clifford-algebrasexterior-algebravector-space-isomorphism
I'm studying the book "Spin Geometry" by Lawson and Michelson.
I don't understand the last line of their proof of the fact that the Clifford Algebra is isomorphic to exterior algebra, this is the proposition:
I understand that $\tilde{f}$ is injective, but why is the direct sum of these maps an isomorphism?
$\newcommand{\Cl}{\mathscr{Cl}(V)}$The problem with your definition of the wedge product in $\Cl$ is that it is an $n$-ary product of vectors. And so it's not clear to me how one would even discuss associativity of it (i.e. it doesn't allow one to distinguish between the product of a bivector and a vector vs the product of a vector and a bivector). So while $\Cl$ with the wedge product is an algebra -- it isn't an associative algebra like the exterior algebra is. Let me instead recommend a different definition based on grade projection (basically the same definition given in the works of Alan Macdonald):
To define the wedge product, we first define it in the case of single-grade elements. Let $A,B$ be an $r$-vector and $s$-vector, respectively. Then
$$A\wedge B := \langle AB\rangle_{r+s}$$
Now using this special case, we define the wedge product for all multivectors. Let $C,D$ be (potentially mixed grade) multivectors in $\Cl$. Then we define
$$C \wedge D := \sum_{j,k} \langle C\rangle_j \wedge \langle D\rangle_k$$
Proof that for all $A,B,C\in \Cl$, we have $A\wedge (B\wedge C) = (A\wedge B)\wedge C$:
Consider a $j$-vector $A$, $k$-vector $B$, and $l$-vector $C$. Then $$(A\wedge B)\wedge C = \langle AB\rangle_{j+k}\wedge C = \langle \langle AB\rangle_{j+k}C\rangle_{j+k+l} = \langle (AB)C\rangle_{j+k+l}$$ where that last step holds because only the \langle AB\rangle_{j+k} part of $AB$ can contribute to $\langle (AB)C\rangle_{j+k+l}$.$^\dagger$.
Then $$(A\wedge B)\wedge C = \langle (AB)C\rangle_{j+k+l} = \langle A(BC)\rangle_{j+k+l} = A\wedge\langle BC\rangle_{k+l} = A\wedge(B\wedge C)$$
To finish the proof by allowing $A,B,C$ to be any multivectors in $\Cl$, just note that our definition of the wedge product is bilinear.$^\ddagger$$\ \ \ \ \square$
Now we'll show that with this definition, your proposition holds. Take $\wedge_C$ to be the wedge product in $\Cl$ and $\wedge_T$ to be the wedge product in $\Lambda V$ (i.e. the coproduct of the antisymmetric tensor powers of $V$).
Let $\{e_1, \dots, e_n\}$ be an orthogonal basis for $V$ wrt the symmetric bilinear form and construct bases for $\Cl$ and $\Lambda V$ in the obvious way from it. Let $\Gamma: \Cl \to \Lambda V$ be given by $$\Gamma(A_0 + A_1e_1 + A_2e_2 + \cdots + A_{12\cdots n}e_1\wedge_C e_2\wedge_C \cdots \wedge_C e_n) = A_0 + A_1e_1 + A_2e_2 + \cdots + A_{12\cdots n}e_1\wedge_T e_2\wedge_T \cdots \wedge_T e_n$$
By definition this is linear. It's also clearly invertible. To see that it's multiplicative, it suffices (due to linearity of $\Gamma$ and bilinearity of $\wedge_C$) to show that $\Gamma$ is multiplicative on blades. Let $A=A_{i_1\cdots i_p}e_{i_1}\wedge_C\cdots\wedge_C e_{i_p}$ be a $p$-blade and $B=B_{j_1\cdots j_r}e_{j_1}\wedge_C\cdots\wedge_C e_{j_r}$ an $r$-vector. Then $$\begin{align}\Gamma(A\wedge_C B) &= \Gamma(A_{i_1\cdots i_p}e_{i_1}\wedge_C\cdots\wedge_C e_{i_p}\wedge_CB_{j_1\cdots j_r}e_{j_1}\wedge_C\cdots\wedge_C e_{j_r}) \\ &= A_{i_1\cdots i_p}e_{i_1}\wedge_T\cdots\wedge_T e_{i_p}\wedge_T(B_{j_1\cdots j_r}e_{j_1})\wedge_T\cdots\wedge_T e_{j_r} \\ &= (A_{i_1\cdots i_p}e_{i_1}\wedge_T\cdots\wedge_T e_{i_p})\wedge_T(B_{j_1\cdots j_r}e_{j_1}\wedge_T\cdots\wedge_T e_{j_r}) \\ &=\Gamma(A_{i_1\cdots i_p}e_{i_1}\wedge_C\cdots\wedge_C e_{i_p})\wedge_T\Gamma(B_{j_1\cdots j_r}e_{j_1}\wedge_C\cdots\wedge_C e_{j_r}) \\ &= \Gamma(A)\wedge_T\Gamma(B)\end{align}$$
A linear, invertible, multiplicative map is by definition an isomorphism of algebras. Hence $\Cl$ is isomorphic to $\Lambda V$, as desired.
