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Thread: Show element is invertible in a Clifford algebra

  1. #1
    Super Member redsoxfan325's Avatar
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    Show element is invertible in a Clifford algebra

    Here is the problem:

    Let $\displaystyle F$ be a field and $\displaystyle V$ a quadratic space with an anisotropic form $\displaystyle q$ not representing $\displaystyle 1$ (i.e. $\displaystyle q(v)\neq1$ for all $\displaystyle v\in V$). Show that if $\displaystyle a\in F$ and $\displaystyle v\in V$ are not both zero, then $\displaystyle a+v$ is an invertible element in the Clifford algebra $\displaystyle C(V,q)$.

    I'm not really sure how to start this problem. I don't fully understand Clifford algebras, so I don't even really know what invertible elements even look like. I understand the definition - $\displaystyle T(V)/I$, where I is the ideal generated by $\displaystyle v\otimes v-q(v)$ - but I don't understand the resulting algebra (except when $\displaystyle q$ is a binary or nonzero unary form).

    Any help is appreciated.
    Last edited by redsoxfan325; Nov 22nd 2011 at 06:49 PM. Reason: clarified "represent"
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  2. #2
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    Re: Show element is invertible in a Clifford algebra

    Quote Originally Posted by redsoxfan325 View Post
    Here is the problem:

    Let $\displaystyle F$ be a field and $\displaystyle V$ a quadratic space with an anisotropic form $\displaystyle q$ not representing $\displaystyle 1$. Show that if $\displaystyle a\in F$ and $\displaystyle v\in V$ are not both zero, then $\displaystyle a+v$ is an invertible element in the Clifford algebra $\displaystyle C(V,q)$.

    I'm not really sure how to start this problem. I don't fully understand Clifford algebras, so I don't even really know what invertible elements even look like. I understand the definition - $\displaystyle T(V)/I$, where I is the ideal generated by $\displaystyle v\otimes v-q(v)$ - but I don't understand the resulting algebra (except when $\displaystyle q$ is a binary or nonzero unary form).

    Any help is appreciated.
    what do u mean by "not representing 1"? do u mean $\displaystyle q(v) \neq 1$ for all $\displaystyle v \in V$?
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  3. #3
    Super Member redsoxfan325's Avatar
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    Re: Show element is invertible in a Clifford algebra

    Quote Originally Posted by NonCommAlg View Post
    what do u mean by "not representing 1"? do u mean $\displaystyle q(v) \neq 1$ for all $\displaystyle v \in V$?
    Yes.
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  4. #4
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    Re: Show element is invertible in a Clifford algebra

    Quote Originally Posted by redsoxfan325 View Post
    Yes.
    ok then, so $\displaystyle q(v) \neq a^2$ because otherwise $\displaystyle q(a^{-1}v)=1.$ let $\displaystyle b = a(a^2 - q(v))^{-1}$ and $\displaystyle c =-ba^{-1}$ and see that $\displaystyle (a+v)(b+cv)=1.$
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