Proof that geometric product is associative The Next CEO of Stack OverflowIn Geometric...
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Proof that geometric product is associative
The Next CEO of Stack OverflowIn Geometric Algebra, is there a geometric product between matrices?Help me evaluate this geometric productinner product of trivector and bivector in geometric algebrageometric product between nonorthogonal basis and its reciprocal basisGeometric ProductVisualizing the geometric product?Different answers to geometric product problemHow does the geometric product work? Inconsistent/circular?Is there a relationship between the trace and the Clifford/geometric product?General expression for geometric product of blades in terms of scalar and exterior products
$begingroup$
Geometric product has nice property since it is a ring and it is associative to multiplication, which is not the case for vector cross product. But besides it is an axiom for geometric product, in the process of actually defining geometric product in a constructive way, is there a proof that it is indeed satisfy the associativity?
i.e., the geometric product of a blade $A_r$ and blade $B_s$ by grade expansion of
$$ A_rB_s = langle A_rB_srangle_{|r-s|} + ... + langle A_rB_srangle_{r+s}$$ is this associative?
geometric-algebras
edited Aug 29 '14 at 17:58
rschwieb
107k12103252
asked Aug 29 '14 at 14:47
ahalaahala
1,10811123
$endgroup$
|
show 1 more comment
$begingroup$
Geometric product has nice property since it is a ring and it is associative to multiplication, which is not the case for vector cross product. But besides it is an axiom for geometric product, in the process of actually defining geometric product in a constructive way, is there a proof that it is indeed satisfy the associativity?
i.e., the geometric product of a blade $A_r$ and blade $B_s$ by grade expansion of
$$ A_rB_s = langle A_rB_srangle_{|r-s|} + ... + langle A_rB_srangle_{r+s}$$ is this associative?
geometric-algebras
edited Aug 29 '14 at 17:58
rschwieb
107k12103252
asked Aug 29 '14 at 14:47
ahalaahala
1,10811123
$endgroup$
$begingroup$
Isn't it one of the three axioms for the geometric product? You can't prove an axiom!
$endgroup$
– user122283
Aug 29 '14 at 15:15
1
$begingroup$
For the main examples it is easy to see that they are associative.
$endgroup$
– Dietrich Burde
Aug 29 '14 at 15:17
1
$begingroup$
The usual approach is to take the fact that the multiplication is associative as an axiom. See Chapter 1 of Hestenes and Sobczyk, Clifford Algebra to Geometric Calculus, Reidel 1984.
$endgroup$
– almagest
Aug 29 '14 at 15:22
$begingroup$
@Sanath, you have to prove the computation rules made up for the geometric product do satisfy the axioms. Otherwise you know there is a thing called geometric product, but do not know whether it is the same thing you are calculating.
$endgroup$
– ahala
Aug 29 '14 at 15:28
$begingroup$
@ahala You're taking as axiomatic that the geometric product is associative, so what "rules" do you think must be shown to be consistent with this associative axiom?
$endgroup$
– Muphrid
Aug 29 '14 at 16:26
|
show 1 more comment
$begingroup$
Geometric product has nice property since it is a ring and it is associative to multiplication, which is not the case for vector cross product. But besides it is an axiom for geometric product, in the process of actually defining geometric product in a constructive way, is there a proof that it is indeed satisfy the associativity?
i.e., the geometric product of a blade $A_r$ and blade $B_s$ by grade expansion of
$$ A_rB_s = langle A_rB_srangle_{|r-s|} + ... + langle A_rB_srangle_{r+s}$$ is this associative?
geometric-algebras
edited Aug 29 '14 at 17:58
rschwieb
107k12103252
asked Aug 29 '14 at 14:47
ahalaahala
1,10811123
$endgroup$
Geometric product has nice property since it is a ring and it is associative to multiplication, which is not the case for vector cross product. But besides it is an axiom for geometric product, in the process of actually defining geometric product in a constructive way, is there a proof that it is indeed satisfy the associativity?
i.e., the geometric product of a blade $A_r$ and blade $B_s$ by grade expansion of
$$ A_rB_s = langle A_rB_srangle_{|r-s|} + ... + langle A_rB_srangle_{r+s}$$ is this associative?
