Homological Algebra

  • Let \(f:M\to N\) be a morphism of \({\mathfrak{g}}{\hbox{-}}\)modules over a field \(k\). Show that the \(k{\hbox{-}}\)modules \(\ker(f), \operatorname{im}(f), \operatorname{coker}(f)\) are the kernel, image, and cokernel respectively of \(f\) in the category \({\mathfrak{g}{\hbox{-}}\mathsf{Mod}}\)

  • Show that a monic (resp. epi) in \({\mathfrak{g}{\hbox{-}}\mathsf{Mod}}\) is also a monic (resp. epi) in \({\mathsf{k}{\hbox{-}}\mathsf{Mod}}\). Use (1) to show that \({\mathfrak{g}{\hbox{-}}\mathsf{Mod}}\) is an abelian category.


Note that there is an inclusion of sets \begin{align*} \mathop{\mathrm{Hom}}_{{\mathfrak{g}{\hbox{-}}\mathsf{Mod}}}(M, N) &\coloneqq\left\{{ f\in \mathop{\mathrm{Hom}}_{{\mathsf{k}{\hbox{-}}\mathsf{Mod}}}(M, N) {~\mathrel{\Big\vert}~}f(gm) = gf(m)\,\, \forall g\in {\mathfrak{g}},\, \forall m\in M }\right\} \\ &\subseteq \mathop{\mathrm{Hom}}_{{\mathsf{k}{\hbox{-}}\mathsf{Mod}}}(M, N) ,\end{align*} where we can regard \(M, N\) as \(k{\hbox{-}}\)modules by applying a forgetful functor \({\mathfrak{g}{\hbox{-}}\mathsf{Mod}}\to {\mathsf{k}{\hbox{-}}\mathsf{Mod}}\). This is in fact a \(k{\hbox{-}}\)submodule: if \(f_1, f_2 \in \mathop{\mathrm{Hom}}_{{\mathfrak{g}{\hbox{-}}\mathsf{Mod}}}(M, N)\) and \(t\in k\), we have \begin{align*} (tf_1 + f_2)(gm) \coloneqq tf_1(gm) + f_2(gm) = g\cdot tf_1(m) + g\cdot f_2(m) = g\cdot ( tf_1(m) + f_2(m) ,\end{align*} which shows \(tf_1 + f_2 \in \mathop{\mathrm{Hom}}_{{\mathfrak{g}{\hbox{-}}\mathsf{Mod}}}(M, N)\) and the one-step submodule test applies.


Moreover, kernels exist in \({\mathfrak{g}{\hbox{-}}\mathsf{Mod}}\) since they exist in \({\mathsf{k}{\hbox{-}}\mathsf{Mod}}\): given \(f\in \mathop{\mathrm{Hom}}_{{\mathfrak{g}{\hbox{-}}\mathsf{Mod}}}(M, N)\), using the submodule structure above we can identify \(f\in \mathop{\mathrm{Hom}}_{{\mathsf{k}{\hbox{-}}\mathsf{Mod}}}(M, N)\) and produce \(\ker f\) as a \(k{\hbox{-}}\)submodule of \(M\). Using the kernels are set inclusions in categories of \(R{\hbox{-}}\)modules, we can define a \({\mathfrak{g}}{\hbox{-}}\)module structure on \(\ker f\leq M\) by restricting the \({\mathfrak{g}}{\hbox{-}}\)action on \(M\). Then the \(k{\hbox{-}}\)module inclusion \(\iota: \ker f\hookrightarrow M\) is a morphism of \({\mathfrak{g}}{\hbox{-}}\)modules, since \(\iota(\ell) = \ell\) for \(\ell\in \ker f\), so it is product-preserving: \begin{align*} g\iota(\ell) = g\ell = \iota(g\ell) .\end{align*}

Similarly, \(\operatorname{im}(f)\) in \({\mathfrak{g}{\hbox{-}}\mathsf{Mod}}\) is gotten by setting \(\operatorname{im}(f) \coloneqq\operatorname{im}_{{\mathsf{k}{\hbox{-}}\mathsf{Mod}}}(f)\) and restricting the \({\mathfrak{g}}{\hbox{-}}\)action from \(N\) to \(\operatorname{im}(f)\), and the cokernel is obtained as the quotient \(\operatorname{coker}(f) \coloneqq N/\operatorname{im}(f)\) with a \({\mathfrak{g}}{\hbox{-}}\)module structure induced by the canonical quotient map.

