# User:Jan A. Sanders/An introduction to Lie algebra cohomology/Lecture 4

## Construction of an extension

We now turn the question around. What is the situation if we have Lie algebras $${\mathfrak{g}}$$ (projective) and $${\mathfrak{a}}$$ (abelian), and $$[a_2]\in H_{\wedge}^2({\mathfrak{g}},{\mathfrak{a}})\ ?$$

If a Lie section (that is, a section which is a Lie algebra morphism) is possible, then one can view $${\mathfrak{k}}$$ as the direct sum of $${\mathfrak{g}}$$ and $${\mathfrak{a}}\ .$$ What is the situation if $$[a_2]\in H_{\wedge}^2({\mathfrak{g}},{\mathfrak{a}})$$ is not zero?

To see how one should define a Lie algebra structure on the sum of $${\mathfrak{g}}$$ and $${\mathfrak{a}}\ ,$$ it pays to have a look at $${\mathfrak{k}}\ .$$

Every element in $$y\in{\mathfrak{k}}$$ can be uniquely written as $$\sigma(x)+\iota(a)\ ,$$ $$x\in{\mathfrak{g}}, a\in{\mathfrak{a}}\ :$$ take $$x=p(y)$$ and$$a=\iota^{-1} (1-\sigma p)(y) \ .$$

Let $$\psi:{\mathfrak{k}}\rightarrow {\mathfrak{a}}\oplus_R{\mathfrak{g}}$$ be defined by $$\psi(\sigma(x))+\iota(a))=(a,x)\ .$$

Then $\psi([\sigma(x)+\iota(a),\sigma(y)+\iota(b)])=\psi([\sigma(x),\sigma(y)]+[\iota(a),\sigma(y)]+[\sigma(x),\iota(b)])$ $=\psi(\sigma([x,y])-\iota(a_2(x,y))+[\iota(a),\sigma(y)]+[\sigma(x),\iota(b)])$ $=\psi(\sigma([x,y])-\iota(a_2(x,y))-\iota d_1(y)a(a)+\iota d_1(x)b)$ $=(d_1(x)b-d_1(y)a-a_2(x,y),[x,y])$

### definition

One now defines for an arbitrary representation $$d_1$$ and $$a_2\in C_{\wedge}^2({\mathfrak{g}},{\mathfrak{a}})\ ,$$ $[(a,x),(b,y)]=(d_1(x)b-d_1(y)a-a_2(x,y),[x,y])$

### lemma

The new bracket $$[\cdot,\cdot]_{a_2}$$ obeys the Jacobi identity as long as $$a_2\in Z_{\wedge}^2({\mathfrak{g}},{\mathfrak{a}})\ .$$

### proof

$[[(a,x),(b,y)],(c.z)]-[(a,x),[(b,x),(c,z)]]+[(b,y),[(a,x),(c,y)]]$ $=[({d_1}(x)b-{d_1}(y)a-a_2(x,y),[x,y]),(c,z)]$ $-[(a,x),({d_1}(y)c-{d_1}(z)b-a_2(y,z),[y,z])]$ $+[(b,y),({d_1}(x)c-{d_1}(z)a-a_2(x,z),[x,z])]$ $=({d_1}([x,y])c-{d_1}(z){d_1}(x)b+{d_1}(z){d_1}(y)a+{d_1}(z)a_2(x,y)-a_2([x,y],z),[[x,y],z])$ $-({d_1}(x){d_1}(y)c-{d_1}(x){d_1}(z)b-{d_1}(x)a_2(y,z)-{d_1}([y,z])a-a_2(x,[y,z],[x,[y,z]])$ $+({d_1}(y){d_1}(x)c-{d_1}(y){d_1}(z)a-{d_1}(y)a_2(x,z)-{d_1}([x,z])b-a_2(y,[x,z],[y,[x,z]])$ $=({d_1}([y,z])a-{d_1}(y){d_1}(z)a+{d_1}(z){d_1}(y)a$ $-{d_1}([x,z])b-{d_1}(z){d_1}(x)b+{d_1}(x){d_1}(z)b$ $+{d_1}([x,y])c-{d_1}(x){d_1}(y)c +{d_1}(y){d_1}(x)c$ $+{d_1}(x)a_2(y,z)-{d_1}(y)a_2(x,z)+{d_1}(z)a_2(x,y)$ $-a_2([x,y],z)-a_2(y,[x,z])+a_2(x,[y,z]),[[x,y],z] -[x,[y,z]] +[y,[x,z]])$ $=(d^{2}a_2(x,y,z),0)$ $=(0,0).$$$\square$$

### antisymmetry

One needs $$a_2\in Z_{\wedge}^2({\mathfrak{g}},{\mathfrak{a}})$$ in order to make the bracket antisymmetric.

### definition

We denote the new Lie algebra by $${\mathfrak{g}}{\oplus}_{a_2} {\mathfrak{a}}\ ,$$ the semidirect product of $${\mathfrak{g}}$$ and $${\mathfrak{a}}$$ induced by $$a_2 \in Z_{\wedge}^2({\mathfrak{g}},{\mathfrak{a}})\ .$$

### theorem

Let $$a_2$$ be as defined in lecture two. Then

$\mathfrak{k}\simeq \mathfrak{a}\oplus_{a_2} \mathfrak{g}\ .$

### proof

Consider $$\mathfrak{a}\oplus_{a_2}\mathfrak{g}\ ,$$ where the representation $$d_1$$ and the form $$a_2\in Z_{\wedge}^2(\mathfrak{g},\mathfrak{a})$$ are constructed as in the second lecture.

