# Tuesday, September 07

## Intro: Historical POV on Brauer Groups

:::{.remark}
Last time we defined $\Br(k) \da H^2(k; k\units)$ and had a SES
\[
1 \to (k^s)\units \to \GL_n(k^s)\to \PGL_n \to 1
.\]
We identified a subset of $\PGL_n\dash$torsors in $H^1(k; \PGL_n(k^s)) \mapsvia{\iota_n} H^2(k; (k^s)\units)$, and alternatively defined $\Br(k) = \union_n \im(\iota_n)$.
We'll now look at geometric interpretations of elements of $H^1$.
:::

:::{.example title="?"}
$\Aut(X) = \PGL_n$ for the following:

- $\PP^{n-1}$
- $\GL_n$
- $\Mat(n\times n)$, by the Skolem-Noether theorem.
:::

:::{.corollary title="?"}
For any of the $X$ above, there is an isomorphism:
\[
H^1(k; \PGL_n(k^s)) 
\mapsvia{\sim} \ts{\text{Forms of $X$}}\slice{\sim}
\mapsvia{\sim} 
\ts{\PGL_n\dash\text{torsors}}\slice{\sim}
.\]

:::

:::{.definition title="Severi-Brauers"}
A **Severi-Brauer** variety over $k$ is a form of $\PP^n\slice k$ for some $n$.
:::

:::{.example title="?"}
\envlist

- $C$ a conic with no rational points, e.g. $x^2 + y^2 + z^2 = 0$ over $\RR$.
- $\Sym^n C$ is a nontrivial Severi-Brauer if $n$ is odd.
  It's difficult to write any down for even $n$, e.g. there are no Severi-Brauer surfaces over $\RR$.

:::

:::{.definition title="CSAs/Azumaya Algebras"}
A finite dimensional **central simple algebra** or **Azumaya algebra** over $k$ is a associative algebra over $k$ with no nontrivial 2-sided ideals with center $k$ which is finite-dimensional as a $k\dash$vector space.
:::

:::{.theorem title="Classification of CSAs"}
Let $A\in \Alg\slice k$, then TFAE:

- $\exists$ a finite separable extension $L \slice k$ where after base-changing to $L$ one obtains $A\tensor_k L \cong \Mat(n\times n, L)$.
- $A\tensor_k k^s \cong \Mat(n\times n, k^s)$.
- $\exists$ a finite (not necessarily separable) extension $L\slice k$ such that $A\tensor_k L \cong \Mat(n\times n, L)$.
- $A$ is a finite dimensional central simple algebra / Azumaya algebra.
- $A$ is a matrix algebra over a finite-dimensional central $k\dash$division algebra.

This is essentially a classification theorem: they're all forms of matrix algebras over division algebras.
Moreover there is a bijection
\[
\ts{\text{Central simple $k\dash$algebras }} \to H^2(k; (k^s)\units)
.\]
:::

:::{.definition title="Opposite algebra"}
If $A \in \CSA\slice k$, then $A\op\in \CSA\slice k$ is an algebra with the same underlying vector space as $A$ with $a \cdot_{\op} b \da ba$.
:::

:::{.definition title="Morita equivalence"}
$A, B$ are Morita equivalent if $A\tensor_k B\op$ is isomorphic to a matrix algebra.
:::

:::{.theorem title="?"}
Given $A, B \in \CSA \slice k$ which correspond to elements $[A], [B]\in H^2$, then

- $[A] = [B] \iff A, B$ are Morita equivalent.
- $[A]\inv = [A\op]$.
- $[A] \cdot [B] = [A\tensor_k B]$.

:::

## The Boundary Map and Twisted Vector Space

:::{.remark}
We'd now like to make the boundary map explicit:
\[
H^1(k; \PGL_n(k^s)) \to H^2(k; (k^s)\units)
.\]
Given $[f] \in H^1$, choose a representable cocycle $f$:

\begin{tikzcd}
	{\Gal(k^s/k)} && {\PGL(k^s)} \\
	\\
	{\Gal(L/k)} && {\PGL_n(L)}
	\arrow["f", from=1-1, to=1-3]
	\arrow[from=1-1, to=3-1]
	\arrow[hook, from=3-3, to=1-3]
	\arrow["{\tilde f}"', from=3-1, to=3-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJcXEdhbChrXnMvaykiXSxbMiwwLCJcXFBHTChrXnMpIl0sWzAsMiwiXFxHYWwoTC9rKSJdLFsyLDIsIlxcUEdMX24oTCkiXSxbMCwxLCJmIl0sWzAsMl0sWzMsMSwiIiwyLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbMiwzLCJcXHRpbGRlIGYiLDJdXQ==)

