Formula (1.18) page 43 in Hestenes book “New foundations for Classical mechanics”
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The formula is $(A_rland b)cdot C_s=A_rcdot (bcdot C_s)$, where $0<r<s$. Hestenes suggests to expand $(A_rb)C_s=A_r(bC_s)$ and extract the $s-r-1$-vector part. But this method requires one to know the formula for the Clifford product of two arbitrary blades, and as far as I can see, the book only gives formulas for the Clifford product of a vector and a blade.
One can prove this formula using formulas for the exterior and interior product in Grassmann algebra. But this is not the method Hestenes suggests. If somebody knows the proof that Hestenes means, please let me know too.
geometric-algebras
asked Sep 26 '18 at 1:41
alexanderyaacovalexanderyaacov
404
$endgroup$
add a comment |
$begingroup$
The formula is $(A_rland b)cdot C_s=A_rcdot (bcdot C_s)$, where $0<r<s$. Hestenes suggests to expand $(A_rb)C_s=A_r(bC_s)$ and extract the $s-r-1$-vector part. But this method requires one to know the formula for the Clifford product of two arbitrary blades, and as far as I can see, the book only gives formulas for the Clifford product of a vector and a blade.
One can prove this formula using formulas for the exterior and interior product in Grassmann algebra. But this is not the method Hestenes suggests. If somebody knows the proof that Hestenes means, please let me know too.
geometric-algebras
asked Sep 26 '18 at 1:41
alexanderyaacovalexanderyaacov
404
$endgroup$
$begingroup$
@mr_e_man What is the symbol that looks like an upside down r?
$endgroup$
– alexanderyaacov
Dec 11 '18 at 10:08
$begingroup$
That $lrcorner$ "lrcorner" is the contraction product, an asymmetric dot product. en.wikipedia.org/wiki/… , av8n.com/physics/clifford-intro.htm#sec-contractions
$endgroup$
– mr_e_man
Dec 15 '18 at 0:19
$begingroup$
I remember this identity by comparing it with the analogous one in logic: $(Aland B)to Cequiv Ato(Bto C)$, where $A,B,C$ are statements which are either true or false. In words, "$A$ and $B$ together imply $C$" is equivalent to "If $A$ is true, then $B$ implies $C$".
$endgroup$
– mr_e_man
Dec 15 '18 at 0:25
$begingroup$
@mr_e_man why is it then a special case? It seems to me to be exactly the same.
$endgroup$
– alexanderyaacov
Dec 16 '18 at 7:24
$begingroup$
It's a special case because the grades are restricted. The identity I gave is always true; $B$ doesn't need to be a vector, nor $A$ lower-grade than $C$.
$endgroup$
– mr_e_man
Dec 16 '18 at 23:43
add a comment |
$begingroup$
The formula is $(A_rland b)cdot C_s=A_rcdot (bcdot C_s)$, where $0<r<s$. Hestenes suggests to expand $(A_rb)C_s=A_r(bC_s)$ and extract the $s-r-1$-vector part. But this method requires one to know the formula for the Clifford product of two arbitrary blades, and as far as I can see, the book only gives formulas for the Clifford product of a vector and a blade.
One can prove this formula using formulas for the exterior and interior product in Grassmann algebra. But this is not the method Hestenes suggests. If somebody knows the proof that Hestenes means, please let me know too.
geometric-algebras
asked Sep 26 '18 at 1:41
alexanderyaacovalexanderyaacov
404
$endgroup$
The formula is $(A_rland b)cdot C_s=A_rcdot (bcdot C_s)$, where $0<r<s$. Hestenes suggests to expand $(A_rb)C_s=A_r(bC_s)$ and extract the $s-r-1$-vector part. But this method requires one to know the formula for the Clifford product of two arbitrary blades, and as far as I can see, the book only gives formulas for the Clifford product of a vector and a blade.
One can prove this formula using formulas for the exterior and interior product in Grassmann algebra. But this is not the method Hestenes suggests. If somebody knows the proof that Hestenes means, please let me know too.
geometric-algebras
geometric-algebras
asked Sep 26 '18 at 1:41
alexanderyaacovalexanderyaacov
404
asked Sep 26 '18 at 1:41
alexanderyaacovalexanderyaacov
404
asked Sep 26 '18 at 1:41
alexanderyaacovalexanderyaacov
404
asked Sep 26 '18 at 1:41
alexanderyaacovalexanderyaacov
404
asked Sep 26 '18 at 1:41
alexanderyaacovalexanderyaacov
404
404
$begingroup$
@mr_e_man What is the symbol that looks like an upside down r?
$endgroup$
– alexanderyaacov
Dec 11 '18 at 10:08
$begingroup$
That $lrcorner$ "lrcorner" is the contraction product, an asymmetric dot product. en.wikipedia.org/wiki/… , av8n.com/physics/clifford-intro.htm#sec-contractions
$endgroup$
– mr_e_man
Dec 15 '18 at 0:19
$begingroup$
I remember this identity by comparing it with the analogous one in logic: $(Aland B)to Cequiv Ato(Bto C)$, where $A,B,C$ are statements which are either true or false. In words, "$A$ and $B$ together imply $C$" is equivalent to "If $A$ is true, then $B$ implies $C$".
$endgroup$
– mr_e_man
Dec 15 '18 at 0:25
$begingroup$
@mr_e_man why is it then a special case? It seems to me to be exactly the same.
$endgroup$
– alexanderyaacov
Dec 16 '18 at 7:24
$begingroup$
It's a special case because the grades are restricted. The identity I gave is always true; $B$ doesn't need to be a vector, nor $A$ lower-grade than $C$.
$endgroup$
– mr_e_man
Dec 16 '18 at 23:43
add a comment |
$begingroup$
@mr_e_man What is the symbol that looks like an upside down r?
$endgroup$
– alexanderyaacov
Dec 11 '18 at 10:08
$begingroup$
That $lrcorner$ "lrcorner" is the contraction product, an asymmetric dot product. en.wikipedia.org/wiki/… , av8n.com/physics/clifford-intro.htm#sec-contractions
$endgroup$
– mr_e_man
Dec 15 '18 at 0:19
$begingroup$
I remember this identity by comparing it with the analogous one in logic: $(Aland B)to Cequiv Ato(Bto C)$, where $A,B,C$ are statements which are either true or false. In words, "$A$ and $B$ together imply $C$" is equivalent to "If $A$ is true, then $B$ implies $C$".
