# Clifford algebra non-zero

This should be a very easy question, but the proof in Lawson/Michelson (Spin geometry) is wrong and I do not find a really correct and complete argument:

Let V be a nonzero real vector space with scalar product: Why is the Clifford-algebra (constructed from the tensor algebra by quotiening out an ideal) non-zero?

• Apparently it is somewhat common knowledge that the proof (of what fact? That the inclusion $V \hookrightarrow CL(V,q)$ is injective?) in Lawson-Michelsohn is incorrect. I am interested to know what the the error is, I am apparently too unskilled in basic algebra to see it. In Atiyah-Bott-Shapiro (Clifford modules) it is stated as an easy to verify fact, together with the vector space isomorphism of $CL(V,q)$ and $Gr(V)$, the Grassman algebra of $V$, for vector spaces over any commutative field, so I am curious to know if there is a simpler proof. Jun 21, 2011 at 15:20
• For the injectivity (p.8) they use a "contraction"; but if the different a_i have different degree (only the sum of degrees of a_i and b_i was fixed), then it is not specified at which place we have to contract. You can construct examples (e.g. for deg a_i + deg b_i = 3) where this cancellation argument fails. Jun 21, 2011 at 15:32
• I retagged your question because a discussion on meta suggested they were trying to get rid of the generic "algebra" tag: tea.mathoverflow.net/discussion/1071/… Jun 28, 2011 at 13:41

To show that an algebra constructed as a quotient of the tensor algebra of a vector space is nonzero, one of the main ways to go is to construct representations. We can do this for the Clifford algebra as follows.

Let $V$ be a vector space over a field $k$ and $(,):V\times V \to k$ a symmetric bilinear form on $V$. The Clifford algebra (for this form) is given by $$Cl(V)= T(V)/\langle v \otimes v - (v,v)\rangle.$$ We will construct a representation of the Clifford algebra on the exterior algebra $\bigwedge (V)$.

For $v \in V$, define two $k$-endomorphisms of $\bigwedge(V)$ by $$l_v(x) = v \wedge x$$ and $$\delta_v(x) = \sum_{j=1}^k (-1)^{j-1}(v,x_j) x_1 \wedge \dots \wedge \widehat{x_j} \wedge \dots \wedge x_k$$ if $x = x_1 \wedge \dots \wedge x_k$.

Then check that $l_v^2 = \delta_v^2 = 0$, and moreover that $l_v \delta_v + \delta_v l_v = (v,v) \cdot \mathrm{id}$.

Extend the map linear $v \mapsto l_v + \delta_v$ to an algebra homomorphism from the tensor algebra $T(V)$ to $\mathrm{End}_k(\bigwedge(V))$. By the previous remark, this descends to a map, let's call it $\phi$, from the Clifford algebra to $\mathrm{End}_k(\bigwedge(V))$.

In particular, $\phi(v)1 = v$, so $V$ injects into the Clifford algebra.

Edit: I believe also that the map $$x \mapsto \phi(x)1$$ gives a linear isomorphism of the Clifford algebra with the exterior algebra.

A great reference for this stuff is Chevalley's monograph, The Algebraic Theory of Clifford Algebras and Spinors.

• Very nice proof, but I don't think it appears in Chevalley's works (however I have read them rather superficially). Jun 21, 2011 at 18:57
• Perhaps I am misremembering. If it's not in Chevalley, it is in Clifford Algebras and Spinor Norms Over a Commutative Ring, by Hyman Bass. Or perhaps a mixture of stuff that I read in those two references.
– MTS
Jun 21, 2011 at 19:13
• Thank you very much for this proof in the spirit of "the universal property for Clifford algebras". This is a nice and direct argument. Jun 21, 2011 at 19:34
• If you're a physicist, you will recognize $a=l_v$ and $a^\dagger=\delta_v$ as the creation and annihilation operators for fermions. The sum $a+a^\dagger$ is a Majorana operator. Jun 22, 2011 at 9:59
• typo: $k-1\to j-1$ in your definition of $\delta_v$. Jun 25, 2011 at 15:09

As Darij said, a way to prove this is to show a PBW theorem for the Clifford algebra.

It is a consequence of a more general PBW theorem for quadratic algebras of Koszul type by Braverman and Gaitsgory (see Theorem 4.1, and $\S 5.3$ for its application to Clifford algebras).

By the way, the result of Braverman and Gaitsgory works for $V$ being a module over a semi-simple ring (actually, I guess it works even if you start with a projective module $V$ over an arbitrary ring), while Darij's result might be a bit more general.

