User:Jan A. Sanders/An introduction to Lie algebra cohomology/Lecture 10

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The Gerstenhaber bracket

The Gerstenhaber bracket was introduced in The cohomology structure of an associative ring and the following is entirely based on this paper, with some input from La renaissance des opérades, followed by examples from Cohomology of infinite-dimensional Lie algebras. We write $\mathfrak{x}_\alpha^\beta$ for $x_\alpha,\cdots,x_{\alpha+\beta}\ .$

definition - Gerstenhaber bracket

Let $a\in C^k(\mathfrak{g},\mathfrak{g})$ and $b\in C^l(\mathfrak{g},\mathfrak{g})\ .$ We write $|a|=k-1\mod 2\ ,$ and then wite $a$ for $|a|\ ,$ since there can be no confusion. Define for $i=0,\dots,a$

$$a\circ_i b (\mathfrak{x}_{s}^{a+b})=a(\mathfrak{x}_{s}^{i-1},b(\mathfrak{x}_{s+i}^{b}),\mathfrak{x}_{s+b+i+1}^{a-i-1}) ,$$

that is, the $b$ is inserted at the position of $x_{s+i}\ ,$ and let

$$a\circ b =\sum_{i=0}^{a} (-1)^{ib}a\circ_{i} b$$

Then define the Gerstenhaber bracket $[\cdot,\cdot]_{\mathcal{G}}$ by

$$[a,b]_{\mathcal{G}}=a\circ b-(-1)^{ab} b\circ a$$

proposition - super Lie algebra

This gives a super Lie algebra structure on $C^\cdot(\mathfrak{g},\mathfrak{g})\ .$

proof

First one shows that it is enough to have (writing $xy$ for $x\circ y$)

$$\tag{1} x[y,z]_{\mathcal{G}}=(x y)z-(-1)^{yz}(x z)y.$$

Indeed,

$$[[x,y]_{\mathcal{G}},z]_{\mathcal{G}}-[x,[y,z]_{\mathcal{G}}]_{\mathcal{G}}+(-1)^{xy}[y,[x,z]_{\mathcal{G}}]_{\mathcal{G}}=$$ $$=[x,y]_{\mathcal{G}}z-(-1)^{(x+y)z}z[x,y]_{\mathcal{G}}-x[y,z]_{\mathcal{G}}+(-1)^{x(y+z)}[y,z]_{\mathcal{G}}x+(-1)^{xy}y[x,z]_{\mathcal{G}}-(-1)^{yz}[x,z]_{\mathcal{G}}y$$ $$=(xy-(-1)^{xy}yx)z-(-1)^{(x+y)z}z[x,y]_{\mathcal{G}}-x[y,z]_{\mathcal{G}}+(-1)^{x(y+z)}(yz-(-1)^{yz}zy)x+(-1)^{xy}y[x,z]_{\mathcal{G}}-(-1)^{yz}(xz-(-1)^{xz}zx)y$$ $$=-(x[y,z]_{\mathcal{G}}-(xy)z+(-1)^{yz}(xz)y)+(-1)^{xy}(y[x,z]_{\mathcal{G}}-(yx)z+(-1)^{xz}(yz)x)-(-1)^{(x+y)z}(z[x,y]_{\mathcal{G}}-(zx)y+(-1)^{xy}(zy)x)$$ $$=0$$ Let the associator $\alpha\in C^3(\mathfrak{g},\mathfrak{g})$ be defined by $\alpha(x,y,z)=x( yz)-(xy)z \ .$ Then Condition (1) is written as

