We denote $M^G:=\{m\in m|gm=m,\forall g\in G\}$, then $M^G$ is an abelian group, and this define a functor $(.)^G:Mod_G\to Ab$ defined by $M\mapsto M^G$. Let
$$0\to M\to N\to P\to 0$$
be an exact sequence of $G$-modules, we then apply the functor $(.)^G$ to obtain the following exact sequence of abelian groups
$$0\to M^G\to N^G\to P^G$$
And this yields $(.)^G$ is a left exact functor.
Proposition 3.1. $Mod_G$ has enough injectives.
Proof. We recall that an abelian category $C$ is said to e enough injective if for all object $M$ in $C$, there exists a monomorphism $M\to I$ where $I$ is an injective object in $C$, i.e. the functor $Hom_C(_,I)$ is exact.
Let $M$ be a $G$-modules, and $M_0$ its underlying abelian group structure. We know that the category of abelian group has enough injectives, so there exists an injective object $I$ in $Ab$ such that $M_0\hookrightarrow I$, and since $Ind^G$ is an exact functor, we have $Ind^G(M_0)\hookrightarrow Ind^G(I)$, which is also an injective object in $Mod_G$, since $Ind^G$ is exact. And by Proposition 2.6, we can embed $M$ into $Ind^G(M_0)$. And this yields $Mod_G$ has enough injectives. (Q.E.D)
Proposition 3.2. Let $M$ be in $Mod_G$, then there exists an injective resolution of $M$, i.e. there exists an exact sequence of $G$-modules
$$0\to M\to I^0\to I^1\to...$$
of $G$-modules, and $I^i$ is injective object in $G$-modules, for all $i$.
Proof. Due to Proposition 3.1, there exists $I^0$: an injective object in $Mod_G$ such that $f_0: M\hookrightarrow I^0$. Let $B^0$ be the cokernel of the map $f_0$, i.e. $B^0=I^0/Im f^0$, then we can embed $B^0\hookrightarrow I^1$, where $I^1$ is an injective object in $Mod_G$. And the composition $I^0\to B^0\to I^1$ has the kernel exactly $Im(f_0)$. Continuing this process, we obtain the desired exact sequence
$$0\to M\to I^0\to I^1\to...$$
(Q.E.D)
Fon Proposition 3.2, assume that
$$0\to M\to I^0\to I^1\to...$$
is an injective resolution of a $G$-module $M$. If we apply the functor $(.)^G$, we then obtain the following complex of abelian groups
$$0\xrightarrow{d^{-1}}(I^0)^G\xrightarrow{d^0}(I^1)^G\xrightarrow{d^1}...$$
And we can define $H^i(G,M):=\ker d^i/Im (d^{i-1})$, which is called the $i$-th cohomology group. The theory of derived functors implies that the $i$-th cohomology group $H^i(G,M)$ does not depend on the choice of injective resolution of $M$. And it also follows that if we begin with a short exact sequence
$$0\to M\to N\to P\to 0$$
of $G$-modules, then we will get a long exact sequence of abelian groups
$$0\to H^0(G,M)\to H^0(G,N)\to H^0(G,P)\to H^1(G,M)\to ...$$
Proposition 3.3. $H^0(G,M)=M^G$ for any $G$-module $M$.
Proof. We have $H^0(G,M)=\ker d^0$, and since $(.)^G$ is a left exact functor, we have
$$0\to M^G\to (I^0)^G\to (I^1)^G$$
is exact, and this yields $\ker d^0=Im(M\to (I^0)^G)=M^G$. (Q.E.D)
We can consider $\mathbb{Z}$ as a $G$-module with trivial action from $G$. Let $M$ be any $G$-module , then any $G$-module homomorphism $f$ from $\mathbb{Z}$ to $M$ is uniquely determined by $f(1)$, and since $f(1)=f(g1)=gf(1)$, we have $f(1)\in M^G$. And it follows that $Hom_G(\mathbb{Z},M)=M^G$. Using this, one can prove the Shapiro's lemma
Proposition 3.4. Let $H$ be a subgroup of $G$, and $N$ an $H$-module, then there is a canonical isomorphism $H^r(G, Ind^G_H(N))\cong H^r(H,N)$, for all $r\ge 0$.