Another Interesting Tidbit:
$\Lambda V$ is actually (or at least can be) defined by a set of axioms like other types of vector spaces.
From nlab:
Suppose $V$ is a vector space over a field $K$. Then the exterior algebra $\Lambda V$ is generated by the elements of $V$ using these operations:
- addition and scalar multiplication
- an associative binary operation $\wedge$ called the exterior product or wedge product,
subject to these identities:
- the identities necessary for $\Lambda V$ to be an associative algebra
- the identity $v\wedge v = 0$ for all $v\in V$.
It is easily confirmed that $\Cl$ satisfies these properties under the definition of $\wedge$ given above (and thus why associativity is so important for $\wedge$). Thus, not only can we construct an isomorphism, but we can even consider $\Cl$ to just be $\Lambda V$ with extra structure (the Clifford product).
$\dagger:$ To prove this, note that it is known that if $\Bbb F$ is not characteristic $2$, then $V$ has an orthogonal basis wrt any symmetric bilinear form. Then you can construct a "standard" basis for $\Cl$ and decompose your multivectors wrt to it.
$\ddagger:$ By the linearity of the grade projection operator.
$ \newcommand\K{\mathbb K} \newcommand\Cl{\mathrm{Cl}} \newcommand\tensor\otimes \newcommand\C{\mathbb C} \newcommand\R{\mathbb R} $
We utilize the fact that $\C^n \cong \R^n\tensor_\R\C$. Define $q\tensor \C := q_\C : \C^n \to \C$ as the unique quadratic form such that $$ q_\C(v\tensor x) := q(v)x^2. $$ It follows easily that $$ q_\C(v\tensor 1 + v'\tensor i) = q(v) - q(v') + 2iB(v, v') $$ where $B(v, v') = \tfrac12(q(v + v') - q(v) - q(v'))$ is the bilinear form associated to $q$.
To show that $$ \Cl(\C^n, q_\C) \cong C := \Cl(\R^n, q)\tensor_\R\C \tag{$*$} $$ as complex algebras, it suffices to show that the RHS satisfies the universal property of complex Clifford algebras. We consider $\C^n \cong \R^n\tensor_\R\C$, and we let the inclusion $\C^n \to C$ be given by $v\tensor x \mapsto v\tensor x$ for $v \in \R^n$ and $x \in \C$ (so we are identifying $\R^n$ as a subset of $\Cl(\R^n, q)$. Then for any complex algebra $A$, let $f : \C^n \to A$ be a homomorphism such that $f(v\tensor x)^2 = q_\C(v\tensor x) = q(v)x^2$. Since $$ \Cl(\R^n, q)\tensor_\R\R \cong \Cl(\R^n, q) $$ as real algebras, the restriction of $f$ to $\R^n \to A$ as a real homomorphism and $A$ gives a unique real homomorphism $\phi : \Cl(\R^n, q) \to A$ such that $\phi(v) = f(v\tensor1)$ by the universal property of $\Cl(\R^n, q)$. Now define $\phi_\C : C \to A$ by $$ \phi_\C(X\tensor x) = \phi(X)x,\quad X \in \Cl(\R^n, q). $$ This is clearly a complex homomorphism. It immediately follows for $v \in \R^n$ that $$ \phi_\C(v\tensor x) = \phi(v)x = f(v\tensor1)x = f(v\tensor x). $$ $\phi_\C$ is automatically unique amongst maps $\phi' : C \to A$ such that $\phi'(v\tensor x) = f(v\tensor x)$ by virtue of being a homomorphism. This completes the proof of ($*$).
Best Answer
$ \newcommand\Ext{{\bigwedge}} $
Index each $\tilde f$ as $\tilde{f_r}$ for clarity, and let $\pi_r : \mathscr F^r \to \mathscr F^{r-1}$ be the mentioned projection. What we need to establish is that $\tilde{f_r}(x) = \tilde{f_s}(y)$ implies $x = 0 = y$ for any $x \in \Ext^r V$ and $y \in \Ext^s V$ when $r \not= s$. Without loss of generality take $r < s$. Applying $\pi_s$ gives $$ \pi_s(\tilde{f_r}(x)) = \pi_s(\tilde{f_s}(y)) \implies 0 = (\pi_s\circ\tilde{f_s})(y), $$ but $\pi_s\circ\tilde{f_s}$ is a bijection so $y = 0$; then $\tilde{f_r}(x) = 0$ and $x = 0$ since $\tilde{f_r}$ is injective.
What this shows is that the images of $\tilde{f_1}, \tilde{f_2}, \dotsc$ do not overlap except trivially; thus the direct sum $F = \tilde{f_1}\oplus\tilde{f_2}\oplus\cdots$ is injective since each $\tilde{f_r}$ is injective. So we have an injection $F : \Ext V \to Cl(V, Q)$.
Proposition 1.2 gives $\Ext V \cong \mathscr G$, where $\mathscr G$ is the associated graded algebra of $Cl(V, Q)$. Though they don't appear to show or claim it, if we use the fact that $\dim\mathscr G = \dim Cl(V, Q)$ then $F$ is an isomorphism and we're done. I don't know how else they are concluding that $F$ is surjective.