$endgroup$
– mr_e_man
Dec 15 '18 at 0:25
$begingroup$
@mr_e_man why is it then a special case? It seems to me to be exactly the same.
$endgroup$
– alexanderyaacov
Dec 16 '18 at 7:24
$begingroup$
It's a special case because the grades are restricted. The identity I gave is always true; $B$ doesn't need to be a vector, nor $A$ lower-grade than $C$.
$endgroup$
– mr_e_man
Dec 16 '18 at 23:43
$begingroup$
@mr_e_man What is the symbol that looks like an upside down r?
$endgroup$
– alexanderyaacov
Dec 11 '18 at 10:08
$begingroup$
@mr_e_man What is the symbol that looks like an upside down r?
$endgroup$
– alexanderyaacov
Dec 11 '18 at 10:08
$begingroup$
That $lrcorner$ "lrcorner" is the contraction product, an asymmetric dot product. en.wikipedia.org/wiki/… , av8n.com/physics/clifford-intro.htm#sec-contractions
$endgroup$
– mr_e_man
Dec 15 '18 at 0:19
$begingroup$
That $lrcorner$ "lrcorner" is the contraction product, an asymmetric dot product. en.wikipedia.org/wiki/… , av8n.com/physics/clifford-intro.htm#sec-contractions
$endgroup$
– mr_e_man
Dec 15 '18 at 0:19
$begingroup$
I remember this identity by comparing it with the analogous one in logic: $(Aland B)to Cequiv Ato(Bto C)$, where $A,B,C$ are statements which are either true or false. In words, "$A$ and $B$ together imply $C$" is equivalent to "If $A$ is true, then $B$ implies $C$".
$endgroup$
– mr_e_man
Dec 15 '18 at 0:25
$begingroup$
I remember this identity by comparing it with the analogous one in logic: $(Aland B)to Cequiv Ato(Bto C)$, where $A,B,C$ are statements which are either true or false. In words, "$A$ and $B$ together imply $C$" is equivalent to "If $A$ is true, then $B$ implies $C$".
$endgroup$
– mr_e_man
Dec 15 '18 at 0:25
$begingroup$
@mr_e_man why is it then a special case? It seems to me to be exactly the same.
$endgroup$
– alexanderyaacov
Dec 16 '18 at 7:24
$begingroup$
@mr_e_man why is it then a special case? It seems to me to be exactly the same.
$endgroup$
– alexanderyaacov
Dec 16 '18 at 7:24
$begingroup$
It's a special case because the grades are restricted. The identity I gave is always true; $B$ doesn't need to be a vector, nor $A$ lower-grade than $C$.
$endgroup$
– mr_e_man
Dec 16 '18 at 23:43
$begingroup$
It's a special case because the grades are restricted. The identity I gave is always true; $B$ doesn't need to be a vector, nor $A$ lower-grade than $C$.
$endgroup$
– mr_e_man
Dec 16 '18 at 23:43
add a comment |
2 Answers
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Here's a proof from the appendix of Geometric Algebra for Electrical Engineers, with the variable names changed to match your question. The order of the variables is reverse of your question, but you can apply the reverse operator to each side and make a change of variables to prove the formula as stated in NFCM.
Theorem: Distribution of inner products
Given two blades $C_s, A_r$ with grades subject to $s > r > 0$, and a vector $mathbf{b}$, the inner product distributes according to
$$C_s cdot left( { mathbf{b} wedge A_r } right) = left( { C_s cdot mathbf{b} } right) cdot A_r.$$
Proof:
The proof is straightforward, relying primarily on grade selection, but also mechanical.
Start by expanding the wedge and dot products within a grade selection operator
$$C_s cdot left( { mathbf{b} wedge A_r } right)={leftlangle{{C_s (mathbf{b} wedge A_r)}}rightrangle}_{{s - (r + 1)}}=frac{1}{{2}} {leftlangle{{C_s left( {mathbf{b} A_r + (-1)^{r} A_r mathbf{b}} right) }}rightrangle}_{{s - (r + 1)}}.$$
Solving for $C_r mathbf{b}$ in
$$2 mathbf{b} cdot A_r = mathbf{b} A_r - (-1)^{r} A_r mathbf{b},$$
we have
$$begin{aligned}C_s cdot left( mathbf{b} wedge A_r right) &= frac{1}{2} {leftlangle C_s mathbf{b} A_r + C_s left( mathbf{b} A_r - 2 mathbf{b} cdot A_r right) rightrangle}_{s - (r + 1)} \ &= {leftlangle C_s mathbf{b} A_r rightrangle}_{s - (r + 1)} - {leftlangle C_s left( mathbf{b} cdot A_r right) rightrangle}_{s - (r + 1)} \ &= {leftlangle C_s mathbf{b} A_r rightrangle}_{s - (r + 1)}.end{aligned}$$
The last term in the second step is zero since we are selecting the $s - r - 1$ grade element of a multivector with grades $s - r + 1$ and $s + r - 1$, which has no terms for $r > 0$.
Now we can expand the $C_s mathbf{b}$ multivector product, for
$$C_s cdot left( { mathbf{b} wedge A_r } right)={leftlangle{{ left( { C_s cdot mathbf{b} + C_s wedge mathbf{b}} right) A_r }}rightrangle}_{{s - (r + 1)}}.$$
The latter multivector (with the wedge product factor) above has grades $s + 1 - r$ and $s + 1 + r$, so this selection operator finds nothing.
This leaves
$$C_s cdot left( { mathbf{b} wedge A_r } right)={leftlangle{{left( { C_s cdot mathbf{b} } right) cdot A_r+ left( { C_s cdot mathbf{b} } right) wedge A_r}}rightrangle}_{{s - (r + 1)}}.$$
The first dot products term has grade $s - 1 - r$ and is selected, whereas the wedge term has grade $s - 1 + r ne s - r - 1$ (for $r > 0$), which completes the proof.
answered Sep 28 '18 at 0:17
Peeter JootPeeter Joot
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This is a special case of the identity $(Awedge B),lrcorner,C=A,lrcorner,(B,lrcorner,C)$. It can be proven first for blades, then extended to all multivectors by linearity.