Proof. When $r=0$, we have
$$H^0(H,N)=N^H=Hom_H(\mathbb{Z},N)=Hom_G(\mathbb{Z},Ind^G_H(N))=Ind^G_H(N)^G=H^0(G,Ind^G(N))$$
where the third identity follows from the Proposition 2.3. Let
$$0\to N\to I^0\to I^1\to...$$
be an injective resolution of $N$ in $Mod_H$, then since $Ind^G_H$ is an exact functor, which preserves injectives, we have
$$0\to Ind^G_H(N)\to Ind^G_H(I_0)\to Ind^G_H(I_1)\to ...$$
is an injective resolution of $Ind^G_H(N)$, and
$$H^r(G, Ind^G_H(N))=H^r(Ind^G_H(I^\bullet)^G)\cong H^r((I^\bullet)^H)=H^r(H,N)$$
where the second identity follows from the previous argument that $Ind^G_H(I^j)^G=(I^j)^H$. (Q.E.D)
As a corollary, we get
Corollary 3.5. If $M$ is an induced $G$-module, then $H^r(G,M)=0$, for all $r>0$.
Proof. We can write $M=Ind^G(N)$, for some abelian group $N$. According to Proposition 3.4, we have
$$H^r(G, Ind^G(N))\cong H^r({1},N)=0\forall r>0$$
(Q.E.D)
To make things more explicit, we need to describe cohomology groups in terms of co-chains. And we will need some more about projective resolution. Because we can identify $Mod_G$ and $Mod_{\mathbb{Z}[G]}$, it follows that if $P$ is a free $\mathbb{Z}[G]$-module, then $P$ is a projective object in $Mod_G$. We recall that $P$ is a projective object in an abelian category $\mathscr{C}$ if $Hom_{\mathscr{C}}(P,_)$ is an exact functor. Equivalently, any morphism from $P$ to a quotient object $M/N$ can be lifted to a morphism from $P$ to $M$.
Proposition 3.6. $Mod_G$ has enough projectives, i.e. for all $M$ in $Mod_G$, there exists $P$: projective in $Mod_G$, such that $P\to M\to 0$ is exact.
Proof. Let $(m_i)_{i\in I}$ be the set of generators of $M$ as $G$-modules, i.e. for all $n\in M$, there exists $(g_i)_{i\in I}$, such that $m=\sum_{i}g_im_i$. Consider the $G$-module $\mathbb{Z}[G]^I$, which is a direct sum of $\mathbb{Z}[G]$, with indexes run on $I$. Then the map $\mathbb{Z}[G]^I\to M$ defined by $\sum_i \gamma_i\mapsto \sum_i\gamma_im_i$, this map is clearly a surjective $G$-module homomorphism. And $\mathbb{Z}[G]^I$ is projective $G$-module. (Q.E.D)
Proposition 3.7. Any $G$-module $M$ has a projective resolution, i.e. there exists $P_0,P_1,...$: projective $G$-modules, such that
$$...\to P_2\to P_1\to P_0\to M\to 0$$
is an exact sequence of $G$-modules.
Proof. From Proposition 3.6, there exists a projective object $P_0$ in $Mod_G$ such that
$$P_0\xrightarrow{f_0} M\to 0$$
is exact. Let $B_0:=\ker (P_0\to M)$, there exists a projective object $P_1$ in $Mod_G$ such that $P_1\to B_0\to 0$ is exact. And it is clear that $P_1\to P_0\to M\to 0$ is exact. Continuing this process, we get that $M$ admits a projective resolution. (Q.E.D)
Now, let $A,B$ be two $G$-modules, let $B\to I^\bullet$ be an injective resolution of $B$, then since $Hom_G(A,_)$ is a left exact functor, we get
$$0\to Hom_G(A,I^0)\to Hom_G(A,I^1)\to... $$
is a complex in $Ab$, and let us denote $Ext^r(A,B):=H^r(Hom_G(A,I^bullet))$
Now, let $A\leftarrow P_\bullet$ be a projective resolution of $A$, then since $Hom_G(_,B)$ is a contravariant left exact functor, we have:
$$0\to Hom_G(A,B)\to Hom_G(P_0,B)\to Hom_G(P_1,B)\to...$$
is a complex in $Ab$, and we denote $Ext'^r(A,B):=H^r(Hom_G(P_\bullet,B))$. And an interesting thing is that $Ext^r(A,B)=Ext'^r(A,B)$ [Milne ...]. Using this, we can deduce an interesting result
Proposition 3.8. Let $M$ be a $G$-module, and $\mathbb{Z}$ with the trivial action from $G$. Let $\mathbb{Z}\leftarrow P_\bullet$ be a projective resolution of $\mathbb{Z}$, then
$$H^r(Hom_G(P_\bullet,M))\cong H^r(G,M)$$
Proof. We have $H^r(Hom_G(P_\bullet,M))=Ext^r(\mathbb{Z},M)$, and $Ext^r$ is the $r$-th derived functor of $Hom_G(\mathbb{Z},-)$, but it follows that $Hom_G(\mathbb{Z},-)\equiv (.)^G$, so their $r$-th derived functor are the same. (Q.E.D)
We can use Proposition 8 to write explicitly elements in a cohomology groups. Let $G$ be a group, we denote $P_r$ the free $\mathbb{Z}$-module generated by tuples $(g_0,...,g_{r})$, where $g_i\in G$. We define a map $d_r: P_r\to P_{r-1}$ defined by $(g_0,...,g_r)\mapsto \sum_{i=1}^r(-1)^i (g_0,...,\hat{g_i},...,g_r)$. And it can be seen that
$$...\xrightarrow{d_{r+1}}P_r\xrightarrow{d_r} P_{r-1}\xrightarrow{d_{r-1}}...\xrightarrow{d_2}P_1\xrightarrow{d_1}P_0\xrightarrow{d_0} \mathbb{Z}\to 0 (*)$$
is a complex of $G$-modules, where $G$ acts trivially on $\mathbb{Z}$.