$$\alpha(x,y,z)=(-1)^{yz}\alpha(x,z,y) ,$$

that is, the associator is symmetric in its last two arguments. Indeed, one has

$$\alpha(x,y,z)-(-1)^{yz}\alpha(x,z,y)=$$ $$=x(yz)-(xy)z-(-1)^{yz}(x(zy)-(xz)y)$$ $$=x[y,z]_\mathcal{G}-(xy)z+(-1)^{yz}(xz)y$$ One computes $$\alpha(a,b,c)=a\circ (b\circ c)-(a\circ b)\circ c$$ $$=\sum_{i=0}^{ {a}} (-1)^{i({b}+{c})} a \circ_{i} (b\circ c)-\sum_{i=0}^{ {a}+{b} } (-1)^{i{c} }(a\circ b)\circ_{i} c$$ $$=\sum_{i=0}^{ {a}}\sum_{j=0}^b (-1)^{jc+i({b}+{c})} a \circ_{i} (b\circ_j c)-\sum_{i=0}^{a+b}\sum_{j=0}^a (-1)^{jb+i{c} }(a\circ_j b)\circ_{i} c$$ $$=\sum_{i=0}^{ {a}}(-1)^{ib}( \sum_{j=0}^b (-1)^{(j+i){c}} a \circ_{i} (b\circ_j c)- \sum_{j=0}^{i-1} (-1)^{j{c} }(a\circ_i b)\circ_{j} c$$ $$- \sum_{j=0}^{b} (-1)^{(i+j){c}}(a\circ_i b)\circ_{i+j} c - \sum_{j=i+1}^{ {a}} (-1)^{(b+j){c} }(a\circ_i b)\circ_{b+j} c )$$

lemma

$$(a\circ_i b)\circ_{j+b} c=(a\circ_j c)\circ_i b , \quad 0\leq i<j \leq a$$

proof

For $0\leq i<j \leq a$ one has

$$(a\circ_i b)\circ_{j+b} c(\mathfrak{x}_{0}^{ {a}+{b}+{c} })=$$ $$=(a\circ_i b)(\mathfrak{x}_{0}^{b+j-1},c(\mathfrak{x}_{b+j}^{c}),\mathfrak{x}_{b+c+j+1}^{a-j-1})$$ $$=a(\mathfrak{x}_{0}^{i-1},b(\mathfrak{x}_{i}^{b}),\mathfrak{x}_{b+i+1}^{j-i-2},c(\mathfrak{x}_{b+j}^{c}),\mathfrak{x}_{b+c+j+1}^{a-j-1})$$ $$=(a\circ_j c)(\mathfrak{x}_{0}^{i-1},b(\mathfrak{x}_{i}^{b}),\mathfrak{x}_{b+i+1}^{a+c-i-1})$$ $$=(a\circ_j c)\circ_i b(\mathfrak{x}_{0}^{ {a}+{b}+{c} })$$

lemma

$$(a\circ_i b)\circ_{i+j} c=a\circ_i( b\circ_j c) , \quad 0\leq i\leq a, 0\leq j\leq b$$