Suppose $A$ has grade $r$, $B$ has grade $k$ (in your case $k=1$), and $C$ has grade $s$. Then
$$AB=langle ABrangle_{r+k}$$
$$+langle ABrangle_{r+k-2}$$
$$+langle ABrangle_{r+k-4}$$
$$+cdots$$
$$+langle ABrangle_{-r+k+2}$$
$$+langle ABrangle_{-r+k}$$
The first term is $Awedge B$, and the last term is $A,lrcorner,B$. We could also write this as $$langle ABrangle_{r+k}+cdots+langle ABrangle_{|r-k|}=Awedge B+cdots+Acdot B$$ because the last term is the same if $rleq k$, and otherwise we're only adding $0$'s after $Acdot B$. (For example, with $r=3$ and $k=1$, the expression is $langle Abrangle_4+langle Abrangle_2+langle Abrangle_0+langle Abrangle_{-2}=langle Abrangle_4+langle Abrangle_2$.)
Multiply by $C$, expanding each term in the same way, to get
$$ABC=langlelangle ABrangle_{r+k}Crangle_{r+k+s}+langlelangle ABrangle_{r+k}Crangle_{r+k+s-2}+cdots+langlelangle ABrangle_{r+k}Crangle_{-(r+k)+s}$$
$$+langlelangle ABrangle_{r+k-2}Crangle_{r+k-2+s}+langlelangle ABrangle_{r+k-2}Crangle_{r+k-2+s-2}+cdots+langlelangle ABrangle_{r+k-2}Crangle_{-(r+k-2)+s}$$
$$+cdots$$
$$+langlelangle ABrangle_{-r+k}Crangle_{-r+k+s}+langlelangle ABrangle_{-r+k}Crangle_{-r+k+s-2}+cdots+langlelangle ABrangle_{-r+k}Crangle_{-(-r+k)+s}$$
Each row has successively fewer terms ($2$ less each time), so the rightmost term should be indented, and the expression above would look something like this trapezoid:
$$begin{matrix} x&x&x&x&x&x&x&x&x \ x&x&x&x&x&x&x \ x&x&x&x&x \ x&x&x end{matrix}$$
The grade decreases going right or going down, so all the terms with a given grade are on a diagonal line. The leftmost diagonal (the highest grade) has only a single term, $langle ABCrangle_{r+k+s}=langlelangle ABrangle_{r+k}Crangle_{r+k+s}=(Awedge B)wedge C$. The rightmost diagonal (the lowest grade) also has only a single term, $langle ABCrangle_{-(r+k)+s}=langlelangle ABrangle_{r+k}Crangle_{-(r+k)+s}=(Awedge B),lrcorner,C$.
Now write it the other way, expanding $BC$ first:
$$BC=langle BCrangle_{k+s}+langle BCrangle_{k+s-2}+cdots+langle BCrangle_{-k+s}$$
and multiplying by $A$:
$$ABC=langle Alangle BCrangle_{k+s}rangle_{r+k+s}+langle Alangle BCrangle_{k+s-2}rangle_{r+k+s-2}+cdots+langle Alangle BCrangle_{-k+s}rangle_{r-k+s}$$
$$+langle Alangle BCrangle_{k+s}rangle_{r-2+k+s}+langle Alangle BCrangle_{k+s-2}rangle_{r-2+k+s-2}+cdots+langle Alangle BCrangle_{-k+s}rangle_{r-2-k+s}$$
$$+cdots$$
$$+langle Alangle BCrangle_{k+s}rangle_{-r+k+s}+langle Alangle BCrangle_{k+s-2}rangle_{-r+k+s-2}+cdots+langle Alangle BCrangle_{-k+s}rangle_{-r-k+s}$$
This time, each row has the same number of terms, so it's just a rectangle instead of a trapezoid. Again, all the terms with a given grade are on a diagonal, and the leftmost diagonal has only $langle ABCrangle_{r+k+s}=langle Alangle BCrangle_{k+s}rangle_{r+k+s}=Awedge(Bwedge C)$. The rightmost diagonal also has only one term, $langle ABCrangle_{-r-k+s}=langle Alangle BCrangle_{-k+s}rangle_{-r-k+s}=A,lrcorner,(B,lrcorner,C)$.
Comparing the two expansions, we can see both identities at once:
$$(Awedge B)wedge C=Awedge(Bwedge C),qquad(Awedge B),lrcorner,C=A,lrcorner,(B,lrcorner, C).$$
answered Dec 17 '18 at 6:40
mr_e_manmr_e_man
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2 Answers
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2 Answers
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$begingroup$
Here's a proof from the appendix of Geometric Algebra for Electrical Engineers, with the variable names changed to match your question. The order of the variables is reverse of your question, but you can apply the reverse operator to each side and make a change of variables to prove the formula as stated in NFCM.
Theorem: Distribution of inner products
Given two blades $C_s, A_r$ with grades subject to $s > r > 0$, and a vector $mathbf{b}$, the inner product distributes according to
$$C_s cdot left( { mathbf{b} wedge A_r } right) = left( { C_s cdot mathbf{b} } right) cdot A_r.$$
Proof:
The proof is straightforward, relying primarily on grade selection, but also mechanical.
Start by expanding the wedge and dot products within a grade selection operator
$$C_s cdot left( { mathbf{b} wedge A_r } right)={leftlangle{{C_s (mathbf{b} wedge A_r)}}rightrangle}_{{s - (r + 1)}}=frac{1}{{2}} {leftlangle{{C_s left( {mathbf{b} A_r + (-1)^{r} A_r mathbf{b}} right) }}rightrangle}_{{s - (r + 1)}}.$$
Solving for $C_r mathbf{b}$ in
$$2 mathbf{b} cdot A_r = mathbf{b} A_r - (-1)^{r} A_r mathbf{b},$$
we have
$$begin{aligned}C_s cdot left( mathbf{b} wedge A_r right) &= frac{1}{2} {leftlangle C_s mathbf{b} A_r + C_s left( mathbf{b} A_r - 2 mathbf{b} cdot A_r right) rightrangle}_{s - (r + 1)} \ &= {leftlangle C_s mathbf{b} A_r rightrangle}_{s - (r + 1)} - {leftlangle C_s left( mathbf{b} cdot A_r right) rightrangle}_{s - (r + 1)} \ &= {leftlangle C_s mathbf{b} A_r rightrangle}_{s - (r + 1)}.end{aligned}$$
The last term in the second step is zero since we are selecting the $s - r - 1$ grade element of a multivector with grades $s - r + 1$ and $s + r - 1$, which has no terms for $r > 0$.