To compute this boundary, we use the original SES:

\begin{tikzcd}
	1 && {k^s} && {\GL_n(k^s)} && {\PGL_n(k^s)} && 1 \\
	\\
	&&&&&& {\Gal(k^s/k)}
	\arrow[from=1-1, to=1-3]
	\arrow[from=1-3, to=1-5]
	\arrow[from=1-5, to=1-7]
	\arrow[from=1-7, to=1-9]
	\arrow["f", from=3-7, to=1-7]
	\arrow["{\text{Choose a set-theoretic lift } \tilde f}"{pos=0.8}, dashed, from=3-7, to=1-5]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMCwwLCIxIl0sWzIsMCwia15zIl0sWzQsMCwiXFxHTF9uKGtecykiXSxbNiwwLCJcXFBHTF9uKGtecykiXSxbOCwwLCIxIl0sWzYsMiwiXFxHYWwoa15zL2spIl0sWzAsMV0sWzEsMl0sWzIsM10sWzMsNF0sWzUsMywiZiJdLFs1LDIsIlxcdGV4dHtDaG9vc2UgYSBzZXQtdGhlb3JldGljIGxpZnQgfSBcXHRpbGRlIGYiLDAseyJsYWJlbF9wb3NpdGlvbiI6ODAsInN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dXQ==)

So $\tilde f: \Gal(k^s/k) \to \GL_n(k^s)$ is a lieft of $f$, and $\delta f$ measures the failure of $\tilde f$ to be a cocycle.
We have 
\[
\delta \tilde f(\sigma, \tau) = \tilde f(\sigma \tau) \qty{ \tilde f( \sigma) \tilde f (\tau)^{ \sigma} }\inv \in (k^s)\units
,\]
using exactness since for $f$ it lands in $\PGL_n$ and is trivial.
:::

:::{.definition title="Twisted vector spaces"}
For $L\slice k$ a separable extension and $\alpha: G\cartpower{2} \to L\units$ a 2-cocycle, so $[\alpha] \in H^2(L\slice k; L\units)$, a **twisted vector space** is a twisted semilinear action of $\Gal(L\slice k)$ on $L^n$.
I.e. it is a map
\[
\tilde f: \Gal(L\slice k) &\to \Aut(L^n) = \GL_n(L) \\
\text{such that } \tilde f( \sigma \tau) &= \tilde f( \sigma) \tilde g( \tau)^{ \sigma} \alpha( \sigma, \tau)
.\]

:::

:::{.remark}
For each \( \sigma\in \Gal(L\slice k) \) we get a $\sigma\dash$semilinear automorphism of $L^n$, i.e. a map 
\[
f_\sigma: L^n &\to L^n \\
\text{where } f_ \sigma(s \cdot v) &= \sigma(s) \cdot f_{ \sigma}(v)
,\]
which is just the definition of semilinearity, and moreover $f_{ \sigma \tau} = f_{ \sigma} f_{ \tau} \alpha( \sigma, \tau)$.
:::

:::{.remark}
If $\alpha = \id$, an $\alpha\dash$twisted vector space is the same as a $k\dash$vector space by Galois descent.
:::

:::{.proposition title="Properties of categories of twisted vector spaces"}
\envlist

1. $\alpha \in \im\qty{ H^1(k; \PGL_n(k^s)) \to H^2(k; (k^s)\units ) } \iff$ there exists an $n\dash$dimensional $\alpha\dash$twisted vector space.
  
    The proof of this is just unwinding definitions, it's literally the same data!

2. The category $\Tw_\alpha$ of $\alpha\dash$twisted vector spaces is abelian -- the only nontrivial thing to check is that there are enough injectives.

