That \(A^G\leq A\) is the maximal trivial \(G{\hbox{-}}\)trivial submodule is clear: if there were any \(A'\) such that \(A^G \leq A' \leq A\) with \(A'\) being \(G{\hbox{-}}\)trivial, we would have \(ga' = a'\) for all \(a' \in A'\). This would imply \(a'\in A^G\) by definition, so \(A'\leq A^g\) and this forces \(A^G = A'\).

Moreover, since right adjoints are left-exact, it suffices to show the adjunction exists. The claim is that there is a natural isomorphism between the sets of morphisms \begin{align*} {\mathsf{Ab}}(A, M^G) &\mathrel{\operatorname*{\rightleftharpoons}_{\Psi}^{\Theta}} \mathsf{G}{\hbox{-}}\mathsf{Mod}(\mathop{\mathrm{Triv}}(A), M)\\ (A \xrightarrow{f} M^G) &\mapsto (A \xrightarrow{\tilde f} M^G \hookrightarrow M) \\ (A \xrightarrow{\tilde w} M^G) &\mapsfrom (\mathop{\mathrm{Triv}}(A) \xrightarrow{g} M) .\end{align*}

Define a map \begin{align*} \mathop{\mathrm{Hom}}_G(M, \mathop{\mathrm{Triv}}(A)) &\to \mathop{\mathrm{Hom}}_{\mathsf{Ab}}(M_G, A) \\ f &\mapsto \mkern 1.5mu\overline{\mkern-1.5muf\mkern-1.5mu}\mkern 1.5mu \\ \tilde h &\mapsfrom h \end{align*} where we realize \(M_G \coloneqq M/ \left\langle{ gm-m }\right\rangle\) and take \(\mkern 1.5mu\overline{\mkern-1.5muf\mkern-1.5mu}\mkern 1.5mu\) to be the natural quotient map:

Link to Diagram

and given \(h: M_G \to A\), we define \(\tilde h = h\circ \pi\), so \begin{align*} \tilde h: M &\to \mathop{\mathrm{Triv}}(A) \\ m &\mapsto h(\pi(m)) .\end{align*}

That \(f\) descends to a well-defined map on the quotient follows from the fact that \(ga = a\) for all \(g\) and all \(a\in \mathop{\mathrm{Triv}}(A)\), so \(f(gm-m) = gf(m) - f(m)\), which is of the form \(ga-a\), and so \(\ker f \subseteq \left\langle{ ga- a }\right\rangle\). The inverse map is a well-defined \(G{\hbox{-}}\)module morphism, using the fact that since \(g\) acts trivially on both sides we have \(gh(m) = h(m) = h(gm)\).

In words, the forward map is descending morphisms to the quotient, and the inverse map is lifting maps from the quotient by composing with the projection. That these compose to the identity in other order and yield a bijection follows from the commutativity of the triangle for the quotient. That this is an adjunction follows from naturality of this isomorphism, for which the proof is similar to the one for the previous problem.

By definition we have \(H^0(G; A) \coloneqq A^G\), and lemma 6.1.1, we also have \(H^0(G; A) \cong \mathop{\mathrm{Hom}}_{{\mathbb{Z}}G}({\mathbb{Z}}, A)\) for any \(A\). Taking \(A \coloneqq{\mathbb{Z}}G\) yields the first isomorphism appearing in (1), so it suffice to show \(({\mathbb{Z}}G)^G = 0\). In this case, we can write \begin{align*} ({\mathbb{Z}}G)^G \coloneqq \left\{{ x\coloneqq \sum_{j \in J} f(g_j) g_{j} {~\mathrel{\Big\vert}~} h \sum_{j\in J} f(g_j) g_{j} = \sum_{j \in J } f(g_j) g_{j}\,\,\forall h\in G }\right\} \subseteq \bigoplus_{g\in G} {\mathbb{Z}} \end{align*} where \(f:G\to {\mathbb{Z}}\) is some coefficient function, and every such sum has only finitely many nonzero terms. Since the \(g_{ \alpha}\) form a basis of the free \({\mathbb{Z}}{\hbox{-}}\)module \({\mathbb{Z}}G\), we can equate coefficients. The coefficient of \(hg_j\) in the first expression is \(f(hg_j)\), and is \(f(g_j)\) in the second, so we must have \(f(g_j) = f(hg_j)\) for all \(h\in G\). Toward a contradiction fix an \(x\in ({\mathbb{Z}}G)^G\) and suppose \(x\neq 0\). Then there is at least one \(j\) for which \(f(g_j) \neq 0\), then noting that \(G\) acts on itself transitively by left-multiplication, we can write any \(g_i\) as \(g_i = h(i) g_j\) for some \(h=h(i)\) depending on \(i\). We then have \begin{align*} x \coloneqq\sum f(g_i) g_i = \sum f(h(i) g_j)h(i)g_j ,\end{align*} and noting that \begin{align*} 0 \neq f(g_j) = f(h(i) g_j) && \forall i ,\end{align*} if \(G\) is infinite then this produces a sum with infinitely terms with nonzero coefficients. \(\contradiction\)

