# Friday, April 23


:::{.remark}
Given $(V, \cdot)$ an inner product space, we defined
\[
\Cl(V) \da { \bigoplus _{n\geq 0} V^{\tensor n} \over \gens{ v\tensor w + w\tensor v = 2v\cdot w } }
.\]
:::

:::{.example title="?"}
We saw that \
\[
\Cl(\RR, \cdot) 
&\cong \RR[e] / e^2 
=-1 \cong \CC\\
\Cl(\RR^2, \cdot) 
&= \RR \gens{ e_1, e_2 } / \gens{ e_1^2 = e_2^2 = -1, e_1e_2 = -e_2 e_1 -} \cong \HH
\]
where $e_1\mapsto i, e_2\mapsto j, e_3 = e_1 e_2 \mapsto k$.
Can we describe $\Cl(\RR^n, \cdot)$ in general?
Choose an orthonormal basis \( \ts{ e_i } \), then 
\[
\Cl(\RR^n, \cdot) = { \RR \gens{ e_1, \cdots, e_n } \over \gens{ e_i^2 = -1, e_i e_j = -e_j e_i \st i\neq j } }
.\]
We saw that replacing $2$ with $\epsilon$ in the defining relation recovers $\Extalg^* V$.

:::

:::{.definition title="Degree Filtration"}
Define the **degree filtration** on $\Cl(V, \cdot)$ as the filtration induced by the degree filtration on $T(V) \da \bigoplus _{n\geq 0} V^{\tensor n}$.
:::

:::{.example title="?"}
Consider $\Cl(\RR^2, \cdot)$. 
Then

- Degree 0: $\RR$.
- Degree 1: $\RR \oplus \RR e_1 \oplus \RR e_2$
- Degree 2: $\RR \oplus \RR e_1 \oplus \RR e_2 \oplus \RR e_1 e_2$
:::

:::{.definition title="Grading and Filtration"}
Recall that there's a distinction between gradings and filtration:

- Gradings: $R^i R^j \subset R^{i+j}$ and $R = \bigoplus_i R^i$.
- Filtrations: $F^1 \subset F^2 \subset \cdots$ with $F^i F^j \subseteq F^{i+j}$

An algebra equipped with a grading is a **graded algebra**, and similarly an algebra equipped with a filtration is a **filtered algebra**.
:::

:::{.remark}
Note that 

- $k[x_1, \cdots, x_n]$ is graded (by monomials of uniform degree) and filtered (by polynomials of a bounded degree)
- $T(V)$ is graded and filtered, since multiplying a pure $p$ tensor with a pure $q$ tensor yields a pure $p+q$ tensor
- $\Cl(V)$ is a quotient of $T(V)$, but one can't simply define $\Cl(V, \cdot)^i = \im T(V)^i$ since the relations have mixed degree: for example $e_1^2 = -1$
  So $\Cl(V)$ isn't graded, but is still filtered: take the filtration $F$ on $T(V)$ defined by $F^i \da \bigoplus _{j\leq  i} V^{\tensor j}$ and descend it through the quotient map.
  The relations can only decrease degree, so this is well defined.
:::

:::{.definition title="Filtration on the Clifford Algebra"}
Define a filtration $F^\wait$ on $\Cl(V)$ by the following:
\[
F^i \Cl(V) \da \spanof \ts{ \elts{e_j}{i} } 
.\]
:::

:::{.definition title="The associated graded"}
The **associated graded** ring $\gr_{F^\wait} R$ is the graded ring defined by 
\[
(\gr_{F^\wait})^i \da F^i R / F^{i-1} R
.\]
This induces a decomposition
\[
\gr_{F^\wait} \cong \bigoplus _{i\geq 0} F^i R/ F^{i-1} R = \bigoplus _{i\geq 0} (\gr_{F^\wait})^i
,\]
which has a multiplicative structure
\[
F^i/F_{i-1} \cdot F^j/F_{j-1} \to F^{i+j} / F^{i+j-1}
.\]
:::

:::{.remark}
Note that if $R \in \gr\Ring$, then $\gr(R) = R$, so taking the associated graded recovers the ring itself.
What's happening: taking the smallest homogeneous ideal.
:::

:::{.fact}
If one has relations of mixed degree, the associated graded also has the top degree part of each relation.
:::

