Deformed super brackets on forms of a manifold

Let $M$ be a differentiable manifold and $\mathrm{T}(M)$ and $\mathrm{T}^{*}(M)$ be the tangent and cotangent bundle of $M$ (or the spaces of the sections). In addition to the typical example of Lie superalgebra $\mathfrak{g}=\sum_{p=1}^{\textrm{dim}M}\Lambda^{p}\mathrm{T}(M)$ with the Schouten bracket, the space of differential forms $\mathfrak{h}=\sum_{q=0}^{\textrm{dim}M}\Lambda^{q}\mathrm{T}^{*}(M)$ with the bracket

(1.1)

$$\{\alpha,\beta\}=(-1)^{a}d(\alpha\wedge\beta)\quad\text{where}\alpha\in\Lambda^{a}\mathrm{T}^{*}(M)\ \text{ and }\ \beta\in\Lambda^{b}\mathrm{T}^{*}(M)$$

becomes a Lie superalgebra, where the grading of $\Lambda^{a}\mathrm{T}^{*}(M)$ is $-a-1$, and is often referred to as $a^{\prime}$ (cf.  [5]). The grading of $\Lambda^{a}\mathrm{T}(M)$ is $a-1$, and is also represented by $a^{\prime}$.

There is a notion of deformation of the exterior differentiation $d$ by a 1-form $\phi$ defined by $d_{t}\alpha=d\alpha+t\phi\wedge\alpha$ where $t$ is a scalar parameter runs at least $[0,1]$ interval, and it is well-known that $d_{t}\circ d_{t}=0$ if $\phi$ is a 1-cocycle. It is natural to expect $d_{\phi}$ defines a Lie superalgebra structure, namely

$$\{\alpha,\beta\}^{t}=(-1)^{a}d_{t}(\alpha\wedge\beta)=\{\alpha,\beta\}+{\alpha}\wedge(t\phi)\wedge{\beta}$$

will be a super bracket for each $t$. Super symmetry holds good. About super Jacobi identity,

	$\displaystyle\quad\{\{\alpha,\beta\}^{t},\gamma\}^{t}$
	$\displaystyle=\{\{\alpha,\beta\},\gamma\}+(-1)^{a+1}d(\alpha\wedge\beta\wedge t\phi\wedge\gamma)+(-1)^{a}d({\alpha}\wedge{\beta})\wedge t\phi\wedge\gamma$
	$\displaystyle=\{\{\alpha,\beta\},\gamma\}+(-1)^{b+c+1}\alpha\wedge\beta\wedge d\gamma\wedge t\phi+(-1)^{b+1}\alpha\wedge\beta\wedge\gamma\wedge d(t\phi)$
	$\displaystyle=\{\{\alpha,\beta\},\gamma\}+(-1)^{b+c+1}\alpha\wedge\beta\wedge d\gamma\wedge t\phi\quad\text{if $\phi$ is closed.}$
	$\displaystyle\quad\mathop{\mathfrak{S}}_{\alpha,\beta,\gamma}(-1)^{a^{\prime}c^{\prime}}\{\{\alpha,\beta\}^{t},\gamma\}^{t}$
	$\displaystyle=(-1)^{a^{\prime}c^{\prime}}\{\{\alpha,\beta\},\gamma\}+(-1)^{a^{\prime}c^{\prime}+b+c+1}\alpha\wedge\beta\wedge d\gamma\wedge t\phi+(-1)^{a^{\prime}c^{\prime}+b+1}\alpha\wedge\beta\wedge\gamma\wedge d(t\phi)$
	$\displaystyle=\mathop{\mathfrak{S}}_{\alpha,\beta,\gamma}(-1)^{ac}d(\alpha)\wedge\beta\wedge d\gamma\wedge t\phi+3(-1)^{ac+a+b+c}\alpha\wedge\beta\wedge\gamma\wedge d(t\phi)$
	$\displaystyle=\mathop{\mathfrak{S}}_{\alpha,\beta,\gamma}(-1)^{ac}d(\alpha)\wedge\beta\wedge d\gamma\wedge t\phi\quad\text{if $\phi$ is closed.}$

So far, there is no affirmative statement in general setting. On the other hand, there is a result of deformation of the Schouten bracket by D. Iglesias and J. C. Marrero in [1]. They say for a 1-cocycle $\phi$,

(1.2)

$$[P,Q]^{\phi}=[P,Q]+(-1)^{p}P(\phi)\wedge(q-1)Q+(p-1)P\wedge Q(\phi)\quad\text{ where }\quad P(\phi)=\iota_{\phi}P$$

satisfies the axioms of bracket of Lie superalgebra.