Now we can expand the $C_s mathbf{b}$ multivector product, for
$$C_s cdot left( { mathbf{b} wedge A_r } right)={leftlangle{{ left( { C_s cdot mathbf{b} + C_s wedge mathbf{b}} right) A_r }}rightrangle}_{{s - (r + 1)}}.$$
The latter multivector (with the wedge product factor) above has grades $s + 1 - r$ and $s + 1 + r$, so this selection operator finds nothing.
This leaves
$$C_s cdot left( { mathbf{b} wedge A_r } right)={leftlangle{{left( { C_s cdot mathbf{b} } right) cdot A_r+ left( { C_s cdot mathbf{b} } right) wedge A_r}}rightrangle}_{{s - (r + 1)}}.$$
The first dot products term has grade $s - 1 - r$ and is selected, whereas the wedge term has grade $s - 1 + r ne s - r - 1$ (for $r > 0$), which completes the proof.
answered Sep 28 '18 at 0:17
Peeter JootPeeter Joot
605410
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add a comment |
$begingroup$
Here's a proof from the appendix of Geometric Algebra for Electrical Engineers, with the variable names changed to match your question. The order of the variables is reverse of your question, but you can apply the reverse operator to each side and make a change of variables to prove the formula as stated in NFCM.
Theorem: Distribution of inner products
Given two blades $C_s, A_r$ with grades subject to $s > r > 0$, and a vector $mathbf{b}$, the inner product distributes according to
$$C_s cdot left( { mathbf{b} wedge A_r } right) = left( { C_s cdot mathbf{b} } right) cdot A_r.$$
Proof:
The proof is straightforward, relying primarily on grade selection, but also mechanical.
Start by expanding the wedge and dot products within a grade selection operator
$$C_s cdot left( { mathbf{b} wedge A_r } right)={leftlangle{{C_s (mathbf{b} wedge A_r)}}rightrangle}_{{s - (r + 1)}}=frac{1}{{2}} {leftlangle{{C_s left( {mathbf{b} A_r + (-1)^{r} A_r mathbf{b}} right) }}rightrangle}_{{s - (r + 1)}}.$$
Solving for $C_r mathbf{b}$ in
$$2 mathbf{b} cdot A_r = mathbf{b} A_r - (-1)^{r} A_r mathbf{b},$$
we have
$$begin{aligned}C_s cdot left( mathbf{b} wedge A_r right) &= frac{1}{2} {leftlangle C_s mathbf{b} A_r + C_s left( mathbf{b} A_r - 2 mathbf{b} cdot A_r right) rightrangle}_{s - (r + 1)} \ &= {leftlangle C_s mathbf{b} A_r rightrangle}_{s - (r + 1)} - {leftlangle C_s left( mathbf{b} cdot A_r right) rightrangle}_{s - (r + 1)} \ &= {leftlangle C_s mathbf{b} A_r rightrangle}_{s - (r + 1)}.end{aligned}$$
The last term in the second step is zero since we are selecting the $s - r - 1$ grade element of a multivector with grades $s - r + 1$ and $s + r - 1$, which has no terms for $r > 0$.
Now we can expand the $C_s mathbf{b}$ multivector product, for
$$C_s cdot left( { mathbf{b} wedge A_r } right)={leftlangle{{ left( { C_s cdot mathbf{b} + C_s wedge mathbf{b}} right) A_r }}rightrangle}_{{s - (r + 1)}}.$$
The latter multivector (with the wedge product factor) above has grades $s + 1 - r$ and $s + 1 + r$, so this selection operator finds nothing.
This leaves
$$C_s cdot left( { mathbf{b} wedge A_r } right)={leftlangle{{left( { C_s cdot mathbf{b} } right) cdot A_r+ left( { C_s cdot mathbf{b} } right) wedge A_r}}rightrangle}_{{s - (r + 1)}}.$$
The first dot products term has grade $s - 1 - r$ and is selected, whereas the wedge term has grade $s - 1 + r ne s - r - 1$ (for $r > 0$), which completes the proof.
answered Sep 28 '18 at 0:17
Peeter JootPeeter Joot
605410
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add a comment |
$begingroup$
Here's a proof from the appendix of Geometric Algebra for Electrical Engineers, with the variable names changed to match your question. The order of the variables is reverse of your question, but you can apply the reverse operator to each side and make a change of variables to prove the formula as stated in NFCM.
Theorem: Distribution of inner products
Given two blades $C_s, A_r$ with grades subject to $s > r > 0$, and a vector $mathbf{b}$, the inner product distributes according to
$$C_s cdot left( { mathbf{b} wedge A_r } right) = left( { C_s cdot mathbf{b} } right) cdot A_r.$$
Proof:
The proof is straightforward, relying primarily on grade selection, but also mechanical.
Start by expanding the wedge and dot products within a grade selection operator
$$C_s cdot left( { mathbf{b} wedge A_r } right)={leftlangle{{C_s (mathbf{b} wedge A_r)}}rightrangle}_{{s - (r + 1)}}=frac{1}{{2}} {leftlangle{{C_s left( {mathbf{b} A_r + (-1)^{r} A_r mathbf{b}} right) }}rightrangle}_{{s - (r + 1)}}.$$
Solving for $C_r mathbf{b}$ in
$$2 mathbf{b} cdot A_r = mathbf{b} A_r - (-1)^{r} A_r mathbf{b},$$
we have
$$begin{aligned}C_s cdot left( mathbf{b} wedge A_r right) &= frac{1}{2} {leftlangle C_s mathbf{b} A_r + C_s left( mathbf{b} A_r - 2 mathbf{b} cdot A_r right) rightrangle}_{s - (r + 1)} \ &= {leftlangle C_s mathbf{b} A_r rightrangle}_{s - (r + 1)} - {leftlangle C_s left( mathbf{b} cdot A_r right) rightrangle}_{s - (r + 1)} \ &= {leftlangle C_s mathbf{b} A_r rightrangle}_{s - (r + 1)}.end{aligned}$$
The last term in the second step is zero since we are selecting the $s - r - 1$ grade element of a multivector with grades $s - r + 1$ and $s + r - 1$, which has no terms for $r > 0$.
