Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $k$ be a field which does not have characteristic 2. Let $M$ be the free $k$-vector space generated by two elements $\{ c, x \}$. Let $T(M)$ be the tensor algebra of $M$ and let $I$ be the ideal generated by the elements $$\{c^2 -1, x^2, cx + xc \}.$$

I am trying to show that $B = \{ 1, c, x, cx\}$ forms a basis of the quotient space $T(M) / I$. I can see why it spans the space but I cannot show that it is linearly independent. Any help would be appreciated.

share|improve this question

1 Answer 1

up vote 1 down vote accepted

As you probably know, this is an example of a Clifford algebra attached to the quadratic form which gives two orthogonal basis elements $c,x$ respective "norms" $1$ and $0$, and $c,x$ anti-commute, etc.

The standard trick to show that the dimension is as large as possible begins with the case of the Clifford algebra of a one-dimensional vector space with bilinear form, and in that case we know that $k[x]/(x^2-a)$ is two-dimensional, in any case, because we have simpler control in the commutative case.

It is perhaps over-kill for the two-dimensional vectorspace case, but the induction step in proving that a Clifford algebra on an $n$-dimensional v.s. $V$ has dimension $2^n$ (rather than accidentally smaller) is to express $V$ as an orthogonal sum $V_1\oplus V_2$ where $V_1$ is one-dimensional. There is the parity grading into odd and even elements, inherited from the universal associative (tensor) algebra, and for monomials define a twisted multiplication $(x\otimes y)\cdot (x'\otimes y') = xx'\otimes yy' \cdot (-1)^{\deg x'\cdot \deg y}$ on $Cl(V_1)\otimes CL(V_2)$. One checks directly that this gives the Clifford multiplication, and we know the dimension of a tensor product (whether or not the multiplication is defined with $-1$'s in it).

(So this is not as serious a result as Poincare-Birkhoff-Witt for universal enveloping algebras, but it's not completely trivial.)

Edit: Actually, knowing the above, one can pretend to have the epiphany that this algebra is inside the $4\times 4$ matrix algebra over the ground-field (it has some degeneracy) so prove the four-dimensionality by mapping this (universal) object surjectively to a four-dimensional subalgebra of the matrix algebra. But this is a bit of a prank... :)

share|improve this answer
    
Thanks for you answer. I wasn't aware it was a Clifford algebra, but I will look into this! –  Paul Slevin Aug 30 '12 at 20:49
    
You're welcome. This smallish case is also some sort of "quaternion algebra", which indeed is the smallest non-commutative Clifford algebra. –  paul garrett Aug 30 '12 at 20:58

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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