To see that monics in \({\mathfrak{g}{\hbox{-}}\mathsf{Mod}}\) are also monics in \({\mathsf{k}{\hbox{-}}\mathsf{Mod}}\), first consider the forgetful functor \begin{align*} F: {\mathfrak{g}{\hbox{-}}\mathsf{Mod}}\to {\mathsf{k}{\hbox{-}}\mathsf{Mod}} .\end{align*} This is adjoint to the trivial \({\mathfrak{g}}{\hbox{-}}\)module functor, yielding an adjunction \begin{align*} \adjunction{F}{\mathop{\mathrm{Triv}}} {{\mathfrak{g}{\hbox{-}}\mathsf{Mod}}}{{\mathsf{k}{\hbox{-}}\mathsf{Mod}}} .\end{align*} We need to check that if \(f: A\to B\) in \({\mathfrak{g}{\hbox{-}}\mathsf{Mod}}\) is monic, then its image \(F(f): F(A) \to F(B)\) is monic in \({\mathsf{k}{\hbox{-}}\mathsf{Mod}}\). Note that being a monomorphism is equivalent to having the following injections on hom sets for all \(Z\in {\mathfrak{g}{\hbox{-}}\mathsf{Mod}}\): \begin{align*} f_*: \mathop{\mathrm{Hom}}_{{\mathfrak{g}{\hbox{-}}\mathsf{Mod}}}(Z, A) &\hookrightarrow\mathop{\mathrm{Hom}}_{{\mathfrak{g}{\hbox{-}}\mathsf{Mod}}}(Z, B) \\ h_i &\mapsto f\circ h_i .\end{align*} So the content of the problem is to check that for all \(W\in {\mathsf{k}{\hbox{-}}\mathsf{Mod}}\), the following map is an injection: \begin{align*} F(f)_*: \mathop{\mathrm{Hom}}_{{\mathsf{k}{\hbox{-}}\mathsf{Mod}}}(W, F(A)) &\hookrightarrow\mathop{\mathrm{Hom}}_{{\mathsf{k}{\hbox{-}}\mathsf{Mod}}}(W, F(B)) \\ g_i &\mapsto F(f)\circ g_i .\end{align*} Using the adjunction, we have natural isomorphism \begin{align*} \mathop{\mathrm{Hom}}_{{\mathsf{k}{\hbox{-}}\mathsf{Mod}}}(W, F(A)) &\xrightarrow{\sim} \mathop{\mathrm{Hom}}_{{\mathfrak{g}{\hbox{-}}\mathsf{Mod}}}(\mathop{\mathrm{Triv}}(W), A) \\ \mathop{\mathrm{Hom}}_{{\mathsf{k}{\hbox{-}}\mathsf{Mod}}}(W, F(B)) &\xrightarrow{\sim} \mathop{\mathrm{Hom}}_{{\mathfrak{g}{\hbox{-}}\mathsf{Mod}}}(\mathop{\mathrm{Triv}}(W), B) ,\end{align*} and by assumption, \begin{align*} f_*: \mathop{\mathrm{Hom}}_{{\mathfrak{g}{\hbox{-}}\mathsf{Mod}}}(\mathop{\mathrm{Triv}}(W), A) \hookrightarrow \mathop{\mathrm{Hom}}_{{\mathfrak{g}{\hbox{-}}\mathsf{Mod}}}(\mathop{\mathrm{Triv}}(W), B) ,\end{align*} since we can take \(Z \coloneqq\mathop{\mathrm{Triv}}(W)\). Since \(f_*\) is an injection, pushing it through the isomorphism shows that \(F(f)_*\) is an isomorphism, and so any monic \({\mathfrak{g}}{\hbox{-}}\)module morphism descends to a monic \(k{\hbox{-}}\)module morphism.