Now we can expand the $C_s mathbf{b}$ multivector product, for
$$C_s cdot left( { mathbf{b} wedge A_r } right)={leftlangle{{ left( { C_s cdot mathbf{b} + C_s wedge mathbf{b}} right) A_r }}rightrangle}_{{s - (r + 1)}}.$$
The latter multivector (with the wedge product factor) above has grades $s + 1 - r$ and $s + 1 + r$, so this selection operator finds nothing.
This leaves
$$C_s cdot left( { mathbf{b} wedge A_r } right)={leftlangle{{left( { C_s cdot mathbf{b} } right) cdot A_r+ left( { C_s cdot mathbf{b} } right) wedge A_r}}rightrangle}_{{s - (r + 1)}}.$$
The first dot products term has grade $s - 1 - r$ and is selected, whereas the wedge term has grade $s - 1 + r ne s - r - 1$ (for $r > 0$), which completes the proof.
answered Sep 28 '18 at 0:17
Peeter JootPeeter Joot
605410
$endgroup$
Here's a proof from the appendix of Geometric Algebra for Electrical Engineers, with the variable names changed to match your question. The order of the variables is reverse of your question, but you can apply the reverse operator to each side and make a change of variables to prove the formula as stated in NFCM.
Theorem: Distribution of inner products
Given two blades $C_s, A_r$ with grades subject to $s > r > 0$, and a vector $mathbf{b}$, the inner product distributes according to
$$C_s cdot left( { mathbf{b} wedge A_r } right) = left( { C_s cdot mathbf{b} } right) cdot A_r.$$
Proof:
The proof is straightforward, relying primarily on grade selection, but also mechanical.
Start by expanding the wedge and dot products within a grade selection operator
$$C_s cdot left( { mathbf{b} wedge A_r } right)={leftlangle{{C_s (mathbf{b} wedge A_r)}}rightrangle}_{{s - (r + 1)}}=frac{1}{{2}} {leftlangle{{C_s left( {mathbf{b} A_r + (-1)^{r} A_r mathbf{b}} right) }}rightrangle}_{{s - (r + 1)}}.$$
Solving for $C_r mathbf{b}$ in
$$2 mathbf{b} cdot A_r = mathbf{b} A_r - (-1)^{r} A_r mathbf{b},$$
we have
$$begin{aligned}C_s cdot left( mathbf{b} wedge A_r right) &= frac{1}{2} {leftlangle C_s mathbf{b} A_r + C_s left( mathbf{b} A_r - 2 mathbf{b} cdot A_r right) rightrangle}_{s - (r + 1)} \ &= {leftlangle C_s mathbf{b} A_r rightrangle}_{s - (r + 1)} - {leftlangle C_s left( mathbf{b} cdot A_r right) rightrangle}_{s - (r + 1)} \ &= {leftlangle C_s mathbf{b} A_r rightrangle}_{s - (r + 1)}.end{aligned}$$
The last term in the second step is zero since we are selecting the $s - r - 1$ grade element of a multivector with grades $s - r + 1$ and $s + r - 1$, which has no terms for $r > 0$.
Now we can expand the $C_s mathbf{b}$ multivector product, for
$$C_s cdot left( { mathbf{b} wedge A_r } right)={leftlangle{{ left( { C_s cdot mathbf{b} + C_s wedge mathbf{b}} right) A_r }}rightrangle}_{{s - (r + 1)}}.$$
The latter multivector (with the wedge product factor) above has grades $s + 1 - r$ and $s + 1 + r$, so this selection operator finds nothing.
This leaves
$$C_s cdot left( { mathbf{b} wedge A_r } right)={leftlangle{{left( { C_s cdot mathbf{b} } right) cdot A_r+ left( { C_s cdot mathbf{b} } right) wedge A_r}}rightrangle}_{{s - (r + 1)}}.$$
The first dot products term has grade $s - 1 - r$ and is selected, whereas the wedge term has grade $s - 1 + r ne s - r - 1$ (for $r > 0$), which completes the proof.
answered Sep 28 '18 at 0:17
Peeter JootPeeter Joot
605410
edited Sep 28 '18 at 1:03
answered Sep 28 '18 at 0:17
Peeter JootPeeter Joot
605410
answered Sep 28 '18 at 0:17
Peeter JootPeeter Joot
605410
answered Sep 28 '18 at 0:17
Peeter JootPeeter Joot
605410
605410
add a comment |
add a comment |
$begingroup$
This is a special case of the identity $(Awedge B),lrcorner,C=A,lrcorner,(B,lrcorner,C)$. It can be proven first for blades, then extended to all multivectors by linearity.
Suppose $A$ has grade $r$, $B$ has grade $k$ (in your case $k=1$), and $C$ has grade $s$. Then
$$AB=langle ABrangle_{r+k}$$
$$+langle ABrangle_{r+k-2}$$
$$+langle ABrangle_{r+k-4}$$
$$+cdots$$
$$+langle ABrangle_{-r+k+2}$$
$$+langle ABrangle_{-r+k}$$
The first term is $Awedge B$, and the last term is $A,lrcorner,B$. We could also write this as $$langle ABrangle_{r+k}+cdots+langle ABrangle_{|r-k|}=Awedge B+cdots+Acdot B$$ because the last term is the same if $rleq k$, and otherwise we're only adding $0$'s after $Acdot B$. (For example, with $r=3$ and $k=1$, the expression is $langle Abrangle_4+langle Abrangle_2+langle Abrangle_0+langle Abrangle_{-2}=langle Abrangle_4+langle Abrangle_2$.)
