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A Brief Introduction to Cohomology of Lie Algebra

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

Affiliation

Fei Ye

QCC-CUNY

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

1 Representations of Lie Algebras

Let $\mathfrak{g}$ be a Lie algebra, $M$ be a vector space $M$, and $\mathfrak{gl}(M)$ be the space of endomorphisms of $M$. The vector space $\mathfrak{gl}(M)$ 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 $\rho :\mathfrak{g}\to \mathfrak{gl}(M)$, 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 $\mathfrak{g}$-module structure if there is a $\mathbb{k}$-bilinear map $\pi :\mathfrak{g}{\otimes }_{\mathbb{k}}M\to M$ 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 $\rho :\mathfrak{g}\to \mathfrak{gl}(M)$ defines a $\mathfrak{g}$-module structure on $M$: $g\cdot m=\rho (g)(m)$. Conversely, a $\mathfrak{g}$-module structure on $M$ defines a representation.

Example 1 (Adjoint representation) Let $\mathfrak{g}$ be a Lie algebra. Consider the map $\mathrm{ad}:\mathfrak{g}\to \mathfrak{gl}(\mathfrak{g})$ defined by $x\mapsto {\mathrm{ad}}_x:=[x,-]$. By Jacobian identity, the map $\mathrm{ad}$ defines representation $\mathfrak{g}$ of $\mathfrak{g}$ called an adjoint representation. Indeed, $$\begin{array}{rl}\mathrm{ad}([x,y])(z)=& [[x,y],z]\\ =& [x,[y,z]]-[y,[x,z]]\\ =& {\mathrm{ad}}_x({\mathrm{ad}}_y(z))-{\mathrm{ad}}_y({\mathrm{ad}}_x(z))\\ =& [{\mathrm{ad}}_x,{\mathrm{ad}}_y](z).\end{array}$$

Note that an adjoint is a $\mathbb{k}$-derivation of $\mathfrak{g}$, that is ${\mathrm{ad}}_x$ satisfies the Leibniz rule for the Lie bracket

$${\mathrm{ad}}_x[g,h]=[{\mathrm{ad}}_x(g),h]+[g,{\mathrm{ad}}_x(h)].$$

Example 2 (Coadjoint representation) Given any Lie algebra $\mathfrak{g}$, denote by ${\mathfrak{g}}^{\ast }$ the dual vector space, the space for $\mathbb{k}$-linear functions on $\mathfrak{g}$. We define $<\xi ,x>=\xi (x)$, where $\xi \in {\mathfrak{g}}^{\ast }$ and $x\in \mathfrak{g}$. Consider the map ${\mathrm{ad}}^{\ast }:\mathfrak{g}\to \mathfrak{gl}({\mathfrak{g}}^{\ast })$ defined by $$x\mapsto {\mathrm{ad}}_x^{\ast }:{\mathfrak{g}}^{\ast }\to {\mathfrak{g}}^{\ast },$$ where ${\mathrm{ad}}_x^{\ast }$ is given by ${\mathrm{ad}}_x^{\ast }(\xi )(y)=-<\xi ,[x,y]>=-\xi ([x,y])$ for any $\xi \in {\mathfrak{g}}^{\ast }$ and $y\in \mathfrak{g}$.

The map ${\mathrm{ad}}^{\ast }$ defines representation of $\mathfrak{g}$ in ${\mathfrak{g}}^{\ast }$ called an coadjoint representation of $\mathfrak{g}$. For any $x,y,z$ in $\mathfrak{g}$ and $\xi$ in ${\mathfrak{g}}^{\ast }$, we see that $$\begin{array}{rl}{\mathrm{ad}}_{[x,y]}^{\ast }(\xi )(z)=& -<\xi ,[[x,y],z]>\\ =& -<\xi ,[x,[y,z]]>+<\xi ,[y,[x,z]]>\\ =& {\mathrm{ad}}_x^{\ast }(\xi )([y,z])-{\mathrm{ad}}_y^{\ast }(\xi )([x,z])\\ =& -{\mathrm{ad}}_y^{\ast }({\mathrm{ad}}_x^{\ast }(\xi ))(z)+{\mathrm{ad}}_x^{\ast }({\mathrm{ad}}_y^{\ast }(\xi ))(z)\\ =& [{\mathrm{ad}}_x^{\ast },{\mathrm{ad}}_y^{\ast }](\xi )(z).\end{array}$$

