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How to compute if a multivector inverse exists in Clifford Algebra
The Next CEO of Stack OverflowCalculating the inverse of a multivectorInverse of a general nonfactorizable multivectorHow do I evaluate the Clifford product in dimensions greater than 3?Clifford Algebra Multiplication IntuitionInverse of a general nonfactorizable multivectorFinding an element in a Clifford algebra that satisfy some specific commutation and anti-commutation relationGeneralizing the dot product to multivectorsWhat happens to Clifford algebra structure and periodicity when the field is weird?Is the reverse really an anti-automorphism of a a Clifford algebra?Transformation of metric tensor under clifford algebra?How to perform wedge productOn the algebraic formulation of the Clifford algebra
$begingroup$
Suppose we have a 4 dimension positive signature clifford algebra. In Calculating the inverse of a multivector and Inverse of a general nonfactorizable multivector, the inverse of a multivector is presented as a solution when vectors/bivectors are present
$B^-1 = fracB^daggerB B^dagger$
but the above is not true for any multivector. For example, how to know if
$(1+e_1234)^-1$
exists and how to compute it?
clifford-algebras geometric-algebras
asked Mar 19 at 13:10
paco gilpaco gil
82
$endgroup$
add a comment |
$begingroup$
Suppose we have a 4 dimension positive signature clifford algebra. In Calculating the inverse of a multivector and Inverse of a general nonfactorizable multivector, the inverse of a multivector is presented as a solution when vectors/bivectors are present
$B^-1 = fracB^daggerB B^dagger$
but the above is not true for any multivector. For example, how to know if
$(1+e_1234)^-1$
exists and how to compute it?
clifford-algebras geometric-algebras
asked Mar 19 at 13:10
paco gilpaco gil
82
$endgroup$
add a comment |
$begingroup$
Suppose we have a 4 dimension positive signature clifford algebra. In Calculating the inverse of a multivector and Inverse of a general nonfactorizable multivector, the inverse of a multivector is presented as a solution when vectors/bivectors are present
$B^-1 = fracB^daggerB B^dagger$
but the above is not true for any multivector. For example, how to know if
$(1+e_1234)^-1$
exists and how to compute it?
clifford-algebras geometric-algebras
asked Mar 19 at 13:10
paco gilpaco gil
82
$endgroup$
Suppose we have a 4 dimension positive signature clifford algebra. In Calculating the inverse of a multivector and Inverse of a general nonfactorizable multivector, the inverse of a multivector is presented as a solution when vectors/bivectors are present
$B^-1 = fracB^daggerB B^dagger$
but the above is not true for any multivector. For example, how to know if
$(1+e_1234)^-1$
exists and how to compute it?
clifford-algebras geometric-algebras
clifford-algebras geometric-algebras
asked Mar 19 at 13:10
paco gilpaco gil
82
asked Mar 19 at 13:10
paco gilpaco gil
82
asked Mar 19 at 13:10
paco gilpaco gil
82
asked Mar 19 at 13:10
paco gilpaco gil
82
asked Mar 19 at 13:10
paco gilpaco gil
82
82
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Naively speaking, the existence of inverses will depend on the signature $(p,q)$ of the quadratic space $mathbbR^p,q=(mathbbR^p+q,g)$, in which for an orthonormal basis $e_i_i=1^n=p+q$ and $v=sum v^ie_i$ we have
$$g(v,v)=(v^1)^2+(v^2)^2+cdots+(v^p)^2-(v^p+1)^2-cdots-(v^p+q)^2.$$
For your example, notice that if $(e_1234)^2=-1$, then
$$(1+e_1234)frac12(1-e_1234)=1,$$
which means that $(1+e_1234)^-1=frac12(1-e_1234)$.
Now, if $(e_1234)^2=1$, then there is no inverse for $(1+e_1234)$, which is due to the fact that $xoverlinex=0$. More specifically, one can derive conditions for which there are inverses for the cases where $p+q=nleq 5$. The discovery of new faster methods for higher dimensions, which do not depend on the signature is a problem still under development as of today. As an example, we could cite this.
answered Mar 20 at 16:33
Aquerman KuczmendaAquerman Kuczmenda
1065
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Naively speaking, the existence of inverses will depend on the signature $(p,q)$ of the quadratic space $mathbbR^p,q=(mathbbR^p+q,g)$, in which for an orthonormal basis $e_i_i=1^n=p+q$ and $v=sum v^ie_i$ we have
$$g(v,v)=(v^1)^2+(v^2)^2+cdots+(v^p)^2-(v^p+1)^2-cdots-(v^p+q)^2.$$
For your example, notice that if $(e_1234)^2=-1$, then
$$(1+e_1234)frac12(1-e_1234)=1,$$
which means that $(1+e_1234)^-1=frac12(1-e_1234)$.