geometric-algebras
geometric-algebras
edited Aug 29 '14 at 17:58
rschwieb
107k12103252
asked Aug 29 '14 at 14:47
ahalaahala
1,10811123
edited Aug 29 '14 at 17:58
rschwieb
107k12103252
asked Aug 29 '14 at 14:47
ahalaahala
1,10811123
edited Aug 29 '14 at 17:58
rschwieb
107k12103252
edited Aug 29 '14 at 17:58
rschwieb
107k12103252
edited Aug 29 '14 at 17:58
rschwieb
107k12103252
107k12103252
asked Aug 29 '14 at 14:47
ahalaahala
1,10811123
asked Aug 29 '14 at 14:47
ahalaahala
1,10811123
asked Aug 29 '14 at 14:47
ahalaahala
1,10811123
1,10811123
$begingroup$
Isn't it one of the three axioms for the geometric product? You can't prove an axiom!
$endgroup$
– user122283
Aug 29 '14 at 15:15
1
$begingroup$
For the main examples it is easy to see that they are associative.
$endgroup$
– Dietrich Burde
Aug 29 '14 at 15:17
1
$begingroup$
The usual approach is to take the fact that the multiplication is associative as an axiom. See Chapter 1 of Hestenes and Sobczyk, Clifford Algebra to Geometric Calculus, Reidel 1984.
$endgroup$
– almagest
Aug 29 '14 at 15:22
$begingroup$
@Sanath, you have to prove the computation rules made up for the geometric product do satisfy the axioms. Otherwise you know there is a thing called geometric product, but do not know whether it is the same thing you are calculating.
$endgroup$
– ahala
Aug 29 '14 at 15:28
$begingroup$
@ahala You're taking as axiomatic that the geometric product is associative, so what "rules" do you think must be shown to be consistent with this associative axiom?
$endgroup$
– Muphrid
Aug 29 '14 at 16:26
|
show 1 more comment
$begingroup$
Isn't it one of the three axioms for the geometric product? You can't prove an axiom!
$endgroup$
– user122283
Aug 29 '14 at 15:15
1
$begingroup$
For the main examples it is easy to see that they are associative.
$endgroup$
– Dietrich Burde
Aug 29 '14 at 15:17
1
$begingroup$
The usual approach is to take the fact that the multiplication is associative as an axiom. See Chapter 1 of Hestenes and Sobczyk, Clifford Algebra to Geometric Calculus, Reidel 1984.
$endgroup$
– almagest
Aug 29 '14 at 15:22
$begingroup$
@Sanath, you have to prove the computation rules made up for the geometric product do satisfy the axioms. Otherwise you know there is a thing called geometric product, but do not know whether it is the same thing you are calculating.
$endgroup$
– ahala
Aug 29 '14 at 15:28
$begingroup$
@ahala You're taking as axiomatic that the geometric product is associative, so what "rules" do you think must be shown to be consistent with this associative axiom?
$endgroup$
– Muphrid
Aug 29 '14 at 16:26
$begingroup$
Isn't it one of the three axioms for the geometric product? You can't prove an axiom!
$endgroup$
– user122283
Aug 29 '14 at 15:15
$begingroup$
Isn't it one of the three axioms for the geometric product? You can't prove an axiom!
$endgroup$
– user122283
Aug 29 '14 at 15:15
1
1
$begingroup$
For the main examples it is easy to see that they are associative.
$endgroup$
– Dietrich Burde
Aug 29 '14 at 15:17
$begingroup$
For the main examples it is easy to see that they are associative.
$endgroup$
– Dietrich Burde
Aug 29 '14 at 15:17
1
1
$begingroup$
The usual approach is to take the fact that the multiplication is associative as an axiom. See Chapter 1 of Hestenes and Sobczyk, Clifford Algebra to Geometric Calculus, Reidel 1984.
$endgroup$
– almagest
Aug 29 '14 at 15:22
$begingroup$
The usual approach is to take the fact that the multiplication is associative as an axiom. See Chapter 1 of Hestenes and Sobczyk, Clifford Algebra to Geometric Calculus, Reidel 1984.
$endgroup$
– almagest
Aug 29 '14 at 15:22
$begingroup$
@Sanath, you have to prove the computation rules made up for the geometric product do satisfy the axioms. Otherwise you know there is a thing called geometric product, but do not know whether it is the same thing you are calculating.
$endgroup$
– ahala
Aug 29 '14 at 15:28
$begingroup$
@Sanath, you have to prove the computation rules made up for the geometric product do satisfy the axioms. Otherwise you know there is a thing called geometric product, but do not know whether it is the same thing you are calculating.
