# 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.

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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.

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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 $$f$$.

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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.