1. 1.

   $[P,Q]^{\phi}=-(-1)^{(p-1)(q-1)}[Q,P]^{\phi}$

2. 2.

   $[P,[Q,R]^{\phi}]^{\phi}=[[P,Q]^{\phi},R]^{\phi}+(-1)^{(p-1)(q-1)}[Q,[P,R]^{\phi}]^{\phi}$   .

Inspired by the work above, in this note for a given 1-form $\phi$, we fix properties of a function $F$ so that

(1.3)

$$(-1)^{a}d(\alpha\wedge\beta)+F(a,b){\alpha}\wedge(t\phi)\wedge{\beta}\quad\quad(\alpha\in\Lambda^{a}\mathrm{T}^{*}(M),\;\beta\in\Lambda^{b}\mathrm{T}^{*}(M))\;$$

becomes super bracket for $t$. The function $F$ should be defined on $\{(a,b)\in\mathbb{Z}^{2}\mid a+b\leqq\textrm{dim}M-1\}$.

Main results in this note is that there are deformations of two super brackets on the space $\mathfrak{h}=\sum_{q=0}^{\textrm{dim}M}\Lambda^{q}\mathrm{T}^{*}(M)$, and there is a natural extension to a subalgebra of $\mathrm{T}(M)$.

Claim 1: A deformation of the trivial bracket: For a given 1-form $\phi$

$$\{\alpha,\beta\}^{t,\phi}=F(a,b){\alpha}\wedge(t\phi)\wedge{\beta}\quad\quad(\alpha\in\Lambda^{a}\mathrm{T}^{*}(M),\;\beta\in\Lambda^{b}\mathrm{T}^{*}(M))\;$$

is a super bracket on $\mathfrak{h}$ when $F$ is a symmetric function.

Claim 2: A deformation of the standard bracket: For a given closed 1-form $\phi$,

$$\{\alpha,\beta\}^{t,\phi}=(-1)^{a}d({\alpha}\wedge{\beta})+\tfrac{a+b+2}{2}\alpha\wedge t\phi\wedge\beta\quad\quad(\alpha\in\Lambda^{a}\mathrm{T}^{*}(M),\;\beta\in\Lambda^{b}\mathrm{T}^{*}(M))\;$$

is a super bracket on $\mathfrak{h}$.

Claim 3: An extension of the deformation of the standard bracket: For a given closed 1-form $\phi$, Claim 2 says $\mathfrak{h},\{\cdot,\cdot\}^{t,\phi}$ are Lie superalgebra. Let ${\mathfrak{g}}_{0}^{\prime}=\{X\in\mathrm{T}(M)\mid\mathit{L}_{X}\phi=0\}$, which is a subalgebra of $\mathrm{T}(M)$. Then $(\mathfrak{h},\{\cdot,\cdot\}^{t,\phi})\oplus{\mathfrak{g}}_{0}^{\prime}$ becomes a Lie superalgebra naturally by the Lie derivative for each $t$.

Based on these results, there are many issues to be studied. We would like to develop homology theory of deformed superalgebra. When a Lie group $G$ acts on $M$, we get $\sum_{p=1}^{\textrm{dim}M}\Lambda^{p}\mathrm{T}_{G}(M)$ of $G$-invariant multivector fields, and $\sum_{q=0}^{\textrm{dim}M}\Lambda^{q}\mathrm{T}^{*}_{G}(M)$ of $G$-invariant differential forms. Since the action preserves the Jacobi-Lie bracket, the Schouten bracket is preserved by the action. Also the action commutes with the differentiation $d$, the Lie superalgebra bracket is preserved by the action. In short, $\sum_{p=1}^{\textrm{dim}M}\Lambda^{p}\mathrm{T}_{G}(M)$ and $\sum_{q=0}^{\textrm{dim}M}\Lambda^{q}\mathrm{T}^{*}_{G}(M)$ have Lie superalgebra structures. The simplest case is a Lie group acts on itself. $\overline{\mathfrak{g}}=\sum_{p=1}^{n}\Lambda^{p}\mathfrak{g}$ where $\mathfrak{g}=$Lie algebra of $G$, has a Lie superalgebra structure by the Schouten bracket (cf.[4]). The differential $d$ gives a Lie superalgebra structure on $\overline{\mathfrak{h}}=\sum_{p=0}^{n}\Lambda^{p}\mathfrak{h}$, where $\mathfrak{h}=\mathfrak{g}^{*}$ (cf.  [5]). Concrete and fancy examples are presented from those superalgebras.