$endgroup$
– ahala
Aug 29 '14 at 15:28
$begingroup$
@ahala You're taking as axiomatic that the geometric product is associative, so what "rules" do you think must be shown to be consistent with this associative axiom?
$endgroup$
– Muphrid
Aug 29 '14 at 16:26
$begingroup$
@ahala You're taking as axiomatic that the geometric product is associative, so what "rules" do you think must be shown to be consistent with this associative axiom?
$endgroup$
– Muphrid
Aug 29 '14 at 16:26
|
show 1 more comment
3 Answers
3
active
oldest
votes
$begingroup$
If you defined the geometric product "from the bottom" on basis elements, then it follows from the fact the product is defined to be associative on the basis elements (See proposition 1 pg 4 in this text by Jacobson for a proof that associativity on basis elements is sufficient for associativity of the ring.) (I think this is basically what you're asking when you're talking about the grade decomposition.)
If you are defining it from the top down as a quotient of the tensor algebra on $V$, then it is associative because the tensor algebra is associative.
answered Aug 29 '14 at 17:45
rschwiebrschwieb
107k12103252
$endgroup$
$begingroup$
thanks. I think that is a valid question since one has to recover vector analysis from an implementation of geometric product, in which it needs a proof to link the axioms to the implementation.
$endgroup$
– ahala
Aug 29 '14 at 18:06
$begingroup$
@ahala I don't follow what you're saying, but that's ok.
$endgroup$
– rschwieb
Aug 29 '14 at 18:18
add a comment |
$begingroup$
There are two approaches:
Write down a list of axioms for geometric algebra. Associativity probably needs to be on the list. Advantage: One can get down to the business of using the algebra right away. Disadvantage: How does one know that the list of axioms does not hide an inconsistency?
For most people "it has been proved consistent" is a good enough answer to the question above. But some will want to see a construction of GA and proofs of its properties, including associativity of the geometric product. There are several proofs. My own is the topic of the paper An elementary construction of the geometric algebra, Adv. Appl. Clif. Alg. 12, 1-6 (2002). A somewhat improved version is available at my website http://faculty.luther.edu/~macdonal/ . The paper cites other proofs.
answered Aug 29 '14 at 18:41
Alan MacdonaldAlan Macdonald
44625
$endgroup$
add a comment |
$begingroup$
Mr. Macdonald, the link for this paper is broken or miss-linked, if I can say that. This is a sad coincidence because I am really interested in this kind of simple constructions.
answered Mar 18 at 2:12
Carlos Manuel Rodríguez RománCarlos Manuel Rodríguez Román
1
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If you defined the geometric product "from the bottom" on basis elements, then it follows from the fact the product is defined to be associative on the basis elements (See proposition 1 pg 4 in this text by Jacobson for a proof that associativity on basis elements is sufficient for associativity of the ring.) (I think this is basically what you're asking when you're talking about the grade decomposition.)
If you are defining it from the top down as a quotient of the tensor algebra on $V$, then it is associative because the tensor algebra is associative.
answered Aug 29 '14 at 17:45
rschwiebrschwieb
107k12103252
$endgroup$
$begingroup$
thanks. I think that is a valid question since one has to recover vector analysis from an implementation of geometric product, in which it needs a proof to link the axioms to the implementation.
$endgroup$
– ahala
Aug 29 '14 at 18:06
$begingroup$
@ahala I don't follow what you're saying, but that's ok.
$endgroup$
– rschwieb
Aug 29 '14 at 18:18
add a comment |
$begingroup$
If you defined the geometric product "from the bottom" on basis elements, then it follows from the fact the product is defined to be associative on the basis elements (See proposition 1 pg 4 in this text by Jacobson for a proof that associativity on basis elements is sufficient for associativity of the ring.) (I think this is basically what you're asking when you're talking about the grade decomposition.)
If you are defining it from the top down as a quotient of the tensor algebra on $V$, then it is associative because the tensor algebra is associative.
answered Aug 29 '14 at 17:45
rschwiebrschwieb
107k12103252
$endgroup$
$begingroup$
thanks. I think that is a valid question since one has to recover vector analysis from an implementation of geometric product, in which it needs a proof to link the axioms to the implementation.
$endgroup$
– ahala
Aug 29 '14 at 18:06
$begingroup$
@ahala I don't follow what you're saying, but that's ok.
