# A Brief Introduction to Cohomology of Lie Algebra

Differential Geometry

Let $$G$$ be a compact simply connected Lie group and $$\mathfrak{g}$$ its Lie algebra. In this posts, we will discuss the motivation of Lie algebra cohomology of $$\mathfrak{g}$$ and its connection with the de Rham cohomology of $$G$$.

Author
Published

January 7, 2021

We fix a field $$\mathbb{k}$$. All Lie algebras and vector spaces will be considered over $$\kk$$.

## 1 Representations of Lie Algebras

Let $$\mathfrak{g}$$ be a Lie algebra, $$M$$ be a vector space $$M$$, and $$\Lgl(M)$$ be the space of endomorphisms of $$M$$. The vector space $$\Lgl(M)$$ admits a Lie algebra structure given by $$[A, B]=A\circ B-B\circ A$$.

Definition 1 (Representations of a Lie Algebra) The vector space $$M$$ is said to be a representation of $$\mathfrak{g}$$ if there is a Lie algebra homomorphism $$\rho: \Lg\to \Lgl(M)$$, i.e. $$\rho$$ is a linear map such that $$\rho([x, y])=[\rho(x),\rho(y)]$$.

Definition 2 (Modules over a Lie Algebra) A vector space $$M$$ is said to admit a $$\Lg$$-module structure if there is a $$\kk$$-bilinear map $$\pi: \Lg\otimes_{\kk}M\to M$$ such that $$[x,y]\cdot m=x\cdot(y\cdot m)-y\cdot (x\cdot m)$$, where $$x\cdot m:=\pi(x, m)$$.

One can check directly that a representation $$\rho:\Lg\to \Lgl(M)$$ defines a $$\Lg$$-module structure on $$M$$: $$g\cdot m=\rho(g)(m)$$. Conversely, a $$\Lg$$-module structure on $$M$$ defines a representation.

Example 1 (Adjoint representation) Let $$\Lg$$ be a Lie algebra. Consider the map $$\ad: \Lg\to \Lgl(\Lg)$$ defined by $$x\mapsto \ad_x:=[x, -]$$. By Jacobian identity, the map $$\ad$$ defines representation $$\Lg$$ of $$\Lg$$ called an adjoint representation. Indeed, \begin{aligned} \ad([x, y])(z)=&[[x, y], z]\\ =&[x, [y, z]]-[y, [x,z]]\\ =&\ad_x(\ad_y(z))-\ad_y(\ad_x(z))\\ =&[\ad_x,\ad_y](z). \end{aligned}

Note that an adjoint is a $$\kk$$-derivation of $$\Lg$$, that is $$\ad_x$$ satisfies the Leibniz rule for the Lie bracket

$\ad_x[g,h]=[\ad_x(g),h]+[g,\ad_x(h)].$

Example 2 (Coadjoint representation) Given any Lie algebra $$\Lg$$, denote by $$\Lg^*$$ the dual vector space, the space for $$\kk$$-linear functions on $$\Lg$$. We define $$<\xi, x>=\xi(x)$$, where $$\xi\in\Lg^*$$ and $$x\in \Lg$$. Consider the map $$\ad^*: \Lg \to \Lgl(\Lg^*)$$ defined by $x \mapsto \ad_x^*: \Lg^* \to \Lg^*,$ where $$\ad_x^*$$ is given by $$\ad_x^*(\xi)(y)=-<\xi, [x, y]>=-\xi([x, y])$$ for any $$\xi\in \Lg^*$$ and $$y\in\Lg$$.

The map $$\ad^*$$ defines representation of $$\Lg$$ in $$\Lg^*$$ called an coadjoint representation of $$\Lg$$. For any $$x, y, z$$ in $$\Lg$$ and $$\xi$$ in $$\Lg^*$$, we see that \begin{aligned} \ad_{[x, y]}^*(\xi)(z)=&-<\xi, [[x, y], z]>\\ =&-<\xi, [x, [y, z]]>+<\xi, [y, [x,z]]>\\ =&\ad_x^*(\xi)([y,z])-\ad_y^*(\xi)([x,z])\\ =&-\ad_y^*(\ad_x^*(\xi))(z)+\ad_x^*(\ad_y^*(\xi))(z)\\ =&[\ad_x^*,\ad_y^*](\xi)(z). \end{aligned}

Note that the negative sign in the definition of $$\ad_x^*$$ is necessary so that $$\ad$$ is a Lie algebra homomorphism.

