Mixing
Equivalent definitions of Clifford algebra, verification
$begingroup$
Let $(V,B)$ be a finite dimensional $k$ vector space $V$ with an associated quadratic form $Q$. $char , k not= 2$.
Let $X:= {e_i }_{i=1}^n$ be a set of basis for $V$. Construct $klangle X rangle$, the free associative unital algebra generated by set $X$. Consider the ideal $I:= langle e_i e_j +e_j e_i + 2B(e_i, e_j) rangle$,
$klangle X rangle / I$ satisfies the universal property of the Clifford algebra.
Or equivalently, as the tensor algebra $T(V)$ is free associative algebra of $V$,
$T(V)/I'$ satisfies the universal property of the Clifford algebra.
$I'$ is defined similarly as $I$.
What I know:
That $Cl(V,Q)$ can be constructed by $T(V)/I''$ where $I'' = langle v otimes v - Q(v), : , v in V rangle.$
I believe I could supply proof for both claims if needed. In sketch: Proof for 1. is to use pg 8 of Thomas Friedrich's Dirac Operators; Proof for 2 is to show that $I=I''$.
I just want to know if these claims are true.
abstract-algebra representation-theory mathematical-physics algebras clifford-algebras
asked Jan 19 at 15:51
CL.CL.
2,2572925
$endgroup$
add a comment |
$begingroup$
Let $(V,B)$ be a finite dimensional $k$ vector space $V$ with an associated quadratic form $Q$. $char , k not= 2$.
Let $X:= {e_i }_{i=1}^n$ be a set of basis for $V$. Construct $klangle X rangle$, the free associative unital algebra generated by set $X$. Consider the ideal $I:= langle e_i e_j +e_j e_i + 2B(e_i, e_j) rangle$,
$klangle X rangle / I$ satisfies the universal property of the Clifford algebra.
Or equivalently, as the tensor algebra $T(V)$ is free associative algebra of $V$,
$T(V)/I'$ satisfies the universal property of the Clifford algebra.
$I'$ is defined similarly as $I$.
What I know:
That $Cl(V,Q)$ can be constructed by $T(V)/I''$ where $I'' = langle v otimes v - Q(v), : , v in V rangle.$
I believe I could supply proof for both claims if needed. In sketch: Proof for 1. is to use pg 8 of Thomas Friedrich's Dirac Operators; Proof for 2 is to show that $I=I''$.
I just want to know if these claims are true.
abstract-algebra representation-theory mathematical-physics algebras clifford-algebras
asked Jan 19 at 15:51
CL.CL.
2,2572925
$endgroup$
$begingroup$
They are the same thing. Lift to free associative algebra level and you can identify the ideal generated.
$endgroup$
– user45765
Jan 19 at 16:51
add a comment |
$begingroup$
Let $(V,B)$ be a finite dimensional $k$ vector space $V$ with an associated quadratic form $Q$. $char , k not= 2$.
Let $X:= {e_i }_{i=1}^n$ be a set of basis for $V$. Construct $klangle X rangle$, the free associative unital algebra generated by set $X$. Consider the ideal $I:= langle e_i e_j +e_j e_i + 2B(e_i, e_j) rangle$,
$klangle X rangle / I$ satisfies the universal property of the Clifford algebra.
Or equivalently, as the tensor algebra $T(V)$ is free associative algebra of $V$,
$T(V)/I'$ satisfies the universal property of the Clifford algebra.
$I'$ is defined similarly as $I$.
What I know:
That $Cl(V,Q)$ can be constructed by $T(V)/I''$ where $I'' = langle v otimes v - Q(v), : , v in V rangle.$
I believe I could supply proof for both claims if needed. In sketch: Proof for 1. is to use pg 8 of Thomas Friedrich's Dirac Operators; Proof for 2 is to show that $I=I''$.
I just want to know if these claims are true.
abstract-algebra representation-theory mathematical-physics algebras clifford-algebras
asked Jan 19 at 15:51
CL.CL.
2,2572925
$endgroup$
Let $(V,B)$ be a finite dimensional $k$ vector space $V$ with an associated quadratic form $Q$. $char , k not= 2$.
Let $X:= {e_i }_{i=1}^n$ be a set of basis for $V$. Construct $klangle X rangle$, the free associative unital algebra generated by set $X$. Consider the ideal $I:= langle e_i e_j +e_j e_i + 2B(e_i, e_j) rangle$,
$klangle X rangle / I$ satisfies the universal property of the Clifford algebra.
Or equivalently, as the tensor algebra $T(V)$ is free associative algebra of $V$,
$T(V)/I'$ satisfies the universal property of the Clifford algebra.
$I'$ is defined similarly as $I$.
What I know:
That $Cl(V,Q)$ can be constructed by $T(V)/I''$ where $I'' = langle v otimes v - Q(v), : , v in V rangle.$
I believe I could supply proof for both claims if needed. In sketch: Proof for 1. is to use pg 8 of Thomas Friedrich's Dirac Operators; Proof for 2 is to show that $I=I''$.
I just want to know if these claims are true.
abstract-algebra representation-theory mathematical-physics algebras clifford-algebras
abstract-algebra representation-theory mathematical-physics algebras clifford-algebras
asked Jan 19 at 15:51
CL.CL.
2,2572925
asked Jan 19 at 15:51
CL.CL.
2,2572925
edited Jan 19 at 16:07
CL.
asked Jan 19 at 15:51
CL.CL.
2,2572925
asked Jan 19 at 15:51
CL.CL.
2,2572925
asked Jan 19 at 15:51
CL.CL.
2,2572925
2,2572925
$begingroup$
They are the same thing. Lift to free associative algebra level and you can identify the ideal generated.
$endgroup$
– user45765
Jan 19 at 16:51
add a comment |
$begingroup$
They are the same thing. Lift to free associative algebra level and you can identify the ideal generated.
$endgroup$
– user45765
Jan 19 at 16:51
$begingroup$
They are the same thing. Lift to free associative algebra level and you can identify the ideal generated.
$endgroup$
– user45765
Jan 19 at 16:51
$begingroup$
They are the same thing. Lift to free associative algebra level and you can identify the ideal generated.
$endgroup$
– user45765
Jan 19 at 16:51
add a comment |
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$begingroup$
They are the same thing. Lift to free associative algebra level and you can identify the ideal generated.
$endgroup$
– user45765
Jan 19 at 16:51