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We define a modified Clifford algebra, we take the free algebra over the vector space $E$ with the relations: $$ e\otimes f + f\otimes e + \alpha (e)f + \alpha (f)e = -2 g(e,f)$$ with $\alpha$ a linear forme over $E$. Could we define a modified Dirac operator $\cal D$ over a spin manifold $M$, such that: $${\cal D}^2 + \nabla_a {\cal D} = -\Delta + \frac {r}{4}$$, with $\nabla$, the Levi-Civita connection, $a= \alpha^*$ and $\Delta$, the Laplacian?

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  • $\begingroup$ Could you introduce some background about such setting. $\endgroup$
    – DLIN
    Aug 4, 2016 at 7:53
  • $\begingroup$ Are you sure this algebra is not zero? Can you say more about its structure (Brauer invariants)? $\endgroup$
    – user1688
    Aug 4, 2016 at 8:48
  • $\begingroup$ @Anton , FWIW, I checked now that for any $\alpha$ the defining relations of this algebra form a Gröbner basis, so it has the same size as the usual Clifford algebra. This seems to imply that in the case of a non-degenerate $g$, this algebra is isomorphic to the usual Clifford algebra (the one for $\alpha=0$), since if the degeneration of an algebra is semisimple (consider the family $t\alpha$ for all $t$), it does not leave much of space for what this algebra can be... $\endgroup$ Aug 4, 2016 at 9:16
  • $\begingroup$ Due to the grading of the algebra, we can be sure that this algebra is not zero. The modified Clifford algebra is isomorphic to an Clifford algebra but the presentation is different... Perhaps, I should not call it modified, but we can construct, I suppose, an adapted Dirac operator over a 'spin' manifold. $\endgroup$ Aug 4, 2016 at 19:34
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    $\begingroup$ I think this is isomorphic to the usual Clifford algebra with a different $g'(e, f) = g(e, f) - \frac{1}{4} \alpha(e) \alpha(f)$, through the isomorphism $e \rightarrow e + \frac{1}{2} \alpha(e)$. Does that allow you to construct your desired operators? $\endgroup$
    – user44191
    Aug 11, 2016 at 2:26

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