For \(M\in {\mathsf{k}{\hbox{-}}\mathsf{Mod}}\), let \(E \coloneqq\mathop{\mathrm{End}}_{{\mathsf{k}{\hbox{-}}\mathsf{Mod}}}(M) \in {\mathsf{Alg}_{/k} }\) be the associative algebra of \(k{\hbox{-}}\)module endomorphisms of \(M\). Show that there is a correspondence \begin{align*} \left\{{\substack{ \text{Maps ${\mathfrak{g}}\otimes M \to M$} \\ \text{making $M$ a ${\mathfrak{g}}{\hbox{-}}$module} }}\right\} &\rightleftharpoons \left\{{\substack{ \text{Lie algebra morphisms}\\ {\mathfrak{g}}\to \mathsf{Lie}(E) }}\right\} \end{align*} Conclude that a \({\mathfrak{g}}{\hbox{-}}\)module may also be described as an \(M\in{\mathsf{k}{\hbox{-}}\mathsf{Mod}}\) together with a morphism of Lie algebras \begin{align*} {\mathfrak{g}}\to \mathsf{Lie}\qty{ \mathop{\mathrm{End}}_{{\mathsf{k}{\hbox{-}}\mathsf{Mod}}}(M) } .\end{align*}

Define maps \begin{align*} \left\{{ f: {\mathfrak{g}}\otimes_k M \to M \in {\mathsf{k}{\hbox{-}}\mathsf{Mod}}}\right\} & \mathrel{\operatorname*{\rightleftharpoons}_{\Psi}^{\Theta}} \left\{{ {\mathfrak{g}}\to \mathsf{Lie}\qty{\mathop{\mathrm{End}}_{{\mathsf{k}{\hbox{-}}\mathsf{Mod}}}(M) } \in {\mathsf{Lie}{\hbox{-}}\mathsf{Alg}}}\right\} \\ f & \mapsto \Theta_f \coloneqq(g \mapsto f(g\otimes{-}) ) \\ \Psi_h \coloneqq(g\otimes m \mapsto h_g (m) ) &\mapsfrom h ,\end{align*} where \(h_g(m) \coloneqq h(g)(m)\), and we have \(\Theta_f(g)(m) \coloneqq f(g, m)\) and \(\Psi_h(g, m) \coloneqq h(g)(m)\).

Let \(f: {\mathfrak{g}}\otimes_k M\to M\) be a \(k{\hbox{-}}\)module morphism. \(\Theta_f\) defines a morphism of Lie algebras.

Write \([{-}, {-}]_{\mathfrak{g}}\) for the bracket on \({\mathfrak{g}}\) and \([{-}, {-}]_{\mathop{\mathrm{End}}}\) for the bracket on \(\mathsf{Lie}(\mathop{\mathrm{End}}_{{\mathsf{k}{\hbox{-}}\mathsf{Mod}}} (M))\) defined by \([x, y]_{\mathop{\mathrm{End}}} \coloneqq x\circ y - y\circ x\), it then suffices to check the following identity: \begin{align*} [\Theta_f(x), \Theta_f(y)]_{\mathop{\mathrm{End}}} = \Theta_f\qty{ [x, y]_{\mathfrak{g}}} .\end{align*} Expanding the right-hand side, we have \begin{align*} [\Theta_f(x), \Theta_f(y)]_{\mathop{\mathrm{End}}} &\coloneqq \Theta_f(x) \circ \Theta_f(y) - \Theta_f(y) \circ \Theta_f(x) \\ &= f(x\otimes{-}) \circ f(y\otimes{-}) - f(y\otimes{-}) \circ f(x\otimes{-}) \\ &= f(x \otimes f(y\otimes{-})) - f(y \otimes f(x\otimes{-})) ,\end{align*} and now using multiplicative notation to write \(gm \coloneqq f(g, m)\), evaluating this on an element \(m\in M\) yields \begin{align*} \qty{ f(x\otimes f(y\otimes{-})) - f(y\otimes f(x\otimes{-})) }(m) = x(ym) - y(xm) .\end{align*} For the right-hand side, we have \begin{align*} \Theta_f \qty{ [x, y]_{\mathfrak{g}}}(m) &\coloneqq f([x, y]_{\mathfrak{g}}\otimes m) \\ &= [x, y]_{\mathfrak{g}}(m) \\ &\coloneqq x(ym) - y(xm) ,\end{align*} where in the last step we’ve used that \(f\) is a structure map that makes \(M\) into a \({\mathfrak{g}}{\hbox{-}}\)module. So the two sides agree on every element of \(M\), and are thus equal as \(k{\hbox{-}}\)module endomorphisms of \(M\).