$endgroup$
– rschwieb
Aug 29 '14 at 18:18
add a comment |
$begingroup$
If you defined the geometric product "from the bottom" on basis elements, then it follows from the fact the product is defined to be associative on the basis elements (See proposition 1 pg 4 in this text by Jacobson for a proof that associativity on basis elements is sufficient for associativity of the ring.) (I think this is basically what you're asking when you're talking about the grade decomposition.)
If you are defining it from the top down as a quotient of the tensor algebra on $V$, then it is associative because the tensor algebra is associative.
answered Aug 29 '14 at 17:45
rschwiebrschwieb
107k12103252
$endgroup$
If you defined the geometric product "from the bottom" on basis elements, then it follows from the fact the product is defined to be associative on the basis elements (See proposition 1 pg 4 in this text by Jacobson for a proof that associativity on basis elements is sufficient for associativity of the ring.) (I think this is basically what you're asking when you're talking about the grade decomposition.)
If you are defining it from the top down as a quotient of the tensor algebra on $V$, then it is associative because the tensor algebra is associative.
answered Aug 29 '14 at 17:45
rschwiebrschwieb
107k12103252
edited Aug 29 '14 at 18:00
answered Aug 29 '14 at 17:45
rschwiebrschwieb
107k12103252
answered Aug 29 '14 at 17:45
rschwiebrschwieb
107k12103252
answered Aug 29 '14 at 17:45
rschwiebrschwieb
107k12103252
107k12103252
$begingroup$
thanks. I think that is a valid question since one has to recover vector analysis from an implementation of geometric product, in which it needs a proof to link the axioms to the implementation.
$endgroup$
– ahala
Aug 29 '14 at 18:06
$begingroup$
@ahala I don't follow what you're saying, but that's ok.
$endgroup$
– rschwieb
Aug 29 '14 at 18:18
add a comment |
$begingroup$
thanks. I think that is a valid question since one has to recover vector analysis from an implementation of geometric product, in which it needs a proof to link the axioms to the implementation.
$endgroup$
– ahala
Aug 29 '14 at 18:06
$begingroup$
@ahala I don't follow what you're saying, but that's ok.
$endgroup$
– rschwieb
Aug 29 '14 at 18:18
$begingroup$
thanks. I think that is a valid question since one has to recover vector analysis from an implementation of geometric product, in which it needs a proof to link the axioms to the implementation.
$endgroup$
– ahala
Aug 29 '14 at 18:06
$begingroup$
thanks. I think that is a valid question since one has to recover vector analysis from an implementation of geometric product, in which it needs a proof to link the axioms to the implementation.
$endgroup$
– ahala
Aug 29 '14 at 18:06
$begingroup$
@ahala I don't follow what you're saying, but that's ok.
$endgroup$
– rschwieb
Aug 29 '14 at 18:18
$begingroup$
@ahala I don't follow what you're saying, but that's ok.
$endgroup$
– rschwieb
Aug 29 '14 at 18:18
add a comment |
$begingroup$
There are two approaches:
Write down a list of axioms for geometric algebra. Associativity probably needs to be on the list. Advantage: One can get down to the business of using the algebra right away. Disadvantage: How does one know that the list of axioms does not hide an inconsistency?
For most people "it has been proved consistent" is a good enough answer to the question above. But some will want to see a construction of GA and proofs of its properties, including associativity of the geometric product. There are several proofs. My own is the topic of the paper An elementary construction of the geometric algebra, Adv. Appl. Clif. Alg. 12, 1-6 (2002). A somewhat improved version is available at my website http://faculty.luther.edu/~macdonal/ . The paper cites other proofs.
answered Aug 29 '14 at 18:41
Alan MacdonaldAlan Macdonald
44625
$endgroup$
add a comment |
$begingroup$
There are two approaches:
Write down a list of axioms for geometric algebra. Associativity probably needs to be on the list. Advantage: One can get down to the business of using the algebra right away. Disadvantage: How does one know that the list of axioms does not hide an inconsistency?
For most people "it has been proved consistent" is a good enough answer to the question above. But some will want to see a construction of GA and proofs of its properties, including associativity of the geometric product. There are several proofs. My own is the topic of the paper An elementary construction of the geometric algebra, Adv. Appl. Clif. Alg. 12, 1-6 (2002). A somewhat improved version is available at my website http://faculty.luther.edu/~macdonal/ . The paper cites other proofs.
answered Aug 29 '14 at 18:41
Alan MacdonaldAlan Macdonald
44625
$endgroup$
add a comment |
$begingroup$
There are two approaches:
Write down a list of axioms for geometric algebra. Associativity probably needs to be on the list. Advantage: One can get down to the business of using the algebra right away. Disadvantage: How does one know that the list of axioms does not hide an inconsistency?