Example 3 (Tensor product of representations) Given two representations $$\varphi: \Lg\to\Lgl(M)$$ and $$\psi: \Lg\to \Lgl(N)$$, there is a natural representation on the tensor product $$M\otimes N$$ given by $(\varphi\otimes\psi)(g)=\varphi(g)\otimes I+I\otimes\psi(g),$ for any $$g$$ in $$\Lg$$. We can check that $$\varphi\otimes\psi$$ is a Lie algebra morphism by direct calculations, i.e. $$\varphi\otimes\psi([g, h])=[(\varphi\otimes\psi)(g), (\varphi\otimes\psi)(h)]$$ for any $$g$$ and $$h$$ in $$\Lg$$. Indeed, we have

\begin{aligned} \varphi\otimes\psi([g, h])=&\varphi([g, h])\otimes I+I\otimes\psi([g, h])\\ =&([\varphi(g), \varphi(h)])\otimes I+I\otimes([\psi(g), \psi(h)])\\ =&(\varphi(g)\varphi(h)-\varphi(h)\varphi(g))\otimes I\\ &+I\otimes(\psi(g)\psi(h)-\psi(h)\psi(g)), \end{aligned} and \begin{aligned} &[(\varphi\otimes\psi)(g), (\varphi\otimes\psi)(h)]\\ =&(\varphi\otimes\psi)(g) (\varphi\otimes\psi)(h)-(\varphi\otimes\psi)(h) (\varphi\otimes\psi)(g)\\ =&(\varphi(g)\otimes I+I\otimes\psi(g))(\varphi(h)\otimes I+I\otimes\psi(h))\\ &-(\varphi(h)\otimes I+I\otimes\psi(h))(\varphi(g)\otimes I+I\otimes\psi(g))\\ =&(\varphi(g)\varphi(h)\otimes I+\varphi(h)\otimes\psi(g)+\varphi(g)\otimes\psi(h)+I\otimes\psi(g)\psi(h))\\ &-(\varphi(h)\varphi(g)\otimes I+\varphi(g)\otimes\psi(h)+\varphi(h)\otimes\psi(g)+I\otimes\psi(h)\psi(g))\\ =&(\varphi(g)\varphi(h)-\varphi(h)\varphi(g))\otimes I\\ &+I\otimes(\psi(g)\psi(h)-\psi(h)\psi(g)). \end{aligned}

## 2 Cohomology of Lie Algebra

One motivation of Cohomology of Lie Algebra is the de Rham cohomology.

Let $$G$$ be a connected Lie group and $$\Lg$$ the associated Lie algebra, i.e. the tangent space $$T_eG$$ equipped with the Lie bracket $$[X, Y](f):=X(Y(f))-Y(X(f))$$. Note that the tangent space $$T_eG$$ can be identified with the space of left-invariant vector fields on $$G$$. Taking the dual, we may identify $$\Lg^*$$ with space of left-invariant differential forms on $$G$$. Then there is a complex $0\to\wedge^0\Lg^*=\RR \to \wedge^1\Lg^*=\Lg^*\to \wedge^2\Lg^*\to \cdots$ which is isomorphic to the subcomplex of the de Rham complex $0\to \RR \to \Omega_L^1(G)^G\to \Omega_L^2(G)^G\to \cdots,$ where $$\Omega_L^k(G)^G$$ are spaces of left-invariant differential $$k$$-forms on $$G$$.

Let $$H^k(\Lg)$$ and $$H^k_L(G)$$ be the cohomology groups of the above complex respectively.

By Theorem 15.1 of , we know that that $H^k_L(G)\cong H^k(\Lg).$ If in addition that $$G$$ is also compact, then by Theorem 15.2 of , we know that $H^k_{dR}(G)\cong H^k_L(G)\cong H^k(\Lg),$ where $$H^k_{dR}(G)$$ is the de Rham cohomology group.

Another motivation is from the study of extensions of Lie algebras. We refer the reader to for detailed explorations.

Viewing $$\wedge^k\Lg^*$$ as the space $$\Hom_{\RR}(\Lg, \RR)$$ of multilinear alternating forms from $$\Lg\to \RR$$, where $$\RR$$ is the $$\Lg$$-module associated to the trivial representation of $$\Lg$$ in $$\RR$$, we can generalized the complex $$(\wedge^*\Lg^*,d)$$ to general $$\Lg$$-modules.