Let \(h: {\mathfrak{g}}\to \mathsf{Lie}( \mathop{\mathrm{End}}_{{\mathsf{k}{\hbox{-}}\mathsf{Mod}}} (M))\) be a morphism of Lie algebras. \(\Psi_h\) defines a morphism of \(k{\hbox{-}}\)modules.

It suffices to check \(\Psi_h(rx+y) = r\qty{ \Psi_h(x) +\Psi_h(y)}\) for \(x, y\) in the domain \({\mathfrak{g}}\otimes M\) and \(r\in k\). This follows from a computation: on elementary tensors, we have \begin{align*} \Psi_h( r(g_1 \otimes m_1) + (g_2 \otimes m_2)) &= \Psi_h(r(g_1 + g_2) \otimes(m_1+ m_2)) \\ &\coloneqq h(r(g_1 + g_2))(m_1 + m_2) \\ &= r\qty{ h(g_1) + h(g_2) } (m_1 + m_2) \quad \text{using $r{\hbox{-}}$linearity of $h$}\\ &= r\qty{ h(g_1) + h(g_2) } (m_1 + m_2) \\ &= r\qty{ h(g_1) + h(g_2) \otimes m_1 + m_2} \\ &= r\qty{ (h(g_1) \otimes m_1) + (h(g_2)\otimes m_2)} \\ &= r\qty{ \Psi_h(g_1 \otimes m_1) + \Psi_h(g_2 \otimes m_2) } ,\end{align*} and extending by linearity shows that \(\Psi_h\) is \(k{\hbox{-}}\)linear.

\(\Psi_h\) makes \(M\) into a \({\mathfrak{g}}{\hbox{-}}\)module.

In multiplicative notation, the condition we need to check is the following: \begin{align*} [x, y]m = x(ym) - y(xm) && \forall x,y\in {\mathfrak{g}},\, m\in M .\end{align*} Writing this explicitly, we want ` \begin{align*} \Theta_h( [xy], m)

\Theta_h(x, \Theta_h(y, m)) - \Theta_h(y, \Theta_h(x, m)) .\end{align*} {=html} This follows from a computation: \begin{align*} \Theta_h([xy]{\mathfrak{g}}, m) &\coloneqq h( [xy]{\mathfrak{g}})(m) \ &= \qty{ [h(x), h(y)]_{\mathop{\mathrm{End}}}} (m) \quad \ast \ &\coloneqq\qty{ h(x) \circ h(y) - h(y) \circ h(x) }(m) \ &= h(x)( h(y)(m)) - h(y)(h(x)(m))\ &\coloneqq\Theta_h(x, \Theta_h(y, m)) - \Theta_h(y, \Theta_h(x, m)) ,\end{align*} `{=html} where in the line marked \(\ast\) we’ve used that \(h\) was a Lie algebra morphism and thus we can commute the brackets.

\(\Theta\) and \(\Psi\) are mutually inverse.

Starting with \(f: {\mathfrak{g}}\otimes M\to M\), we obtain \(\Theta_f\coloneqq(g\mapsto f(g\otimes{-}))\). Applying \(\Psi\) yields \begin{align*} \Psi_{\Theta_f} &\coloneqq(g \otimes m \mapsto \Theta_f(g)(m)) \\ &\coloneqq(g\otimes m \mapsto f(g \otimes{-})(m) ) \\ &\coloneqq(g\otimes m \mapsto f(g\otimes m)) ,\end{align*} and so this recovers \(f\).


Similarly, starting now with \(h: {\mathfrak{g}}\to {\mathfrak{gl}}(M)\) and letting \(h_g \coloneqq h(g)\), applying \(\Psi\) yields \(\Psi_h \coloneqq(g\otimes m \mapsto h_g(m))\), and applying \(\Theta\) yields \begin{align*} \Theta_{\Psi_h} &\coloneqq(g \mapsto \Psi_h(g \otimes{-})) \\ &\coloneqq(g\mapsto (g\otimes{-}\mapsto h_g({-}) ) ) \\ &\coloneqq(g\mapsto h_g) ,\end{align*} which recovers \(h\).

