1
$\begingroup$

Let $R$be a commutative ring with unit element, and $M$ be an $R-$module. Let $f:M \times M \to R$ be a nondegenerate symmetric bilinear quadratic form, and $C(f)$ be the corresponding Clifford algebra with $\mathbb{Z}_2$ grading $C(f)=C(f)^{0} \oplus C(f)^1$. Now let $P=P^0 \oplus P^1 $ be a graded $C(f)$-module which is irreducible. Thus there is a natural multiplication map $\psi : C(f) \otimes P^0 \to P $. My question is :

Is this map necessarily an isomorphism ?

I am able to see that this map is epimorphic (using the irreducibility assumption), but not able to see why this map must be injective.

Edit :

Motivation for this problem comes from trying to prove proposition 12(6.3) of Fibre Bundles by Husemoller. Pic attached : enter image description here

$\endgroup$
1
$\begingroup$

What's the motivation for the question? If you pick $R = M = k$ a field and $f = 0$, then $C := C(f) \cong k[\varepsilon]/(\varepsilon^2)$ with the ${\mathbb Z}/2{\mathbb Z}$-grading coming from the natural $\mathbb Z$-grading. Then the residue field of $C$ is an irreducible graded $C$-module for which $\psi$ is not an isomorphism.

$\endgroup$
  • $\begingroup$ Thanks, I think you are right. I have updated the question so as to include the motivation. I think Husemoller is assuming $f$ to be non-degenerate somewhere. I have accordingly added this condition. Could you advise some example with non-degenerate $f$ ? $\endgroup$ – user90041 May 28 '16 at 18:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.