Multiply by $C$, expanding each term in the same way, to get
$$ABC=langlelangle ABrangle_{r+k}Crangle_{r+k+s}+langlelangle ABrangle_{r+k}Crangle_{r+k+s-2}+cdots+langlelangle ABrangle_{r+k}Crangle_{-(r+k)+s}$$
$$+langlelangle ABrangle_{r+k-2}Crangle_{r+k-2+s}+langlelangle ABrangle_{r+k-2}Crangle_{r+k-2+s-2}+cdots+langlelangle ABrangle_{r+k-2}Crangle_{-(r+k-2)+s}$$
$$+cdots$$
$$+langlelangle ABrangle_{-r+k}Crangle_{-r+k+s}+langlelangle ABrangle_{-r+k}Crangle_{-r+k+s-2}+cdots+langlelangle ABrangle_{-r+k}Crangle_{-(-r+k)+s}$$
Each row has successively fewer terms ($2$ less each time), so the rightmost term should be indented, and the expression above would look something like this trapezoid:
$$begin{matrix} x&x&x&x&x&x&x&x&x \ x&x&x&x&x&x&x \ x&x&x&x&x \ x&x&x end{matrix}$$
The grade decreases going right or going down, so all the terms with a given grade are on a diagonal line. The leftmost diagonal (the highest grade) has only a single term, $langle ABCrangle_{r+k+s}=langlelangle ABrangle_{r+k}Crangle_{r+k+s}=(Awedge B)wedge C$. The rightmost diagonal (the lowest grade) also has only a single term, $langle ABCrangle_{-(r+k)+s}=langlelangle ABrangle_{r+k}Crangle_{-(r+k)+s}=(Awedge B),lrcorner,C$.
Now write it the other way, expanding $BC$ first:
$$BC=langle BCrangle_{k+s}+langle BCrangle_{k+s-2}+cdots+langle BCrangle_{-k+s}$$
and multiplying by $A$:
$$ABC=langle Alangle BCrangle_{k+s}rangle_{r+k+s}+langle Alangle BCrangle_{k+s-2}rangle_{r+k+s-2}+cdots+langle Alangle BCrangle_{-k+s}rangle_{r-k+s}$$
$$+langle Alangle BCrangle_{k+s}rangle_{r-2+k+s}+langle Alangle BCrangle_{k+s-2}rangle_{r-2+k+s-2}+cdots+langle Alangle BCrangle_{-k+s}rangle_{r-2-k+s}$$
$$+cdots$$
$$+langle Alangle BCrangle_{k+s}rangle_{-r+k+s}+langle Alangle BCrangle_{k+s-2}rangle_{-r+k+s-2}+cdots+langle Alangle BCrangle_{-k+s}rangle_{-r-k+s}$$
This time, each row has the same number of terms, so it's just a rectangle instead of a trapezoid. Again, all the terms with a given grade are on a diagonal, and the leftmost diagonal has only $langle ABCrangle_{r+k+s}=langle Alangle BCrangle_{k+s}rangle_{r+k+s}=Awedge(Bwedge C)$. The rightmost diagonal also has only one term, $langle ABCrangle_{-r-k+s}=langle Alangle BCrangle_{-k+s}rangle_{-r-k+s}=A,lrcorner,(B,lrcorner,C)$.
Comparing the two expansions, we can see both identities at once:
$$(Awedge B)wedge C=Awedge(Bwedge C),qquad(Awedge B),lrcorner,C=A,lrcorner,(B,lrcorner, C).$$
answered Dec 17 '18 at 6:40
mr_e_manmr_e_man
1,1101424
$endgroup$
add a comment |
$begingroup$
This is a special case of the identity $(Awedge B),lrcorner,C=A,lrcorner,(B,lrcorner,C)$. It can be proven first for blades, then extended to all multivectors by linearity.
Suppose $A$ has grade $r$, $B$ has grade $k$ (in your case $k=1$), and $C$ has grade $s$. Then
$$AB=langle ABrangle_{r+k}$$
$$+langle ABrangle_{r+k-2}$$
$$+langle ABrangle_{r+k-4}$$
$$+cdots$$
$$+langle ABrangle_{-r+k+2}$$
$$+langle ABrangle_{-r+k}$$
The first term is $Awedge B$, and the last term is $A,lrcorner,B$. We could also write this as $$langle ABrangle_{r+k}+cdots+langle ABrangle_{|r-k|}=Awedge B+cdots+Acdot B$$ because the last term is the same if $rleq k$, and otherwise we're only adding $0$'s after $Acdot B$. (For example, with $r=3$ and $k=1$, the expression is $langle Abrangle_4+langle Abrangle_2+langle Abrangle_0+langle Abrangle_{-2}=langle Abrangle_4+langle Abrangle_2$.)
Multiply by $C$, expanding each term in the same way, to get
$$ABC=langlelangle ABrangle_{r+k}Crangle_{r+k+s}+langlelangle ABrangle_{r+k}Crangle_{r+k+s-2}+cdots+langlelangle ABrangle_{r+k}Crangle_{-(r+k)+s}$$
$$+langlelangle ABrangle_{r+k-2}Crangle_{r+k-2+s}+langlelangle ABrangle_{r+k-2}Crangle_{r+k-2+s-2}+cdots+langlelangle ABrangle_{r+k-2}Crangle_{-(r+k-2)+s}$$
$$+cdots$$
$$+langlelangle ABrangle_{-r+k}Crangle_{-r+k+s}+langlelangle ABrangle_{-r+k}Crangle_{-r+k+s-2}+cdots+langlelangle ABrangle_{-r+k}Crangle_{-(-r+k)+s}$$
Each row has successively fewer terms ($2$ less each time), so the rightmost term should be indented, and the expression above would look something like this trapezoid:
$$begin{matrix} x&x&x&x&x&x&x&x&x \ x&x&x&x&x&x&x \ x&x&x&x&x \ x&x&x end{matrix}$$
The grade decreases going right or going down, so all the terms with a given grade are on a diagonal line. The leftmost diagonal (the highest grade) has only a single term, $langle ABCrangle_{r+k+s}=langlelangle ABrangle_{r+k}Crangle_{r+k+s}=(Awedge B)wedge C$. The rightmost diagonal (the lowest grade) also has only a single term, $langle ABCrangle_{-(r+k)+s}=langlelangle ABrangle_{r+k}Crangle_{-(r+k)+s}=(Awedge B),lrcorner,C$.
