The *Bianchi identity* is a differential equation satisfied by curvature data.

It can be thought of as generalizing the equation $d (d A) = 0$ for a real-valued 1-form $A$ to higher degree and nonabelian forms.

Generally it applies to the curvature of ∞-Lie algebroid valued differential forms.

Let $U$ be a smooth manifold.

For $A \in \Omega^1(U)$ a differential 1-form, its curvature 2-form is the de Rham differential $F_A = d A$. The Bianchi identity in this case is the equation

$d F = 0
\,.$

More generally, for $\mathfrak{g}$ an arbitrary Lie algebra and $A \in \Omega^1(U,\mathfrak{g})$ a Lie-algebra valued 1-form, its curvature is the 2-form $F_A = d A + [A \wedge A]$. The Bianchi identity in this case is the equation

$d F_A + [A\wedge F_A] = 0$

satisfied by these curvature 2-forms.

We may reformulate the above identities as follows.

For $\mathfrak{g}$ a Lie algebra we have naturally associated two dg-algebras: the Chevalley-Eilenberg algebra $CE(\mathfrak{g})$ and the Weil algebra $W(\mathfrak{g})$.

The dg-algebra morphisms

$\Omega^\bullet(U) \leftarrow W(\mathfrak{g}) : (A,F_A)$

are precisely in bijection with Lie-algebra valued 1-forms as follows: the Weil algebra is of the form

$W(\mathfrak{g}) = \wedge^\bullet (\mathfrak{g}^* \oplus \mathfrak{g}^*[1]), d_{W(\mathfrak{g})}$

with one copy of $\mathfrak{g}^*$ in degree 1, the other in degree 2. By the free nature of the Weil algebra, dg-algebra morphisms $\Omega^\bullet(U) \leftarrow W(\mathfrak{g})$ are in bijection to their underlying morphisms of vector spaces of generators

$\Omega^1(U) \leftarrow \mathfrak{g}^* : A
\,.$

This identifies the 1-form $A \in \Omega^1(U,\mathfrak{g})$. This extends uniquely to a morphism of dg-algebras and thereby fixes the image of the shifted generators

$\Omega^2(U) \leftarrow \mathfrak{g}^*[1] : F_A
\,.$

The *Bianchi identity* is precisely the statement that these linear maps, extended to morphisms of graded algebra, are compatible with the differentials and hence do constitute dg-algebra morphisms.

Concretely, if $\{t^a\}$ is a dual basis for $\mathfrak{g}^*$ and $\{r^a\}$ the corresponding dual basis for $\mathfrak{g}^*[1]$ and $\{C^a{}_{b c}\}$ the structure constants of the Lie bracket $[-,-]$ on $\mathfrak{g}$, then the differential $d_{W(\mathfrak{g})}$ on the Weil algebra is defined on generators by

$d_{W(\mathfrak{g})} t^a = - \frac{1}{2} C^a{}_{b c} t^b \wedge t^c + r^a$

and

$d_{W(\mathfrak{g})} r^a = C^a{}_{b c} t^b \wedge r^c
\,.$

The image of $t^a$ under $\Omega^\bullet(U) \leftarrow W(\mathfrak{g}) : (A,F_A)$ is the component $A^a$. The image of $r^a$ is therefore, by respect for the differential on $t^a$

$r^a \mapsto (F_A)^a := d A^a + \frac{1}{2}C^a{}_{b c} A^b \wedge A^c
\,.$

Respect for the differential on $r^a$ then implies

$d (F_A)^a + C^a{}_{b c} A^a \wedge (F_A)^c = 0
\,.$

This is the Bianchi identity.

Let now $\mathfrak{g}$ be an arbitrary ∞-Lie-algebra and $W(\mathfrak{g})$ its Weil algebra. Then a collection of ∞-Lie algebra valued differential forms is a dg-algebra morphism

$\Omega^\bullet(U) \leftarrow W(\mathfrak{g}) : A
,.$

It curvature is the composite of morphism of graded vector space

$\Omega^\bullet(U)
\stackrel{A}{\leftarrow}
W(\mathfrak{g})
\stackrel{F_{(-)}}{\leftarrow}
\mathfrak{g}^*[1]
:
F_A
\,.$

Since $A$ is a homomorphism of dg-algebras, this satisfies

$d_{dR} F_A + A(d_{W(\mathfrak{g})}(-)) = 0
\,.$

This identity is the **Bianchi identity for $\infty$-Lie algebra valued forms**.

The Bianchi identity for ∞-Lie algebroid valued differential forms is discussed in

- SSSI (web)

Last revised on July 17, 2022 at 14:48:16. See the history of this page for a list of all contributions to it.