We first identify \({\mathbb{Z}}G^G = {\mathbb{Z}}N\) and \({\mathbb{Z}}G_G \cong {\mathbb{Z}}\otimes_{{\mathbb{Z}}G} {\mathbb{Z}}G\) where we regard \({\mathbb{Z}}\) as a trivial \({\mathbb{Z}}G{\hbox{-}}\)module. We then identify the natural map \({\mathbb{Z}}G^G \to {\mathbb{Z}}\) as the composition \begin{align*} F: {\mathbb{Z}}G^G &\to {\mathbb{Z}}\otimes_{{\mathbb{Z}G}} {\mathbb{Z}G}&&\to {\mathbb{Z}}\\ nN &\mapsto n \otimes N &&\mapsto N \curvearrowright n \coloneqq\sum_{g\in G} n ,\end{align*} where we’ve used the trivial action \({\mathbb{Z}G}\curvearrowright{\mathbb{Z}}\) to write \((g+h)\curvearrowright n = gn + hn \coloneqq n + n = 2n\) and extended this over the sum. So we have \begin{align*} n N \mapsto \sum_{g\in G} n = n \sum_{g\in G} 1 = n (# G) ,\end{align*} and taking \(n=1\) sends \(N\mapsto # G\). From this construction, we find that \(\ker F = \left\{{n\in {\mathbb{Z}}{~\mathrel{\Big\vert}~}n (# G) = 0}\right\} = \left\{{0}\right\}\), so \(F\) is injective.

It suffices to show \(Nx = 0 \iff {\varepsilon}(x) = 0\), since this yields both subset containments.

\(\implies\): Let \(N: {\mathbb{Z}G}\to {\mathbb{Z}G}\) by multiplication by \(N\), then \(\ker N \coloneqq\left\{{ \sum n_g g \in {\mathbb{Z}G}{~\mathrel{\Big\vert}~}N\sum n_g g = 0 }\right\}\). Letting \(x\coloneqq\sum n_g g\in \ker N\), supposing that \(Nx = 0\), we can expand this product using the multiplication in \({\mathbb{Z}G}\): \begin{align*} \qty{\sum_h n_h h } \qty{ \sum_g n_g g} = \sum_{g, h} n_h n_g hg ,\end{align*} and so \begin{align*} Nx &\coloneqq\qty{ \sum_h h } \qty{ \sum_g n_g g} = \sum_{h,g} n_g hg = 0 .\end{align*} We can now consider the coefficient of the identity \(1_G\). For any fixed \(g\), there is some \(h\) such that \(h=g^{-1}\), and this will contribute a term of the form \(n_g \cdot 1_G\). Letting \(g\) run over \(G\) yields \begin{align*} Nx = \qty{ \sum_g n_g}\cdot 1_G + \cdots = 0 .\end{align*} Equating coefficients, we must have \(\sum_g n_g = 0\), and noting that \({\varepsilon}(x) = {\varepsilon}(\sum n_g g) = \sum n_g = 0\), this means that \(x\in \ker {\varepsilon}\coloneqq\mathcal{J}\).

\(\impliedby\): Conversely, if \({\varepsilon}(x) = 0\), then we can use an alternative characterization of the product: \begin{align*} \qty{ \sum_h n_h h } \qty{ \sum _g n_g g} = \sum_g \qty{ \sum_k n_{k^{-1}g} } g .\end{align*} Applying this to \(N \coloneqq\sum_h h\), we have \begin{align*} Nx \coloneqq\qty{ \sum_h h} \qty{\sum_g n_g g} = \sum_g \qty{ \sum_k n_{k^{-1}g} }g .\end{align*} But the inner sum runs over all elements of \(G\), so for each fixed \(g\) it is equal to \(\sum n_g \coloneqq{\varepsilon}(x)\) and we in fact have \begin{align*} Nx = \sum_g {\varepsilon}(x) g = \sum_g 0\cdot g = 0 .\end{align*}

Following an argument from here

Without loss of generality, reindex \(E\) so that \(p-1=0\) and \(p=1\). Then since \(n\) is fixed, we have \(q = n-p\) so that \(q=n, n-1\); we then want to show that there is a SES \begin{align*} 0 \to E_{0,n} \to H_n T \to E_{1, n-1} \to 0 .\end{align*} Noting that the differentials on \(E_r\) have bidegree \((-r, r-1)\) since this is a homological spectral sequence, when \(r\geq 2\), fix an element on \(E_r\) in column 1 or 2 and consider the following two cases for the differentials:

• For outgoing differentials, the source is in column 0 or 1, so the target is in column \(j\leq -1\)
• For incoming differentials, the target is in column 0 or 1, so the source is in column \(j\geq 2\).