The main result now follows.

Given an \(M\in {\mathsf{k}{\hbox{-}}\mathsf{Mod}}\), consider the Lie algebra \(\mathsf{Lie}(T(M))\) underlying the tensor algebra \(T(M)\). Let \(\mathfrak{f}\) denote the Lie subalgebra generated by \(M\), so elements of \(\mathfrak{f}\) are iterated brackets of elements: \begin{align*} x \in \mathfrak{f} \implies x = \sum [x_1\, [x_2\, [ \cdots x_n]]] && x_i \in M .\end{align*} Show that \(\mathfrak{f}\) satisfies the universal property of a free Lie algebra of \(M\) (see 7.1.5).

It suffices to show that the following map is an isomorphism of hom sets: \begin{align*} \mathop{\mathrm{Hom}}_{{\mathsf{k}{\hbox{-}}\mathsf{Mod}}}(M, \mathop{\mathrm{Forget}}({\mathfrak{g}})) &\xrightarrow{\sim} \mathop{\mathrm{Hom}}_{{\mathsf{Lie}{\hbox{-}}\mathsf{Alg}}}({\mathfrak{f}}, {\mathfrak{g}}) \\ f &\mapsto \tilde f \\ \mkern 1.5mu\overline{\mkern-1.5mug\mkern-1.5mu}\mkern 1.5mu &\mapsfrom g ,\end{align*} where \(\mkern 1.5mu\overline{\mkern-1.5mug\mkern-1.5mu}\mkern 1.5mu\) is the restriction of \(g\) to only those sums with a single term and no iterated brackets. To define \(\tilde f\), first write \([{-}, {-}] \coloneqq[{-}, {-}]_{\mathfrak{g}}\) for the bracket on \({\mathfrak{g}}\). Using the compatibility of \(f\) with sums and brackets on \({\mathfrak{g}}\), given an element \(x \coloneqq\sum [x_1, [x_2, \cdots, x_n]] \in {\mathfrak{f}}\), we can pass \(f\) through each iterated bracket inductively: \begin{align*} f(x) &\coloneqq\tilde f\qty{ \sum [x_1, [x_2, \cdots, x_n]] } \\ &= \sum f \qty{ [x_1, [x_2, \cdots, x_n]] } \\ &= \sum [f(x_1), f\qty{ [x_2, \cdots, x_n]} ] \\ &= \sum [f(x_1), [ f(x_2), f \qty{ \cdots, x_n]} ] \\ &= \sum [f(x_1), [f(x_2), \cdots, f(x_n)]] .\end{align*} So we can extend \(f: M\to {\mathfrak{g}}\) to \(\tilde f: {\mathfrak{f}}\to {\mathfrak{g}}\) by linearity and bracketing to make the following definition: \begin{align*} \tilde f(x) \coloneqq\tilde f\qty{ \sum [x_1, [x_2, \cdots, x_n]] } &\coloneqq \sum [f(x_1), [f(x_2), \cdots, f(x_n)]] .\end{align*}

Since \(\tilde f: {\mathfrak{f}}\to {\mathfrak{g}}\) is a morphism on the underlying \(k{\hbox{-}}\)modules, it remains to check that it defines a morphism of Lie algebras. For this to be true, we need that \begin{align*} \tilde f \qty{ [x, y]_{\mathfrak{f}}} = [\tilde f(x), \tilde f(y)] ,\end{align*} where since \({\mathfrak{f}}\leq \mathsf{Lie}(T(M))\) is a subalgebra of the tensor algebra, its bracket is defined by \([x, y]_{{\mathfrak{f}}} \coloneqq[xy-yx]\). Since both brackets are bilinear, it suffices to check this on sums with a single term and extend by linearity. Moreover since \(\tilde f\) is defined inductively, it suffices to check on a single iteration of bracketing, in which case we have\

Note: I don’t see a clear way to get this result. Since \(f\) is a morphism of \(k{\hbox{-}}\)modules, one can easily get something like \begin{align*} f( [x, y]_{\mathfrak{f}}) \coloneqq f(xy-yx) = f(x)f(y) - f(y)f(x) = [f(x), f(y) ]_{{\mathfrak{g}}'} ,\end{align*} where \({\mathfrak{g}}'\) is the same underlying \(k{\hbox{-}}\)module as \({\mathfrak{g}}\) but with the bracket defined as \([a,b]_{{\mathfrak{g}}'} \coloneqq ab-ba\). However, this isn’t a priori related to the original bracket \([a, b]_{{\mathfrak{g}}}\).

