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Math Help - 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 F be a field and V a quadratic space with an anisotropic form q not representing 1 (i.e. q(v)\neq1 for all v\in V). Show that if a\in F and v\in V are not both zero, then a+v is an invertible element in the Clifford algebra 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 - T(V)/I, where I is the ideal generated by v\otimes v-q(v) - but I don't understand the resulting algebra (except when q is a binary or nonzero unary form).

    Any help is appreciated.
    Last edited by redsoxfan325; November 22nd 2011 at 07:49 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 F be a field and V a quadratic space with an anisotropic form q not representing 1. Show that if a\in F and v\in V are not both zero, then a+v is an invertible element in the Clifford algebra 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 - T(V)/I, where I is the ideal generated by v\otimes v-q(v) - but I don't understand the resulting algebra (except when q is a binary or nonzero unary form).

    Any help is appreciated.
    what do u mean by "not representing 1"? do u mean q(v) \neq 1 for all 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 q(v) \neq 1 for all 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 q(v) \neq a^2 because otherwise q(a^{-1}v)=1. let b = a(a^2 - q(v))^{-1} and c =-ba^{-1} and see that (a+v)(b+cv)=1.
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