We already know that the superalgebra $\mathfrak{h}=\sum_{i=0}^{\textrm{dim}M}\Lambda^{i}\mathrm{T}^{*}(M)$ with the bracket $\{\alpha,\beta\}=(-1)^{a}d(\alpha\wedge\beta)$ has an extension by ${\mathfrak{g}}_{0}=\mathrm{T}(M)$ through Lie derivative in [5]. Here we study the deformed superalgebra given by the bracket (2.1) has an extension by a subalgebra of $\mathrm{T}(M)$ through the Lie derivative $\mathit{L}_{X}=\iota_{X}\circ d+d\circ\iota_{X}$ with respect to $X$.

Some concrete example will appear in the tail of the next section.

It is known in [5] that $\{\alpha,\beta\}=(-1)^{a}d(\alpha\wedge\beta)$ defines a super bracket on $\sum_{i=0}^{\textrm{dim}M}\Lambda^{i}\mathrm{T}^{*}(M)$. Looking at the deformed Schouten bracket, the bracket we expect is of form $\{\alpha,\beta\}^{t,\phi}=\{\alpha,\beta\}+F(a,b)\alpha\wedge t\phi\wedge\beta$ for some function $F$ on $\{(a,b)\in\mathbb{Z}^{2}\mid a+b\leqq\textrm{dim}M-1\}$.

About super symmetric property, we have

(3.1)

$$\{\alpha,\beta\}^{t,\phi}+(-1)^{a^{\prime}b^{\prime}}\{\beta,\alpha\}^{t,\phi}=(F(a,b)-F(b,a))\alpha\wedge t\phi\wedge\beta$$

About Jacobi identity, we see

	$\displaystyle\quad\mathop{\mathfrak{S}}_{\alpha,\beta,\gamma}(-1)^{a^{\prime}c^{\prime}}\left(\{\{\alpha,\beta\}^{t,\phi},\gamma\}^{t,\phi}-\{\{\alpha,\beta\},\gamma\}\right)$
	$\displaystyle=\left(F(1+b+c,a)-F(a,b)-F(b,c)-F(c,a)+F(1+c+a,b)\right)\left(-1\right)^{ac+a+c}\left(\alpha\wedge t\phi\wedge\beta\wedge d\left(\gamma\right)\right)$
	$\displaystyle-\left(-1\right)^{ac+a+c}\left(F(1+a+b,c)+F\left(1+b+c,a\right)-F(a,b)-F\left(b,c\right)-F\left(c,a\right)\right)\left(\alpha\wedge d\left(\beta\right)\wedge t\phi\wedge\gamma\right)$
	$\displaystyle+\left(F(a,b)+F(b,c)+F(c,a)\right)\left(-1\right)^{ac+a+b+c}\left(\alpha\wedge d\left(t\phi\right)\wedge\beta\wedge\gamma\right)$
	$\displaystyle-\left(-1\right)^{ac+c}\left(F(1+a+b,c)-F(a,b)-F(b,c)-F(c,a)+F(1+c+a,b)\right)\left(d\left(\alpha\right)\wedge\beta\wedge t\phi\wedge\gamma\right)\;.$

Thus, if $\phi$ is a 1-cocycle, then we get the following sufficient conditions for super Jacobi identity

(3.2)		$\displaystyle 0$	$\displaystyle=F(1+b+c,a)+F(1+c+a,b)-F(a,b)-F(b,c)-F(c,a)\;,$
(3.3)		$\displaystyle 0$	$\displaystyle=F(1+a+b,c)+F(1+b+c,a)-F(a,b)-F(b,c)-F(c,a)\;,$
(3.4)		$\displaystyle 0$	$\displaystyle=F(1+c+a,b)+F(1+a+b,c)-F(a,b)-F(b,c)-F(c,a)\;.$

The difference $\eqref{eq:x:1}-\eqref{eq:x:2}=0$ implies $F(1+c+a,b)=F(1+a+b,c)$. Putting $c=0$, we have $F(1+a,b)=F(1+a+b,0)$ and $\displaystyle F(a,0)=F(0,0)(\frac{a}{2}+1)$, we see that the symmetric function $F$ satisfying the above 3 conditions is

(3.5)

$$F(a,b)=\kappa(a+b+2)\ \ \text{where}\ \kappa\text{ is a constant.}$$