Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Update: now with a question 2 which is much more elementary (and should be well-known!).

Let $k$ be a commutative ring with $1$. Let $L$ be a $k$-module, and $g:L\to k$ be a quadratic form, i. e., a map for which the map $L\times L\to k,\ \left(x,y\right)\mapsto g\left(x+y\right)-g\left(x\right)-g\left(y\right)$ is $k$-bilinear and which satisfies

$g\left(\lambda x\right)=\lambda^2 g\left(x\right)$ for all $x\in L$ and $\lambda\in k$.

We define the Clifford algebra $\mathrm{Cl}\left(L,g\right)$ as the tensor algebra $\otimes L$ (all tensors are over $k$) divided by the two-sided ideal generated by $x\otimes x-g\left(x\right)$ for all $x\in L$.

Question 1: Is $\mathrm{Cl}\left(L,g\right)$ isomorphic to $\wedge L$ as a $k$-module? Is the associated graded object of $\mathrm{Cl}\left(L,g\right)$ isomorphic to $\wedge L$ as a $k$-algebra?

Remark: The answer is yes if the quadratic form $g$ comes from a bilinear (not necessarily symmetric!) form. As a consequence, the answer is yes if $L$ is a free $k$-module or, more generally, a direct sum of quotient $k$-modules of $k$ (in fact, it is easy to see that in this case, every quadratic form on $L$ comes from a bilinear form). (I think that in the case when $L$ is a finite-free $k$-module, the answer "yes" can also be proven by the diamond lemma, though I have not checked.) I am interested in the "perverse" cases when none of these holds, but I suffer from a lack of perversion: I can't name any such case. So here is the question which obviously needs to be addressed first:

Question 2: Find a commutative ring $k$ with $1$ and a $k$-module $L$ with a quadratic form (this is defined as above) which doesn't come from any bilinear (symmetric or not) form on $L$. (A quadratic form $g$ is said to come from a bilinear form $h$ if we have $g\left(v\right)=h\left(v,v\right)$ for every $v\in L$.)

Note: The Clifford algebra of a quadratic form is, in some sense, a "little sister" of the universal enveloping algebra of a Lie or pseudo-Lie algebra. ("Little sister" not in the historical sense, but in the sense of partly having the same properties, but them being easier to prove in the Clifford case than in the universal enveloping algebra case.) The above question asks for a kind of Poincaré-Birkhoff-Witt (PBW) theorem for the Clifford algebra. (Note that the PBW theorem itself requires some niceness conditions such as $L$ being finite-free or $k$ being a $\mathbb Q$-algebra, so it wouldn't surprise me if the Clifford case also doesn't work in full generality. But the opposite case wouldn't surprise me either, because PBW is substantially harder than PBW for Clifford algebras, even in characteristic $0$.)

share|improve this question
    
If it is true in the finite-free case it should work in the free case, no? –  Mariano Suárez-Alvarez Apr 4 '11 at 19:38
    
Probably. I just haven't checked. I don't know much about the diamond lemma in infinite cases. –  darij grinberg Apr 4 '11 at 19:39
add comment

2 Answers 2

What follows is incorrect, as pointed out by Darij and Theo. In particular, it doesn't seem that anywhere I am using the fact that $g$ is a quadratic form. I am leaving it here in the hope that it can be somehow rescued by additional assumptions, or some adaptation of the method will work.

Original answer follows:

I believe this follows from the general statement that taking the associated graded object is an exact functor from the category of filtered $k$-algebras to the category of graded $k$-algebras. Denote the tensor algebra of $L$ by $T(L)$, denote the relevant ideal by $$ I = \langle x \otimes x - g(x) \rangle $$ and the inclusion by $\iota: I \to T(L)$.

Then the morphism $\mathrm{gr} (\iota)$ of graded $k$-algebras maps $x \otimes x - g(x)$ (or, more pedantically, its image in $\mathrm{gr}(I)$) to $x \otimes x$ (after identifying $\mathrm{gr}(T(L))$ with $T(L)$).

This shows that the associated graded algebra of $\mathrm{Cl}(L,g)$ is $\bigwedge(L)$, and this lifts to an isomorphism of $k$-modules between $\mathrm{Cl}(L,g)$ and $\bigwedge(L)$.

Edit: I have expanded my original answer.

Let $A = \cup_{n=0}^\infty A_n$ be any filtered $k$-algebra and let $I$ be an ideal. Denote $I_n = I \cap A_n$, so $I$ is filtered by $I = \cup_{n=0}^\infty I_n$. Let $$ q: A \to A/I $$ be the quotient map. Then $A/I$ is filtered by $$ A/I = q(A) = \cup_{n=0}^\infty q(A_n).$$ Let the inclusion map be denoted $\iota : I \to A$. Then we have the short exact sequence $$ 0 \to I \overset{\iota}{\longrightarrow} A \overset{q}{\longrightarrow} A/I \to 0$$ in the category of filtered $k$-algebras.

My claim is that the sequence $$ 0 \to \mathrm{gr}(I) \overset{\mathrm{gr}(\iota)}{\longrightarrow} \mathrm{gr}(A) \overset{\mathrm{gr}(q)}{\longrightarrow} \mathrm{gr}(A/I) \to 0$$ remains exact in the category of graded $k$-algebras. This is a bit tedious to check but not difficult.