Let $$\phi: \mathfrak{a}\oplus_{a_2}\mathfrak{g}\rightarrow \mathfrak{k}$$ be defined by $$\phi((a,x))=\iota(a)+\sigma(x)\ .$$

Then $\phi([(a,x),(b,y)])=\phi((d_1(x)b-d_1(y)a-a_2(x,y),[x,y]))\ :$ $=\iota(d_1(x)b-d_1(y)a-a_2(x,y))+\sigma([x,y])\ :$ $=[\sigma(x),\iota(b)]+[\iota(a),\sigma(y)]+[\sigma(x),\sigma(y)]\ :$ $=[\iota(a)+\sigma(x),\iota(b)+\sigma(y)]\ :$ $=[\phi((a,x)),\phi((b,y))].$

Let now $$\psi:\mathfrak{k}\rightarrow \mathfrak{a}\oplus_{a_2}\mathfrak{g}$$ be defined by $\psi(x)=(\iota^{-1}(x-\sigma(p(x))),p(x))\ .$

Then $\psi([x,y])=(\iota^{-1}([x,y]-\sigma(p([x,y]))),p([x,y]))\ :$ $=(\iota^{-1}([x,y]-\sigma([p(x),p(y)])),[p(x),p(y)])\ :$ $=(\iota^{-1}([x,y]-[\sigma(p(x)),\sigma(p(y))]-\iota a_2(p(x),p(y))),[p(x),p(y))])\ :$ $=(\iota^{-1}([x-\sigma(p(x)),\sigma(p(y))]+[\sigma(p(x)),y-\sigma(p(y))]-\iota a_2(p(x),p(y))),[p(x),p(y))])\ :$ $=(\iota^{-1}([\sigma(p(x)),y-\sigma(p(y))])+\iota^{-1}([x-\sigma(p(x),\sigma(p(y))])-a_2(p(x),p(y)),[p(x),p(y)])\ :$ $=(d_1(p(x))\iota^{-1}(y-\sigma(p(y)))-d_1(p(y))\iota^{-1}(x-\sigma(p(x))-a_2(p(x),p(y)),[p(x),p(y)])\ :$ $=[(\iota^{-1}(x-\sigma(p(x))),p(x)),(\iota^{-1}(y-\sigma(p(y))),p(y))]\ :$ $=[\psi(x),\psi(y)]$ This implies that $$\phi$$ and $$\psi$$ are Lie algebra homomorphisms.

Furthermore, $\phi(\psi(x))=\phi((\iota^{-1}(x-\sigma(p(x))),p(x)))\ :$ $= x-\sigma(p(x))+\sigma(p(x)\ :$ $=x$ and $\psi(\phi(a,x))=\psi(\iota(a)+\sigma(x))\ :$ $=(\iota^{-1}(\iota(a)+\sigma(x)-\sigma(p(\iota(a)+\sigma(x))),\sigma(p(\iota(a)+\sigma(x)))\ :$ $=(\iota^{-1}(\iota(a)+\sigma(x)-\sigma(x)),\sigma(x))\ :$ $=(a,\sigma(x))$ This $$\phi$$ and $$\psi$$ are Lie algebra isomorphisms. One has

$\mathfrak{k}\simeq \mathfrak{a}\oplus_{a_2} \mathfrak{g}\quad\square$

What if one now applies the construction in the second lecture to $$\mathfrak{a}\oplus_{a_2} \mathfrak{g}\ ?$$

The maps $$\iota$$ and $$p$$ are given by $\iota(a)=(a,0)$ $p((a,x))=x$ From the definition of the bracket it follows that $$p$$ is a Lie algebra homomorphism, for $$\iota$$ this is trivial since $$\mathfrak{a}$$ is abelian.

One chooses a section $$\tilde{\sigma}$$ as follows. $\tilde{\sigma}(x)=(a_1(x),x),\quad a_1\in C^1(\mathfrak{g},\mathfrak{a})\ .$

Then $\iota(\tilde{d}_1(x)a=[\tilde{\sigma}(x),\iota(a)]\ :$ $=[(a_1(x),x),(a,0)]\ :$ $=(d_1(x)a,0)$ and $\iota(\tilde{d}_1(x)a=-[\iota(a),\tilde{\sigma}(x)]\ :$ $=-[((a,0),a_1(x),x)]\ :$ $=(d_1(x)a,0)$ It follows that $$\tilde{d}_1=d_1\ ,$$ which implies that the induced coboundary operators will be the same.

Now $\iota(\tilde{a}_2(x,y))=\tilde{\sigma}([x,y])-[\tilde{\sigma}(x),\tilde{\sigma}(y)]\ :$ $=(a_1([x,y]),[x,y])-[(a_1(x),x),(a_1(y),y)]\ :$ $=(a_1([x,y]),[x,y])-(d_1(x)a_1(y)-d_1(y)a_1(x)-a_2(x,y),[x,y])\ :$ $=(a_2(x,y)-d^1 a_1(x,y),0)$ and this implies $\tilde{a}_2=a_2-d^1 a_1\ .$ Furthermore

$\mathfrak{k}\simeq \mathfrak{a}\oplus_{a_2} \mathfrak{g}\simeq \mathfrak{a}\oplus_{\tilde{a}_2} \mathfrak{g}$

and if $$a_1\in Z^1(\mathfrak{g},\mathfrak{a})$$ one has

$\mathfrak{a}\oplus_{a_2} \mathfrak{g}= \mathfrak{a}\oplus_{\tilde{a}_2} \mathfrak{g}$

### theorem

To the short exact sequence of Lie algebras $0\rightarrow\mathfrak{a}\rightarrow\mathfrak{k}\rightarrow\mathfrak{g}\rightarrow 0$ is associated an $[a_2]\in H_{\wedge}^2(\mathfrak{g},\mathfrak{a})$ such that for any $$a_2\in[a_2]$$ one has

$\mathfrak{k}\simeq \mathfrak{a}\oplus_{a_2} \mathfrak{g}\ .$

### example - de Rham continued

Take $$a_2=0\ .$$ Then

$\mathfrak{k}\simeq \mathfrak{a}\oplus \mathfrak{g}$

can be identified with the Lie algebra of vector fields of the form $$\sum_{i=1}^n f_i \frac{\partial}{\partial x_i}\ .$$

#### exercise

What is the Lie bracket?$$\square$$

There is an induced representation $$d_1$$ of $$\mathfrak{k}$$ on $$\mathfrak{a}\ .$$

Observe that $$d g(\sum_{i=1}^n f_i \frac{\partial}{\partial x_i})=\sum_{i=1}^n f_i\frac{\partial g}{\partial x_i}=\sum_{i=1}^n f_i d g(\frac{\partial}{\partial x_i})\ ,$$ that is, $$d g$$ is an $$\mathfrak{a}$$-linear one-form.

Is the same true for $$d^1\ ?$$

Before we answer this question, we first observe that $$d x_j(\sum_{i=1}^n f_i \frac{\partial}{\partial x_i})= f_j\ ,$$ that is, we can write $$d g=\sum_{i=1}^n \frac{\partial g}{\partial x_i} d x_i\ .$$

Let $$a_1=\sum_{i=1}^n a_1^i d x_i\in C^1(\mathfrak{k},\mathfrak{a})$$ be an $$\mathfrak{a}$$-linear one-form.

