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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: Proposition 1.3 Proof

I understand that $\tilde{f}$ is injective, but why is the direct sum of these maps an isomorphism?

marc
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$ \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.

Nicholas Todoroff
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  • Thank you, the only thing i don't understand is the fact that $\mathcal{G} \simeq C\ell(V,q)$. Is this a general fact of filtered vector spaces? If so, can you give me some references? – marc Nov 15 '22 at 18:31
  • @MarcoTavano Yes this is a general fact, though the isomorphism is not necessarily canonical which is why I phrased it in terms of dimension. I don't know of a reference, but the argument goes something like this: Define a filtration on $\mathscr G$ by $$\mathscr H^r = \mathscr G^0\oplus\mathscr G^1\oplus\cdots\oplus\mathscr G^r.$$ Calculating $\dim\mathscr H^r$, we get a telescoping sum and $\dim\mathscr H^r = \dim\mathscr F^r$. Since $Cl(V, Q) = \bigcup_r\mathscr F^r$ and $\mathscr G = \bigcup_r\mathscr H^r$ the result follows. – Nicholas Todoroff Nov 15 '22 at 23:48