Then apply this to $A = T(L)$, $I = \langle x \otimes x - g(x) \rangle$. Then $A/I$ is your Clifford algebra, and I am saying that you can figure out its associated graded object as $$ \mathrm{gr}(A/I) \simeq \mathrm{gr}(A)/ \mathrm{gr}(\iota)(\mathrm{gr}(I)).$$ Finally, what does the map $\mathrm{gr}(\iota)$ do to the ideal? It turns out that $$ \mathrm{gr}(\iota) (x \otimes x - g(x)) = x \otimes x,$$ so the image of $\mathrm{gr}(I)$ under $\mathrm{gr}(\iota)$ is exactly the ideal you quotient by to get the exterior algebra $\bigwedge(L)$.

share|improve this answer
    
What would stop your argument (which I don't really understand) from showing the same for ANY map $g$, which is clearly wrong (any sufficiently random $g$ would collapse the degree $0$-part of $\otimes L$)? –  darij grinberg Apr 4 '11 at 20:52
2  
OK, I think your argument breaks at your claim that "the image of gr(I) under gr(ι) is exactly the ideal you quotient by to get the exterior algebra ⋀(L)". This does NOT follow from $(\mathrm{gr}(\iota))(x\otimes x - g(x)) = x\otimes x$. In fact, you might have wild cancellations (= a sum of tensors where the highest degrees cancel out but the lower don't). –  darij grinberg Apr 4 '11 at 21:40
2  
An example might help. Assume that $g$ is random. Take two generic vectors $x$ and $y$. We know that $I$ contains the tensors $x\otimes x - g(x)$, $y\otimes y - g(y)$, $(x+y)\otimes (x+y) - g(x+y)$ and $(x-y)\otimes (x-y) - g(x-y)$. Thus, $I$ also contains $((x+y)\otimes (x+y) - g(x+y)) + ((x-y)\otimes (x-y) - g(x-y)) - 2 (x\otimes x - g(x)) - 2 (y\otimes y - g(y))$. Now, this turns out to be a tensor of degree $0$, namely $-g(x+y)-g(x-y)+2g(x)+2g(y)$, and for a random $g$ this has no reason to be $0$. **If we treat it as an element of $I_2$*, then its image under $\mathrm{gr}\iota$ is ... –  darij grinberg Apr 4 '11 at 21:50
1  
... zero and thus, of course, lies in the ideal you quotient by to get the exterior algebra. But it happens to be an element of $I_0$, and we have no reason to assume that it lies in that ideal. –  darij grinberg Apr 4 '11 at 21:51
1  
This answer can't be right, because it would prove PBW for arbitrary Lie algebras over arbitrary rings. But PBW fails: Cohn, P. M. A remark on the Birkhoff-Witt theorem. J. London Math. Soc. 38 1963 197--203. MR0148717 . –  Theo Johnson-Freyd Apr 5 '11 at 14:26
show 5 more comments
up vote 1 down vote accepted

If someone could check the below I'd be very indebted.

Found the counterexample (to Question 1 and thus also to Question 2). It is inspired by the counterexample to ring-theoretical PBW in P. M. Cohn, A remark on the Birkhoff-Witt theorem. J. London Math. Soc. 38 1963 pp. 197-203, MR0148717 (though I still don't know whether Cohn's counterexample is true for all primes $p$ - but here I only need $p=2$).

Let $k$ be the commutative ring $\mathbb F_2 \left[\alpha,\beta,\gamma\right] / \left(\alpha^2,\beta^2,\gamma^2\right)$.

Let $L$ be the $k$-module $\left\langle x,y,z\right\rangle / \left\langle \alpha x - \beta y - \gamma z\right\rangle$.

Define a map $q:L\to k$ by $q\left(\overline{ax+by+cz}\right) = a^2+b^2+c^2$ for all $a,b,c\in k$. This map $q$ is easily seen to be well-defined and a quadratic form; it also satisfies $q\left(\overline x\right)=q\left(\overline y\right)=q\left(\overline z\right)=1$.

In the Clifford algebra $\mathrm{Cl}\left(L,q\right)$, we have $\left(\overline{\alpha x-\beta y-\gamma z}\right)\cdot\overline x = \alpha \underbrace{\overline{x}\cdot\overline{x}}_{=q\left(\overline x\right)=1} - \beta \overline y \overline x - \gamma \overline z \overline x = \alpha - \beta \overline y \overline x - \gamma \overline z \overline x$. Since the left hand side of this equality is $0$ (because $\overline{\alpha x-\beta y-\gamma z}=0$), we thus have $ \alpha - \beta \overline y \overline x - \gamma \overline z \overline x = 0$. Multiplied with $\beta\gamma$, this becomes $\alpha\beta\gamma = 0$ (because $\beta^2=0$ and $\gamma^2=0$ make the other two terms disappear). This must hence hold in $\mathrm{Cl}\left(L,q\right)$. But this does not hold in $k$, and thus not in $\wedge L$ either, so we cannot have an isomorphism.

share|improve this answer
add comment

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.