Cocommutator map

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In mathematics a Lie coalgebra is the dual structure to a Lie algebra.

In finite dimensions, these are dual objects: the dual vector space to a Lie algebra naturally has the structure of a Lie coalgebra, and conversely.

Definition

Let E {\displaystyle E} {\displaystyle E} be a vector space over a field k {\displaystyle \mathbb {k} } {\displaystyle \mathbb {k} } equipped with a linear mapping d : E → E ∧ E {\displaystyle d\colon E\to E\wedge E} {\displaystyle d\colon E\to E\wedge E} from E {\displaystyle E} {\displaystyle E} to the exterior product of E {\displaystyle E} {\displaystyle E} with itself. It is possible to extend d {\displaystyle d} {\displaystyle d} uniquely to a graded derivation (this means that, for any a , b ∈ E {\displaystyle a,b\in E} {\displaystyle a,b\in E} which are homogeneous elements, d ( a ∧ b ) = ( d a ) ∧ b + ( − 1 ) deg ⁡ a a ∧ ( d b ) {\displaystyle d(a\wedge b)=(da)\wedge b+(-1)^{\deg a}a\wedge (db)} {\displaystyle d(a\wedge b)=(da)\wedge b+(-1)^{\deg a}a\wedge (db)}) of degree 1 on the exterior algebra of E {\displaystyle E} {\displaystyle E}:

d : ⋀ ∙ E → ⋀ ∙ + 1 E . {\displaystyle d\colon \bigwedge ^{\bullet }E\rightarrow \bigwedge ^{\bullet +1}E.} {\displaystyle d\colon \bigwedge ^{\bullet }E\rightarrow \bigwedge ^{\bullet +1}E.}

Then the pair ( E , d ) {\displaystyle (E,d)} {\displaystyle (E,d)} is said to be a Lie coalgebra if d 2 = 0 {\displaystyle d^{2}=0} {\displaystyle d^{2}=0}, i.e., if the graded components of the exterior algebra with derivation ( ⋀ ∗ E , d ) {\textstyle (\bigwedge ^{*}E,d)} {\textstyle (\bigwedge ^{*}E,d)} form a cochain complex:

E   → d   E ∧ E   → d   ⋀ 3 E → d   ⋯ {\displaystyle E\ \xrightarrow {d} \ E\wedge E\ \xrightarrow {d} \ \bigwedge ^{3}E\xrightarrow {d} \ \cdots } {\displaystyle E\ \xrightarrow {d} \ E\wedge E\ \xrightarrow {d} \ \bigwedge ^{3}E\xrightarrow {d} \ \cdots }

Relation to de Rham complex

Just as the exterior algebra (and tensor algebra) of vector fields on a manifold form a Lie algebra (over the base field k {\displaystyle \mathbb {k} } {\displaystyle \mathbb {k} }), the de Rham complex of differential forms on a manifold form a Lie coalgebra (over the base field k {\displaystyle \mathbb {k} } {\displaystyle \mathbb {k} }). Further, there is a pairing between vector fields and differential forms.

However, the situation is subtler: the Lie bracket is not linear over the algebra of smooth functions C ∞ ( M ) {\displaystyle C^{\infty }(M)} {\displaystyle C^{\infty }(M)} (the error is the Lie derivative), nor is the exterior derivative: d ( f g ) = ( d f ) g + f ( d g ) ≠ f ( d g ) {\displaystyle d(fg)=(df)g+f(dg)\neq f(dg)} {\displaystyle d(fg)=(df)g+f(dg)\neq f(dg)} (it is a derivation, not linear over functions): they are not tensors. They are not linear over functions, but they behave in a consistent way, which is not captured simply by the notion of Lie algebra and Lie coalgebra.

Further, in the de Rham complex, the derivation is not only defined for Ω 1 → Ω 2 {\displaystyle \Omega ^{1}\to \Omega ^{2}} {\displaystyle \Omega ^{1}\to \Omega ^{2}}, but is also defined for C ∞ ( M ) → Ω 1 ( M ) {\displaystyle C^{\infty }(M)\to \Omega ^{1}(M)} {\displaystyle C^{\infty }(M)\to \Omega ^{1}(M)}.