Then

$d^1 a_1(\sum_{i=1}^n f_i\frac{\partial }{\partial x_i},\sum_{j=1}^n g_j\frac{\partial }{\partial x_j})=\sum_{i=1}^n f_i\frac{\partial}{\partial x_i}a_1(\sum_{j=1}^n g_j\frac{\partial }{\partial x_j}) -\sum_{j=1}^n g_j\frac{\partial }{\partial x_j}a_1(\sum_{i=1}^n f_i\frac{\partial}{\partial x_i})-a_1(\sum_{i=1}^n\sum_{j=1}^n \left(f_i\frac{\partial g_j}{\partial x_i}-g_i\frac{\partial f_j}{\partial x_i}\right) \frac{\partial }{\partial x_j}) \ :$

$=\sum_{i=1}^n f_i\frac{\partial}{\partial x_i}\sum_{j=1}^n g_j a_1(\frac{\partial }{\partial x_j}) -\sum_{j=1}^n g_j\frac{\partial }{\partial x_j}\sum_{i=1}^n f_ia_1(\frac{\partial}{\partial x_i})-\sum_{i=1}^n\sum_{j=1}^n\left(f_i\frac{\partial g_j}{\partial x_i}-g_i\frac{\partial f_j}{\partial x_i}\right) a_1(\frac{\partial }{\partial x_j}) \ :$ $=\sum_{i=1}^n f_i\sum_{j=1}^n g_j \frac{\partial}{\partial x_i}a_1(\frac{\partial }{\partial x_j}) -\sum_{j=1}^n g_j\sum_{i=1}^n f_i\frac{\partial }{\partial x_j}a_1(\frac{\partial}{\partial x_i}) \ :$ $=\sum_{i=1}^n \sum_{j=1}^n f_i g_j d^1 a_1(\frac{\partial}{\partial x_i},\frac{\partial }{\partial x_j}),$ that is, $$d^1 a_1$$ is an $$\mathfrak{a}$$-linear two-form.

Remark that $$d^1 a_1 (\frac{\partial}{\partial x_i},\frac{\partial }{\partial x_j})=\frac{\partial a_1^j}{\partial x_i}-\frac{\partial a_1^i}{\partial x_j}\ .$$

### exactness (Poincaré lemma)

Observe that $$d x_j(\sum_{i=1}^n f_i \frac{\partial}{\partial x_i})= f_j\ ,$$ that is, we can write $$d g=\sum_{i=1}^n \frac{\partial g}{\partial x_i} d x_i\ .$$

Given $$a_1=\sum_{i=1}^n a_1^i d x_i\in Z^1(\mathfrak{k},\mathfrak{a})\ ,$$ can we find $$a$$ such that $$a_1=d a\ ?$$

Consider the Euler vector field $$E=\sum_{i=1}^n x_i \frac{\partial}{\partial x_i}\ .$$

If the functions $$a_1^i$$ are homogeneous of degree $$k\ ,$$ then $$Ea_1(\frac{\partial}{\partial x_j})=\sum_{i=1}^n (E a_1^i)dx_i(\frac{\partial}{\partial x_j})-a_1^id x_i([E,\frac{\partial}{\partial x_j}]) =\sum_{i=1}^n k a_1^i d x_i(\frac{\partial}{\partial x_j}) +a_1^i d x_i( \frac{\partial}{\partial x_j})=(k+1)a_1(\frac{\partial}{\partial x_j})\ .$$

In other words, linear combinations of terms with degrees $$\neq -1$$ are exact by the homotopy method.

The general formula was $$a_n=d^{n-1}\lambda(a_n)^{-1}\iota^n(s)a_n\in B^n(\mathfrak{k},\mathfrak{a})\ ,$$ where in our case $$n=1, \lambda(a_1)=k+1, s=E\ .$$

In this particular case ($$a_1^i$$ are homogeneous of degree $$k$$) the formula reduces to $$a_1=d\frac{1}{k+1}a_1(E)=\frac{1}{k+1}d \sum_{i=1}^n x_i a_1^i \ .$$

### exercise

Is $$d^2 a_2$$ an $$\mathfrak{a}$$-linear three-form if $$a_2$$ is an $$\mathfrak{a}$$-linear two-form?

### solution

We see that

$\tag{1} d^2 a_2(f_i\frac{\partial}{\partial x_i},g_j\frac{\partial}{\partial x_j},h_k\frac{\partial}{\partial x_k})=\ :$

$=d_1(f_i\frac{\partial}{\partial x_i})a_2(g_j\frac{\partial}{\partial x_j},h_k\frac{\partial}{\partial x_k}) -d_1(g_j\frac{\partial}{\partial x_j})a_2(f_i\frac{\partial}{\partial x_i},h_k\frac{\partial}{\partial x_k}) +d_1(h_k\frac{\partial}{\partial x_k})a_2(f_i\frac{\partial}{\partial x_i},g_j\frac{\partial}{\partial x_j}) -a_2([f_i\frac{\partial}{\partial x_i},g_j\frac{\partial}{\partial x_j}],h_k\frac{\partial}{\partial x_k}) -a_2(g_j\frac{\partial}{\partial x_j},[f_i\frac{\partial}{\partial x_i},h_k\frac{\partial}{\partial x_k}]) +a_2(f_i\frac{\partial}{\partial x_i},[g_j\frac{\partial}{\partial x_j},h_k\frac{\partial}{\partial x_k}])\ .$

$=f_i\frac{\partial}{\partial x_i}g_jh_ka_2(\frac{\partial}{\partial x_j},\frac{\partial}{\partial x_k}) -g_j\frac{\partial}{\partial x_j}f_i h_k a_2(\frac{\partial}{\partial x_i},\frac{\partial}{\partial x_k}) +h_k\frac{\partial}{\partial x_k}f_i g_j a_2(\frac{\partial}{\partial x_i},\frac{\partial}{\partial x_j}) -h_k a_2([f_i\frac{\partial}{\partial x_i},g_j\frac{\partial}{\partial x_j}],\frac{\partial}{\partial x_k}) -g_j a_2(\frac{\partial}{\partial x_j},[f_i\frac{\partial}{\partial x_i},h_k\frac{\partial}{\partial x_k}]) +f_i a_2(\frac{\partial}{\partial x_i},[g_j\frac{\partial}{\partial x_j},h_k\frac{\partial}{\partial x_k}]) \ :$

