0
$\begingroup$

In Bourbaki Lie Groups and Lie algebras chapter 6 section 4 excercise 1(c), they have used the word pseudo-discriminant. The reference is given to be Algebra chapter IX which I can't find a English translation of. Here the following definition is given. Since I don't know French I can't understand the definition. Any help will be appreciated. Sorry if this kind of question is not for this site. enter image description here

$\endgroup$
1
$\begingroup$

Here is a translation in English. I added few things into parentheses.

Assume that $A$ is a field of characteristic $2$, $E$ is a vector space on $A$ of even dimension $2r$, and $Q$ is a non-degenerate quadratic form on $E$.

a) Let $(e_i)$ a symplectic basis of $E$ for the alternating bilinear fomr $\Phi$ associated to $Q$ . Show that the element $z=e_1e_1+e_3e_4+\cdots+e_{2r-1}e_{2r}\in C^+(Q)$ (the even Clifford algebra) forms, together with the unit element, a basis of the center $Z$ of $C(Q) (the Clifford algbra).

Moreover, $Z$ is the direct sum of two fields of and only if th element $\Delta(Q)=Q(e_1)Q(e_2)+Q(e_3)Q(e_4)+\cdots+ Q(e_{2r-1})Q(e_{2r})$, called the pseudo-discriminant of $Q$ with respect to the symplectic basis $(e_i)$, has the form $\lambda^2+\lambda$ for some $\lambda\in A$.

Some comments on this notion. Even if this is not explicitely asked by the OP, I think it would be nice to comment on this notion.

Recall that the determinant $\det(Q)$ of a non degenerate quadratic form $Q$ over a field $A$ (to stick with the previous notation) is the square class of the determinant of any representative matrix of the associated bilinear form. It is an element $A^\times/A^{\times 2}$, which is an invariant of the isometry class of $Q$ (two isometric quadratic forms have equal determinants). If $A$ has characteristic different from two, this invariant is quite useful, and may be use to obtain classification results.

If $A$ has characteristic two, then a representative matrix of the associated bilinear form is alternating, so its determinant is always a square. Hence, in this case , $\det(Q)$ is always the trivial square class.

Set $\wp(A)=\{ \lambda^2+\lambda\mid \lambda\in A\}$. This is a subgroup of $A$. One may show that the class of $\Delta(Q)$ in $A/\wp(A)$ does not depend on the choice of the symplectic basis, so we can set ${\rm Arf}(Q)$ to be the class of $\Delta(Q)$ in $A/\wp(A)$. The class ${\rm Arf}(Q)$ is called the Arf invariant of $Q$. One may show this is an anvariant of the isometry class of $Q$. It plays the same role as the determinant.

$\endgroup$
  • $\begingroup$ Thanks a lot.!! $\endgroup$ – justanothermathstudent Sep 11 at 12:00
  • $\begingroup$ I've added some comments about this notion in my answer. Hope this helps. $\endgroup$ – GreginGre Sep 11 at 12:02
  • $\begingroup$ Thanks again, sorry for late reply. Doesn't the existence of symplectic basis require the characteristic to be not equal to 2? $\endgroup$ – justanothermathstudent Sep 12 at 14:08
  • $\begingroup$ No I was mistaken sorry. $\endgroup$ – justanothermathstudent Sep 12 at 14:10

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.