proof

For $0\leq i<j \leq a$ one has

$$(a\circ_i b)\circ_{i+j} c(\mathfrak{x}_{0}^{ {a}+{b}+{c} })=$$ $$=(a\circ_i b)(\mathfrak{x}_{0}^{i+j-1},c(\mathfrak{x}_{i+j}^{c}),\mathfrak{x}_{c+i+j+1}^{{a}+{b}-i-j-1})$$ $$=(a(\mathfrak{x}_{0}^{i-1},b(\mathfrak{x}_{i}^{j-1},c(\mathfrak{x}_{i+j}^{c}),\mathfrak{x}_{c+i+j+1}^{b-j-1}),\mathfrak{x}_{b+c+i+1}^{{a}-i-1})$$ $$=a\circ_i( b\circ_j c)(\mathfrak{x}_{0}^{ {a}+{b}+{c} })$$ It follows that $$\alpha(a,b,c)=a\circ (b\circ c)-(a\circ b)\circ c=$$ $$=\sum_{i=0}^{ {a}}(-1)^{ib}( \sum_{j=0}^b (-1)^{(j+i){c}} a \circ_{i} (b\circ_j c)- \sum_{j=0}^{i-1} (-1)^{j{c}}(a\circ_i b)\circ_{j} c- \sum_{j=0}^{b} (-1)^{(i+j){c}}a\circ_i( b\circ_{j} c) - \sum_{j=i+1}^{ {a} } (-1)^{(b+j){c} }(a\circ_j c)\circ_{i} b )$$ $$=-\sum_{i=0}^{ {a}}(\sum_{j=0}^{i-1} (-1)^{ib}(-1)^{j{c}}(a\circ_i b)\circ_{j} c+ \sum_{j=i+1}^{ {a} } (-1)^{ib}(-1)^{(b+j){c} }(a\circ_j c)\circ_{i} b )$$ $$=-\sum_{i=0}^{ {a}} \sum_{j=0}^{i-1} (-1)^{ib}(-1)^{j{c}}(a\circ_i b)\circ_{j} c -\sum_{j=0}^{ {a}} \sum_{i=j+1}^{{a} } (-1)^{jb}(-1)^{(b+i){c} }(a\circ_i c)\circ_{j} b$$ $$=-\sum_{i=0}^{ {a}} \sum_{j=0}^{i-1} (-1)^{ib}(-1)^{j{c}}(a\circ_i b)\circ_{j} c -\sum_{i=0}^{ {a} } \sum_{j=0}^{i-1} (-1)^{jb}(-1)^{(b+i){c} }(a\circ_i c)\circ_{j} b$$ $$=(-1)^{bc} \alpha(a,c,b)$$

remark

Observe that the Lie algebra is not necessarily a Poisson-Lie algebra, at least not with the present multiplication $\circ\ .$ For this to happen we would need

$$[a\circ b,c]=a\circ [b,c]-[a,c]\circ b$$

but one finds

$$[a\circ b,c]=a\circ [b,c]-[a,c]\circ b+\alpha(c,a,b)$$

example: associative structure

Let $a^2\in C^2(\mathfrak{g},\mathfrak{g})$ be an associative structure, that is,

$$a^2(a^2(x,y),z)=a^2(x,a^2(y,z))$$ Then $$[a^2,a^2]_{\mathcal{G}}(x_1,x_2,x_3)=2(a^2\circ a^2)(x_1,x_2,x_3)\ :$$ $$= 2a^2\circ_1 a^2 (x_1,x_2,x_3)-2a^2\circ_2 a^2 (x_1,x_2,x_3)\ :$$ $$= 2a^2(a^2(x_1,x_2),x_3)-2a^2(x_1,a^2(x^2,x^3))\ :$$ $$=0$$ The equation $[a^2,a^2]_{\mathcal{G}}=0$ characterizes an associative structure.

Let

$$d^m b^m=[a^2,b^m]_{\mathcal{G}}$$ Then $$d^{m+1} d^m b^m =d^{m+1} [a^2,b^m]_{\mathcal{G}} = [a^2,[a^2,b^m]_{\mathcal{G}}]_{\mathcal{G}} = \frac{1}{2} [[a^2,a^2]_{\mathcal{G}},b^m]_{\mathcal{G}} = 0$$ The complex $(C^\cdot(\mathfrak{g},\mathfrak{g}),d^\cdot)$ is called the Hochschild complex. One observes that $$d^{k+l-1}[b^k,c^l]_{\mathcal{G}}=[a^2,[b^k,c^l]_{\mathcal{G}}]_{\mathcal{G}}\ :$$ $$=[[a^2,b^k]_{\mathcal{G}},c^l]_{\mathcal{G}}+(-1)^{k-1}[b^k,[a^2,c^l]_{\mathcal{G}}]_{\mathcal{G}}\ :$$ $$=[d^k b^k,c^l]_{\mathcal{G}}+(-1)^{k-1}[b^k,d^l c^l]_{\mathcal{G}}$$ This implies that the Gerstenhaber bracket can be lifted to $H^\cdot(\mathfrak{g},\mathfrak{g})\ ,$ since $Z^\cdot(\mathfrak{g},\mathfrak{g})$ is a Lie subalgebra and $B^\cdot(\mathfrak{g},\mathfrak{g})$ is an ideal in $Z^\cdot(\mathfrak{g},\mathfrak{g})\ .$

