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+\input{preamble.tex}
+\title{Galois Cohomology of Algebraic Groups}
+\author[Ayushi Tsydendorzhiev]{Ayushi Tsydendorzhiev}
+\usepackage{enumitem}
+\usepackage{stmaryrd}
+\usepackage{xhfill}
+\usepackage{array}
+\usepackage{outlines}
+\usepackage{faktor}
+\usepackage{float}
+\usepackage{mathtools}
+\usepackage{tikz-cd}
+\usepackage{centernot}
+\usepackage{cancel}
+\newcommand{\ra}{\rightarrow}
+\newcommand{\lra}{\longrightarrow}
+\newcommand{\eps}{\varepsilon}
+\renewcommand{\P}{\mathbb{P}}
+\newcommand{\acts}{\curvearrowright}
+\newcommand\Gal[1]{\operatorname{Gal}({#1})}
+\renewcommand{\phi}{\varphi}
+
+\begin{document}
+
+\maketitle
+
+\tableofcontents
+\clearpage
+
+\section{Introduction}
+
+\subsection{Galois group actions}
+\marginnote{Lecture 1, 10.10.2024}
+\begin{outline}
+\0 Let $L/K$ be a Galois extension and $G=\Gal{L/K}$ its Galois group. The Galois group $G$ acts on $L$ via field automorphisms:
+    \1 Action on the field extension $L$: For $\Q(\sqrt{2})$ its Galois group $\Gal{\Q(\sqrt{2})/\Q}$ acts either by identity or by sending $\sqrt{2}$ to $-\sqrt{2}$.
+    \1 Action on the dual of the field extension $L^*$: For $\Q(\sqrt{2})^*$ its Galois group acts on $f(x_1,x_2)=x_1\cdot 1 + x_2 \cdot \sqrt{2}$ either by identity or by sending $f$ to $f'=x_1\cdot 1 -x_2 \cdot \sqrt{2}$.
+    \1 Action on the group of $n$th roots of unity $\mu_n(L)$: 
+        \2 In $\Q(\sqrt{2})$, the $n$th roots of unity consist of $\{-1,1\}$ if $n$ is even and $\{1\}$ if $n$ is odd. Both automorphisms in $\Gal{\Q(\sqrt{2})/\Q}$ leave $\mu_n(\Q)$ fixed, so this tells us that they all belong to the base field (are rational, in this case). 
+        \2 A more interesting example is the $n$th cyclotomic field $\Q(\zeta_n)$.In this field $\mu_n (Q(\zeta_n))=\langle \zeta_n \rangle$, the cyclic group generated by $\zeta_n$. The Galois group $\Gal{Q(\zeta_n)/Q}$ is isomorphic to $(\Z/n\Z)^*$. For $n=5$ (prime), the Galois group is cyclic and consists of $\{1, \zeta_5, \zeta_5^2, \zeta_5^3, \zeta_5^4\}$. The action of the Galois group then permutes the $5$th roots of unity. For $n=8$, the Galois group $\Gal{\Q(\zeta_8)/\Q}$ is isomorphic to $(\Z/8\Z)^*=\{1,3,5,7\}$ and is cyclic of order $4$. The basis of $\Q(\zeta_8)$ over $\Q$ is given by $\{1,\zeta_8,\zeta_8^2, \zeta_8^3\}$. The actions is given as: $\sigma_1$ acts trivially, $\sigma_3$ maps $\zeta_8$ to $\zeta_8^3$, $\sigma_5$ acts by multiplication by $-1$ and $\sigma_7$ maps $\zeta_8$ to $\zeta^7_8$.
+    \1 Action on the cyclic group $(\Z/n\Z)^*$: same as above.
+    \1 Action on a finite abelian group $M$: trivial action.
+    \1 Action on the general linear group $\GL_n(L)$ over a field $L$ of characteristic $0$: $\GL_n(L)$ consists of $n\times n$ invertible matrices over $L$. We have a Galois extension $L/K$. The Galois group acts by applying the field automorphisms to the entries of the matrices, so $\sigma(A)=\sigma(a_{ij}) \forall 1\leq ij \leq n$. The fixed points contain $\GL_n(K)$.
+        \2 Backstory: The determinant of a $n\times n$ matrix $A$ is defined as 
+        \marginnote{$\sgn(\pi)$ is either even or odd. $+1$ if even and $-1$ if odd.}
+        $$\det(A) = \sum_{\pi \in S_n} \left( \sgn(\pi) \prod_{i=1}^n a_{i,\pi(i)} \right)$$
+        Consider $\sigma(\det (A))$, where $\sigma \in \Gal{L/K}$ is a field automorphism. It distributes over addition and multiplication:
+        $$\sigma(\det(A)) = \sum_{\pi \in S_n} \left( \sgn(\pi) \prod_{i=1}^n \sigma(a_{i,\pi(i)}) \right)$$
+        The signum is either $+1$ or $-1$, so it is always in the base field $K$ and is fixed by $\sigma$. Thus $\sigma(\det(A))=\det(\sigma(A))$. So the action of the Galois group preserves determinants.