$=\left( f_i\frac{\partial}{\partial x_i}g_jh_k -h_k f_i \frac{\partial g_j}{\partial x_i} -g_j f_i\frac{\partial h_k}{\partial x_i} \right) a_2^{j,k} +\left( f_i g_j \frac{\partial h_k}{\partial x_j} -g_j\frac{\partial}{\partial x_j}f_i h_k +h_k g_j \frac{\partial f_i}{\partial x_j} \right) a_2^{i,k} +g_j h_k \frac{\partial f_i}{\partial x_k} a_2^{j,i} +\left( h_k\frac{\partial}{\partial x_k}f_i g_j -f_i h_k \frac{\partial g_j}{\partial x_k} \right) a_2^{i,j} \ :$ $= f_ig_jh_k\frac{\partial}{\partial x_i} a_2^{j,k} -g_jf_i h_k \frac{\partial}{\partial x_j} a_2^{i,k} +g_j h_k \frac{\partial f_i}{\partial x_k} (a_{j,i}^2+a_2^{i,j}) +h_kf_i g_j\frac{\partial}{\partial x_k} a_2^{i,j} \ :$ $=f_ig_jh_k d^2 a_2(\frac{\partial}{\partial x_i},\frac{\partial}{\partial x_j},\frac{\partial}{\partial x_k}) +g_j h_k \frac{\partial f_i}{\partial x_k} (a_2^{j,i}+a_2^{i,j})$

and it follows that we need to require that $$a_2$$ is antisymmetric for $$d^2 a_2$$ to inherit the $$\mathfrak{a}$$-linearity.

This is the reason why Lie algebra cohomology is usually done with antisymmetric forms; apart from this historical reason there seems to be no good reason why this should be done.

The main results, Whitehead's first and second Lemma, remain valid in this more general framework, as well as other important constructions to be treated in the following lectures.

## Derivations

### definition - derivation

One says that $$\mathfrak{d}\in \mathrm{End}(\mathfrak{g})$$ is a derivation if $\mathfrak{d}[x,y]=[\mathfrak{d}x,y]+[x,\mathfrak{d}y].$ The space of derivations on $$\mathfrak{g}$$ is denoted by $$\mathfrak{der}(\mathfrak{g})\ .$$

### lemma - $$\mathfrak{der}(\mathfrak{g})$$ Lie algebra

$$\mathfrak{der}(\mathfrak{g})$$ is a Lie algebra with bracket $$[\mathfrak{d},\mathfrak{e}]=\mathfrak{d} \mathfrak{e} -\mathfrak{e} \mathfrak{d} \ .$$

### proof

Let $$\mathfrak{d} ,\mathfrak{e}\in \mathfrak{der}(\mathfrak{g})\ .$$ Then $$[\mathfrak{d},\mathfrak{e}]\in\mathfrak{der}(\mathfrak{g})$$ since $[\mathfrak{d},\mathfrak{e}][x,y]=\mathfrak{d} \mathfrak{e} [x,y]-\mathfrak{e} \mathfrak{d} [x,y]\ :$ $= \mathfrak{d} [\mathfrak{e} x , y]+ \mathfrak{d} [x,\mathfrak{e}y]- \mathfrak{e} [ \mathfrak{d} x , y ]-\mathfrak{e} [x,\mathfrak{d} y]\ :$ $=[\mathfrak{d} \mathfrak{e} x , y ]+ [\mathfrak{e} x , \mathfrak{d} y] + [\mathfrak{d} x, \mathfrak{e} y]+[x,\mathfrak{d} \mathfrak{e} y]-[\mathfrak{e} \mathfrak{d} x,y]-[\mathfrak{d} x, \mathfrak{e} y]-[\mathfrak{e} x , \mathfrak{d} y]- [x, \mathfrak{e} \mathfrak{d} y]\ :$ $=[(\mathfrak{d} \mathfrak{e}-\mathfrak{e} \mathfrak{d})x,y]+[x,(\mathfrak{d} \mathfrak{e}-\mathfrak{e} \mathfrak{d})y]\ :$ $=[[\mathfrak{d},\mathfrak{e}]x,y]+[x,[\mathfrak{d},\mathfrak{e}]y].$

### definition - inner derivations

For $$x\in\mathfrak{g}$$ one has $$\mathrm{ad}(x)\in\mathfrak{der}(\mathfrak{g})\ .$$

Such derivations are called inner derivations, denoted by $$\mathfrak{inn}(\mathfrak{g})\ .$$

### lemma - ideal

The inner derivations form an ideal in $$\mathfrak{der}(\mathfrak{g})\ .$$

### proof

One has, with $$x,y\in\mathfrak{g}\ ,$$ $[\mathfrak{d},\mathrm{ad}(x)]y=\mathfrak{d}[x,y]-[x,\mathfrak{d}y]\ :$ $=[\mathfrak{d}x,y]\ :$ $=\mathrm{ad}(\mathfrak{d}x)y\ .$

### definition - outer derivations

The factor space $$\mathfrak{out}(\mathfrak{g})= \mathfrak{der}(\mathfrak{g})/\mathfrak{inn}(\mathfrak{g})$$ consists of the outer derivations.

## Crossed Modules

Material in this section is taken from Wagemann (2006).

### definition - crossed modules

A crossed module of Lie algebras is a homomorphism $$\mu:\mathfrak{m}\rightarrow \mathfrak{n}$$ and an action $$\eta:\mathfrak{n}\rightarrow\mathfrak{der}(\mathfrak{m})$$ of such that

• $\mu(\eta(n)\cdot m)=[n,\mu(m)],\quad n\in\mathfrak{n}, m\in \mathfrak{m}\ ,$
• $\eta(\mu(m))\cdot m'=[m,m'],\quad m,m'\in\mathfrak{m}\ .$

The aim of this lecture is to show the existence of a $${[\gamma_3]}\in H^3(\mathfrak{n}/\mathrm{im\ }\mu,\ker \mu)\ .$$

Remark that $$\eta$$ has to be an element of $$\mathfrak{der}(\mathfrak{m})$$ in order to be compatible with the Jacobi identity in $$\mathfrak{m}\ :$$

$\eta(\mu(m))\cdot [m',m'']=[m,[m',m'']]=[m',[m,m'']]+[[m,m'],m'']=[m',\eta(\mu(m))m'']+[\eta(\mu(m))m',m'']\ .$

### definition - exact sequence

To a crossed module one associates an exact sequence $0\rightarrow \mathfrak{a}\stackrel{\iota}{\rightarrow}\mathfrak{m}\stackrel{\mu}{\rightarrow}\mathfrak{n}\stackrel{\pi}{\rightarrow}\mathfrak{g}\rightarrow 0.$ with $$\mathfrak{a}=\ker \mu$$ and $$\mathfrak{g}=\mathrm{coker\ } \mu\ .$$

### proposition

$$\mathfrak{a}$$ is in the center of $$\mathfrak{m}$$ (and therefore abelian).

