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Why is clifford group a group?

Center of Clifford Algebra depending on the parity of $dim V$?Why is the radical of a Clifford algebra generated by the kernel of the associated symmetric form?Injection of vector space in clifford algebraLie group structures inside of Clifford algebrasWhy are (S)pinor representations *restrictions* of Clifford algebra representations?Question about Clifford algebras (reference Matrix Groups: An Introduction to Lie Group Theory, Baker)Representation of Clifford AlgebrasFinding an element in a Clifford algebra that satisfy some specific commutation and anti-commutation relationClifford algebra vs 'universal' clifford algebra?A confusing formula in Clifford algebra

2

0

$begingroup$

Let $C(Q)$ denote the clifford algebra of vector space $Q$ with respect to a quadratic form $q:V rightarrow Bbb R$. Hence we have the relation $w^2 = Q(w) cdot 1$ for $w in V$.

Let $alpha:C(Q) rightarrow C(Q)$ be the canonical automoprhism $alpha^2=id, alpha=-x$.

The Clifford group of $Q$ is
$$Gamma (Q) = x in C(Q)^* , ; $$

How is this set closed under inverses?

abstract-algebra clifford-algebras

share|cite|improve this question

asked Feb 12 at 5:28

CL.CL.

2,2982925

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can rewrite the condition as $alpha(x)Vx^-1=V$.
  $endgroup$
  – Lord Shark the Unknown
  Feb 12 at 5:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok it doesn't seem clear to me how the equality holds: the way is define $f_x :V rightarrow V$, then as $alpha(x) cdot v cdot x^-1$. This map is injective, $V$ finite dimensinoal so bijective. Is there an easier way?
  $endgroup$
  – CL.
  Feb 12 at 5:51
  

  
  
  
  

add a comment | 

2

0

$begingroup$

Let $C(Q)$ denote the clifford algebra of vector space $Q$ with respect to a quadratic form $q:V rightarrow Bbb R$. Hence we have the relation $w^2 = Q(w) cdot 1$ for $w in V$.

Let $alpha:C(Q) rightarrow C(Q)$ be the canonical automoprhism $alpha^2=id, alpha=-x$.

The Clifford group of $Q$ is
$$Gamma (Q) = x in C(Q)^* , ; $$

How is this set closed under inverses?

abstract-algebra clifford-algebras

share|cite|improve this question

asked Feb 12 at 5:28

CL.CL.

2,2982925

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can rewrite the condition as $alpha(x)Vx^-1=V$.
  $endgroup$
  – Lord Shark the Unknown
  Feb 12 at 5:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok it doesn't seem clear to me how the equality holds: the way is define $f_x :V rightarrow V$, then as $alpha(x) cdot v cdot x^-1$. This map is injective, $V$ finite dimensinoal so bijective. Is there an easier way?
  $endgroup$
  – CL.
  Feb 12 at 5:51
  

  
  
  
  

add a comment | 

$begingroup$

Let $C(Q)$ denote the clifford algebra of vector space $Q$ with respect to a quadratic form $q:V rightarrow Bbb R$. Hence we have the relation $w^2 = Q(w) cdot 1$ for $w in V$.

Let $alpha:C(Q) rightarrow C(Q)$ be the canonical automoprhism $alpha^2=id, alpha=-x$.

The Clifford group of $Q$ is
$$Gamma (Q) = x in C(Q)^* , ; $$

How is this set closed under inverses?

abstract-algebra clifford-algebras

share|cite|improve this question

asked Feb 12 at 5:28

CL.CL.

2,2982925

$endgroup$

share|cite|improve this question

asked Feb 12 at 5:28

CL.CL.

2,2982925

share|cite|improve this question

asked Feb 12 at 5:28

CL.CL.

2,2982925

asked Feb 12 at 5:28

CL.CL.

2,2982925

asked Feb 12 at 5:28

CL.CL.

2,2982925

1 Answer
1

active

oldest

votes

0

$begingroup$

I will assume, as it is usually defined, that $C(Q)^*$ is the set of invertible elements in $C(Q)$. Then, the existence of an element $x^-1in C(Q)$ such that $xx^-1=x^-1x=1$ is, by definition, guaranteed for each $xin C(Q)^*$. What we need to show is that $x^-1$ is in fact an element in $Gamma(Q)$.

First, for each $xin Gamma(Q)$, the function $sigma(x):Vrightarrow V$ given by $sigma(x)(v)=alpha(x)vx^-1$ is a vector space isomorphism. It is clear that it is a linear transformation, since the Clifford product is bilinear. Moreover,
$$ sigma(x)(v)=0 ;;Rightarrow;; alpha(x)vx^-1=0;;Rightarrow;; v=0,$$
where we simply multiplied (using the Clifford product) the right side by $x^-1$ and the left side by $alpha(x)^-1$.