+\end{outline}
+
+\subsection{The fixed point functor and exact sequences}
+
+\begin{outline}
+\0 All of these examples are special cases of a more general concept: a group $G$ acting on an algebraic group $\bG \subseteq \GL_n$.
+\marginnote{An algebraic group is a matrix group defined by polynomial conditions, at least this is what \enquote{The theory of group schemes of
+finite type over a field.} by Milne says. I guess this is the consequence of Chevalley theorem?}
+
+When studying group actions, we're often interested in fixed points 
+$$A^G =\{a\in A \mid \forall \sigma \in G: \sigma a = a \}$$ 
+Here, $A^G$ represents the set of all elements in $A$ that are fixed by every element of $G$. To study fixed points more systematically, we introduce the fixed point functor $-^G$. This functor takes a $\Z G$-module  and returns its fixed points. We're particularly interested in how this functor behaves with respect to exact sequences.
+
+\begin{note}
+~\\
+\textbf{Group action perspective}: A $\Z G$-module is an abelian group $A$ endowed with a (left) action $(\sigma, a) \mapsto \sigma a$ of $G$ on $A$ such that for all $\sigma\in G$ the map $\phi_\sigma : a \mapsto \sigma a$ from $A$ to $A$ is a morphism of abelian groups. This implies that the action of $G$ is distributive, $\phi_\sigma (ab) = \phi_\sigma (a) + \phi_\sigma (b)$.\\
+\textbf{Ring module perspective}: Equivalently, a $\Z G$-module is a module over the group ring $\Z[G]$, where elements consist of formal linear combinations of elements from group $G$ with integer coefficients, so something like $3g_1+4g_2+10g_3 \in \Z[G]$. It contains both $\Z$ and $G$ as subrings.\\
+The $\Z[G]$-module structure encapsulates both the abelian group structure of $A$ and the $G$-action on $A$, which leads to the key insight:
+$$\{\text{module over } \Z[G]\} \leftrightarrow \{\text{abelian group $A$ with $G$-action}\}$$
+\end{note}
+
+\begin{lemma}
+    Consider an exact sequence of $\Z G$-modules:
+    $$\begin{tikzcd}
+    0\arrow[r] & A \arrow[r, "f"] & B\arrow[r, "g"] & C\arrow[r, "h"] & 0
+    \end{tikzcd}$$
+    Applying the fixed point functor $-^G$ to this sequence yields:
+    $$\begin{tikzcd}
+    0\arrow[r] & A^G \arrow[r, "f^G"] & B^G \arrow[r, "g^G"] & C^G
+    \end{tikzcd}$$
+    This new sequence is exact in Ab (the category of abelian groups). Thus the functor $-^G$ is left-exact, meaning it preserves exactness at the left end of the sequence.
+\end{lemma}
+
+    \1 A natural question arises: Is the fixed point functor also right-exact? If such a lifting always exists, then the fixed point functor preserves exactness at $C$, making it right-exact.
+    If not, we've discovered an obstruction that tells us something about the Galois action and the structure of our groups.
+    \1 To investigate this, we need to check if $\ker h^G = \im g^G$, or equivalently, if $\im g^G = C^G$. Breaking this down:
+        \2 Take any $c \in C^G$. 
+        \2 Since $C^G \subseteq C$, there exists a $b \in B$ such that $g(b) = c$.
+        \2 If $b$ were fixed by $G$, we'd be done. But it might not be. 
+        \marginnote{Why $\sigma b = b$?}
+            \3 Consider $\sigma b - b$ for any $\sigma \in G$. We have $g(\sigma b - b) = g(\sigma b) - g(b)= \sigma g(b)-g(b)=\sigma c - c$. 
+            \3 Since $c\in C^G$, $\sigma c - c = 0$ and $(\sigma b-b)\in \ker g$.
+            \3 By exactness, $\ker g = \im f$, so $\sigma b - b \in \im f$.
+            \3 We can view this as an element of $A$ (considering $f$ as an inclusion $A \subseteq B$). 
+    \1[] So the question of right-exactness boils down to whether or not every $G$-invariant element of $C$ can be lifted to a $G$-invariant element of $B$ and the obstruction to it lives inside $A$.
