MathOverflow is a question and answer site for professional mathematicians. It only takes a minute to sign up.
Sign up to join this community
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?
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.
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.
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).
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.)
$ \newcommand\Ext{\mathop{\textstyle\bigwedge}} \newcommand\Cl{\mathrm{Cl}} \newcommand\gtensor{\mathbin{\hat\otimes}} \newcommand\gen[1]{\langle#1\rangle} \newcommand\K{\mathbb K} $Let $V$ be a vector space over a field $\K$ and let $Q$ be a quadratic form on $V$.
I'd like to add a proof I gave in this MSE post that $V \mapsto \Cl(V,Q)$ is injective. This is similar to what Darij Grinberg mentions, making use of the fact that $\Cl(V_1\oplus V_2,Q) \cong \Cl(V_1,Q)\gtensor\Cl(V_2,Q)$ when $V_1 \perp V_2$, but I fail to see how this relies on $V$ being finite dimensional. I also make use of the fact that $V = V_1 \oplus V^\perp$ where $V_1$ is non-degenerate and $V^\perp$ is the radical; as I understand this decomposition should work in infinite dimensions so long as we have choice.
We write $T(V)$ for the tensor algebra of $V$ and assume $\K,V \subseteq T(V)$, we write $Q(\cdot,\cdot)$ for the bilinear form associated to $Q$, and we write $$ V^\perp = \{v \in V \;:\; Q(v) = 0\text{ and }\forall w \in V.\,Q(v,w) = 0\} $$ for the radical of $V$.
We define $\Cl(V, Q) = T(V)/I$ where $T(V)$ is the tensor algebra and $I = \gen{v\otimes v - Q(v) \;:\; v \in V}$, and we write $\pi T(V) \to \Cl(V,Q)$ for the canonical projection. $\pi|_\K$ is injective by linearity, since $\pi(1) \ne 0$, and since $\K$ is a field so we take the map $\pi|_\K : \K \to \pi(\K)$ to be the identity. Then for any $v \in V$ $$ 0 =\pi(v\otimes v - Q(v)) = \pi(v)^2 - Q(v) \implies \pi(v)^2 = Q(v). $$ This definition of $\Cl(V,Q)$ gives the following universal property:
We now proceed by showing that $\pi|_V$ is injective when $Q$ is nondegenerate; then we prove the decomposition $\Cl(V, Q) \cong \Cl(V_1,Q)\gtensor\Cl(V_2,Q)$ when $V = V_1\oplus V_2$ with $V_1\perp V_2$. We always have the decomposition $V = V_1\oplus V^\perp$ with $V_1$ nondegenerate, and from here it follows that $\pi|_V$ is injective.
Lemma 1. $\pi|_V$ is injective for $\Ext V := \Cl(V, 0)$.
Proof. Note that in this case every element of $I$ has degree at least 2. If $v \in V\cap I$ then $v$ must have degree 1 and degree at least 2, so $v = 0$. $\quad\Box$
Lemma 2. If $Q$ is nondegenerate (i.e. $V^\perp = \{0\}$) then $\pi|_V$ is injective.
Proof. Let $v, w \in V$ and assume $\pi(v) = 0$. Write $Q(\cdot,\cdot)$ for the bilinear form associated with $Q$. Then $Q(v) = \pi(v)^2 = 0$ and $$ Q(v, w) = \pi(v + w)^2 - \pi(v)^2 - \pi(w)^2 = \pi(v)\pi(w) + \pi(w)\pi(v) = 0, $$ Hence $v \in V^\perp = \{0\}$. $\quad\Box$
Theorem. If $V = V_1 \oplus V_2$ with $V_1\perp V_2$ then there is an algebra isomorphism $$ \psi : \Cl(V,Q) \cong \Cl(V_1,Q)\gtensor\Cl(V_2,Q) $$ (where on the RHS $Q$ is restricted as appropriate) such that $\psi\circ\pi_1|_{V_1} = \pi|_{V_1}$ and $\psi\circ\pi_2|_{V_2} = \pi|_{V_2}$ where $\pi_i$ is the canonical projection $T(V_i) \to \Cl(V_i,Q)$.
Proof. Let $C = \Cl(V_1,Q)\gtensor\Cl(V_2,Q)$ and identify $\Cl(V_1,Q)$ and $\Cl(V_2,Q)$ as subalgebras in the obvious way. For each $i = 1,2$, by the universal property of $\Cl(V_i, Q)$ we get a homomorphism $\psi_i : \Cl(V_i,Q) \to \Cl(V,Q)$ such that $\psi_i(\pi_i(w)) = \pi(w)$ for $w \in V_i$. Define $\psi = \psi_1\otimes\psi_2 : C \to \Cl(V,Q)$. By the universal property of $\Cl(V,Q)$ we get $\phi : \Cl(V,Q) \to C$ such that $$ \phi(\pi(w_i)) = \pi_i(w_i)\quad\text{when }w_i \in V_i. $$ If $X_1 \in T(V_1)$ and $X_2 \in T(V_2)$ it now easily follows that $$\begin{aligned} \phi\Bigl(\psi(\pi_1(X_1)\otimes\pi_2(X_2))\Bigr) &= \phi\Bigl(\psi_1\bigl(\pi_1(X_1))\otimes\psi_2(\pi_2(X_2)\bigr)\Bigr) \\ &= \phi(\pi(X_1))\otimes\phi(\pi(X_2)) \\ &= \pi_1(X_1)\otimes\pi_2(X_2). \end{aligned}$$ Every element of $C$ is a sum of elements of the form $\pi_1(X_1)\otimes\pi_2(X_2)$, so by linearity $\psi = \phi^{-1}$ is an isomorphism with the requisite properties. $\quad\Box$
Corollary. $\pi$ is injective on $V$ for any $Q$.
Proof. We choose $V_2 = V^\perp$ whence $\Cl(V_2, Q) = \Ext V^\perp$. We can choose $V_1$ to be any complement of $V^\perp$. By Lemma 1 $\pi_2|_{V^\perp}$ is injective so $\pi|_{V^\perp} = \psi\circ\pi_2|_{V^\perp}$ is injective. Similarly $\pi_1|_{V_1}$ is injective by Lemma 2 and so $\pi|_{V_1}$ is injective. Hence $\pi|_V = \pi|_{V_1} \oplus \pi|_{V^\perp}$ is injective. $\quad\Box$
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.
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.
