$. Only one of them is going to produce a reflection that maps $v$ onto $w$. You’ve chosen a different one.
– amd
Jul 22 at 4:52
How do I know which one ive chosen?
– Emilio Minichiello
Jul 22 at 5:26
How do I know which one ive chosen?
– Emilio Minichiello
Jul 22 at 5:26
1
1
To make the comment by @amd clear, we might not be dealing with hyperplanes through the origin. See Theorem A.3 on p. 15 of math.uconn.edu/~kconrad/blurbs/grouptheory/isometryRn.pdf, which uses notation from Lemma A.1.
– KCd
Jul 22 at 5:31
To make the comment by @amd clear, we might not be dealing with hyperplanes through the origin. See Theorem A.3 on p. 15 of math.uconn.edu/~kconrad/blurbs/grouptheory/isometryRn.pdf, which uses notation from Lemma A.1.
– KCd
Jul 22 at 5:31
 |Â
show 1 more comment
2 Answers
2
active
oldest
votes
up vote
2
down vote
Here's a less convoluted version of your calculation that should make your error more immediately obvious:
$$mathrmRef_v-w(v) = −(v−w)v(v−w)^−1 = -frac(v−w)vv−w = -v$$
In short, you've tricked yourself by using fraction notation in a non-commutative algebra. Since geometric algebra is non-commutative, there's a difference between $ab^-1$ and $b^-1a$ (if that were not the case, you could simply exchange the last two terms in the reflection formula, and then cancel $(v-w)$ with $(v-w)^-1$).
Now if you choose to use fraction notation $fracab$ (which is a very bad idea for non-commutative algebra to begin with), you have to decide on a meaning for it. Now in the step
$$-(v-w)v(v-w)^-1 = frac-(v-w)v(v-w)(v-w)^2$$
you used the definition $fracab = ab^-1$ (it's the only definition under which that equation would hold), but in the step
$$fracwv - v^2v-w = -v$$
you used the definition $fracab = b^-1a$ (again, that's the only definition where that step would be valid).
So essentially, in those two steps together, you exchanged the factors in $v(v-w)^-1$; all the other stuff only serves to obfuscate this. And thus your result is easily explained, as indeed
$$-(v-w)(v-w)^-1v = -v,$$
it's just no longer the reflection of $v$ on a hyperplane.
Bottom line: When dealing with non-commutative algebra, don't use fraction notation. It only invites errors such as this one.
answered Jul 22 at 11:42
celtschk
28.1k65495
Ah okay. So my calculation should be $-frac(v-w)v(v-w)^2 = - fracv^2 = - frac(^2$ right? What can I do now? I should be getting $w$ right?
– Emilio Minichiello
Jul 22 at 12:25
add a comment |Â
up vote
1
down vote
accepted
I found the answer in Guide to Geometric Algebra in Practice by Dorst and Lasenby. As celtschk explains, my use of fraction notation messed me up. This is the correct derivation.
$$-(v-w)v(v-w)^-1 = (-(v-w)v)(v-w)^-1 = (-(v-w)((v+w)-w))(v-w)^-1=(-(v^2 - w^2) + w(v-w))(v-w)^-1 = (w(v-w) - (v^2-w^2))(v-w)^-1 = w - (v^2 - w^2)(v-w)^-1 = w$$
Because $v^2 = |v|^2 = |w|^2 = w^2$ since $f(v) = w$ is an isometry.
answered Jul 23 at 18:49
Emilio Minichiello
1917
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Here's a less convoluted version of your calculation that should make your error more immediately obvious:
$$mathrmRef_v-w(v) = −(v−w)v(v−w)^−1 = -frac(v−w)vv−w = -v$$
In short, you've tricked yourself by using fraction notation in a non-commutative algebra. Since geometric algebra is non-commutative, there's a difference between $ab^-1$ and $b^-1a$ (if that were not the case, you could simply exchange the last two terms in the reflection formula, and then cancel $(v-w)$ with $(v-w)^-1$).
Now if you choose to use fraction notation $fracab$ (which is a very bad idea for non-commutative algebra to begin with), you have to decide on a meaning for it. Now in the step
$$-(v-w)v(v-w)^-1 = frac-(v-w)v(v-w)(v-w)^2$$
you used the definition $fracab = ab^-1$ (it's the only definition under which that equation would hold), but in the step
$$fracwv - v^2v-w = -v$$
you used the definition $fracab = b^-1a$ (again, that's the only definition where that step would be valid).
So essentially, in those two steps together, you exchanged the factors in $v(v-w)^-1$; all the other stuff only serves to obfuscate this. And thus your result is easily explained, as indeed
$$-(v-w)(v-w)^-1v = -v,$$
it's just no longer the reflection of $v$ on a hyperplane.
Bottom line: When dealing with non-commutative algebra, don't use fraction notation. It only invites errors such as this one.
answered Jul 22 at 11:42
celtschk
28.1k65495
Ah okay. So my calculation should be $-frac(v-w)v(v-w)^2 = - fracv^2 = - frac(^2$ right? What can I do now? I should be getting $w$ right?