Now write it the other way, expanding $BC$ first:
$$BC=langle BCrangle_{k+s}+langle BCrangle_{k+s-2}+cdots+langle BCrangle_{-k+s}$$
and multiplying by $A$:
$$ABC=langle Alangle BCrangle_{k+s}rangle_{r+k+s}+langle Alangle BCrangle_{k+s-2}rangle_{r+k+s-2}+cdots+langle Alangle BCrangle_{-k+s}rangle_{r-k+s}$$
$$+langle Alangle BCrangle_{k+s}rangle_{r-2+k+s}+langle Alangle BCrangle_{k+s-2}rangle_{r-2+k+s-2}+cdots+langle Alangle BCrangle_{-k+s}rangle_{r-2-k+s}$$
$$+cdots$$
$$+langle Alangle BCrangle_{k+s}rangle_{-r+k+s}+langle Alangle BCrangle_{k+s-2}rangle_{-r+k+s-2}+cdots+langle Alangle BCrangle_{-k+s}rangle_{-r-k+s}$$
This time, each row has the same number of terms, so it's just a rectangle instead of a trapezoid. Again, all the terms with a given grade are on a diagonal, and the leftmost diagonal has only $langle ABCrangle_{r+k+s}=langle Alangle BCrangle_{k+s}rangle_{r+k+s}=Awedge(Bwedge C)$. The rightmost diagonal also has only one term, $langle ABCrangle_{-r-k+s}=langle Alangle BCrangle_{-k+s}rangle_{-r-k+s}=A,lrcorner,(B,lrcorner,C)$.
Comparing the two expansions, we can see both identities at once:
$$(Awedge B)wedge C=Awedge(Bwedge C),qquad(Awedge B),lrcorner,C=A,lrcorner,(B,lrcorner, C).$$
answered Dec 17 '18 at 6:40
mr_e_manmr_e_man
1,1101424
$endgroup$
add a comment |
$begingroup$
This is a special case of the identity $(Awedge B),lrcorner,C=A,lrcorner,(B,lrcorner,C)$. It can be proven first for blades, then extended to all multivectors by linearity.
Suppose $A$ has grade $r$, $B$ has grade $k$ (in your case $k=1$), and $C$ has grade $s$. Then
$$AB=langle ABrangle_{r+k}$$
$$+langle ABrangle_{r+k-2}$$
$$+langle ABrangle_{r+k-4}$$
$$+cdots$$
$$+langle ABrangle_{-r+k+2}$$
$$+langle ABrangle_{-r+k}$$
The first term is $Awedge B$, and the last term is $A,lrcorner,B$. We could also write this as $$langle ABrangle_{r+k}+cdots+langle ABrangle_{|r-k|}=Awedge B+cdots+Acdot B$$ because the last term is the same if $rleq k$, and otherwise we're only adding $0$'s after $Acdot B$. (For example, with $r=3$ and $k=1$, the expression is $langle Abrangle_4+langle Abrangle_2+langle Abrangle_0+langle Abrangle_{-2}=langle Abrangle_4+langle Abrangle_2$.)
Multiply by $C$, expanding each term in the same way, to get
$$ABC=langlelangle ABrangle_{r+k}Crangle_{r+k+s}+langlelangle ABrangle_{r+k}Crangle_{r+k+s-2}+cdots+langlelangle ABrangle_{r+k}Crangle_{-(r+k)+s}$$
$$+langlelangle ABrangle_{r+k-2}Crangle_{r+k-2+s}+langlelangle ABrangle_{r+k-2}Crangle_{r+k-2+s-2}+cdots+langlelangle ABrangle_{r+k-2}Crangle_{-(r+k-2)+s}$$
$$+cdots$$
$$+langlelangle ABrangle_{-r+k}Crangle_{-r+k+s}+langlelangle ABrangle_{-r+k}Crangle_{-r+k+s-2}+cdots+langlelangle ABrangle_{-r+k}Crangle_{-(-r+k)+s}$$
Each row has successively fewer terms ($2$ less each time), so the rightmost term should be indented, and the expression above would look something like this trapezoid:
$$begin{matrix} x&x&x&x&x&x&x&x&x \ x&x&x&x&x&x&x \ x&x&x&x&x \ x&x&x end{matrix}$$
The grade decreases going right or going down, so all the terms with a given grade are on a diagonal line. The leftmost diagonal (the highest grade) has only a single term, $langle ABCrangle_{r+k+s}=langlelangle ABrangle_{r+k}Crangle_{r+k+s}=(Awedge B)wedge C$. The rightmost diagonal (the lowest grade) also has only a single term, $langle ABCrangle_{-(r+k)+s}=langlelangle ABrangle_{r+k}Crangle_{-(r+k)+s}=(Awedge B),lrcorner,C$.
Now write it the other way, expanding $BC$ first:
$$BC=langle BCrangle_{k+s}+langle BCrangle_{k+s-2}+cdots+langle BCrangle_{-k+s}$$
and multiplying by $A$:
$$ABC=langle Alangle BCrangle_{k+s}rangle_{r+k+s}+langle Alangle BCrangle_{k+s-2}rangle_{r+k+s-2}+cdots+langle Alangle BCrangle_{-k+s}rangle_{r-k+s}$$
$$+langle Alangle BCrangle_{k+s}rangle_{r-2+k+s}+langle Alangle BCrangle_{k+s-2}rangle_{r-2+k+s-2}+cdots+langle Alangle BCrangle_{-k+s}rangle_{r-2-k+s}$$
$$+cdots$$
$$+langle Alangle BCrangle_{k+s}rangle_{-r+k+s}+langle Alangle BCrangle_{k+s-2}rangle_{-r+k+s-2}+cdots+langle Alangle BCrangle_{-k+s}rangle_{-r-k+s}$$
This time, each row has the same number of terms, so it's just a rectangle instead of a trapezoid. Again, all the terms with a given grade are on a diagonal, and the leftmost diagonal has only $langle ABCrangle_{r+k+s}=langle Alangle BCrangle_{k+s}rangle_{r+k+s}=Awedge(Bwedge C)$. The rightmost diagonal also has only one term, $langle ABCrangle_{-r-k+s}=langle Alangle BCrangle_{-k+s}rangle_{-r-k+s}=A,lrcorner,(B,lrcorner,C)$.
Comparing the two expansions, we can see both identities at once:
$$(Awedge B)wedge C=Awedge(Bwedge C),qquad(Awedge B),lrcorner,C=A,lrcorner,(B,lrcorner, C).$$
answered Dec 17 '18 at 6:40
mr_e_manmr_e_man
1,1101424
$endgroup$
This is a special case of the identity $(Awedge B),lrcorner,C=A,lrcorner,(B,lrcorner,C)$. It can be proven first for blades, then extended to all multivectors by linearity.