the shuffled Gerstenhaber bracket

The following variation is possible. Define for $i=1,\dots,k$

$$a\hat{\circ}_i b =\sum_{\sigma\in\mathcal{S}_a^b}(-1)^{|\sigma|}( a\circ_i b)^\sigma$$

where $\mathcal{S}_a^b$ is the group of permutations of the arguments, in such a way that the order within $a$ and $b$ is preserved (shuffle) and the last argument is fixed, and where $|\sigma|$ is the sign of the permutation. So from the set $x_0,\dots,x_{a+b-1}$ we take out $b$ elements if $i<a+b-1$ and $b-1$ elements if $i=a+b-1$ and put those in $b\ .$ An example, with $a.b \in C^2(\mathfrak{g},\mathfrak{g})\ :$

$$a\hat{\circ}_0 b(x_0,x_1,x_2)=a \circ_0 b (x_0,x_1,x_2) = a(b(x_0,x_1),x_2)$$

$$a\hat{\circ}_1 b(x_0,x_1,x_2)=a \circ_1 b (x_0,x_1,x_2)-a \circ_1 b (x_1,x_0,x_2)$$ $$=a(x_0,b(x_1,x_2))-a(x_1,b(x_0,x_2))$$ Let

$$a\hat{\circ} b =\sum_{i=0}^b (-1)^{ib}a\hat{\circ}_i b$$

Then define the shuffled Gerstenhaber bracket $[\cdot,\cdot]_{\hat{\mathcal{G}}}$ by

$$[a,b]_{\hat{\mathcal{G}}}=a\hat{\circ} b-(-1)^{ab} b\hat{\circ} a$$

This again defines a super Lie algebra structure, but this needs proof.

lemma

$$(a\hat{\circ}_i b)\hat{\circ}_{j+b} c=(a\hat{\circ}_j c)\hat{\circ}_i b , \quad 0\leq i<j \leq a$$

proof

$$(a\hat{\circ}_i b)\hat{\circ}_{j+b} c)=$$ $$= \sum_{\sigma\in\mathcal{S}_{a+b}^c}(-1)^{|\sigma|}( (a\hat{\circ}_i b) \circ_{j+b} c)^\sigma$$ $$= \sum_{\sigma\in\mathcal{S}_{a+b}^c}\sum_{\pi\in\mathcal{S}_{a}^b}(-1)^{|\sigma|+|\pi|}( (a\circ_i b) ^\pi \circ_{j+b} c)^\sigma$$ The permutations $\pi$ and $\sigma$ give us a partition of $(A|B|C)$ of $(0,\dots,a+b+c)\ .$ Using uniquely determined permutations $\hat{\pi}\in S_a^c$ (the inverse of joining $A$ and $C$) and $\hat{\sigma}\in S_{a+c}^b$ (the inverse of joining the join of $A$ and $C$ with $B$) we can obtain the partition $(A|C|B)\ .$ Thus we can write, using Lemma, $$(a\hat{\circ}_i b)\hat{\circ}_{j+b} c)=$$ $$= \sum_{\sigma\in\mathcal{S}_{a+b}^c}\sum_{\pi\in\mathcal{S}_{a}^b}(-1)^{|\sigma|+|\pi|}( (a\circ_i b) ^\pi \circ_{j+b} c)^\sigma$$ $$= \sum_{\hat{\sigma}\in\mathcal{S}_{a+c}^b}\sum_{\hat{\pi}\in\mathcal{S}_{a}^c}(-1)^{|\hat{\sigma}|+|\hat{\pi}|}( (a\circ_j c)^{\hat{\pi}} \circ_{i} b)^{\hat{\sigma }}$$ $$= (a\hat{\circ}_j c)\hat{\circ}_i b$$

lemma

$$(a\hat{\circ}_i b)\hat{\circ}_{i+j} c=a\hat{\circ}_i( b\hat{\circ}_j c) , \quad 0\leq i\leq a, 0\leq j \leq b$$

proof

The permutations $\pi\in S_a^b$ and $\sigma\in S_{a+b}^c$ give us a partition of $(A|B|C)$ of $(0,\dots,a+b+c)\ .$ Using uniquely determined permutations $\hat{\pi}\in S_b^c$ (the inverse of joining $B$ and $C$) and $\hat{\sigma}\in S_{a}^{b+c}$ (the inverse of joining $A$ with the join of $B$ and $C$) we can obtain the same partition $(A|B|C)\ .$ The result follows from the Lemma.