– Emilio Minichiello
Jul 22 at 12:25
add a comment |Â
up vote
2
down vote
Here's a less convoluted version of your calculation that should make your error more immediately obvious:
$$mathrmRef_v-w(v) = −(v−w)v(v−w)^−1 = -frac(v−w)vv−w = -v$$
In short, you've tricked yourself by using fraction notation in a non-commutative algebra. Since geometric algebra is non-commutative, there's a difference between $ab^-1$ and $b^-1a$ (if that were not the case, you could simply exchange the last two terms in the reflection formula, and then cancel $(v-w)$ with $(v-w)^-1$).
Now if you choose to use fraction notation $fracab$ (which is a very bad idea for non-commutative algebra to begin with), you have to decide on a meaning for it. Now in the step
$$-(v-w)v(v-w)^-1 = frac-(v-w)v(v-w)(v-w)^2$$
you used the definition $fracab = ab^-1$ (it's the only definition under which that equation would hold), but in the step
$$fracwv - v^2v-w = -v$$
you used the definition $fracab = b^-1a$ (again, that's the only definition where that step would be valid).
So essentially, in those two steps together, you exchanged the factors in $v(v-w)^-1$; all the other stuff only serves to obfuscate this. And thus your result is easily explained, as indeed
$$-(v-w)(v-w)^-1v = -v,$$
it's just no longer the reflection of $v$ on a hyperplane.
Bottom line: When dealing with non-commutative algebra, don't use fraction notation. It only invites errors such as this one.
answered Jul 22 at 11:42
celtschk
28.1k65495
Ah okay. So my calculation should be $-frac(v-w)v(v-w)^2 = - fracv^2 = - frac(^2$ right? What can I do now? I should be getting $w$ right?
– Emilio Minichiello
Jul 22 at 12:25
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Here's a less convoluted version of your calculation that should make your error more immediately obvious:
$$mathrmRef_v-w(v) = −(v−w)v(v−w)^−1 = -frac(v−w)vv−w = -v$$
In short, you've tricked yourself by using fraction notation in a non-commutative algebra. Since geometric algebra is non-commutative, there's a difference between $ab^-1$ and $b^-1a$ (if that were not the case, you could simply exchange the last two terms in the reflection formula, and then cancel $(v-w)$ with $(v-w)^-1$).
Now if you choose to use fraction notation $fracab$ (which is a very bad idea for non-commutative algebra to begin with), you have to decide on a meaning for it. Now in the step
$$-(v-w)v(v-w)^-1 = frac-(v-w)v(v-w)(v-w)^2$$
you used the definition $fracab = ab^-1$ (it's the only definition under which that equation would hold), but in the step
$$fracwv - v^2v-w = -v$$
you used the definition $fracab = b^-1a$ (again, that's the only definition where that step would be valid).
So essentially, in those two steps together, you exchanged the factors in $v(v-w)^-1$; all the other stuff only serves to obfuscate this. And thus your result is easily explained, as indeed
$$-(v-w)(v-w)^-1v = -v,$$
it's just no longer the reflection of $v$ on a hyperplane.
Bottom line: When dealing with non-commutative algebra, don't use fraction notation. It only invites errors such as this one.
answered Jul 22 at 11:42
celtschk
28.1k65495
Here's a less convoluted version of your calculation that should make your error more immediately obvious:
$$mathrmRef_v-w(v) = −(v−w)v(v−w)^−1 = -frac(v−w)vv−w = -v$$
In short, you've tricked yourself by using fraction notation in a non-commutative algebra. Since geometric algebra is non-commutative, there's a difference between $ab^-1$ and $b^-1a$ (if that were not the case, you could simply exchange the last two terms in the reflection formula, and then cancel $(v-w)$ with $(v-w)^-1$).
Now if you choose to use fraction notation $fracab$ (which is a very bad idea for non-commutative algebra to begin with), you have to decide on a meaning for it. Now in the step
$$-(v-w)v(v-w)^-1 = frac-(v-w)v(v-w)(v-w)^2$$
you used the definition $fracab = ab^-1$ (it's the only definition under which that equation would hold), but in the step
$$fracwv - v^2v-w = -v$$
you used the definition $fracab = b^-1a$ (again, that's the only definition where that step would be valid).
So essentially, in those two steps together, you exchanged the factors in $v(v-w)^-1$; all the other stuff only serves to obfuscate this. And thus your result is easily explained, as indeed
$$-(v-w)(v-w)^-1v = -v,$$
it's just no longer the reflection of $v$ on a hyperplane.
Bottom line: When dealing with non-commutative algebra, don't use fraction notation. It only invites errors such as this one.
answered Jul 22 at 11:42
celtschk
28.1k65495
answered Jul 22 at 11:42
celtschk
28.1k65495
answered Jul 22 at 11:42
celtschk
28.1k65495
answered Jul 22 at 11:42
celtschk
28.1k65495
28.1k65495
Ah okay. So my calculation should be $-frac(v-w)v(v-w)^2 = - fracv^2 = - frac(^2$ right? What can I do now? I should be getting $w$ right?