For most people "it has been proved consistent" is a good enough answer to the question above. But some will want to see a construction of GA and proofs of its properties, including associativity of the geometric product. There are several proofs. My own is the topic of the paper An elementary construction of the geometric algebra, Adv. Appl. Clif. Alg. 12, 1-6 (2002). A somewhat improved version is available at my website http://faculty.luther.edu/~macdonal/ . The paper cites other proofs.
answered Aug 29 '14 at 18:41
Alan MacdonaldAlan Macdonald
44625
$endgroup$
There are two approaches:
Write down a list of axioms for geometric algebra. Associativity probably needs to be on the list. Advantage: One can get down to the business of using the algebra right away. Disadvantage: How does one know that the list of axioms does not hide an inconsistency?
For most people "it has been proved consistent" is a good enough answer to the question above. But some will want to see a construction of GA and proofs of its properties, including associativity of the geometric product. There are several proofs. My own is the topic of the paper An elementary construction of the geometric algebra, Adv. Appl. Clif. Alg. 12, 1-6 (2002). A somewhat improved version is available at my website http://faculty.luther.edu/~macdonal/ . The paper cites other proofs.
answered Aug 29 '14 at 18:41
Alan MacdonaldAlan Macdonald
44625
answered Aug 29 '14 at 18:41
Alan MacdonaldAlan Macdonald
44625
answered Aug 29 '14 at 18:41
Alan MacdonaldAlan Macdonald
44625
answered Aug 29 '14 at 18:41
Alan MacdonaldAlan Macdonald
44625
44625
add a comment |
add a comment |
$begingroup$
Mr. Macdonald, the link for this paper is broken or miss-linked, if I can say that. This is a sad coincidence because I am really interested in this kind of simple constructions.
answered Mar 18 at 2:12
Carlos Manuel Rodríguez RománCarlos Manuel Rodríguez Román
1
$endgroup$
add a comment |
$begingroup$
Mr. Macdonald, the link for this paper is broken or miss-linked, if I can say that. This is a sad coincidence because I am really interested in this kind of simple constructions.
answered Mar 18 at 2:12
Carlos Manuel Rodríguez RománCarlos Manuel Rodríguez Román
1
$endgroup$
add a comment |
$begingroup$
Mr. Macdonald, the link for this paper is broken or miss-linked, if I can say that. This is a sad coincidence because I am really interested in this kind of simple constructions.
answered Mar 18 at 2:12
Carlos Manuel Rodríguez RománCarlos Manuel Rodríguez Román
1
$endgroup$
Mr. Macdonald, the link for this paper is broken or miss-linked, if I can say that. This is a sad coincidence because I am really interested in this kind of simple constructions.
answered Mar 18 at 2:12
Carlos Manuel Rodríguez RománCarlos Manuel Rodríguez Román
1
answered Mar 18 at 2:12
Carlos Manuel Rodríguez RománCarlos Manuel Rodríguez Román
1
answered Mar 18 at 2:12
Carlos Manuel Rodríguez RománCarlos Manuel Rodríguez Román
1
answered Mar 18 at 2:12
Carlos Manuel Rodríguez RománCarlos Manuel Rodríguez Román
1
1
add a comment |
add a comment |
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$begingroup$
Isn't it one of the three axioms for the geometric product? You can't prove an axiom!
$endgroup$
– user122283
Aug 29 '14 at 15:15
1
$begingroup$
For the main examples it is easy to see that they are associative.
$endgroup$
– Dietrich Burde
Aug 29 '14 at 15:17
1
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The usual approach is to take the fact that the multiplication is associative as an axiom. See Chapter 1 of Hestenes and Sobczyk, Clifford Algebra to Geometric Calculus, Reidel 1984.
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– almagest
Aug 29 '14 at 15:22
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@Sanath, you have to prove the computation rules made up for the geometric product do satisfy the axioms. Otherwise you know there is a thing called geometric product, but do not know whether it is the same thing you are calculating.
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– ahala
Aug 29 '14 at 15:28
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@ahala You're taking as axiomatic that the geometric product is associative, so what "rules" do you think must be shown to be consistent with this associative axiom?
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– Muphrid
Aug 29 '14 at 16:26