### proof

Let $$a\in\mathfrak{a}, m\in\mathfrak{m}\ .$$ Then $[\iota a , m]=\eta(\mu \iota a) \cdot m=0.\square$

Let $$\alpha_1$$ be a linear section of $$\pi\ .$$

Let $$\alpha_2(x,y)=[\alpha_1(x),\alpha_1(y)]-\alpha_1([x,y]), x,y \in \mathfrak{g}, \alpha_2\in C_\wedge^2(\mathfrak{g},\mathfrak{n})\ .$$

Then $$\pi\alpha_2(x,y)=0$$ and $$\alpha_2(x,y)\in\ker\pi=\mathrm{im\ }\mu\ .$$

At some point we need to show that the construction is independent of our choices.

Let $$\alpha_1+\delta\alpha_1$$ be a different linear section to $$\pi\ .$$

One sees that $$0=\pi\delta\alpha_1(x)\ ,$$ that is $$\delta\alpha_1(x)=\mu(\kappa_1(x)), \kappa_1\in C^1(\mathfrak{g},\mathfrak{m})\ .$$

Then $\delta\alpha_2(x,y)=[\alpha_1(x)+\delta\alpha_1(x),\alpha_1(y)+\delta\alpha_1(y)]-\alpha_1([x,y])-\delta\alpha_1([x,y])-[\alpha_1(x),\alpha_1(y)]+\alpha_1([x,y])\ :$ $=[\alpha_1(x),\delta\alpha_1(y)]+[\delta\alpha_1(x),\alpha_1(y)]+[\delta\alpha_1(x),\delta\alpha_1(y)]-\delta\alpha_1([x,y])\ :$ $=[\alpha_1(x),\mu(\kappa_1(y))]+[\mu(\kappa_1(x)),\alpha_1(y)]+[\mu(\kappa_1(x)),\mu(\kappa_1(y))]-\delta\alpha_1([x,y])\ :$ $=\mu(\eta(\alpha_1(x))\cdot \kappa_1(y)-\eta(\alpha_1(y))\cdot\kappa_1(x)-\kappa_1([x,y])+[\kappa_1(x),\kappa_1(y)])$ One can choose a linear section $$\sigma: \mathfrak{n}\rightarrow\mathfrak{m}$$ of $$\mu$$ and take $$\beta_2(x,y)=\sigma\alpha_2(x,y)\ .$$

Then $$\mu\beta_2(x,y)=\alpha_2(x,y)$$ and $$\beta_2\in C_\wedge^2(\mathfrak{g},\mathfrak{m})\ .$$

Let $$\partial_1(x) m=\eta(\alpha_1(x))m\ .$$

This is not in general an action, as long as $$\alpha_1$$ is not a Lie algebra section.

Indeed, $\partial_2(x,y)=[\partial_1(x),\partial_1(y)]m-\partial_1([x,y])m\ :$ $=[\eta(\alpha_1(x)),\eta(\alpha_1(y))]m-\eta(\alpha_1([x,y]))m\ :$ $=\eta([\alpha_1(x),\alpha_1(y)])m-\eta([\alpha_1(x),\alpha_1(y)])m+\eta(\alpha_2(x,y))m\ :$ $=\eta(\mu\beta_2(x,y))m\ :$ $=[\beta_2(x,y),m]\ :$ $=\mathrm{ad}(\beta_2(x,y))m$ or, $$\partial_2(x,y)=\mathrm{ad}(\beta_2(x,y))\ .$$

This implies $\partial^2\partial^1\kappa_1(x,y,z)=\partial_1(x)\partial^1\kappa_1(y,z)-\partial_1(y)\partial^1\kappa_1(x,z)+\partial_1(z)\partial^1\kappa_1(x,y)-\partial^1\kappa_1([x,y],z) -\partial^1\kappa_1(y,[x,z])+\partial^1\kappa_1(x,[y,z])\ :$ $=\partial_1(x)\partial_1(y)\kappa_1(z)-\partial_1(x)\partial_1(z)\kappa_1(y)-\partial_1(x)\kappa_1([y,z]) -\partial_1(y)\partial_1(x)\kappa_1(z)+\partial_1(y)\partial_1(z)\kappa_1(x)+\partial_1(y)\kappa_1([x,z])\ :$

$+\partial_1(z)\partial_1(x)\kappa_1(y)-\partial_1(z)\partial_1(y)\kappa_1(x)-\partial_1(z)\kappa_1([x,y]) -\partial_1([x,y])\kappa_1(z)+\partial_1(z)\kappa_1([x,y])+\kappa_1([[x,y],z])\ :$
$-\partial_1(y)\kappa_1([x,z])+\partial_1([x,z])\kappa_1(y)+\kappa_1([y,[x,z]]) +\partial_1(x)\kappa_1([y,z])-\partial_1([y,z])\kappa_1(x)-\kappa_1([x,[y,z]])\ :$

$= \partial_1(x)\partial_1(y)\kappa_1(z)-\partial_1(y)\partial_1(x)\kappa_1(z)-\partial_1([x,y])\kappa_1(z)\ :$

$+\partial_1(z)\partial_1(x)\kappa_1(y)-\partial_1(x)\partial_1(z)\kappa_1(y)+\partial_1([x,z])\kappa_1(y)\ :$
$+\partial_1(y)\partial_1(z)\kappa_1(x)-\partial_1(z)\partial_1(y)\kappa_1(x)-\partial_1([y,z])\kappa_1(x) \ :$

$=[\beta_2(x,y),\kappa_1(z)]-\beta_2(x,z),\kappa_1(y)+[\beta_2(y,z),\kappa_1(x)]\ .$ Ignoring this fact, we write (using the usual formula for the coboundary operator in terms of the representation) $\delta\alpha_2(x,y)=\mu(\partial^1 \kappa_1(x,y)+[\kappa_1(x),\kappa_1(y)])$ or $\delta\beta_2(x,y)=\partial^1 \kappa_1(x,y)+[\kappa_1(x),\kappa_1(y)]\ .$ Compute $$\partial^2 \beta_2\in C_\wedge^3(\mathfrak{g},\mathfrak{m})\ ,$$ using the same formulas as one would when $$\partial_1$$ was an action.