+    \1 This analysis leads us to define a map (for a given $c\in C^G$):
+    $$\phi: G \to A, \quad \sigma \mapsto \sigma b - b =: a_\sigma$$
+    This map is called a crossed homomorphism (also known as a derivation or 1-cocycle). It measures how far $b$ is from being $G$-invariant. If $b$ were $G$-invariant, this map would be identically $0$.
+
+\begin{proposition}
+    The map $\sigma \mapsto a_\sigma$ satisfies:
+    $$a_{\sigma\tau} = a_\sigma + \sigma a_\tau$$
+    This property is what defines a crossed homomorphism.
+\end{proposition}
+
+    \1 \textbf{In the abelian case}, we define 
+        \2 $Z^1(G,A)=\{a':G\ra A \mid a'_{\sigma \tau} = a'_{\sigma}+\sigma a'_{\tau}\}$, the set of all crossed homomorphisms from $G$ to $A$.
+        \2 $B^1(G,A)=\{a:\sigma \in Z^1(G,A) \mid \exists a'\in A : a_\sigma = \sigma a' - a'\}$. 
+        \2 The quotient $H^1(G,A) = Z^1(G,A) / B^1(G,A)$ is called the \textbf{first cohomology group} of $G$ with coefficients in $A$. It measures the obstruction to the right-exactness of the fixed point functor.
+        \marginnote{The functor $A\mapsto H^1(G,A)$ is a derived functor of the $A\mapsto A^G$ functor.}
+
+    \1[] The obstructions for right-exactness: find $\sigma b -b \in A$ such that it is $0$ under projection in $Z^1(G,A)/B^1(G,A)$. It is given by $\delta(c) = [a_\sigma]\in H^1(G,A) = Z^1(G,A)/B^1(G,A)$. We can extend our original sequence to a longer exact sequence:
+    $$\begin{tikzcd}[column sep=small]
+    0 \arrow[r] & A^G \arrow[r] & B^G \arrow[r] & C^G \arrow[r, "\delta"] & H^1(G,A) \arrow[r] & H^1(G,B) \arrow[r] & H^1(G,C) \arrow[r] & 0
+    \end{tikzcd}$$
+    This sequence is exact in Ab, and the map $\delta$ (called the connecting homomorphism) measures the failure of right-exactness of the fixed point functor.
+
+\begin{exercise}
+    Show that $H^1(G,-)$ is functorial and 
+    $$\begin{tikzcd}[column sep=small]
+    0 \arrow[r] & A^G \arrow[r] & B^G \arrow[r] & C^G \arrow[r] & H^1(G,A) \arrow[r] & H^1(G,B) \arrow[r] & H^1(G,C) \arrow[r] & 0
+    \end{tikzcd}$$
+    is exact. Find example with $\delta \neq 0$.
+\end{exercise}
+
+    \1 \textbf{In the non-abelian case}, we define 
+        \2 $H^0(G,A)=A^G$, the fixed points as before. 
+        \2 $H^1(G,A)=Z^1(F,A)/\sim$, where $\sim$ is an equivalence relation defined by: $a_\sigma \sim b_\sigma \iff \exists a' \in A : b_\sigma = (a')^{-1} \cdot a_\sigma \cdot \prescript{\sigma}{}{a'}$. 
+        \marginnote{We cannot expect $B^1(G,A)$ to be a subgroup. Why?}
+        \marginnote{$\prescript{\sigma}{}{a}$ denotes the action of $\sigma$ on $a$.}
+    \1[] In this case, $H^1(G,A)$ doesn't have a group structure, but it's a pointed set (a set with a distinguished element). We can still define a notion of exactness for sequences of pointed sets.
+\marginnote{Exactness in pointed sets $(A,*)$ is defined as $\im f = \ker g = g^{-1}(*)$}
+
+\marginnote{$A\leq_G B$ is $G$-equivariant inclusion.}
+\begin{proposition}
+    For $A\leq_G B$, we obtain $G\acts B/A$ and 
+    $$\begin{tikzcd}[column sep=small]
+    1 \arrow[r] & H^0(G,A) \arrow[r] & H^0(G,B) \arrow[r] & H^0(G,C) \arrow[r] & H^1(G,A) \arrow[r] & H^1(G,B)
+    \end{tikzcd}$$
+    is exact.
+\end{proposition}
+
+This is the \textbf{Galois cohomology}. Why do we care? In the non-commutative case $H^1(G,A)$ classifies \enquote{K-objects}. In our lecture we will use this to classify simple and simply connected linear algebraic $k$-groups $\bG$.
+
+\end{outline}
+
+\marginnote{Lecture 2, 17.10.24}
+
+\end{document}
+

preamble.tex

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