3. There are natural functors
\[
(\wait) \tensor (\wait) &: 
\Tw_\alpha \times \Tw_{\alpha'} \to \Tw_{\alpha \alpha'} \\
\Hom(\wait, \wait)&:
(\Tw_\alpha)^{\op} \times \Tw_{\alpha'} \to \Tw_{\alpha' \alpha\inv} \\
\Sym^n, \Extalg^n &:
\Tw_\alpha \to \Tw_{\alpha^n}
.\]

4. If $F\slice k$ is a separable field extension, then 
\[
(\wait)\tensor F: \Tw_{\alpha}{}\slice k \to \Tw_{\alpha}{}\slice F
.\]

5. There is an equivalence of categories
\[
\Tw_{\id}{}\slice k \mapsvia{\sim} \mods{k}
.\]


:::

:::{.proposition title="?"}
There is a 1-dimensional $\alpha\dash$twisted vector space iff $[\alpha] = 1\in H^1(k; (k^s)\units)$.
:::

:::{.proof title="?"}
$\impliedby$:
First suppose $\alpha \equiv 1$, then $\Tw_\alpha \mapsvia{\sim} \Vect\slice k$, so just take the vector space $k$.
If $\alpha = \delta g$ for some $g: \Gal(k^s/k) \to (k^s)\units$.
Then the action $\Gal(k^s/k)\actson k^s$ where $f_\sigma = g(\sigma)$ is a 1-dimensional $\alpha\dash$twisted vector space by sending $1\to g(\sigma)$ and extending semilinearly.

$\implies$:
Let $V$ be a 1-dimensional $\alpha\dash$twisted vector space.
Choose an isomorphism $V \mapsvia{\sim} k^s$
For each \( \sigma\in \Gal(k^s/k) \) set $g(\sigma) = g(1)$ and $g( \sigma \tau) = g( \sigma) g( \tau)^{ \sigma} \alpha( \sigma, \tau)$, then 
\[ 
\alpha= \delta g = g( \sigma \tau )\qty{g( \sigma) g( \tau)^{ \sigma} }\inv  
.\]
:::

:::{.theorem title="?"}
Suppose \( \alpha\in H^2(k; (k^s)\units) \) is in \( \im \qty{ H^1(k; \PGL_n) \to H^2(k; (k^s)\units )} \), then $\alpha^n = 1$. 
:::

:::{.proof title="?"}
If $\alpha$ is in the image, there exists an $n\dash$dimensional $\alpha\dash$twisted vector space $V\in \Tw_\alpha$, and so $\Extpower^n V \in \Tw_{\alpha^n}$.
:::

:::{.definition title="Index and period"}
Given $H^2(k; (k^s)\units) = \Br(k)$ (which we'll prove soon), the **period** of $\alpha$ is the order of $\alpha$, and the **index** is defined the minimal $n$ such that $\alpha$ is in the above image.
I.e.,
\[
\period(\alpha) &\da \ord(\alpha) \\
\mathrm{index}(\alpha) &\da \min\ts{ n \st \alpha \in \im( H^1 \to H^2)}
.\]
:::

:::{.corollary title="?"}
Period divides index.
:::

:::{.question}
An open question: how different are the period and index?
See the period-index problem.
:::

:::{.remark}
There are some maps between the categories $\Tw_\alpha$, $\SB$ (Severi-Brauers), and $\CSA$:

\begin{tikzcd}
	&&& V \\
	&&& {\Tw_\alpha} \\
	{\PP(V)} &&&&& {\Endo_k(V)} \\
	{} & \SB &&& \CSA
	\arrow[from=2-4, to=4-5]
	\arrow[from=2-4, to=4-2]
	\arrow[dashed, maps to, from=1-4, to=3-1]
	\arrow[dashed, maps to, from=1-4, to=3-6]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNyxbMywxLCJcXFR3X1xcYWxwaGEiXSxbMSwzLCJcXFNCIl0sWzQsMywiXFxDU0EiXSxbMywwLCJWIl0sWzAsM10sWzAsMiwiXFxQUChWKSJdLFs1LDIsIlxcRW5kX2soVikiXSxbMCwyXSxbMCwxXSxbMyw1LCIiLDIseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJtYXBzIHRvIn0sImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dLFszLDYsIiIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Im1hcHMgdG8ifSwiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV1d)

An analogy is that in vector spaces, $\PP^n$ is to $\Endo(V)$ as $\SB$ is to $\CSA$ in twisted vector spaces.
Note that $\Gal(L\slice k) "\actson" V$, which isn't a true action but only fails to be one up to a scalar.
Thus projectivizing yields a semilinear action $\Gal(L\slice k)\actson \PP(V)$, and Galois descent yields forms of $\PP(V)\slice k$.
:::

:::{.remark}
Why is $\Endo(V)$ a form of $\Mat(n\times n)$?
Since $V\in \Tw_\alpha$, split it: choose an $L$ such that $\ro{ \alpha}{L}$ is trivial.
Then $\Tw_{\ro \alpha L}= \Vect\slice L$.
:::