a Form [1] program to compute the shuffled Gerstenhaber bracket

Declaration of the global parameters

• #define k "2";
• #define l "2";
• #define N "{`k'+`l'-1}";

Declaration of the variables

• F f g h a b;
• S m n x1,...,x`N';
• Set var:x1,...,x`N';

Declaration of the sets

• Set prevar:x1,...,x{`N'-1};
• Set fun: g h;
• Set ffun: a b;

Start of the actual program

• L F=f(x1,...,x`N');

Now shuffle

• id f(?a)=distrib_(-1,`l',g,h,?a)-(-1)^((`k'-1)*(`l'-1))*distrib_(-1,`k',h,g,?a);

Insert a into b and move it to all different positions

• id g?fun(?a)*h?fun(?b)=h(g(?a),?b);
• repeat ;
  • id h?fun[n?](?a,g?fun[m?](?b),x1?var,?c)=ffun[n](?a,ffun[m](?b),x1,?c)-(-1)^(nargs_(?b))*fun[n](?a,x1,fun[m](?b),?c);
  • id h?fun[n?](?a,g?fun[m?](?b))=ffun[n](?a,ffun[m](?b));
  • id h?ffun(?a,a?ffun(?b,x1?prevar))=0;
  • id h?ffun(?a,a?ffun(?b),?c,x1?prevar)=0;
• endrepeat;
• .sort
• print;
• .end

The output of this run (with $k=l=2$) is

• F = a(b(x1,x2),x3)- a(x1,b(x2,x3))+ a(x2,b(x1,x3))+ b(a(x1,x2),x3)- b(x1,a(x2,x3))+ b(x2,a(x1,x3));

The computation with, for instance, $k=l=9$ takes a few seconds, but the answer is a bit too long to give here with its 208780 terms.

example: Lie algebra structure

Let $a^2\in C_\wedge^2(\mathfrak{g},\mathfrak{g})$ be a Lie algebra structure, that is,

$$a^2(a^2(x,y),z)=a^2(x,a^2(y,z))-a^2(y,a^2(x,z)$$ and $a^2$ is antisymmetric. Then $$[a^2,a^2]_{\hat{\mathcal{G}}}(x_1,x_2,x_3)=2(a^2\hat{\circ} a^2)(x_1,x_2,x_3)\ :$$ $$= 2a^2\hat{\circ}_1 a^2 (x_1,x_2,x_3)-2a^2\hat{\circ}_2 a^2 (x_1,x_2,x_3)\ :$$ $$= 2a^2(a^2(x_1,x_2),x_3)-2a^2(x_1,a^2(x_2,x_3))+2a^2(x_2,a^2(x_1,x_3))\ :$$ $$=0$$ The equation $[a^2,a^2]_{\hat{\mathcal{G}}}=0$ characterizes a Lie algebra structure.