Lemma 3.9. The complex $(*)$ is exact.
Proof. Fix $g\in G$, let us define the map $e_r: P_r\to P_{r+1}$ defined by $(g_0,...,g_r)\mapsto (g,g_0,...,g_r)$, then it can be seen easily that $d_{r+1}e_r+e_{r-1}d_r=Id_{P_r}$. And so, $\alpha\in \ker d_r$ implies that $d_{r+1}\circ e_r(\alpha)=\alpha$, i.e. $\alpha\in Im(d_{r+1})$. So, the sequence above is exact. (Q.E.D)
From this lemma, we can see that, in fact, $P_\bullet\to \mathbb{Z}$ is a projective resolution of $\mathbb{Z}$. And from Proposition 3.8, we have $H^r(Hom_G(P_\bullet,M))\cong H^r(G,M)$, where the first group begins with the following complex
$$0\to Hom_G(P_0,M)\to Hom_G(P_1,M)\to...$$
Basically, $\tilde{\varphi}\in Hom_G(P_r,M)$ iff $g\tilde{\varphi}(g_0,...,g_r)=\tilde{\varphi}(gg_0,...,gg_r)$. And the map induced from $Hom_G(P_r,M)$ to $Hom_G(P_{r+1},M)$ is defined to be ${d^r}\tilde{\varphi}:=\tilde{\varphi}\circ d_r$, and
$${d^r}\tilde{\varphi}(g_0,...,g_{r+1})=(\tilde{\varphi}\circ d_r)(g_0,...,g_{r+1})=\sum_{i=1}^{r+1}(-1)^i\tilde{\varphi}(g_0,...,\hat{g_i},...,g_{r+1})$$
And we denote
$$\widetilde{C^r}(G,M):=Hom_{G}(P_r,M)=\{\tilde{\varphi}:G^{r+1}\to M|g\tilde{\varphi}(g_0,...,g_r)=\tilde{\varphi}(gg_0,...,gg_r)\}$$
And the map $d^r: \widetilde{C^r}(G,M)\to \widetilde{C^{r+1}}(G,M)$ is defined to be $\widetilde{\varphi}\mapsto d^r\circ \widetilde{\varphi}$. On the other hand, we will prove that
Lemma 3.10. There are a bijective map between $\widetilde{C^r}(G,M)$ and $C^r(G,M):=\{G^r\to M\}$ defined by $\widetilde{\varphi}(1,g_1,g_1g_2,...,g_1...g_n)=:\varphi(g_1,...,g_n)$, for all $g_1,...,g_r\in G$.