The Lie algebra on the dual

A Lie algebra structure on a vector space is a map [ ⋅ , ⋅ ] : g × g → g {\displaystyle [\cdot ,\cdot ]\colon {\mathfrak {g}}\times {\mathfrak {g}}\to {\mathfrak {g}}} {\displaystyle [\cdot ,\cdot ]\colon {\mathfrak {g}}\times {\mathfrak {g}}\to {\mathfrak {g}}} which is skew-symmetric, and satisfies the Jacobi identity. Equivalently, a map [ ⋅ , ⋅ ] : g ∧ g → g {\displaystyle [\cdot ,\cdot ]\colon {\mathfrak {g}}\wedge {\mathfrak {g}}\to {\mathfrak {g}}} {\displaystyle [\cdot ,\cdot ]\colon {\mathfrak {g}}\wedge {\mathfrak {g}}\to {\mathfrak {g}}} that satisfies the Jacobi identity.

Dually, a Lie coalgebra structure on a vector space E is a linear map d : E → E ⊗ E {\displaystyle d\colon E\to E\otimes E} {\displaystyle d\colon E\to E\otimes E} which is antisymmetric (this means that it satisfies τ ∘ d = − d {\displaystyle \tau \circ d=-d} {\displaystyle \tau \circ d=-d}, where τ {\displaystyle \tau } {\displaystyle \tau } is the canonical flip E ⊗ E → E ⊗ E {\displaystyle E\otimes E\to E\otimes E} {\displaystyle E\otimes E\to E\otimes E}) and satisfies the so-called cocycle condition (also known as the co-Leibniz rule)

( d ⊗ i d ) ∘ d = ( i d ⊗ d ) ∘ d + ( i d ⊗ τ ) ∘ ( d ⊗ i d ) ∘ d {\displaystyle \left(d\otimes \mathrm {id} \right)\circ d=\left(\mathrm {id} \otimes d\right)\circ d+\left(\mathrm {id} \otimes \tau \right)\circ \left(d\otimes \mathrm {id} \right)\circ d} {\displaystyle \left(d\otimes \mathrm {id} \right)\circ d=\left(\mathrm {id} \otimes d\right)\circ d+\left(\mathrm {id} \otimes \tau \right)\circ \left(d\otimes \mathrm {id} \right)\circ d}.

Due to the antisymmetry condition, the map d : E → E ⊗ E {\displaystyle d\colon E\to E\otimes E} {\displaystyle d\colon E\to E\otimes E} can be also written as a map d : E → E ∧ E {\displaystyle d\colon E\to E\wedge E} {\displaystyle d\colon E\to E\wedge E}.

The dual of the Lie bracket of a Lie algebra g {\displaystyle {\mathfrak {g}}} {\displaystyle {\mathfrak {g}}} yields a map (the cocommutator)

[ ⋅ , ⋅ ] ∗ : g ∗ → ( g ∧ g ) ∗ ≅ g ∗ ∧ g ∗ {\displaystyle [\cdot ,\cdot ]^{*}\colon {\mathfrak {g}}^{*}\to ({\mathfrak {g}}\wedge {\mathfrak {g}})^{*}\cong {\mathfrak {g}}^{*}\wedge {\mathfrak {g}}^{*}} {\displaystyle [\cdot ,\cdot ]^{*}\colon {\mathfrak {g}}^{*}\to ({\mathfrak {g}}\wedge {\mathfrak {g}})^{*}\cong {\mathfrak {g}}^{*}\wedge {\mathfrak {g}}^{*}}

where the isomorphism ≅ {\displaystyle \cong } {\displaystyle \cong } holds in finite dimension; dually for the dual of Lie comultiplication. In this context, the Jacobi identity corresponds to the cocycle condition.