Let \(M, N \in \mathsf{{\mathfrak{g}}}{\hbox{-}}\mathsf{Mod}\) and make \(\mathop{\mathrm{Hom}}_{{\mathsf{k}{\hbox{-}}\mathsf{Mod}}}(M, N)\) into a \({\mathfrak{g}}{\hbox{-}}\)module via the action \begin{align*} (xf)(m) \coloneqq xf(m) - f(xm) && x\in {\mathfrak{g}}, m\in M .\end{align*} Show that there is a natural isomorphism \begin{align*} \mathop{\mathrm{Hom}}_{{\mathfrak{g}{\hbox{-}}\mathsf{Mod}}}(M, N) \xrightarrow{\sim} \mathop{\mathrm{Hom}}_{{\mathsf{k}{\hbox{-}}\mathsf{Mod}}}(M, N)^{{\mathfrak{g}}} && \in{\mathfrak{g}{\hbox{-}}\mathsf{Mod}} .\end{align*}

The first claim is that these are equal as sets. This follows because for all \(x\in {\mathfrak{g}}\), for \(f\) to be a morphism of \({\mathfrak{g}}{\hbox{-}}\)modules it must be product-preserving, and so we have: \begin{align*} f\in \mathop{\mathrm{Hom}}_{{\mathfrak{g}{\hbox{-}}\mathsf{Mod}}}(M, N) &\iff xf(m) = f(xm) && \forall m\in M \\ &\iff xf(m) - f(xm) = 0_M && \forall m\in M \\ &\iff x\cdot f = 0 \\ &\iff f\in \mathop{\mathrm{Hom}}_{{\mathsf{k}{\hbox{-}}\mathsf{Mod}}}(M, N)^{{\mathfrak{g}}} ,\end{align*} where we’ve used the definition \(A^{{\mathfrak{g}}} \coloneqq\left\{{ a\in A\ {~\mathrel{\Big\vert}~}g a = 0_A\,\, \forall g\in{\mathfrak{g}}}\right\}\). So if we define a map \begin{align*} \tilde F: \mathop{\mathrm{Hom}}_{{\mathfrak{g}{\hbox{-}}\mathsf{Mod}}}(M, N) &\to \mathop{\mathrm{Hom}}_{{\mathsf{k}{\hbox{-}}\mathsf{Mod}}}(M, N)^{{\mathfrak{g}}} \\ f & \mapsto \tilde F_f \coloneqq F(f) \end{align*} where \(F: {\mathfrak{g}{\hbox{-}}\mathsf{Mod}}\to {\mathsf{k}{\hbox{-}}\mathsf{Mod}}\) is the forgetful functor, the above argument shows that this is a bijection. It only remains to check that \(\tilde F\) is a morphism of \({\mathfrak{g}}{\hbox{-}}\)modules, but this follows from the fact that \begin{align*} f(m) = F(f)(m) \coloneqq\tilde F_f(m) ,\end{align*} i.e. \(f\) and \(F(f)\) are pointwise defined in precisely the same way on the underlying sets. So \(\tilde F(xf) = x\tilde F(f)\), since this already holds for \(f\), making \(\tilde F\) product-preserving.

In the following construction, verify that \(d^2 = 0\) and conclude that \(V_* \in \mathsf{Ch}({\mathfrak{g}{\hbox{-}}\mathsf{Mod}})\).




Hint: write \(d(\theta_{i}) = \theta_{i, 1} + \theta_{i, 2}\) and show that \(-\theta_{i, 1}\) is the \(i=1\) part of \(\theta_{2, 1}\) and \(\theta_{2, 2} = 0\).

Then show that \(-\theta_{1, 2}\) is the \(i>1\) part of \(\theta_{2, 1}\).

Todo: start writing calculation.