Now, if $(e_1234)^2=1$, then there is no inverse for $(1+e_1234)$, which is due to the fact that $xoverlinex=0$. More specifically, one can derive conditions for which there are inverses for the cases where $p+q=nleq 5$. The discovery of new faster methods for higher dimensions, which do not depend on the signature is a problem still under development as of today. As an example, we could cite this.
answered Mar 20 at 16:33
Aquerman KuczmendaAquerman Kuczmenda
1065
$endgroup$
add a comment |
$begingroup$
Naively speaking, the existence of inverses will depend on the signature $(p,q)$ of the quadratic space $mathbbR^p,q=(mathbbR^p+q,g)$, in which for an orthonormal basis $e_i_i=1^n=p+q$ and $v=sum v^ie_i$ we have
$$g(v,v)=(v^1)^2+(v^2)^2+cdots+(v^p)^2-(v^p+1)^2-cdots-(v^p+q)^2.$$
For your example, notice that if $(e_1234)^2=-1$, then
$$(1+e_1234)frac12(1-e_1234)=1,$$
which means that $(1+e_1234)^-1=frac12(1-e_1234)$.
Now, if $(e_1234)^2=1$, then there is no inverse for $(1+e_1234)$, which is due to the fact that $xoverlinex=0$. More specifically, one can derive conditions for which there are inverses for the cases where $p+q=nleq 5$. The discovery of new faster methods for higher dimensions, which do not depend on the signature is a problem still under development as of today. As an example, we could cite this.
answered Mar 20 at 16:33
Aquerman KuczmendaAquerman Kuczmenda
1065
$endgroup$
add a comment |
$begingroup$
Naively speaking, the existence of inverses will depend on the signature $(p,q)$ of the quadratic space $mathbbR^p,q=(mathbbR^p+q,g)$, in which for an orthonormal basis $e_i_i=1^n=p+q$ and $v=sum v^ie_i$ we have
$$g(v,v)=(v^1)^2+(v^2)^2+cdots+(v^p)^2-(v^p+1)^2-cdots-(v^p+q)^2.$$
For your example, notice that if $(e_1234)^2=-1$, then
$$(1+e_1234)frac12(1-e_1234)=1,$$
which means that $(1+e_1234)^-1=frac12(1-e_1234)$.
Now, if $(e_1234)^2=1$, then there is no inverse for $(1+e_1234)$, which is due to the fact that $xoverlinex=0$. More specifically, one can derive conditions for which there are inverses for the cases where $p+q=nleq 5$. The discovery of new faster methods for higher dimensions, which do not depend on the signature is a problem still under development as of today. As an example, we could cite this.
answered Mar 20 at 16:33
Aquerman KuczmendaAquerman Kuczmenda
1065
$endgroup$
Naively speaking, the existence of inverses will depend on the signature $(p,q)$ of the quadratic space $mathbbR^p,q=(mathbbR^p+q,g)$, in which for an orthonormal basis $e_i_i=1^n=p+q$ and $v=sum v^ie_i$ we have
$$g(v,v)=(v^1)^2+(v^2)^2+cdots+(v^p)^2-(v^p+1)^2-cdots-(v^p+q)^2.$$
For your example, notice that if $(e_1234)^2=-1$, then
$$(1+e_1234)frac12(1-e_1234)=1,$$
which means that $(1+e_1234)^-1=frac12(1-e_1234)$.
Now, if $(e_1234)^2=1$, then there is no inverse for $(1+e_1234)$, which is due to the fact that $xoverlinex=0$. More specifically, one can derive conditions for which there are inverses for the cases where $p+q=nleq 5$. The discovery of new faster methods for higher dimensions, which do not depend on the signature is a problem still under development as of today. As an example, we could cite this.
answered Mar 20 at 16:33
Aquerman KuczmendaAquerman Kuczmenda
1065
edited Mar 20 at 17:39
answered Mar 20 at 16:33
Aquerman KuczmendaAquerman Kuczmenda
1065
answered Mar 20 at 16:33
Aquerman KuczmendaAquerman Kuczmenda
1065
answered Mar 20 at 16:33
Aquerman KuczmendaAquerman Kuczmenda
1065
1065
add a comment |
add a comment |
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