Note that the negative sign in the definition of ${\mathrm{ad}}_x^{\ast }$ is necessary so that $\mathrm{ad}$ is a Lie algebra homomorphism.

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

$$\begin{array}{rl}\phi \otimes \psi ([g,h])=& \phi ([g,h])\otimes I+I\otimes \psi ([g,h])\\ =& ([\phi (g),\phi (h)])\otimes I+I\otimes ([\psi (g),\psi (h)])\\ =& (\phi (g)\phi (h)-\phi (h)\phi (g))\otimes I\\ & +I\otimes (\psi (g)\psi (h)-\psi (h)\psi (g)),\end{array}$$ and $$\begin{array}{rl}& [(\phi \otimes \psi )(g),(\phi \otimes \psi )(h)]\\ =& (\phi \otimes \psi )(g)(\phi \otimes \psi )(h)-(\phi \otimes \psi )(h)(\phi \otimes \psi )(g)\\ =& (\phi (g)\otimes I+I\otimes \psi (g))(\phi (h)\otimes I+I\otimes \psi (h))\\ & -(\phi (h)\otimes I+I\otimes \psi (h))(\phi (g)\otimes I+I\otimes \psi (g))\\ =& (\phi (g)\phi (h)\otimes I+\phi (h)\otimes \psi (g)+\phi (g)\otimes \psi (h)+I\otimes \psi (g)\psi (h))\\ & -(\phi (h)\phi (g)\otimes I+\phi (g)\otimes \psi (h)+\phi (h)\otimes \psi (g)+I\otimes \psi (h)\psi (g))\\ =& (\phi (g)\phi (h)-\phi (h)\phi (g))\otimes I\\ & +I\otimes (\psi (g)\psi (h)-\psi (h)\psi (g)).\end{array}$$

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 $\mathfrak{g}$ the associated Lie algebra, i.e. the tangent space $T_{e}G$ equipped with the Lie bracket $[X,Y](f):=X(Y(f))-Y(X(f))$. Note that the tangent space $T_{e}G$ can be identified with the space of left-invariant vector fields on $G$. Taking the dual, we may identify ${\mathfrak{g}}^{\ast }$ with space of left-invariant differential forms on $G$. Then there is a complex $$0\to {\wedge }^0{\mathfrak{g}}^{\ast }=\mathbb{R}\to {\wedge }^1{\mathfrak{g}}^{\ast }={\mathfrak{g}}^{\ast }\to {\wedge }^2{\mathfrak{g}}^{\ast }\to \cdots$$ which is isomorphic to the subcomplex of the de Rham complex $$0\to \mathbb{R}\to {\mathrm{\Omega }}_L^1(G{)}^G\to {\mathrm{\Omega }}_L^2(G{)}^G\to \cdots ,$$ where ${\mathrm{\Omega }}_L^k(G{)}^G$ are spaces of left-invariant differential $k$-forms on $G$.

Let $H^k(\mathfrak{g})$ and $H_L^k(G)$ be the cohomology groups of the above complex respectively.

By Theorem 15.1 of (Chevalley and Eilenberg 1948), we know that that $$H_L^k(G)\cong H^k(\mathfrak{g}).$$ If in addition that $G$ is also compact, then by Theorem 15.2 of (Chevalley and Eilenberg 1948), we know that $$H_{dR}^k(G)\cong H_L^k(G)\cong H^k(\mathfrak{g}),$$ where $H_{dR}^k(G)$ is the de Rham cohomology group.