Let $$M$$ be a $$\Lg$$-module. For $$k\geq 1$$, the space $$C^k(\Lg, M)$$ of $$k$$-cochains on $$\Lg$$ with values in $$M$$ is defined to be the space $$\Hom_{\kk}(\wedge^k\Lg, M)$$ of multilinear alternating maps from $$\Lg\to M$$. The space $$C^0(\Lg, M)$$ of $$0$$-cochains is defined to be $$M$$.

There is a coboundary operator $$d: C^k(\Lg, M)\to C^{k+1}(\Lg, M)$$ defined by \begin{aligned} d \omega\left(X_1 \wedge \cdots \wedge X_{k}\right) &=\sum_{j=0}^{k}(-1)^{j+1} X_{j}\cdot\left(\omega\left(X_1 \wedge \cdots \wedge \hat{X}_{j} \wedge \cdots \wedge X_{k}\right)\right) \\ &+\sum_{r<s} (-1)^{r+s}\omega\left(\left[X_{r}, X_{s}\right] \wedge X_1 \wedge \cdots \wedge \hat{X}_{r} \wedge \cdots \wedge \hat{X}_{s} \wedge \cdots \wedge X_{k}\right). \end{aligned}

The definition of $$d$$ is determined by the conditions $$\d f=\sum (X_if)\omega_i$$, $$d^2=0$$, and the Leibniz rule $$\d(\xi\wedge\eta)=\d\xi\wedge\eta+(-1)^{\deg \xi}\xi\wedge\d\eta$$, where $$f\in \Hom_{\kk}(\Lg, M)$$, $$\{X_i\}$$ is a basis of $$\Lg$$ and $$\omega_i\in\Lg^*$$ is the dual of $$X_i$$. Note that the sign $$(-1)^{\deg\xi}$$ is necessary so that $$\d^2=0$$.

To see that $$d$$ is determined by those conditions, assume that $$\Lg$$ is of dimension 2 with a basis $$X$$ and $$Y$$. Denote by $$\omega$$ and $$\mu$$ the dual basis. Suppose that $$[X, Y]=aX+bY$$. From the conditions $$\d f=\sum (X_if)\omega_i$$ and $$d^2=0$$, for any $$f\in \Hom_{\kk}(\Lg, M)$$, we get \begin{aligned} 0=&\d^2f\\ =&\d((Xf)\omega+(Yf)\mu)\\ =&\d(Xf)\wedge\omega+\d(Yf)\wedge\mu+(Xf)\d\omega+(Yf)\d\mu\\ =&Y(Xf)\mu\wedge\omega+X(Y(f))\omega\wedge\mu+(Xf)\d\omega+(Yf)\d\mu\\ =&[X, Y](f)\omega\wedge\mu+(Xf)\d\omega+(Yf)\d\mu\\ =&X(f)(a\omega\wedge\mu+\d\omega)+Y(f)(b\omega\wedge\mu+\d\mu). \end{aligned} Since $$f$$ is arbitrary, we see that $\d\omega(X\wedge Y)=-a\omega\wedge\mu(X\wedge Y)=-\omega([X, Y]),$ and $\d\mu(X\wedge Y)=-b\omega\wedge\mu(X\wedge Y)=-\mu([X, Y).$

The Leibniz rule is necessary for defining higher degree coboundary maps.

The cochain complex $$(C^∗(\Lg, M), d)$$ is called the Chevalley-Eilenberg complex.

Definition 3 The space of $$k$$-cocycles is defined to be $Z^{k}(\mathfrak{g}, M):=\ker\d=\left\{\omega \in C^{k}(\mathfrak{g}, M) \mid \d\omega=0\right\}.$

The space of $$k$$-coboundaries is defined to be
$B^{k}(\mathfrak{g}, M):=\im \d=\left\{\d\omega \mid \omega \in C^{k-1}(\mathfrak{g}, M)\right\}.$

The $$k$$-th cohomology space of $$\mathfrak{g}$$ with values in $$M$$ is defined as the quotient vector space $H^{k}(\mathfrak{g}, M):=Z^{k}(\mathfrak{g}, M)/B^{k}(\mathfrak{g}, M)$

Definition 4 (The Universal Enveloping Algebra) Let $$\Lg$$ be a lie algebra. The quotient algebra $$U(\Lg)$$ defined as $U(\Lg)=T(\Lg)/([x, y]-x\otimes y+y\otimes)$ is called the universal enveloping algebra of $$\Lg$$.

