Merge overleaf-2024-11-08-1441 into main

main.tex

@@ -133,6 +133,13 @@ \subsection{The fixed point functor and exact sequences}
     is exact. Find example with $\delta \neq 0$.
 \end{exercise}
 
+    \1 \textit{Solution:} Consider $\Z_2 = \{e,\sigma\}$. For a function $f:\Z_2\ra \Z$ the cocycle condition states $f(\sigma\tau)=f(\sigma)+\sigma f(\tau)$ for $\sigma,\tau \in \Z_2$. When $\sigma = \tau = e$, we get $f(e)=f(e e)=f(e)+ef(e)= 2f(e)$, implying $f(e)=0$. When $\sigma=\tau$ we get $f(\sigma \sigma)=f(\sigma)+\sigma f(\sigma)$. Since $\sigma^2 = e$, we get $f(\sigma)=-\sigma(f(\sigma))$. Since $\sigma$ acts by negation, we get $f(\sigma)=f(\sigma)$, so there is really no condition on $\sigma$. Each integer gives a different cocycle. Lets calculate coboundaries now. We have the coboundary condition $f(g)=g(a)-a$ for some $a\in\Z$. So we have $f(e)=e(a)-a=a-a=0$ and $f(\sigma)=\sigma(a)-a=-a-a=-2a$. So every coboundary has form $e\mapsto 0$, $\sigma \mapsto -2a$. This implies that $H^1(\Z_2;\Z)=\Z/\Z_2$. Alternatively we can look at the left resolution of $\Z/2\Z$ and compute by hand.
+
+    \1 \textit{Solution 2:} Let's consider a simple elliptic curve $E$ over $\Q$, $f(x,y)=y^2=x^3-x$. it has an obvious $2$-torsion point $(0,0)$. For an elliptic curve $E$, its quadratic twist $E^d$ is another elliptic curve that becomes isomoprhic to $E$ over the quadratic extension $\Q(\sqrt{d})$ but is not isomorphic to $E$ over $\Q$. It is given by $dy^2=x^3-x$. The isomorphism is given by $E\ra E^d, (x,y)\mapsto (x,\sqrt{d}y)$ as we have $dy^2=x^3-x \mapsto d(\frac{y}{\sqrt{d}})^2=x^3-x$ which is equivalent to $y^2=x^3-x$ over $\Q$ (if we can multiply by $\sqrt{d}$, we can transform one equation into the other). The practical use is that over $\Q(\sqrt{d})$ we might get new torsion points, and Galois group $\{1,\sigma\}$ acts on these points by sending $\sqrt{d}$ to $-\sqrt{d}$. This tells us about how different the twist is from the original curve. 
+        \2 We have $E(\Q(\sqrt{2}))^G=E(\Q)$, the fixed points on $E(K)$ are precisely the $\Q$-rational points (both coordinates in $\Q$). For any elliptic curve $E$ we have a short exact sequence $0\ra E[n] \ra E \xrightarrow[]{\times n} E\ra 0$. Applying the fixed point functor $(-)^G$ to it gives us the long exact sequence 
+        $$0\ra E[2]^G \ra E(K)^G \xrightarrow[]{\times n} E(K)^G \ra H^1(G,E[2])\ra H^1(G,E(K))\xrightarrow[]{\times 2} H^1(G,E(K))\ra 0$$
+        Since $E[2]=\{(0,0),(1,0),(-1,0),\infty\}$, we have $E[2]^G=E[2]$.
+
     \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: $a_\sigma \sim b_\sigma \iff \exists a' \in A : b_\sigma = (a')^{-1} \cdot a_\sigma \cdot \prescript{\sigma}{}{a'}$.