### proposition

$\mu\partial^2 \beta_2(x,y,z)=0\ ,$ that is, $$\partial^2 \beta_2(x,y,z)\in\ker\mu=\mathrm{im\ }\iota\ .$$

### proof

Notice that we do use the antisymmetry of the Lie bracket here. $\mu\partial^2 \beta_2(x,y,z)=\mu\partial_1(x)\beta_2(y,z)-\mu\partial_1(y)\beta_2(x,z)+\mu\partial_1(z)\beta_2(x,y)-\mu\beta_2([x,y],z)-\mu\beta_2(y,[x,z])+\mu\beta_2(x,[y,z])\ :$ $=\mu\eta(\rho(x))\cdot\beta_2(y,z)-\mu\eta(\rho(y)\cdot\beta_2(x,z)+\mu\eta(\rho(z)\cdot\beta_2(x,y)-\alpha_2([x,y],z))-\alpha_2(y,[x,z])+\alpha_2(x,[y,z])\ :$ $=[\rho(x),\alpha_2(y,z)]-[\rho(y),\alpha_2(x,z)]+[\rho(z),\alpha_2(x,y)]-\alpha_2([x,y],z))-\alpha_2(y,[x,z])+\alpha_2(x,[y,z])\ :$ $=[\rho(x),[\rho(y),\rho(z)]]-[\rho(x),\rho([y,z]])]-[\rho(y),[\rho(x),\rho(z)]]+[\rho(y),\rho([x,z])]+[\rho(z),[\rho(x),\rho(y)]]-[\rho(z),\rho([x,y])] -[\rho([x,y]),\rho(z)]-\rho([[x,y],z])-[\rho(y),\rho([x,z])]+\rho([y,[x,z]])+[\rho(x),\rho([y,z])]-\rho([x,[y,z]])\ :$ $=[\rho(x),[\rho(y),\rho(z)]]-[\rho(y),[\rho(x),\rho(z)]]+[\rho(z),[\rho(x),\rho(y)]] +\rho(-[[x,y],z]+[y,[x,z]]-[x,[y,z]])\ :$ $=0.$

### corollary

Let $$\tau:\mathfrak{m}\rightarrow \mathfrak{a}$$ be a linear section of $$\iota\ .$$

Define $$\gamma_3(x,y,z)=\tau\partial^2 \beta_2(x,y,z)\ .$$

Then $$\gamma_3\in C_\wedge^3(\mathfrak{g},\mathfrak{a})\ .$$

### remark - independence of choice

If $$\tau + \delta\tau$$ is another section, leading to $$\gamma_3+\delta\gamma_3\ ,$$ then $\delta\gamma_3=\delta\tau\partial^2 \beta_2\in\ker\iota=0\ ,$ that is, the construction of $$\gamma_3$$ is independent of the choice of $$\tau\ .$$

### lemma

Let $$\iota d^{(0)}(x)a=\eta(\rho(x))\iota a\ .$$ Then $$d^{(0)}$$ is a representation of $$\mathfrak{g}$$ on $$\mathfrak{a}\ .$$

### proof

First of all, $\mu \eta(\rho(x))\iota a=[\rho(x),\mu\iota a]=0\ .$ Since $$\eta(\mu(m))\iota(a)=[m,\iota a]=0\ ,$$ ($$\mathfrak{a}$$ is in the center of $$\mathfrak{m}$$) the definition does not depend on the choice of $$\rho\ .$$

This means that $$d_1(x)a$$ is well defined. Furthermore $\iota d_1([x,y])a=\eta(\rho([x,y]))\iota a\ :$ $=\eta([\rho(x),\rho(y)])\iota a-\eta(\alpha_2(x,y))\iota a\ :$ $=[\eta(\rho(x)),\eta(\rho(y)])\iota a-\eta(\mu(\beta_2(x,y)))\iota a\ :$ $=\eta(\rho(x))\eta(\rho(y)\iota a-\eta(\rho(y))\eta(\rho(x)\iota a-[\beta_2(x,y),\iota a]\ :$ $=\eta(\rho(x))\iota d_1(y)a -\eta(\rho(y))\iota d_1(x)a\ :$ $=\iota d_1(x)d_1(y)a-\iota d_1(y)d_1(x)a\ :$ $=\iota[d_1(x),d_1(y)]a,$ that is, $$d_2(x,y)=0\ .$$

### theorem - closed form

$d^3 \gamma_3=0\ .$

### proof

$\iota d^3 \gamma_3(x,y,z,w)=\iota d_1(x)\gamma_3(y,z,w)-\iota d_1(y)\gamma_3(x,z,w)+\iota d_1(z)\gamma_3(x,y,w)-\iota d_1(w)\gamma_3(x,y,z) -\iota \gamma_3([x,y],z,w)-\iota \gamma_3(,y,[x,z],w)-\iota \gamma_3(y,z,[x,w])+\iota \gamma_3(x,[y,z],w)+\iota \gamma_3(x,z,[y,w])-\iota\gamma_3(x,y,[z,w])\ :$ $=\eta(\rho(x))\iota\gamma_3(y,z,w)-\eta(\rho(y))\iota\gamma_3(x,z,w)+\eta(\rho((z)))\iota\gamma_3(x,y,w)-\eta(\rho(w))\iota\gamma_3(x,y,z) -\iota \gamma_3([x,y],z,w)-\iota \gamma_3(,y,[x,z],w)-\iota \gamma_3(y,z,[x,w])+\iota \gamma_3(x,[y,z],w)+\iota \gamma_3(x,z,[y,w])-\iota\gamma_3(x,y,[z,w])\ :$ $=\eta(\rho(x))\partial^2\beta_2(y,z,w)-\eta(\rho(y))\partial^2\beta_2(x,z,w)+\eta(\rho((z)))\partial^2\beta_2(x,y,w)-\eta(\rho(w))\partial^2\beta_2(x,y,z) -\partial^2\beta_2([x,y],z,w)-\partial^2\beta_2(y,[x,z],w)-\partial^2\beta_2(y,z,[x,w])+\partial^2\beta_2(x,[y,z],w)+\partial^2\beta_2(x,z,[y,w])-\partial^2\beta_2(x,y,[z,w])\ :$ $= \eta(\rho(x))\partial_1(y)\beta_2(z,w)-\eta(\rho(x))\partial_1(z)\beta_2(y,w)+\eta(\rho(x))\partial_1(w)\beta_2(y,z) -\eta(\rho(x))\beta_2([y,z],w)-\eta(\rho(x))\beta_2(z,[y,w])+\eta(\rho(x))\beta_2(y,[z,w])\ :$

