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Clifford algebra Cliff(0) as reals, Cliff(1) as complex…
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In this paper by John Baez: http://math.ucr.edu/home/baez/octonions/node6.html he says that assuming a Clifford algebra with $vw + wv = -2<v, w>$, you can see that Cliff(0) = $mathbbR$, Cliff(1) = $mathbbC$ and Cliff(2) = $mathbbH$...
How is that possible? The minus sign would spoil everything... Thanks in advance.
clifford-algebras
asked Jul 22 at 18:12
David
347110
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In this paper by John Baez: http://math.ucr.edu/home/baez/octonions/node6.html he says that assuming a Clifford algebra with $vw + wv = -2<v, w>$, you can see that Cliff(0) = $mathbbR$, Cliff(1) = $mathbbC$ and Cliff(2) = $mathbbH$...
How is that possible? The minus sign would spoil everything... Thanks in advance.
clifford-algebras
asked Jul 22 at 18:12
David
347110
1
The minus sign means that $i^2 = -1$ in $operatornameCliff(1)$ and that $i^2 = j^2 = k^2 = -1$ in $operatornameCliff(2)$, which is presumably what you want?
– Branimir Ćaćić
Jul 22 at 19:57
Thanks, but that doesn't explain how you recover Cliff(0). And for Cliff(1) you have $v, w in mathbbR$, so I can't see how the $i$ appears... Could you elaborate more, please?
– David
Jul 22 at 20:19
The Clifford algebra $operatornameCliff(V)$ of an inner product space $V$ is the unital algebra generated by elements of $V$, subject to the relation $vw + wv = -2langle v,wrangle 1_operatornameCliff(V)$; in particular, it is generated as an algebra by the algebra's unit element $1_operatornameCliff(V)$ together with by any basis for $V$. So, this vacuously gives you $operatornameCliff(0) = mathbbR$, while $operatornameCliff(1)$ is generated by the unit $1_operatornameCliff(1)$ together with the basis $i$ for the $1$-dim'l inner product space $mathbbR$
– Branimir Ćaćić
Jul 22 at 21:36
Ok, sure I'm missing something, but let's put values: we are in $mathbbR$, and for instance $v = 2, w = 3$, and the unit and basis for $mathbbR$ is 1. Then the equality $vw + wv = -2langle v, wrangle$ is not verified...
– David
Jul 22 at 23:56
add a comment |Â
up vote
0
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up vote
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down vote
favorite
In this paper by John Baez: http://math.ucr.edu/home/baez/octonions/node6.html he says that assuming a Clifford algebra with $vw + wv = -2<v, w>$, you can see that Cliff(0) = $mathbbR$, Cliff(1) = $mathbbC$ and Cliff(2) = $mathbbH$...
How is that possible? The minus sign would spoil everything... Thanks in advance.
clifford-algebras
asked Jul 22 at 18:12
David
347110
In this paper by John Baez: http://math.ucr.edu/home/baez/octonions/node6.html he says that assuming a Clifford algebra with $vw + wv = -2<v, w>$, you can see that Cliff(0) = $mathbbR$, Cliff(1) = $mathbbC$ and Cliff(2) = $mathbbH$...
How is that possible? The minus sign would spoil everything... Thanks in advance.
clifford-algebras
asked Jul 22 at 18:12
David
347110
asked Jul 22 at 18:12
David
347110
asked Jul 22 at 18:12
David
347110
asked Jul 22 at 18:12
David
347110
347110
1
The minus sign means that $i^2 = -1$ in $operatornameCliff(1)$ and that $i^2 = j^2 = k^2 = -1$ in $operatornameCliff(2)$, which is presumably what you want?
– Branimir Ćaćić
Jul 22 at 19:57
Thanks, but that doesn't explain how you recover Cliff(0). And for Cliff(1) you have $v, w in mathbbR$, so I can't see how the $i$ appears... Could you elaborate more, please?
– David
Jul 22 at 20:19
The Clifford algebra $operatornameCliff(V)$ of an inner product space $V$ is the unital algebra generated by elements of $V$, subject to the relation $vw + wv = -2langle v,wrangle 1_operatornameCliff(V)$; in particular, it is generated as an algebra by the algebra's unit element $1_operatornameCliff(V)$ together with by any basis for $V$. So, this vacuously gives you $operatornameCliff(0) = mathbbR$, while $operatornameCliff(1)$ is generated by the unit $1_operatornameCliff(1)$ together with the basis $i$ for the $1$-dim'l inner product space $mathbbR$
– Branimir Ćaćić
Jul 22 at 21:36
Ok, sure I'm missing something, but let's put values: we are in $mathbbR$, and for instance $v = 2, w = 3$, and the unit and basis for $mathbbR$ is 1. Then the equality $vw + wv = -2langle v, wrangle$ is not verified...