A $$\kk$$-module $$M$$ is a $$\Lg$$-module if and only if $$M$$ is a $$U(\Lg)$$-module. This result provides another approach to compute Lie algebra cohomology in terms of free resolution of $$U(\Lg)$$-modules. For details, we refer the reader to .

### 2.1 Cohomology of $$\Lg$$ in degree 0

For any $$\Lg$$-module $$M$$, the $$0$$-th cohomology space of $$\Lg$$ with values in $$M$$ is $H^0(\Lg, M)=Z^0(\Lg, M)=\{m\in M\mid X\cdot m=0 \text{ for all } X \text{ in }\Lg\}.$

### 2.2 Cohomology of $$\Lg$$ in degree 1

In degree 1, the space of cochains is $$\Hom_{\kk}(\Lg, M)$$. Let $$\omega$$ be a 1-cochain. Then $\d\omega(X, Y)=X\cdot \omega (Y) - Y\cdot \omega(X)-\omega([X, Y]).$

The space of cocycles is $Z^1(\Lg, M)=\{\omega \mid \omega([X, Y])=X\cdot \omega (Y) - Y\cdot \omega(X)\text{ for all } X, Y \in \Lg\}.$

The space of coboundaries is $B^1(\Lg, M)=\{\omega \mid \omega(X)=X\cdot m \text{ for some } m\in M\}.$

Example 4 Consider $$\Lg$$ as a $$\Lg$$-module via the adjoint representation. Then $Z^1(\Lg, M)=\{D: \Lg\to \Lg\mid D([X, Y])=[X, D(Y)]+[D(X), Y]\}$ is the space of derivations of $$\Lg$$ and $B^1(\Lg, M)=\{D:\Lg\to \Lg\mid D(X)=[X, Y_D]\}=\{\ad_Y\mid Y\in \Lg\}$ is the space of inner derivations of $$\Lg$$. The cohomology space $$H^1(\Lg, \Lg)$$ is known as the space of outer derivations.

Note that those spaces of derivations admit Lie algebra structure.

Example 5 Consider $$\Lg^*$$ as a $$\Lg$$-module via the coadjoint representation. Then $Z^1(\Lg, M)=\{\rho: \Lg\to \Lg^*\mid \rho([X, Y])=\ad_X^*(\rho(Y))-\ad_Y^*(\rho(X))\}$ and \begin{aligned} B^1(\Lg, M)=&\{\rho:\Lg\to \Lg^*\mid \rho(X)=\ad_X^*(\xi) \text{ for some }\xi\in\Lg^*\}\\ =&\{\ad^*(\xi):\Lg\to \Lg^*\mid \xi\in\Lg^*\}. \end{aligned}

Example 6 Consider $$\Lsl(2, \mathbb{C})=<e,f,h>$$ with the relations $$[e, f]=h$$, $$[h, e]=2e$$ and $$[h, f]=-2f$$. Let $$D:\Lg\to \Lg$$ be a linear map such that $D\begin{pmatrix} e\\ f\\ h \end{pmatrix} =\begin{pmatrix} x_e& x_f &x_h\\ y_e&y_f&y_h\\ z_e&z_f&z_h \end{pmatrix}\begin{pmatrix} e\\ f\\ h \end{pmatrix}.$ A direct calculation shows that $$D$$ is a Lie algebra derivation of $$\Lg$$ is and only if $$D=\ad_X$$, where $$X=y_h e-x_h f+\frac12x_e$$, and $$x_f=y_e=z_h=0$$.

Consequently, $$H^1(\Lsl(2,\mathbb{C}, \Lsl(2,\mathbb{C}))=0$$, where the $$\Lsl(2,\mathbb{C})$$-module structure on itself is defined by the adjoint representation.

## References

Chevalley, Claude, and Samuel Eilenberg. 1948. “Cohomology Theory of Lie Groups and Lie Algebras.” Transactions of the American Mathematical Society 63: 85–124. https://doi.org/d58jp5.
Knapp, Anthony W. 1988. Lie Groups, Lie Algebras, and Cohomology. Princeton, N.J: Princeton University Press.