$-\eta(\rho(y))\partial_1(x)\beta_2(z,w)+\eta(\rho(y))\partial_1(z)\beta_2(x,w)-\eta(\rho(y))\partial_1(w)\beta_2(x,z) +\eta(\rho(y))\beta_2([x,z],w)+\eta(\rho(y))\beta_2(z,[x,w])-\eta(\rho(y))\beta_2(x,[z,w]) \ :$
$+\eta(\rho(z))\partial_1(x)\beta(y,w)-\eta(\rho(z))\partial_1(y)\beta(x,w)+\eta(\rho(z))\partial_1(w)\beta_2(x,y) -\eta(\rho(z))\beta_2([x,y],w)-\eta(\rho(z))\beta_2(y,[x,w])+\eta(\rho(z))\beta_2(x,[y,w])\ :$
$-\eta(\rho(w))\partial_1(x)\beta_2(y,z)+\eta(\rho(w))\partial_1(y)\beta_2(x,z)-\eta(\rho(w))\partial_1(z)\beta_2(x,y) +\eta(\rho(w))\beta_2([x,y],z)+\eta(\rho(w))\beta_2(y,[x,z])-\eta(\rho(w))\beta_2(x,[y,z])\ :$
$-\partial_1([x,y])\beta_2(z,w)+\partial_1(z)\beta_2([x,y],w)-\partial_1(w)\beta_2([x,y],z) +\beta_2([[x,y],z],w)+\beta_2(z,[[x,y],w])-\beta_2([x,y],[z,w])\ :$
$-\partial_1(y)\beta_2([x,z],w)+\partial_1([x,z])\beta_2(y,w)-\partial_1(w)\beta_2(y,[x,z]) +\beta_2([y,[x,z]],w)+\beta_2([x,z],[y,w])-\beta_2(y,[[x,z],w])\ :$
$-\partial_1(y)\beta_2(z,[x,w])+\partial_1(z)\beta_2(y,[x,w])-\partial_1([x,w])\beta_2(y,z) +\beta_2([y,z],[x,w])+\beta_2(z,[y,[x,w]])-\beta_2(y,[z,[x,w]])\ :$
$+\partial_1(x)\beta_2([y,z],w)-\partial_1([y,z])\beta_2(x,w)+\partial_1(w)\beta_2(x,[y,z]) -\beta_2([x,[y,z]],w)-\beta_2([y,z],[x,w])+\beta_2(x,[[y,z],w])\ :$
$+\partial_1(x)\beta_2(z,[y,w])-\partial_1(z)\beta_2(x,[y,w)+\partial_1([y,w])\beta_2(x,z) -\beta_2([x,z],[y,w])-\beta_2(z,[x,[y,w]])+\beta_2(x,[z,[y,w]])\ :$
$-\partial_1(x)\beta_2(y,[z,w])+\partial_1(y)\beta_2(x,[z,w])-\partial_1([z,w])\beta_2(x,y) +\beta_2([x,y],[z,w])+\beta_2(y,[x,[z,w]])-\beta_2(x,[y,[z,w]])\ :$

$= \eta([\rho(x),\rho(y)])\beta_2(z,w)-\eta(\rho([x,y]))\beta_2(z,w)\ :$

$+\eta([\rho(x),\rho(w)])\beta_2(y,z)-\eta(\rho([x,w]))\beta_2(y,z) \ :$
$-\eta([\rho(x),\rho(z)])\beta_2(y,w)+\eta(\rho([x,z]))\beta_2(y,w)\ :$
$-\eta([\rho(y),\rho(w)])\beta_2(x,z)+\eta(\rho([y,w]))\beta_2(x,z)\ :$
$+\eta([\rho(y),\rho(z)])\beta_2(x,w)-\eta(\rho([y,z]))\beta_2(x,w)\ :$
$+\eta([\rho(z),\rho(w)])\beta_2(x,y)-\eta(\rho([z,w]))\beta_2(x,y)\ :$

$=\eta(\alpha_2(x,y))\beta_2(z,w)+\eta(\alpha_2(z,w))\beta_2(x,y))-\eta(\alpha_2(x,z))\beta_2(y,w)-\eta(\alpha_2(y,w))\beta_2(x,z)+\eta(\alpha_2(y,z))\beta_2(x,w)+\eta(\alpha_2(x,w))\beta_2(y,z)\ :$ $=\eta(\mu\beta_2(x,y))\beta_2(z,w)+\eta(\mu\beta_2(z,w))\beta_2(x,y))-\eta(\mu\beta_2(x,z))\beta_2(y,w)-\eta(\mu\beta_2(y,w))\beta_2(x,z)+\eta(\mu\beta_2(y,z))\beta_2(x,w)+\eta(\mu\beta_2(x,w))\beta_2(y,z)\ :$ $=[\beta_2(x,y),\beta_2(z,w)]+[\beta_2(z,w),\beta_2(x,y)]-[\beta_2(x,z),\beta_2(y,w)]-[\beta_2(y,w),\beta_2(x,z)]+[\beta_2(y,z),\beta_2(x,w)]+[\beta_2(x,w),\beta_2(y,z)]\ :$ $=0.$

### lemma - independent of choice

The construction of $$\gamma_3$$ is independent of the choice of $$\rho\ .$$

### proof

Recall that $\delta\beta_2(x,y)=(\eta(\rho(x))\cdot \kappa_1(y)-\eta(\rho(y))\cdot\kappa_1(x)-\kappa_1([x,y])+[\kappa_1(x),\kappa_1(y)])$ Then $\delta\gamma_3(x,y,z)=(\tau\eta(\rho(x)+\delta\rho(x))(\beta_2(y,z)+\delta\beta_2(y,z))-\tau\eta(\rho(y)+\delta\rho(y))(\beta_2(x,z)+\delta\beta_2(x,z))+\tau\eta(\rho(z)+\delta\rho(z))(\beta_2(x,y)+\delta\beta_2(x,y)-\tau(\beta_2+\delta\beta_2)([x,y],z)-\tau(\beta_2+\delta\beta_2)(y,[x,z])+\tau(\beta_2+\delta\beta_2)(x,[y,z]))$ $-(\tau\eta(\rho(x))\beta_2(y,z)-\tau\eta(\rho(y))\beta_2(x,z)+\tau\eta(\rho(z)\beta_2(x,y)-\tau\beta_2([x,y],z)-\tau\beta_2(y,[x,z])+\tau\beta_2(x,[y,z]))$