• Yes, my result is general... but if Braverman-Gaitsgory's ones generalize to relative homology, they will probably achieve the same generality as mine and even more. I am still trying to understand their argument, but I am hitting brick walls. Do you happen to know why (i) is equivalent to $d\alpha = 0$ in §4.3? Jun 21, 2011 at 14:54
• $\alpha:K^2=R\to V\subset A$ is viewed as an $A$-bimodule map $\tilde{\alpha}:\tilde{K}^2=A\otimes R\otimes A\to A$. In the very same way you view $\alpha\otimes id-id\otimes\alpha$ as an $A$-bimodule map $\tilde{\gamma}:\tilde{K}^3\to A$. It happens (computation) that $d(\tilde{\alpha})=\tilde{\gamma}$. Now $\alpha\otimes id-id\otimes\alpha$ takes values in $R$ iff $\gamma=0$. Jun 21, 2011 at 15:18
• How is $\alpha\otimes id-id\otimes\alpha$ viewed as an $A$-bimodule map to $A$? I'd expect it to go to $A\otimes A$ (if not to something larger), or do you compose it with multiplication? Also, how exactly is $d$ defined in $d\left(\widetilde{\alpha}\right)$ ? Is it the usual coboundary in the sense of $d\left(\widetilde{\alpha}\right)=\widetilde{\alpha}\circ\left(\mu\otimes id\otimes id\otimes id - id\otimes\mu\otimes id\otimes id + id\otimes id\otimes\mu\otimes id - id\otimes id\otimes id\otimes\mu\right)$ (where $\mu$ is the multiplication map in $A$)? Jun 21, 2011 at 17:13
• $\alpha\otimes id-id\otimes\alpha$ defines a map from $K^3$ to $T(V)$, that induces (by composing with the projection $T(V)\to A$) a map $K^3\to A$, that extend to an $A$-bimodule map $A\otimes K^3\otimes A=\tilde{K}^3\to A$. Jun 21, 2011 at 17:55
• Sorry for necromancing this up now, but my exams are over and I finally have some real time. It seems that the point where I stop understanding Braverman/Gaitsgory is earlier: in §3.6, why is $\widetilde{K}^{\cdot}\left(A\right)$ a subcomplex of $B^{\cdot}\left(A\right)$ ? As far as I understand, the differential of $B^{\cdot}\left(A\right)$ involves multiplying together adjacent tensorands, which leads us out of the submodule $\widetilde{K}^i\left(A\right)$... What am I doing wrong? Jul 6, 2011 at 15:17

Extremely long answer: The Clifford algebra and the Chevalley map - a computational approach Theorem 1 (or 2, or even 3). Actually the reason for writing this text was my disappointment with the wrong proof in Lawson-Michelson, and with the not sufficiently general proofs in the rest of literature. The heuristics for finding the proof are explained in §6.3 of my diploma thesis.

Yes, 90% of the proof are computations.

Now chances are high that you prefer a 1-page proof that works over $$\mathbb R$$ to a 40-pages one that works over any commutative ring, so you might be interested in the proof that uses orthogonal decomposition of the quadratic form (i. e., finding an orthogonal basis using Gram-Schmidt) and application of the fact that $$\operatorname{Cl}\left(V\oplus W\right)\cong \left(\operatorname{Cl}V\right)\hat{\otimes}\left(\operatorname{Cl}W\right)$$ (super-tensor product of superalgebras) for any two quadratic spaces $$V$$ and $$W$$. Such a proof can be found in Milne's ALA Theorem 18.18, and probably in many other places. As Guntram pointed out in the comment below, this proof only works for $$V$$ being finite-dimensional; however, the infinite-dimensional case follows from the fact that (unless I am mistaken) taking the Clifford algebra commutes with the direct limit. (Here we are using the fact that if $$V$$ and $$W$$ are two finite-dimensional quadratic spaces such that $$V\subseteq W$$, then the canonical map $$\operatorname{Cl}V\to \operatorname{Cl}W$$ is injective. This easily follows from the basis theorem for Clifford algebras of finite-dimensional quadratic spaces.)

• It doesn't help to introduce new errors while criticizing others: As given in the source, the tensor product needs to be equipped with a hat - and then this doesn't help in case that $V$ is infinite-dimensional, the only case not explicitly covered in Fulton-Harris Lemma 20.3, where a collection of three proofs is given. Jun 21, 2011 at 15:25
• Thanks for the corrections, they are appreciated. But is the second proof in Fulton-Harris even close to being a proof? Jun 21, 2011 at 15:41

Here is a braindead way to generate PBW theorems like this one:

One is given a presentation of an algebra, which allows one to put words in the generators $x_1,x_2,\dots,x_n$ into a "normal form" (in the case of the Clifford algebra, if $x_1,x_2,\dots,x_n$ is a basis of $V$, then a normal form might be words $x_{i_1} x_{i_2} \cdots x_{i_p}$ with $1 \leq i_1 < i_2 < \cdots < i_p \leq n$). One suspects that the set of words in normal form is actually a basis for the algebra. So one constructs the regular representation of the algebra in question. If the theorem is to be true, there is no choice about this: it is the vector space spanned by words in normal form, and the left (and right) multiplication operators are determined by the relations. Usually the easiest way to write down the formulas is recursively. One is then left to check that these operators satisfy the defining relations. This is "just linear algebra", but the computations in any particular example (or class of examples) may get messy. When it works, it usually works in arbitrary characteristic (and even integrally).