Finally, since $sigma(x)$ is an isomorphism, for each $vin V$ we have a $win V$ such that $alpha(x)wx^-1=v$. Here, we remember that since $alpha$ is an automorphism, we have $alpha(x)^-1=alpha(x^-1)$. It then follows that
$$
v=alpha(x)wx^-1 ;;Rightarrow;; vx=alpha(x)w ;; Rightarrow;; alpha(x^-1)vx=win V,
$$
that is, $x^-1in Gamma(Q)$.

share|cite|improve this answer

edited Mar 13 at 0:59

answered Mar 12 at 20:44

Aquerman KuczmendaAquerman Kuczmenda

665

$endgroup$

add a comment | 

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0

$begingroup$

I will assume, as it is usually defined, that $C(Q)^*$ is the set of invertible elements in $C(Q)$. Then, the existence of an element $x^-1in C(Q)$ such that $xx^-1=x^-1x=1$ is, by definition, guaranteed for each $xin C(Q)^*$. What we need to show is that $x^-1$ is in fact an element in $Gamma(Q)$.

First, for each $xin Gamma(Q)$, the function $sigma(x):Vrightarrow V$ given by $sigma(x)(v)=alpha(x)vx^-1$ is a vector space isomorphism. It is clear that it is a linear transformation, since the Clifford product is bilinear. Moreover,
$$ sigma(x)(v)=0 ;;Rightarrow;; alpha(x)vx^-1=0;;Rightarrow;; v=0,$$
where we simply multiplied (using the Clifford product) the right side by $x^-1$ and the left side by $alpha(x)^-1$.

Finally, since $sigma(x)$ is an isomorphism, for each $vin V$ we have a $win V$ such that $alpha(x)wx^-1=v$. Here, we remember that since $alpha$ is an automorphism, we have $alpha(x)^-1=alpha(x^-1)$. It then follows that
$$
v=alpha(x)wx^-1 ;;Rightarrow;; vx=alpha(x)w ;; Rightarrow;; alpha(x^-1)vx=win V,
$$
that is, $x^-1in Gamma(Q)$.

share|cite|improve this answer

edited Mar 13 at 0:59

answered Mar 12 at 20:44

Aquerman KuczmendaAquerman Kuczmenda

665

$endgroup$

add a comment | 

0

$begingroup$

I will assume, as it is usually defined, that $C(Q)^*$ is the set of invertible elements in $C(Q)$. Then, the existence of an element $x^-1in C(Q)$ such that $xx^-1=x^-1x=1$ is, by definition, guaranteed for each $xin C(Q)^*$. What we need to show is that $x^-1$ is in fact an element in $Gamma(Q)$.

First, for each $xin Gamma(Q)$, the function $sigma(x):Vrightarrow V$ given by $sigma(x)(v)=alpha(x)vx^-1$ is a vector space isomorphism. It is clear that it is a linear transformation, since the Clifford product is bilinear. Moreover,
$$ sigma(x)(v)=0 ;;Rightarrow;; alpha(x)vx^-1=0;;Rightarrow;; v=0,$$
where we simply multiplied (using the Clifford product) the right side by $x^-1$ and the left side by $alpha(x)^-1$.

Finally, since $sigma(x)$ is an isomorphism, for each $vin V$ we have a $win V$ such that $alpha(x)wx^-1=v$. Here, we remember that since $alpha$ is an automorphism, we have $alpha(x)^-1=alpha(x^-1)$. It then follows that
$$
v=alpha(x)wx^-1 ;;Rightarrow;; vx=alpha(x)w ;; Rightarrow;; alpha(x^-1)vx=win V,
$$
that is, $x^-1in Gamma(Q)$.

share|cite|improve this answer

edited Mar 13 at 0:59

answered Mar 12 at 20:44

Aquerman KuczmendaAquerman Kuczmenda

665

$endgroup$

add a comment | 

$begingroup$

I will assume, as it is usually defined, that $C(Q)^*$ is the set of invertible elements in $C(Q)$. Then, the existence of an element $x^-1in C(Q)$ such that $xx^-1=x^-1x=1$ is, by definition, guaranteed for each $xin C(Q)^*$. What we need to show is that $x^-1$ is in fact an element in $Gamma(Q)$.

First, for each $xin Gamma(Q)$, the function $sigma(x):Vrightarrow V$ given by $sigma(x)(v)=alpha(x)vx^-1$ is a vector space isomorphism. It is clear that it is a linear transformation, since the Clifford product is bilinear. Moreover,
$$ sigma(x)(v)=0 ;;Rightarrow;; alpha(x)vx^-1=0;;Rightarrow;; v=0,$$
where we simply multiplied (using the Clifford product) the right side by $x^-1$ and the left side by $alpha(x)^-1$.

Finally, since $sigma(x)$ is an isomorphism, for each $vin V$ we have a $win V$ such that $alpha(x)wx^-1=v$. Here, we remember that since $alpha$ is an automorphism, we have $alpha(x)^-1=alpha(x^-1)$. It then follows that
$$
v=alpha(x)wx^-1 ;;Rightarrow;; vx=alpha(x)w ;; Rightarrow;; alpha(x^-1)vx=win V,
$$
that is, $x^-1in Gamma(Q)$.

share|cite|improve this answer

edited Mar 13 at 0:59

answered Mar 12 at 20:44

Aquerman KuczmendaAquerman Kuczmenda

665

$endgroup$

share|cite|improve this answer

edited Mar 13 at 0:59

answered Mar 12 at 20:44

Aquerman KuczmendaAquerman Kuczmenda

665

answered Mar 12 at 20:44

Aquerman KuczmendaAquerman Kuczmenda

665

answered Mar 12 at 20:44

Aquerman KuczmendaAquerman Kuczmenda

665