– David
Jul 22 at 23:56
add a comment |Â
1
The minus sign means that $i^2 = -1$ in $operatornameCliff(1)$ and that $i^2 = j^2 = k^2 = -1$ in $operatornameCliff(2)$, which is presumably what you want?
– Branimir Ćaćić
Jul 22 at 19:57
Thanks, but that doesn't explain how you recover Cliff(0). And for Cliff(1) you have $v, w in mathbbR$, so I can't see how the $i$ appears... Could you elaborate more, please?
– David
Jul 22 at 20:19
The Clifford algebra $operatornameCliff(V)$ of an inner product space $V$ is the unital algebra generated by elements of $V$, subject to the relation $vw + wv = -2langle v,wrangle 1_operatornameCliff(V)$; in particular, it is generated as an algebra by the algebra's unit element $1_operatornameCliff(V)$ together with by any basis for $V$. So, this vacuously gives you $operatornameCliff(0) = mathbbR$, while $operatornameCliff(1)$ is generated by the unit $1_operatornameCliff(1)$ together with the basis $i$ for the $1$-dim'l inner product space $mathbbR$
– Branimir Ćaćić
Jul 22 at 21:36
Ok, sure I'm missing something, but let's put values: we are in $mathbbR$, and for instance $v = 2, w = 3$, and the unit and basis for $mathbbR$ is 1. Then the equality $vw + wv = -2langle v, wrangle$ is not verified...
– David
Jul 22 at 23:56
1
1
The minus sign means that $i^2 = -1$ in $operatornameCliff(1)$ and that $i^2 = j^2 = k^2 = -1$ in $operatornameCliff(2)$, which is presumably what you want?
– Branimir Ćaćić
Jul 22 at 19:57
The minus sign means that $i^2 = -1$ in $operatornameCliff(1)$ and that $i^2 = j^2 = k^2 = -1$ in $operatornameCliff(2)$, which is presumably what you want?
– Branimir Ćaćić
Jul 22 at 19:57
Thanks, but that doesn't explain how you recover Cliff(0). And for Cliff(1) you have $v, w in mathbbR$, so I can't see how the $i$ appears... Could you elaborate more, please?
– David
Jul 22 at 20:19
Thanks, but that doesn't explain how you recover Cliff(0). And for Cliff(1) you have $v, w in mathbbR$, so I can't see how the $i$ appears... Could you elaborate more, please?
– David
Jul 22 at 20:19
The Clifford algebra $operatornameCliff(V)$ of an inner product space $V$ is the unital algebra generated by elements of $V$, subject to the relation $vw + wv = -2langle v,wrangle 1_operatornameCliff(V)$; in particular, it is generated as an algebra by the algebra's unit element $1_operatornameCliff(V)$ together with by any basis for $V$. So, this vacuously gives you $operatornameCliff(0) = mathbbR$, while $operatornameCliff(1)$ is generated by the unit $1_operatornameCliff(1)$ together with the basis $i$ for the $1$-dim'l inner product space $mathbbR$
– Branimir Ćaćić
Jul 22 at 21:36
The Clifford algebra $operatornameCliff(V)$ of an inner product space $V$ is the unital algebra generated by elements of $V$, subject to the relation $vw + wv = -2langle v,wrangle 1_operatornameCliff(V)$; in particular, it is generated as an algebra by the algebra's unit element $1_operatornameCliff(V)$ together with by any basis for $V$. So, this vacuously gives you $operatornameCliff(0) = mathbbR$, while $operatornameCliff(1)$ is generated by the unit $1_operatornameCliff(1)$ together with the basis $i$ for the $1$-dim'l inner product space $mathbbR$
– Branimir Ćaćić
Jul 22 at 21:36
Ok, sure I'm missing something, but let's put values: we are in $mathbbR$, and for instance $v = 2, w = 3$, and the unit and basis for $mathbbR$ is 1. Then the equality $vw + wv = -2langle v, wrangle$ is not verified...
– David
Jul 22 at 23:56
Ok, sure I'm missing something, but let's put values: we are in $mathbbR$, and for instance $v = 2, w = 3$, and the unit and basis for $mathbbR$ is 1. Then the equality $vw + wv = -2langle v, wrangle$ is not verified...