Thus the spectral sequence collapses at \(E^2\), and we have \(E^{\infty }_{p, q} = E^2_{p, q}\) for all \(p, q\) Since this converges to \(E^2_{p, q} \Rightarrow H_{p+q}T \coloneqq H_n(T)\), there is increasing exhaustive filtration \(F_\bullet\) on \(H_n(T)\) such that \begin{align*} E^{\infty }_{p, q} = {F_p H_n(T) \over F_{p-1}H_n(T)} .\end{align*} Thus the filtration has the form \begin{align*} 0 \subseteq F_0 H_n(T) \subseteq F_1 H_n(T) = F_2 H_n(T) = \cdots = H_n(T) ,\end{align*} where we’ve used that if \(i\neq 0, 1\) we have

\begin{align*} E^{\infty }_{i, q} = 0 = { F_i H_n(T) \over F_{i-1}H_n(T)} \implies F_i H_n(T) = F_{i-1}H_n(T) .\end{align*} Setting \(i=0, 1\), we have \begin{align*} E^2_{0, q} &\cong F_0 H_n(T) \\ E^2_{1, q-1} &\cong F_1 H_n(T) / F_0 H_n(T) ,\end{align*}

and so if we write the following exact sequence \begin{align*} 0 \to F_0 H_n(T) \to F_1 H_n(T) \to {F_1 H_n(T) \over F_0 H_n(T)} \to 0 \end{align*} we can replace each term by an isomorphic term to get an isomorphic exact sequence \begin{align*} 0 \to E^2_{0, q} \to H_n(T) \to E^2_{1, q-1} \to 0 .\end{align*}

\todo[inline]{
This was solved in the problem above -- which probably indicates that the previous problem should have been solved another way!
}

\todo[inline]{Todo}

Consider passing from \(E^2\) to \(E^3\). On the 2nd page, we have the following situation:

\begin{sseqdata}[name = ex1, homological Serre grading, classes = {draw = none }, xscale=4, yscale=3]
\class["E_{0, 2}^2"](0,2)
\class["\color{blue}E_{0, 0}^2"](0,0)
\class["E_{0, 1}^2"](0,1)
\class["\color{blue} E_{1, 0}^2"](1,0)
\class["E_{1, 1}^2"](1,1)
\class["E_{2, 0}^2"](2,0)
\class["E_{3, 0}^2"](3,0)

\d["d_{2, 0}^2"]2(2,0)
\replacetarget["\color{blue}  \operatorname{coker}(d_{2, 0}^2)", xshift = 2em]
\replacesource["\color{blue} \ker(d_{2, 0}^2)"]

\d["d_{3, 0}^2"]2(3,0)
\replacetarget["\color{blue}  \operatorname{coker}(d_{3, 0}^2)"]
\replacesource["\ker(d_{3, 0}^2)", xshift=-0.5em]
\d["d_{3, 0}^3"]3(3,0)
\replacetarget["\color{blue}  E"]

\end{sseqdata}
\printpage[ name = ex1, page = 2 ] \newline

Here the blue terms have stabilized, since either their domain or codomain is a zero. If a first quadrant spectral sequence converges, for each \(n\) there will be a filtration \begin{align*} 0 = F_{-1}H_n(T) \subseteq F_0H_n(T) \subseteq F_1 H_n(T) \subseteq \cdots \subseteq H_n(T) ,\end{align*} where \(E_{p, q}^{\infty } \cong F_pH_{p+q}(T) / F_{p-1} H_{p+q}(T)\). Thus the \(E^{\infty}\) page is the following:

Link to Diagram

We note that for \(H_0(T)\), since \(E\) is first-quadrant, this filtration must be of the form \begin{align*} 0 = F_{-1}H_0(T) \subseteq F_0 H_0(T) = F_1 H_0(T) = \cdots = H_0(T) ,\end{align*} i.e. there are no nontrivial terms. This follows because the degree zero part of the associated graded complex will be given by \(\bigoplus_{p+q=0}E_{p, q}^\infty\), so \(p=q=0\) will contribute a nonzero term, and every other term must involve either \(p<0\) or \(q<0\) and thus be zero, and \(F_j / F_{j-1} = 0\) for some \(j\) forces \(F_j \cong F_{j-1}\). We thus have \begin{align*} E_{0, 0}^2 &= E_{0, 0}^{\infty}\\ &\cong {F_{0} H_{0+0}(T) \over F_{-1} H_{0 + 0}(T)}\\ &= F_0 H_0(T) \\ &= H_0(T) ,\end{align*} which yields part 1.