In the case of the Clifford algebra and similar algebras (e.g. enveloping algebras of Lie algebras, symplectic reflection algebras and their generalizations) this all works without too much difficulty. It also works for the Hecke algebra attached to a Coxeter system, though to make the calculations manageable in a case-free fashion it's good to use the Bourbaki trick of employing both the right and left representations simultaneously. There is even a general theorem here, which usually goes by the name "Bergman diamond lemma" (but in the cases I'd care most about, checking that its conditions are satisfied is just about the same level of difficulty as doing the work directly).

• Speaking of diamonds, Chapter 4 of math.unice.fr/~brunov/Operads.pdf has some extensions of the diamond lemma which do some of one's work. Jun 22, 2011 at 6:32

If $V$ is a vector space and $q$, $q'$ are quadratic forms on $V$ then $Cl(V,q)$ and $Cl(V,q')$ are isomorphic as vector spaces (though obviously not as algebras). Moreover the Clifford algebra of $V$ with $q = 0$ is precisely the exterior algebra.

Added: In the finite dimensional case and when the ground field is $\mathbb{R}$ or $\mathbb{C}$, the isomorphism can be proved without too much difficulty by diagonalizing the quadratic form, though I suppose one does need a PBW-style description of the tensor algebra to prove that the exterior algebra is nonzero. I see now that darij grinberg's answer already included these ideas.

In the infinite dimensional case or in the case where the ground field is more interesting, I suppose there really is something to this. But in the most basic cases (and the ones most relevant to differential geometry) I think the argument is fairly elementary.

• The exterior algebra is obviously non-zero, that is no problem; but how do I prove this isomorphism? Jun 21, 2011 at 15:24
• That is precisely the content of the PBW theorem. Jun 21, 2011 at 15:36

If the symmetric bilinear form $$(\cdot,\cdot)$$ is nondegenerate, then the proof is quick. Let $$i:V\to\text{Cl}(V)$$ be the canonical projection. We must show that $$i$$ is injective.

Let $$x\in\ker(i)$$. Then $$2(x,y)1=i(x)i(y)+i(y)i(x)=0$$ hence $$x=0$$ by nondegeneracy. $$\square$$

I took this from Greub's Multilinear Algebra, § 10.4. He extends the argument to the general case by decomposing $$V$$ as a direct sum of spaces where $$(\cdot,\cdot)$$ is completely degenerate and nondegenerate.

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• This appears to already assume that 1 is not 0 in Cl(V) though? Feb 2 at 18:36
• @kiran You're right, I missed that. I guess Greub missed it too, or just assumes that $1\neq0$? Feb 2 at 20:48

What about: Cl(-) is functorial, the Clifford algebra of a 1-dim real vector space is nonzero-checkable by hand, any nonzero real vector space w/-symmetric-bilinear form has a 1-dimensional summand-w/-symmetric-bilinear-form.

Consider the tensor algebra $T(V)$ as a filtered algebra with $T_{\le k} = \oplus_{i \le k} V^{\otimes i}$. This filtration induces a filtration on the Clifford algebra, $Cl_{\le k}(V) = \pi(T_{\le k}(V))$, where $\pi:T(V) \to Cl(V)$ is the canonical projection. Now it is easy to check that the associated graded algebra $gr Cl(V) := \oplus_k Cl_{\le k}(V)/Cl_{\le k-1}(V)$ is the quotient of the associated graded algebra $gr T(V) = T(V)$ by the ideal generated by the leading terms of the relations defining $Cl(V)$, that is by $v\otimes v$ for all $v \in V$. Thus $gr Cl(V) = \Lambda (V)$, hence $Cl(V)$ is nonzero.

• "Now it is easy to check": no, it is not. Jun 21, 2011 at 18:37
• This is in general not true that for a given subspace of relations $R\subset T(V)$, the associated graded of $A:=T(V)/R$ is the quotient of $T(V)$ by the leading term of $R$. Jun 21, 2011 at 18:41
• See mathoverflow.net/questions/60596/… (in which I made the same mistake) for a fuller discussion of why it isn't easy to show that the associated graded of the Clifford algebra is the exterior algebra. This is part of the content of the PBW theorem mentioned in DamienC's example.
– MTS
Jun 21, 2011 at 18:42