Suppose $A$ has grade $r$, $B$ has grade $k$ (in your case $k=1$), and $C$ has grade $s$. Then
$$AB=langle ABrangle_{r+k}$$
$$+langle ABrangle_{r+k-2}$$
$$+langle ABrangle_{r+k-4}$$
$$+cdots$$
$$+langle ABrangle_{-r+k+2}$$
$$+langle ABrangle_{-r+k}$$
The first term is $Awedge B$, and the last term is $A,lrcorner,B$. We could also write this as $$langle ABrangle_{r+k}+cdots+langle ABrangle_{|r-k|}=Awedge B+cdots+Acdot B$$ because the last term is the same if $rleq k$, and otherwise we're only adding $0$'s after $Acdot B$. (For example, with $r=3$ and $k=1$, the expression is $langle Abrangle_4+langle Abrangle_2+langle Abrangle_0+langle Abrangle_{-2}=langle Abrangle_4+langle Abrangle_2$.)
Multiply by $C$, expanding each term in the same way, to get
$$ABC=langlelangle ABrangle_{r+k}Crangle_{r+k+s}+langlelangle ABrangle_{r+k}Crangle_{r+k+s-2}+cdots+langlelangle ABrangle_{r+k}Crangle_{-(r+k)+s}$$
$$+langlelangle ABrangle_{r+k-2}Crangle_{r+k-2+s}+langlelangle ABrangle_{r+k-2}Crangle_{r+k-2+s-2}+cdots+langlelangle ABrangle_{r+k-2}Crangle_{-(r+k-2)+s}$$
$$+cdots$$
$$+langlelangle ABrangle_{-r+k}Crangle_{-r+k+s}+langlelangle ABrangle_{-r+k}Crangle_{-r+k+s-2}+cdots+langlelangle ABrangle_{-r+k}Crangle_{-(-r+k)+s}$$
Each row has successively fewer terms ($2$ less each time), so the rightmost term should be indented, and the expression above would look something like this trapezoid:
$$begin{matrix} x&x&x&x&x&x&x&x&x \ x&x&x&x&x&x&x \ x&x&x&x&x \ x&x&x end{matrix}$$
The grade decreases going right or going down, so all the terms with a given grade are on a diagonal line. The leftmost diagonal (the highest grade) has only a single term, $langle ABCrangle_{r+k+s}=langlelangle ABrangle_{r+k}Crangle_{r+k+s}=(Awedge B)wedge C$. The rightmost diagonal (the lowest grade) also has only a single term, $langle ABCrangle_{-(r+k)+s}=langlelangle ABrangle_{r+k}Crangle_{-(r+k)+s}=(Awedge B),lrcorner,C$.
Now write it the other way, expanding $BC$ first:
$$BC=langle BCrangle_{k+s}+langle BCrangle_{k+s-2}+cdots+langle BCrangle_{-k+s}$$
and multiplying by $A$:
$$ABC=langle Alangle BCrangle_{k+s}rangle_{r+k+s}+langle Alangle BCrangle_{k+s-2}rangle_{r+k+s-2}+cdots+langle Alangle BCrangle_{-k+s}rangle_{r-k+s}$$
$$+langle Alangle BCrangle_{k+s}rangle_{r-2+k+s}+langle Alangle BCrangle_{k+s-2}rangle_{r-2+k+s-2}+cdots+langle Alangle BCrangle_{-k+s}rangle_{r-2-k+s}$$
$$+cdots$$
$$+langle Alangle BCrangle_{k+s}rangle_{-r+k+s}+langle Alangle BCrangle_{k+s-2}rangle_{-r+k+s-2}+cdots+langle Alangle BCrangle_{-k+s}rangle_{-r-k+s}$$
This time, each row has the same number of terms, so it's just a rectangle instead of a trapezoid. Again, all the terms with a given grade are on a diagonal, and the leftmost diagonal has only $langle ABCrangle_{r+k+s}=langle Alangle BCrangle_{k+s}rangle_{r+k+s}=Awedge(Bwedge C)$. The rightmost diagonal also has only one term, $langle ABCrangle_{-r-k+s}=langle Alangle BCrangle_{-k+s}rangle_{-r-k+s}=A,lrcorner,(B,lrcorner,C)$.
Comparing the two expansions, we can see both identities at once:
$$(Awedge B)wedge C=Awedge(Bwedge C),qquad(Awedge B),lrcorner,C=A,lrcorner,(B,lrcorner, C).$$
answered Dec 17 '18 at 6:40
mr_e_manmr_e_man
1,1101424
answered Dec 17 '18 at 6:40
mr_e_manmr_e_man
1,1101424
answered Dec 17 '18 at 6:40
mr_e_manmr_e_man
1,1101424
answered Dec 17 '18 at 6:40
mr_e_manmr_e_man
1,1101424
1,1101424
add a comment |
add a comment |
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$begingroup$
@mr_e_man What is the symbol that looks like an upside down r?
$endgroup$
– alexanderyaacov
Dec 11 '18 at 10:08
$begingroup$
That $lrcorner$ "lrcorner" is the contraction product, an asymmetric dot product. en.wikipedia.org/wiki/… , av8n.com/physics/clifford-intro.htm#sec-contractions
$endgroup$
– mr_e_man
Dec 15 '18 at 0:19
$begingroup$
I remember this identity by comparing it with the analogous one in logic: $(Aland B)to Cequiv Ato(Bto C)$, where $A,B,C$ are statements which are either true or false. In words, "$A$ and $B$ together imply $C$" is equivalent to "If $A$ is true, then $B$ implies $C$".
$endgroup$
– mr_e_man
Dec 15 '18 at 0:25
$begingroup$
@mr_e_man why is it then a special case? It seems to me to be exactly the same.
$endgroup$
– alexanderyaacov
Dec 16 '18 at 7:24
$begingroup$
It's a special case because the grades are restricted. The identity I gave is always true; $B$ doesn't need to be a vector, nor $A$ lower-grade than $C$.
$endgroup$
– mr_e_man
Dec 16 '18 at 23:43