Now passing to the 3rd page yields the following, where several new terms have stabilized:

\printpage[ name = ex1, page = 3]

Starting in total degree 2, we can identify

• \(E_{0, 1}^{\infty } = F_0 H_1\), and by stabilization this is isomorphic to \(\operatorname{coker}(d_{2, 0}^2)\).

• \(E_{1, 0}^{\infty } = {F_1 H_1 \over F_0 H_1} = {H_1 \over F_0 H_1}\), and by stabilization this is isomorphic to \(E_{1, 0}^2\).

From the second fact, we can extract the canonical quotient SES \begin{align*} 0 \to F_0 H_1 \to H_1 \to {H_1 \over F_0 H_1} \to 0 ,\end{align*} and using the above identifications, this is isomorphic to a SES of the form \begin{align*} 0 \to E_{0, 1}^{\infty } \to H_1 \to E_{1, 0}^{\infty } \to 0 ,\end{align*} which we will splice to form the desired 5-term exact sequence.

In total degree 2, we can identify

• \(E_{0, 2}^{\infty} = F_0 H_2\): this is unknown at this stage, but we won’t need it.

• \(E_{1, 1}^{\infty } = {F_1 H_2 \over F_0 H_2} \cong \operatorname{coker}(d_{3, 0}^2: E_{3, 0}^2 \to E_{1, 1}^2 )\) by stabilization, and

• \(E_{2, 0}^{\infty } \cong {F_2 H_2 \over F_1 H_2} = {H_2 \over F_1 H_2} \cong \ker(d_{2, 0}^2: E_{2, 0}^2 \to E_{0, 1}^2)\) by stabilization.

We will only need this last fact: we can construct the composition \begin{align*} \ker(d_{2, 0}^2: E_{2, 0}^2 \to E_{0, 1}^2) \hookrightarrow E_{2, 0}^2 \xrightarrow{d_{2, 0}^2} E_{0, 1}^2 \twoheadrightarrow \operatorname{coker}(d_{2, 0}^2: E_{2, 0}^2 \to E_{0, 1}^2) .\end{align*} Noting that the first term is isomorphic to \(E_{2, 0}^{\infty }\) by the analysis in degree 2 and the last is isomorphic to \(E_{0, 1}^{\infty }\) by the analysis in degree 1, there is a sequence \begin{align*} E_{2, 0}^{\infty } \hookrightarrow E_{2, 0}^2 \xrightarrow{d_{2, 0}^2} E_{0, 1}^2 \twoheadrightarrow E_{0, 1}^{\infty } .\end{align*} We can now use that \(E_{2, 0}^{\infty }\) is a quotient of \(H_2\) to splice onto the left, and splice the previous SES on the right to obtain

\begin{align*} H_2 {\color{red} \twoheadrightarrow} E_{2, 0}^{\infty } {\color{red} \hookrightarrow} E_{2, 0}^2 \xrightarrow{d_{2, 0}^2} E_{0, 1}^2 {\color{green} \twoheadrightarrow} E_{0, 1}^{\infty } {\color{green} \to} 0 {\color{green} \to} H_1 \to E_{1, 0}^{\infty } \to 0 .\end{align*} This isn’t an exact sequence as written, so we collapse several maps to get a new sequence \begin{align*} H_2 {\color{red} \to } E_{2, 0}^2 \xrightarrow{d_{2, 0}^2} E_{0, 1}^2 {\color{green} \to } H_1 \to E_{1, 0}^{\infty } \to 0 .\end{align*} This new sequence is exact:

• Exactness at the 1st term \(E_{2, 0}^2\): this follows because in the first sequence, \(H_2\) surjected onto \(E_{2, 0}^{\infty}\) which was isomorphic to the kernel of \(d_{2, 0}^2\), which is precisely the next map in the sequence.

• Exactness at the 2nd term \(E_{0, 1}^2\): this follows because \(E_{0, 1}^{\infty } \cong \operatorname{coker}d_{2, 0}^2\), which is precisely the previous map in the sequence. So applying this \(d\) maps onto its image, which is quotiented out in \(E_{0, 1}^{\infty }\) before mapping to \(H_1\).

• Exactness at the 3rd term \(H_1\): this follows because we’ve factored the composite map through zero by construction.

• Exactness at the remaining terms: this follows from exactness of the original SES.