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# 数学代写|同调代数代写Homological Algebra代考|MATH4301 Degree One Group Cohomology

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## 数学代写|同调代数代写Homological Algebra代考|Degree One Group Cohomology

Already for $H^1(G, M)$ one needs some computation. An obvious starting point for the projective resolution of $R$ is
$$0 \longrightarrow I(R G) \stackrel{\iota}{\longrightarrow} R G \longrightarrow R \longrightarrow 0$$
and the rightmost mapping sends all $g \in G$ to 1 . The kernel of this mapping is $I(R G)$, the augmentation ideal, generated by $1-g \in R G$ for all $g \in G$, as a left ideal. Put
$$\operatorname{Der}(G, M)={f: G \longrightarrow M \mid f(g h)=g f(h)+f(g) \forall g, h \in G},$$
the derivations of $G$ with values in $M$, and
$$\operatorname{Inn} \operatorname{Der}(G, M):={f: G \longrightarrow M \mid \exists m \in M: f(g)=g m-m},$$
the inner derivations of $G$ with values in $M$. Then
$$H^1(G, M)=\operatorname{Der}(G, M) / \operatorname{Inn} \operatorname{Der}(G, M) .$$
Indeed, if
$$\cdots \longrightarrow P_2 \stackrel{\partial_1}{\longrightarrow} P_1 \stackrel{\partial_0}{\longrightarrow} R G \longrightarrow R$$
is a projective resolution, then
$$\operatorname{ker}\left(\operatorname{Hom}{R G}\left(\partial_1, M\right)\right)=\operatorname{Hom}{R G}(I(R G), M)$$
since the morphisms $\varphi: P_1 \longrightarrow M$ with $\varphi \circ \partial_1=0$ factor through
$$\operatorname{coker}\left(\partial_1\right)=\operatorname{ker}\left(\partial_0\right)=I(R G) .$$
Hence, since $g(h-1)=g h-g=(g h-1)-(g-1)$, we have
\begin{aligned} \operatorname{ker}\left(\operatorname{Hom}{R G}\left(\partial_1, M\right)\right)=& H m{R G}(I(R G), M) \ =&\left{\varphi \in \operatorname{Hom}_R(I(R G), M) \mid\right.\ &\forall g, h \in G: g \varphi(h-1)=\varphi(g h-1)-\varphi(g-1)} \ =&{f: G \longrightarrow M \mid\ &\forall g, h \in G: g \cdot f(h)=f(g h)-f(g)} \ =& \operatorname{Der}(G, M) . \end{aligned}

## 数学代写|同调代数代写Homological Algebra代考|Degree Two Group Cohomology

In order to compute $H^2(G, M)=E x t_{\mathbb{Z} G}^2(\mathbb{Z}, M)$ we need to give a projective resolution of the trivial module $\mathbb{Z}$ in a more explicit way, and in particular in a way that works for all groups $G$. For algebras in general this is the so-called bar resolution, which will be given in full detail and complete generality in Definition 3.6.4. However, up to degree 2 the resolution is sufficiently simple to be given immediately, in particular for group rings.
Using the fact that
$$0 \longrightarrow I(R G) \longrightarrow R G \longrightarrow R \longrightarrow 0$$
is an exact sequence where $R G$ is free, we need to find a projective $R G$-module $P$ and an epimorphism $P \longrightarrow I(R G)$. Let $P:=R G \otimes_R R G$ with the action of $R G$ by multiplication on the left term of the tensor product. This module is clearly projective, even free. We define
\begin{aligned} R G \otimes_R R G & \stackrel{\partial_0}{\longrightarrow} R G \ g_1 \otimes g_2 & \mapsto g_1\left(g_2-1\right) \end{aligned}
for all $g_1, g_2 \in G$. It is clear that this map is well-defined and surjective since $I(R G)$ is generated by $g-1$ for $g \in G$. Let $Q=R G \otimes_R R G \otimes_R R G$. This is a projective $R G$-module by multiplication on the left-most term. Let
\begin{aligned} R G \otimes_R R G \otimes_R R G & \stackrel{\partial_1}{\longrightarrow} R G \otimes_R R G \ g_1 \otimes g_2 \otimes g_3 \mapsto g_1 g_2 \otimes g_3-g_1 \otimes g_2 g_3+g_1 \otimes g_2 \end{aligned}

## 数学代写|同调代数代写同调代数代考|一次群同调

$$0 \longrightarrow I(R G) \stackrel{\iota}{\longrightarrow} R G \longrightarrow R \longrightarrow 0$$
，最右边的映射将所有$g \in G$发送到1。这个映射的内核是$I(R G)$，这是一个扩展理想，由$1-g \in R G$为所有$g \in G$生成，作为一个左理想。将
$$\operatorname{Der}(G, M)={f: G \longrightarrow M \mid f(g h)=g f(h)+f(g) \forall g, h \in G},$$

$$\operatorname{Inn} \operatorname{Der}(G, M):={f: G \longrightarrow M \mid \exists m \in M: f(g)=g m-m},$$

$$H^1(G, M)=\operatorname{Der}(G, M) / \operatorname{Inn} \operatorname{Der}(G, M) .$$

$$\cdots \longrightarrow P_2 \stackrel{\partial_1}{\longrightarrow} P_1 \stackrel{\partial_0}{\longrightarrow} R G \longrightarrow R$$

$$\operatorname{ker}\left(\operatorname{Hom}{R G}\left(\partial_1, M\right)\right)=\operatorname{Hom}{R G}(I(R G), M)$$

$$\operatorname{coker}\left(\partial_1\right)=\operatorname{ker}\left(\partial_0\right)=I(R G) .$$

\begin{aligned} \operatorname{ker}\left(\operatorname{Hom}{R G}\left(\partial_1, M\right)\right)=& H m{R G}(I(R G), M) \ =&\left{\varphi \in \operatorname{Hom}_R(I(R G), M) \mid\right.\ &\forall g, h \in G: g \varphi(h-1)=\varphi(g h-1)-\varphi(g-1)} \ =&{f: G \longrightarrow M \mid\ &\forall g, h \in G: g \cdot f(h)=f(g h)-f(g)} \ =& \operatorname{Der}(G, M) . \end{aligned}

## 数学代写|同调代数代写同调代数代考|度二群同调

$$0 \longrightarrow I(R G) \longrightarrow R G \longrightarrow R \longrightarrow 0$$

\begin{aligned} R G \otimes_R R G & \stackrel{\partial_0}{\longrightarrow} R G \ g_1 \otimes g_2 & \mapsto g_1\left(g_2-1\right) \end{aligned}
。很明显，这个映射是定义良好且满射的，因为$I(R G)$是由$g-1$为$g \in G$生成的。让$Q=R G \otimes_R R G \otimes_R R G$。这是一个投影$R G$模块，通过对最左边的项进行乘法运算。Let
\begin{aligned} R G \otimes_R R G \otimes_R R G & \stackrel{\partial_1}{\longrightarrow} R G \otimes_R R G \ g_1 \otimes g_2 \otimes g_3 \mapsto g_1 g_2 \otimes g_3-g_1 \otimes g_2 g_3+g_1 \otimes g_2 \end{aligned}

## MATLAB代写

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