– David
Jul 22 at 23:56
add a comment |Â
1 Answer
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Let $(V,langlecdot,cdotrangle)$ be an inner product space over $mathbbK$. First, recall that the tensor algebra generated by $V$ is the vector space
$$T(V)=mathbbKoplus Voplus(Votimes V)oplus(Votimes Votimes V)oplusdots$$
In the text, Baez defines the Clifford algebra as an associative algebra generated by $V$ modulo the relations
$$vv=-|v|,$$
or equivalently
$$vw+wv=-2langle v,wrangle.$$
When he says "generated by $V$ modulo", it means that the product given by juxtaposition of elements $v,win V$ in the equations is defined as a product satisfying those equations. More precisely, since the product must be bilinear, the Clifford algebra will be the quotient of the tensor algebra $T(V)$ by an equivalence relation that will make those relations true.
Aiming to be concise, let $V=mathbbR^n$ and $mathbbK=mathbbR$. Now, if we set $n=0$, we would have $V=0$. Notice that since the juxtaposition product is bilinear, the equations above are automatically satisfied. Therefore, the Clifford algebra will be exactly the tensor algebra. However, $T(V)=mathbbKoplus0oplus(0otimes0)oplusdots=mathbbK=mathbbR$. It yields
$$textCliff(0)=mathbbR.$$
Remark: the symbols $v$ and $w$ in the equations are elements of $V$, not from $mathbbK$. Therefore, the example given in the comments of the question does not provide a counter-example.
answered Jul 26 at 20:00
Aquerman Kuczmenda
214
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Let $(V,langlecdot,cdotrangle)$ be an inner product space over $mathbbK$. First, recall that the tensor algebra generated by $V$ is the vector space
$$T(V)=mathbbKoplus Voplus(Votimes V)oplus(Votimes Votimes V)oplusdots$$
In the text, Baez defines the Clifford algebra as an associative algebra generated by $V$ modulo the relations
$$vv=-|v|,$$
or equivalently
$$vw+wv=-2langle v,wrangle.$$
When he says "generated by $V$ modulo", it means that the product given by juxtaposition of elements $v,win V$ in the equations is defined as a product satisfying those equations. More precisely, since the product must be bilinear, the Clifford algebra will be the quotient of the tensor algebra $T(V)$ by an equivalence relation that will make those relations true.
Aiming to be concise, let $V=mathbbR^n$ and $mathbbK=mathbbR$. Now, if we set $n=0$, we would have $V=0$. Notice that since the juxtaposition product is bilinear, the equations above are automatically satisfied. Therefore, the Clifford algebra will be exactly the tensor algebra. However, $T(V)=mathbbKoplus0oplus(0otimes0)oplusdots=mathbbK=mathbbR$. It yields
$$textCliff(0)=mathbbR.$$
Remark: the symbols $v$ and $w$ in the equations are elements of $V$, not from $mathbbK$. Therefore, the example given in the comments of the question does not provide a counter-example.
answered Jul 26 at 20:00
Aquerman Kuczmenda
214
add a comment |Â
up vote
1
down vote
Let $(V,langlecdot,cdotrangle)$ be an inner product space over $mathbbK$. First, recall that the tensor algebra generated by $V$ is the vector space
$$T(V)=mathbbKoplus Voplus(Votimes V)oplus(Votimes Votimes V)oplusdots$$
In the text, Baez defines the Clifford algebra as an associative algebra generated by $V$ modulo the relations
$$vv=-|v|,$$
or equivalently
$$vw+wv=-2langle v,wrangle.$$
When he says "generated by $V$ modulo", it means that the product given by juxtaposition of elements $v,win V$ in the equations is defined as a product satisfying those equations. More precisely, since the product must be bilinear, the Clifford algebra will be the quotient of the tensor algebra $T(V)$ by an equivalence relation that will make those relations true.
Aiming to be concise, let $V=mathbbR^n$ and $mathbbK=mathbbR$. Now, if we set $n=0$, we would have $V=0$. Notice that since the juxtaposition product is bilinear, the equations above are automatically satisfied. Therefore, the Clifford algebra will be exactly the tensor algebra. However, $T(V)=mathbbKoplus0oplus(0otimes0)oplusdots=mathbbK=mathbbR$. It yields
$$textCliff(0)=mathbbR.$$
Remark: the symbols $v$ and $w$ in the equations are elements of $V$, not from $mathbbK$. Therefore, the example given in the comments of the question does not provide a counter-example.
answered Jul 26 at 20:00
Aquerman Kuczmenda
214
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Let $(V,langlecdot,cdotrangle)$ be an inner product space over $mathbbK$. First, recall that the tensor algebra generated by $V$ is the vector space
$$T(V)=mathbbKoplus Voplus(Votimes V)oplus(Votimes Votimes V)oplusdots$$
In the text, Baez defines the Clifford algebra as an associative algebra generated by $V$ modulo the relations
$$vv=-|v|,$$
or equivalently
$$vw+wv=-2langle v,wrangle.$$
When he says "generated by $V$ modulo", it means that the product given by juxtaposition of elements $v,win V$ in the equations is defined as a product satisfying those equations. More precisely, since the product must be bilinear, the Clifford algebra will be the quotient of the tensor algebra $T(V)$ by an equivalence relation that will make those relations true.