Proof. Assume that $\widetilde{\varphi_1}(1,g_1,...,g_1...g_r)=\varphi(g_1,...,g_r)=\widetilde{\varphi_2}(1,g_1,...,g_1...g_r)$, for all $g_i\in G$, for all $g_i\in G$. If we let $h_i:=g_{i-1}^{-1}g_i$, then
$$\widetilde{\varphi_1}(g_0,...,g_r)=g_0\widetilde{\varphi_1}(1,g_0^{-1}g_1,...,g_0^{-1}g_r)=g_0\widetilde{\varphi_1}(1,h_1,...,h_1...h_r)=$$
$$=g_0\varphi(g_0^{-1}g_1,g_1^{-1}g_2,...,g_{r-1}^{-1}g_r)=g_0\widetilde{\varphi_2}(1, g_0^{-1}g_1,...,g_{r-1}^{-1}g_r)=\widetilde{\varphi_2}(g_0,...,g_r)$$
So we get $\widetilde{\varphi_1}=\widetilde{\varphi_2}$. Now, take any $\varphi: G\to M$, we have to define $\widetilde{\varphi}\in \widetilde{C^r}(G,M)$, such that $\widetilde{\varphi}(1,g_1,...,g_1...g_r)=\varphi(g_1,...,g_r)$. We define
$$\widetilde{\varphi}(g_0,...,g_r):=g_0\widetilde{\varphi}(g_0^{-1}g_1,...,g_{r-1}^{-1}g_r)$$
And it is easy to check that $\widetilde{\varphi}(1,g_1,g_1g_2,...,g_1...g_r)=\varphi(g_1,...,g_r)$. And also,
$$g\widetilde{\varphi}(g_0,...,g_r)=gg_0\widetilde{\varphi}(g_0^{-1}g_1,...,g_{r-1}^{-1}g_r)$$
And $\widetilde{\varphi}(gg_0,...,gg_r)=gg_0\varphi(g_0^{-1}g_1,...,g_{r-1}^{-1}g_r)$. And hence, $\widetilde{\varphi}\in \widetilde{C^{r+1}}(G,M)$. (Q.E.D)
We can see that the map in Lemma 3.10 defines an isomorphism between $\widetilde{C^r}$ and $C^r$. And we can build a map $d^r: C^r(G,M)\to C^{r+1}(G,M)$ such that it is compatible with the map $d^r: \widetilde{C^r}(G,M)\to \widetilde{C^{r+1}}(G,M)$.
For the case $r=0$, for any $\widetilde{\varphi}\in \widetilde{C^0}(G,M)$, we have in this case $\widetilde{\varphi}(1)=\varphi(1)$. And
$$(d^0\widetilde{\varphi})(g_0,g_1)=\widetilde{\varphi}(g_1)-\widetilde{\varphi}(g_0)=g_1\widetilde{\varphi}(1)-g_0\widetilde{\varphi}(1)=g_1\varphi(1)-g_0\varphi(1)$$
And
$$(d^0\widetilde{\varphi})(g_0,g_1)=g_0(d^0\widetilde{\varphi})(1,g_0^{-1}g_1)=g_0(d^0\varphi)(g_0^{-1}g_1)$$
Hence, $g_0(d^0\varphi)(g_0^{-1}g_1)=g_1\varphi(1)-g_0\varphi(1)$, i.e. $(d^0\varphi)(g_0^{-1}g_1)=g_0^{-1}g_1\varphi(1)-\varphi(1)$. If we denote $h:=g_0^{-1}g_1$, we then get $(d^0\varphi)(h):=h\varphi(1)-\varphi(1)$.
When $r\ge 1$, by similar argument, for any $\varphi\in C^r(G,M)$, we have
$$(d^r\varphi)(g_1,...,g_{r+1})=g_1\varphi(g_2,...,g_{r+1})-\sum_{i=1}^r(-1)^i\varphi(g_1,...g_{i}g_{i+1},...,g_r)+(-1)^{r+1}\varphi(g_1,...,g_r)$$
And this yields $H^r(G,M)=H^r(\widetilde{C^\bullet}(G,M))=H^r(C^\bullet(G,M))=\ker d^r/Im (d^{r-1})$. We can now write things explicitly to $H^2$. For $H^0(G,M)$, we already know that it is $M^G$. And for $H^1$, take any $\varphi: G\to M$, we have $\varphi\in \ker d^1$ iff $\varphi(g_1g_2)=g_1\varphi(g_2)+\varphi(g_1)$, and such $\varphi$ is said to be a crossed homomorphism. And $\varphi\in Im(d^0)$ iff $\varphi(g)=g\varphi(1)-\varphi(1)$, i.e. $\varphi(g)=gm-m$, for some $m\in M$, and such $\varphi$ is said to be a principal homomorphism. So, we get
$$H^1(G,M)=\frac{\text{crossed homomorphisms from G to M}}{\text{principal homomorphisms from G to M}}$$
For $H^2$, we have $\varphi\in \ker d^2$ iff $\varphi:G\times G\to M$ satisfying
$$\varphi(g_1g_2,g_3)+\varphi(g_1,g_2)=g_1\varphi(g_2,g_3)+\varphi(g_1,g_2g_3)$$
And $\varphi\in Im(d^1)$ iff $\varphi(g_1,g_2)=h_1\varphi'(h_2)-\varphi'(h_1h_2)+\varphi'(h_1)$, for some $\varphi': G\to M$. And the quotient of them yields the group $H^2(G,M)$.
No comments:
Post a Comment