Another motivation is from the study of extensions of Lie algebras. We refer the reader to (Knapp 1988) for detailed explorations.

Viewing ${\wedge }^k{\mathfrak{g}}^{\ast }$ as the space ${\mathrm{Hom}}_{\mathbb{R}}(\mathfrak{g},\mathbb{R})$ of multilinear alternating forms from $\mathfrak{g}\to \mathbb{R}$, where $\mathbb{R}$ is the $\mathfrak{g}$-module associated to the trivial representation of $\mathfrak{g}$ in $\mathbb{R}$, we can generalized the complex $({\wedge }^{\ast }{\mathfrak{g}}^{\ast },d)$ to general $\mathfrak{g}$-modules.

Let $M$ be a $\mathfrak{g}$-module. For $k\ge 1$, the space $C^k(\mathfrak{g},M)$ of $k$-cochains on $\mathfrak{g}$ with values in $M$ is defined to be the space ${\mathrm{Hom}}_{\mathbb{k}}({\wedge }^k\mathfrak{g},M)$ of multilinear alternating maps from $\mathfrak{g}\to M$. The space $C^0(\mathfrak{g},M)$ of $0$-cochains is defined to be $M$.

There is a coboundary operator $d:C^k(\mathfrak{g},M)\to C^{k+1}(\mathfrak{g},M)$ defined by $$\begin{array}{rl}d\omega (X_1\wedge \cdots \wedge X_k)& =\sum _{j=0}^k(-1{)}^{j+1}{X}_j\cdot \left(\omega (X_1\wedge \cdots \wedge {\hat{X}}_j\wedge \cdots \wedge X_k)\right)\\ & +\sum _{r<s}(-1{)}^{r+s}\omega ([X_r,X_s]\wedge X_1\wedge \cdots \wedge {\hat{X}}_r\wedge \cdots \wedge {\hat{X}}_s\wedge \cdots \wedge X_k).\end{array}$$

The definition of $d$ is determined by the conditions $\mathrm{d}f=\sum (X_{i}f){\omega }_i$, $d^2=0$, and the Leibniz rule $\mathrm{d}(\xi \wedge \eta )=\mathrm{d}\xi \wedge \eta +(-1{)}^{\mathrm{deg}\xi }\xi \wedge \mathrm{d}\eta$, where $f\in {\mathrm{Hom}}_{\mathbb{k}}(\mathfrak{g},M)$, $\{X_i\}$ is a basis of $\mathfrak{g}$ and ${\omega }_i\in {\mathfrak{g}}^{\ast }$ is the dual of $X_i$. Note that the sign $(-1{)}^{\mathrm{deg}\xi }$ is necessary so that ${\mathrm{d}}^2=0$.

To see that $d$ is determined by those conditions, assume that $\mathfrak{g}$ 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 $\mathrm{d}f=\sum (X_{i}f){\omega }_i$ and $d^2=0$, for any $f\in {\mathrm{Hom}}_{\mathbb{k}}(\mathfrak{g},M)$, we get $$\begin{array}{rl}0=& {\mathrm{d}}^{2}f\\ =& \mathrm{d}((Xf)\omega +(Yf)\mu )\\ =& \mathrm{d}(Xf)\wedge \omega +\mathrm{d}(Yf)\wedge \mu +(Xf)\mathrm{d}\omega +(Yf)\mathrm{d}\mu \\ =& Y(Xf)\mu \wedge \omega +X(Y(f))\omega \wedge \mu +(Xf)\mathrm{d}\omega +(Yf)\mathrm{d}\mu \\ =& [X,Y](f)\omega \wedge \mu +(Xf)\mathrm{d}\omega +(Yf)\mathrm{d}\mu \\ =& X(f)(a\omega \wedge \mu +\mathrm{d}\omega )+Y(f)(b\omega \wedge \mu +\mathrm{d}\mu ).\end{array}$$ Since $f$ is arbitrary, we see that $$\mathrm{d}\omega (X\wedge Y)=-a\omega \wedge \mu (X\wedge Y)=-\omega ([X,Y]),$$ and $$\mathrm{d}\mu (X\wedge Y)=-b\omega \wedge \mu (X\wedge Y)=-\mu ([X,Y).$$