– Emilio Minichiello
Jul 22 at 12:25
add a comment |Â
Ah okay. So my calculation should be $-frac(v-w)v(v-w)^2 = - fracv^2 = - frac(^2$ right? What can I do now? I should be getting $w$ right?
– Emilio Minichiello
Jul 22 at 12:25
Ah okay. So my calculation should be $-frac(v-w)v(v-w)^2 = - fracv^2 = - frac(^2$ right? What can I do now? I should be getting $w$ right?
– Emilio Minichiello
Jul 22 at 12:25
Ah okay. So my calculation should be $-frac(v-w)v(v-w)^2 = - fracv^2 = - frac(^2$ right? What can I do now? I should be getting $w$ right?
– Emilio Minichiello
Jul 22 at 12:25
add a comment |Â
up vote
1
down vote
accepted
I found the answer in Guide to Geometric Algebra in Practice by Dorst and Lasenby. As celtschk explains, my use of fraction notation messed me up. This is the correct derivation.
$$-(v-w)v(v-w)^-1 = (-(v-w)v)(v-w)^-1 = (-(v-w)((v+w)-w))(v-w)^-1=(-(v^2 - w^2) + w(v-w))(v-w)^-1 = (w(v-w) - (v^2-w^2))(v-w)^-1 = w - (v^2 - w^2)(v-w)^-1 = w$$
Because $v^2 = |v|^2 = |w|^2 = w^2$ since $f(v) = w$ is an isometry.
answered Jul 23 at 18:49
Emilio Minichiello
1917
add a comment |Â
up vote
1
down vote
accepted
I found the answer in Guide to Geometric Algebra in Practice by Dorst and Lasenby. As celtschk explains, my use of fraction notation messed me up. This is the correct derivation.
$$-(v-w)v(v-w)^-1 = (-(v-w)v)(v-w)^-1 = (-(v-w)((v+w)-w))(v-w)^-1=(-(v^2 - w^2) + w(v-w))(v-w)^-1 = (w(v-w) - (v^2-w^2))(v-w)^-1 = w - (v^2 - w^2)(v-w)^-1 = w$$
Because $v^2 = |v|^2 = |w|^2 = w^2$ since $f(v) = w$ is an isometry.
answered Jul 23 at 18:49
Emilio Minichiello
1917
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
I found the answer in Guide to Geometric Algebra in Practice by Dorst and Lasenby. As celtschk explains, my use of fraction notation messed me up. This is the correct derivation.
$$-(v-w)v(v-w)^-1 = (-(v-w)v)(v-w)^-1 = (-(v-w)((v+w)-w))(v-w)^-1=(-(v^2 - w^2) + w(v-w))(v-w)^-1 = (w(v-w) - (v^2-w^2))(v-w)^-1 = w - (v^2 - w^2)(v-w)^-1 = w$$
Because $v^2 = |v|^2 = |w|^2 = w^2$ since $f(v) = w$ is an isometry.
answered Jul 23 at 18:49
Emilio Minichiello
1917
I found the answer in Guide to Geometric Algebra in Practice by Dorst and Lasenby. As celtschk explains, my use of fraction notation messed me up. This is the correct derivation.
$$-(v-w)v(v-w)^-1 = (-(v-w)v)(v-w)^-1 = (-(v-w)((v+w)-w))(v-w)^-1=(-(v^2 - w^2) + w(v-w))(v-w)^-1 = (w(v-w) - (v^2-w^2))(v-w)^-1 = w - (v^2 - w^2)(v-w)^-1 = w$$
Because $v^2 = |v|^2 = |w|^2 = w^2$ since $f(v) = w$ is an isometry.
answered Jul 23 at 18:49
Emilio Minichiello
1917
answered Jul 23 at 18:49
Emilio Minichiello
1917
answered Jul 23 at 18:49
Emilio Minichiello
1917
answered Jul 23 at 18:49
Emilio Minichiello
1917
1917
add a comment |Â
add a comment |Â
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1
Take $w=0$: should the reflection of $v$ through the hyperplane orthogonal to $v$ be $0$? Your belief of what you are "supposed" to get is a source of error.
– KCd
Jul 22 at 3:35
But what about this result math.stackexchange.com/questions/465304/… ? This is where this is coming from. I'm trying to translate it into Geometric Algebra and don't understand what I'm doing wrong.
– Emilio Minichiello
Jul 22 at 3:48
1
There are many hyperplanes orthogonal to $v-w$. Only one of them is going to produce a reflection that maps $v$ onto $w$. You’ve chosen a different one.
– amd
Jul 22 at 4:52
How do I know which one ive chosen?
– Emilio Minichiello
Jul 22 at 5:26
1
To make the comment by @amd clear, we might not be dealing with hyperplanes through the origin. See Theorem A.3 on p. 15 of math.uconn.edu/~kconrad/blurbs/grouptheory/isometryRn.pdf, which uses notation from Lemma A.1.
– KCd
Jul 22 at 5:31