More explicitly, let E {\displaystyle E} {\displaystyle E} be a Lie coalgebra over a field of characteristic neither 2 nor 3. The dual space E ∗ {\displaystyle E^{*}} {\displaystyle E^{*}} carries the structure of a bracket defined by

α ( [ x , y ] ) = d α ( x ∧ y ) {\displaystyle \alpha ([x,y])=d\alpha (x\wedge y)} {\displaystyle \alpha ([x,y])=d\alpha (x\wedge y)}, for all α ∈ E {\displaystyle \alpha \in E} {\displaystyle \alpha \in E} and x , y ∈ E ∗ {\displaystyle x,y\in E^{*}} {\displaystyle x,y\in E^{*}}.

We show that this endows E ∗ {\displaystyle E^{*}} {\displaystyle E^{*}} with a Lie bracket. It suffices to check the Jacobi identity. For any x , y , z ∈ E ∗ {\displaystyle x,y,z\in E^{*}} {\displaystyle x,y,z\in E^{*}} and α ∈ E {\displaystyle \alpha \in E} {\displaystyle \alpha \in E},

d 2 α ( x ∧ y ∧ z ) = 1 3 d 2 α ( x ∧ y ∧ z + y ∧ z ∧ x + z ∧ x ∧ y ) = 1 3 ( d α ( [ x , y ] ∧ z ) + d α ( [ y , z ] ∧ x ) + d α ( [ z , x ] ∧ y ) ) , {\displaystyle {\begin{aligned}d^{2}\alpha (x\wedge y\wedge z)&={\frac {1}{3}}d^{2}\alpha (x\wedge y\wedge z+y\wedge z\wedge x+z\wedge x\wedge y)\\&={\frac {1}{3}}\left(d\alpha ([x,y]\wedge z)+d\alpha ([y,z]\wedge x)+d\alpha ([z,x]\wedge y)\right),\end{aligned}}} {\displaystyle {\begin{aligned}d^{2}\alpha (x\wedge y\wedge z)&={\frac {1}{3}}d^{2}\alpha (x\wedge y\wedge z+y\wedge z\wedge x+z\wedge x\wedge y)\\&={\frac {1}{3}}\left(d\alpha ([x,y]\wedge z)+d\alpha ([y,z]\wedge x)+d\alpha ([z,x]\wedge y)\right),\end{aligned}}}

where the latter step follows from the standard identification of the dual of a wedge product with the wedge product of the duals. Finally, this gives

d 2 α ( x ∧ y ∧ z ) = 1 3 ( α ( [ [ x , y ] , z ] ) + α ( [ [ y , z ] , x ] ) + α ( [ [ z , x ] , y ] ) ) . {\displaystyle d^{2}\alpha (x\wedge y\wedge z)={\frac {1}{3}}\left(\alpha ([[x,y],z])+\alpha ([[y,z],x])+\alpha ([[z,x],y])\right).} {\displaystyle d^{2}\alpha (x\wedge y\wedge z)={\frac {1}{3}}\left(\alpha ([[x,y],z])+\alpha ([[y,z],x])+\alpha ([[z,x],y])\right).}

Since d 2 = 0 {\displaystyle d^{2}=0} {\displaystyle d^{2}=0}, it follows that

α ( [ [ x , y ] , z ] + [ [ y , z ] , x ] + [ [ z , x ] , y ] ) = 0 {\displaystyle \alpha ([[x,y],z]+[[y,z],x]+[[z,x],y])=0} {\displaystyle \alpha ([[x,y],z]+[[y,z],x]+[[z,x],y])=0}, for any α {\displaystyle \alpha } {\displaystyle \alpha }, x {\displaystyle x} {\displaystyle x}, y {\displaystyle y} {\displaystyle y}, and z {\displaystyle z} {\displaystyle z}.

Thus, by the double-duality isomorphism (more precisely, by the double-duality monomorphism, since the vector space needs not be finite-dimensional), the Jacobi identity is satisfied.

In particular, note that this proof demonstrates that the cocycle condition d 2 = 0 {\displaystyle d^{2}=0} {\displaystyle d^{2}=0} is in a sense dual to the Jacobi identity.

References