:::{.remark}
In our case, the Clifford relation relates degree $k$ pieces to degree $k-2$ pieces, so we obtain
\[
\gr_{F^\wait} \Cl(V) \cong T(V) / \gens{ v\tensor w + w\tensor v = 0 } \da \Extalg^* V
.\]
There is an isomorphism of $k\dash$vector spaces
\[
\Cl(V) & \mapsvia{\sim} \gr \Cl(V) \\
x\in F^i &\mapsto \bar{x} \in F^i / F^{i-1}
.\]
This is because $F^0 \subseteq \cdots \subseteq \cdots$ with $\union_i F^i = \Cl(V)$.
We can conclude $\dim_\RR \Cl(V) = \dim_\RR \Extalg^* V = 2^{\dim_k V}$ and use this to construct a basis for $\Cl(V)$.
The relevant map is 
\[
\elts{e_j}{i} &\mapsto e_{j_1} \wedge \cdots \wedge e_{j_i}
.\]

:::

:::{.corollary title="of the fact"}
The following set forms an $\RR\dash$basis for $\Cl(\RR^n, \cdot)$:
\[
\ts{ \elts{e_j}{i} \st j_1 < j_2 < \cdots < j_i,\, i\leq n } 
.\]
:::

:::{.example title="?"}
Consider
\[
\Cl(\RR^3, \cdot) \cong \spanof_\RR \ts{ 1, e_1, e_2, e_3, e_1e_2, e_1 e_3, e_1 e_2 e_3 } 
.\]
Then
\[
e_1 e_2 \cdot e_1 e_3 
&= -e_1 e_1 e_3 e_3 
  && e_2 e_1 = -e_1 e_2 \\
&= e_2 e_3 
  && e_1^2 = - 1
.\]
:::

:::{.exercise title="?"}
Show that $\Cl(\RR^3) \cong \HH \oplus \HH$.
:::

:::{.definition title="Even and odd parts of the Clifford algebra"}
$\Cl(V)$ has a $\ZZ/2$ ("super") grading, so
\[
\Cl(V) \circ \Cl_0(V) \oplus \Cl_1(V) && \Cl_i(V) \cdot \Cl_j(V) \subset \Cl_{i+j\mod 2}(V)
.\]
The **even** subalgebra is given by 
\[
\Cl_0(V) = \spanof_k \ts{ \elts{e_i}{2k} \st 2k\leq n } 
,\]
where we take an even number of basis elements, which makes sense because the Clifford relation $vw + 2v = -2v\cdot w$ preserves degree mod 2.
This is still an algebra.
The **odd** sub-vector space (not an algebra) is given by
\[
\Cl_1(V) = \spanof_k \ts{ \elts{e_i}{2k+1} \st 2k+1\leq n } 
.\]

:::

:::{.example title="?"}
\[
\Cl(\RR^3) = \spanof_\RR \ts{ 1, e_1 e_2, e_1 e_3, e_2 e_3 } 
,\]
and we saw $e_1 e_2 = e_1 e_3 = e_2 e_3$.
This product has degree 4, and when we applied the relation $e_1^2=1$ we dropped the degree by 2.
For the odd part, $e_3 \in Cl_1(\RR^3)$ and $e_1 e_2 \in \Cl_0(\RR^3)$, and we have
\[
e_3 \cdot (e_1 e_2) = -e_1 e_3 e_2 = e_1 e_2 e_3 \in \Cl_1(\RR^3)
.\]
:::

:::{.proposition title="Decomposing the Clifford algebra of V"}
\[
\Cl(V) \cong \Cl_0(V \oplus \RR)
.\]
:::

:::{.proof title="?"}
Let $e\in \RR$ be a unit vector.
Given $x\in \Cl(V)$, decompose $x = x_0 + x_1 \in \Cl_0(V) \oplus \Cl_1(V)$.
Define an isomorphism
\[
\phi: \Cl(V) &\to \Cl_0(V \oplus \RR) \\
x &\mapsto x_0 + x_1 e
,\]
which is well-defined since $x_0$ was odd degree, and both $x_1, e$ were odd degree and thus $x_1 e$ is even.
One checks that this preserves multiplication:
\[
x\cdot y = (x_0 + x_1) \cdot (y_0 + y_1) = (x_0 y_0 + x_1 y_1) + (x_0 y_1 + x_1 y_0) \in \Cl_0(V) \oplus \Cl_1(V)
,\]
and so
\[
\phi(x) \cdot \phi(y) 
&= (x_0 +x_1 e)(y_0 + y_1 e) \\
&= x_0 y_0 + x_0 y_1 e + x_1 e y_0 + x_1 e y_1 e_1
.\]
The question is if this equals
\[
\phi(xy) \da (x_0 y_0 + x_1 y_1) + (x_0 y_1 + x_1 y_0)e
.\]
But for example, $x_1 e y_0 = (-1)^{|y_0|} x_1 y_0 e$, and $y_0$ is even.
Similarly, $x_1 e y_1 e = -x_1 y_1 e^2 = x_1 y_1$.
:::