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

The cochain complex $(C^{\ast }(\mathfrak{g},M),d)$ is called the Chevalley-Eilenberg complex.

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

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

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 $\mathfrak{g}$ be a lie algebra. The quotient algebra $U(\mathfrak{g})$ defined as $$U(\mathfrak{g})=T(\mathfrak{g})/([x,y]-x\otimes y+y\otimes )$$ is called the universal enveloping algebra of $\mathfrak{g}$.

A $\mathbb{k}$-module $M$ is a $\mathfrak{g}$-module if and only if $M$ is a $U(\mathfrak{g})$-module. This result provides another approach to compute Lie algebra cohomology in terms of free resolution of $U(\mathfrak{g})$-modules. For details, we refer the reader to (Knapp 1988).

2.1 Cohomology of $\mathfrak{g}$ in degree 0

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

2.2 Cohomology of $\mathfrak{g}$ in degree 1

In degree 1, the space of cochains is ${\mathrm{Hom}}_{\mathbb{k}}(\mathfrak{g},M)$. Let $\omega$ be a 1-cochain. Then $$\mathrm{d}\omega (X,Y)=X\cdot \omega (Y)-Y\cdot \omega (X)-\omega ([X,Y]).$$

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

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

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

Note that those spaces of derivations admit Lie algebra structure.

Example 5 Consider ${\mathfrak{g}}^{\ast }$ as a $\mathfrak{g}$-module via the coadjoint representation. Then $$Z^1(\mathfrak{g},M)=\{\rho :\mathfrak{g}\to {\mathfrak{g}}^{\ast }\mid \rho ([X,Y])={\mathrm{ad}}_X^{\ast }(\rho (Y))-{\mathrm{ad}}_Y^{\ast }(\rho (X))\}$$ and $$\begin{array}{rl}{B}^1(\mathfrak{g},M)=& \{\rho :\mathfrak{g}\to {\mathfrak{g}}^{\ast }\mid \rho (X)={\mathrm{ad}}_X^{\ast }(\xi )\text{ for some }\xi \in {\mathfrak{g}}^{\ast }\}\\ =& \{{\mathrm{ad}}^{\ast }(\xi ):\mathfrak{g}\to {\mathfrak{g}}^{\ast }\mid \xi \in {\mathfrak{g}}^{\ast }\}.\end{array}$$

Example 6 Consider $\mathfrak{sl}(2,\mathbb{C})=<e,f,h>$ with the relations $[e,f]=h$, $[h,e]=2e$ and $[h,f]=-2f$. Let $D:\mathfrak{g}\to \mathfrak{g}$ be a linear map such that $$D\left(\begin{array}{c}e\\ f\\ h\end{array}\right)=\left(\begin{array}{ccc}{x}_e& x_f& x_h\\ y_e& y_f& y_h\\ z_e& z_f& z_h\end{array}\right)\left(\begin{array}{c}e\\ f\\ h\end{array}\right).$$ A direct calculation shows that $D$ is a Lie algebra derivation of $\mathfrak{g}$ is and only if $D={\mathrm{ad}}_X$, where $X=y_{h}e-x_{h}f+\frac{1}{2}{x}_e$, and $x_f=y_e=z_h=0$.

Consequently, $H^1(\mathfrak{sl}(2,\mathbb{C},\mathfrak{sl}(2,\mathbb{C}))=0$, where the $\mathfrak{sl}(2,\mathbb{C})$-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.