$= +\tau\eta(\delta\rho(x))\beta_2(y,z) -\tau\eta(\delta\rho(y))\beta_2(x,z) +\tau\eta(\delta\rho(z))\beta_2(x,y) +\tau\eta(\delta\rho(x))\delta\beta_2(y,z)-\tau\eta(\delta\rho(y))\delta\beta_2(x,z)+\tau\eta(\delta\rho(z))\delta\beta_2(x,y)+\tau\eta(\rho(x))\delta\beta_2(y,z)-\tau\eta(\rho(y))\delta\beta_2(x,z)+\tau\eta(\rho(z))\delta\beta_2(x,y) -\tau\delta\beta_2([x,y],z)-\tau\delta\beta_2(y,[x,z])+\tau\delta\beta_2(x,[y,z])$

$= +\tau\eta(\mu\kappa_1(x))\beta_2(y,z) -\tau\eta(\mu\kappa_1(y))\beta_2(x,z) +\tau\eta(\mu\kappa_1(z))\beta_2(x,y) \ :$ $+\tau\eta(\mu\kappa_1(x))\partial^1\kappa_1(y,z)+\tau\eta(\mu\kappa_1(x))[\kappa_1(y),\kappa_1(z)] -\tau\eta(\mu\kappa_1(y))\partial^1\kappa_1(x,z)-\tau\eta(\mu\kappa_1(y))[\kappa_1(x),\kappa_1(z)] +\tau\eta(\mu\kappa_1(z))\partial^1\kappa_1(x,y)+\tau\eta(\mu\kappa_1(z))[\kappa_1(x),\kappa_1(y)] \ :$ $+\tau\eta(\rho(x))\partial^1\kappa_1(y,z)+\tau\eta(\rho(x))[\kappa_1(y),\kappa_1(z)] -\tau\eta(\rho(y))\partial^1\kappa_1(x,z)-\tau\eta(\rho(y))[\kappa_1(x),\kappa_1(z)] +\tau\eta(\rho(z))\partial^1\kappa_1(x,y)+\tau\eta(\rho(z)[\kappa_1(x),\kappa_1(y)] \ :$ $-\tau\partial^1\kappa_1([x,y],z)-\tau[\kappa_1([x,y])\kappa_1(z)] -\tau\partial^1\kappa_1(y,[x,z])-\tau[\kappa_1(y),\kappa_1([x,z])] +\tau\partial^1\kappa_1(x,[y,z])+\tau[\kappa_1(x),\kappa_1([y,z])]$

$= +\tau[\kappa_1(x),\beta_2(y,z)] -\tau[\kappa_1(y),\beta_2(x,z)] +\tau[\kappa_1(z),\beta_2(x,y)] \ :$ $+\tau[\kappa_1(x),\partial^1\kappa_1(y,z)]+\tau[\kappa_1(x),[\kappa_1(y),\kappa_1(z)]] -\tau[\kappa_1(y),\partial^1\kappa_1(x,z)]-\tau[\kappa_1(y),[\kappa_1(x),\kappa_1(z)]] +\tau[\kappa_1(z),\partial^1\kappa_1(x,y)]+\tau[\kappa_1(z),[\kappa_1(x),\kappa_1(y)]] \ :$ $+\tau\eta(\rho(x))\partial^1\kappa_1(y,z)+\tau\eta(\rho(x))[\kappa_1(y),\kappa_1(z)] -\tau\eta(\rho(y))\partial^1\kappa_1(x,z)-\tau\eta(\rho(y))[\kappa_1(x),\kappa_1(z)] +\tau\eta(\rho(z))\partial^1\kappa_1(x,y)+\tau\eta(\rho(z)[\kappa_1(x),\kappa_1(y)] \ :$ $-\tau\partial^1\kappa_1([x,y],z)-\tau[\kappa_1([x,y]),\kappa_1(z)] -\tau\partial^1\kappa_1(y,[x,z])-\tau[\kappa_1(y),\kappa_1([x,z])] +\tau\partial^1\kappa_1(x,[y,z])+\tau[\kappa_1(x),\kappa_1([y,z])]$

$= +\tau[\kappa_1(x),\beta_2(y,z)]+\tau[\kappa_1(x),\partial^1\kappa_1(y,z)]+\tau[\kappa_1(x),\kappa_1([y,z])]-\tau[\kappa_1(x),\eta(\rho(y))\kappa_1(z)]+\tau[\kappa_1(x),\eta(\rho(z)\kappa_1(y)]\ :$ $-\tau[\kappa_1(y),\beta_2(x,z)]-\tau[\kappa_1(y),\partial^1\kappa_1(x,z)]-\tau[\kappa_1(y),\kappa_1([x,z])]+\tau[\kappa_1(y),\eta(\rho(x))\kappa_1(z)]+\tau[\eta(\rho(z)\kappa_1(x),\kappa_1(y)]\ :$ $+\tau[\kappa_1(z),\beta_2(x,y)]+\tau[\kappa_1(z),\partial^1\kappa_1(x,y)]-\tau[\kappa_1([x,y]),\kappa_1(z)]+\tau[\eta(\rho(x))\kappa_1(y),\kappa_1(z)]-\tau[\eta(\rho(y))\kappa_1(x),\kappa_1(z)] \ :$ $+\tau\eta(\rho(x))\partial^1\kappa_1(y,z) -\tau\eta(\rho(y))\partial^1\kappa_1(x,z) +\tau\eta(\rho(z))\partial^1\kappa_1(x,y) \ :$ $-\tau\partial^1\kappa_1([x,y],z) -\tau\partial^1\kappa_1(y,[x,z]) +\tau\partial^1\kappa_1(x,[y,z])$

$= \tau[\kappa_1(x),\beta_2(y,z)]-\tau[\kappa_1(y),\beta_2(x,z)]+\tau[\kappa_1(z),\beta_2(x,y)] +\tau\partial^2\partial^1\kappa_1(x,y,z)$

$=0.\quad\square$

### crossed modules - literature

• Wagemann, Friedrich . On Lie algebra crossed modules. Comm. Algebra 34 (2006), no. 5, 1699--1722.
• Mori, Mitsuya . On the three-dimensional cohomology group of Lie algebras. J. Math. Soc. Japan 5, (1953). 171--183.
• Hochschild, G. Lie algebra kernels and cohomology. Amer. J. Math. 76, (1954). 698--716.