Aiming to be concise, let $V=mathbbR^n$ and $mathbbK=mathbbR$. Now, if we set $n=0$, we would have $V=0$. Notice that since the juxtaposition product is bilinear, the equations above are automatically satisfied. Therefore, the Clifford algebra will be exactly the tensor algebra. However, $T(V)=mathbbKoplus0oplus(0otimes0)oplusdots=mathbbK=mathbbR$. It yields
$$textCliff(0)=mathbbR.$$
Remark: the symbols $v$ and $w$ in the equations are elements of $V$, not from $mathbbK$. Therefore, the example given in the comments of the question does not provide a counter-example.
answered Jul 26 at 20:00
Aquerman Kuczmenda
214
Let $(V,langlecdot,cdotrangle)$ be an inner product space over $mathbbK$. First, recall that the tensor algebra generated by $V$ is the vector space
$$T(V)=mathbbKoplus Voplus(Votimes V)oplus(Votimes Votimes V)oplusdots$$
In the text, Baez defines the Clifford algebra as an associative algebra generated by $V$ modulo the relations
$$vv=-|v|,$$
or equivalently
$$vw+wv=-2langle v,wrangle.$$
When he says "generated by $V$ modulo", it means that the product given by juxtaposition of elements $v,win V$ in the equations is defined as a product satisfying those equations. More precisely, since the product must be bilinear, the Clifford algebra will be the quotient of the tensor algebra $T(V)$ by an equivalence relation that will make those relations true.
Aiming to be concise, let $V=mathbbR^n$ and $mathbbK=mathbbR$. Now, if we set $n=0$, we would have $V=0$. Notice that since the juxtaposition product is bilinear, the equations above are automatically satisfied. Therefore, the Clifford algebra will be exactly the tensor algebra. However, $T(V)=mathbbKoplus0oplus(0otimes0)oplusdots=mathbbK=mathbbR$. It yields
$$textCliff(0)=mathbbR.$$
Remark: the symbols $v$ and $w$ in the equations are elements of $V$, not from $mathbbK$. Therefore, the example given in the comments of the question does not provide a counter-example.
answered Jul 26 at 20:00
Aquerman Kuczmenda
214
edited Jul 26 at 20:11
answered Jul 26 at 20:00
Aquerman Kuczmenda
214
answered Jul 26 at 20:00
Aquerman Kuczmenda
214
answered Jul 26 at 20:00
Aquerman Kuczmenda
214
214
add a comment |Â
add a comment |Â
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1
The minus sign means that $i^2 = -1$ in $operatornameCliff(1)$ and that $i^2 = j^2 = k^2 = -1$ in $operatornameCliff(2)$, which is presumably what you want?
– Branimir Ćaćić
Jul 22 at 19:57
Thanks, but that doesn't explain how you recover Cliff(0). And for Cliff(1) you have $v, w in mathbbR$, so I can't see how the $i$ appears... Could you elaborate more, please?
– David
Jul 22 at 20:19
The Clifford algebra $operatornameCliff(V)$ of an inner product space $V$ is the unital algebra generated by elements of $V$, subject to the relation $vw + wv = -2langle v,wrangle 1_operatornameCliff(V)$; in particular, it is generated as an algebra by the algebra's unit element $1_operatornameCliff(V)$ together with by any basis for $V$. So, this vacuously gives you $operatornameCliff(0) = mathbbR$, while $operatornameCliff(1)$ is generated by the unit $1_operatornameCliff(1)$ together with the basis $i$ for the $1$-dim'l inner product space $mathbbR$
– Branimir Ćaćić
Jul 22 at 21:36
Ok, sure I'm missing something, but let's put values: we are in $mathbbR$, and for instance $v = 2, w = 3$, and the unit and basis for $mathbbR$ is 1. Then the equality $vw + wv = -2langle v, wrangle$ is not verified...
– David
Jul 22 at 23:56