Let

$$d^m b^m=[a^2,b^m]_{\hat{\mathcal{G}}}$$ Then $$d^{m+1} d^m b^m =d^{m+1} [a^2,b^m]_{\hat{\mathcal{G}}}=[a^2,[a^2,b^m]_{\hat{\mathcal{G}}}]_{\hat{\mathcal{G}}}=\frac{1}{2}[[a^2,a^2]_{\hat{\mathcal{G}}},b^m]_{\hat{\mathcal{G}}}=0$$ The complex $(C_\wedge^\cdot(\mathfrak{g},\mathfrak{g}),d^\cdot)$ is called the Lie algebra complex and $d^m$ is identical to the previous $d^m\ .$ One observes that $$d^{k+l-1}[b^k,c^l]_{\hat{\mathcal{G}}}=[a^2,[b^k,c^l]_{\hat{\mathcal{G}}}]_{\hat{\mathcal{G}}}\ :$$ $$=[[a^2,b^k]_{\hat{\mathcal{G}}},c^l]_{\hat{\mathcal{G}}}+(-1)^{k-1}[b^k,[a^2,c^l]_{\hat{\mathcal{G}}}]_{\hat{\mathcal{G}}}\ :$$ $$=[d^k b^k,c^l]_{\hat{\mathcal{G}}}+(-1)^{k-1}[b^k,d^l c^l]_{\hat{\mathcal{G}}}$$ This implies that the shuffled Gerstenhaber bracket can be lifted to $H_\wedge^\cdot(\mathfrak{g},\mathfrak{g})\ .$

Deformations

deformation of a Lie algebra

Let $a_0^2$ denote the bracket on $\mathfrak{g}\ ,$ that is,

$$a_0^2(x_1,x_2)=[x_1,x_2],\quad x_1,x_2\in\mathfrak{g}$$ A formal deformation of a Lie algebra structure is a formal expression $$a_\varepsilon^2=\sum_{i=0}^\infty \varepsilon^i a_i^2$$ such that $a_\varepsilon^2(x_1,x_2)=[x_1,x_2]_\varepsilon$ defines a Lie structure. Writing out the terms of the Jacobi identity one obtains $$\tag{2} 0=[[x_1,x_2]_\varepsilon,x_3]_\varepsilon- [x_1,[x_2,x_3]_\varepsilon]_\varepsilon+ [x_2,[x_1,x_3]_\varepsilon]_\varepsilon\ :$$

$$=\sum_{i=0}^\infty\sum_{j=0}^\infty \varepsilon^{i+j} \left( a_i^2(a_j^2(x_1,x_2),x_3)-a_i^2(x_1,a_j^2(x_2,x_3))+a_i^2(x_2,a_j^2(x_1,x_3))\right)$$ At order zero in $\varepsilon$ this is the Jacobi identity for $a_0^2\ ,$ at order one the expression reduces to $$0=a_0^2(a_1^2(x_1,x_2),x_3)-a_0^2(x_1,a_1^2(x_2,x_3))+a_0^2(x_2,a_1^2(x_1,x_3))+ a_1^2(a_0^2(x_1,x_2),x_3)-a_1^2(x_1,a_0^2(x_2,x_3))+a_1^2(x_2,a_0^2(x_1,x_3))\ :$$ $$=d^{(0)}(x_3)a_1^2(x_1,x_2)-d^{(0)}(x_1)a_1^2(x_2,x_3)+d^{(0)}(x_2)a_1^2(x_1,x_2) +a_1^2([x_1,x_2],x_3)+a_1^2(x_2,[x_1,x_3])-a_1^2(x_1,[x_2,x_3])\ :$$ $$=-d^2 a_1^2(x_1,x_2,x_3)$$ So one has to require that $a_1^2\in Z_\wedge^2(\mathfrak{g},\mathfrak{g})\ .$ A rather trivial way of obtaining a new bracket is by applying a formal transformation to $\mathfrak{g}\ .$ This leads to $$\phi_\varepsilon(x)=\sum_{i=0}^\infty \varepsilon^i b_i^1(x),\quad b_i^1\in C^1(\mathfrak{g},\mathfrak{g})$$ with $b_0^1(x)=x\ ,$ and $$\phi_\varepsilon^{-1} ([\phi_\varepsilon(x_1),\phi_\varepsilon(x_2)])\ :$$ $$=[x_1,x_2]+\varepsilon\left(d^{(0)}(x_1)b_1^1(x_2)-d^{(0)}(x_2)b_1^1(x_1)-b_1^1([x_1,x_2])\right)+O(\varepsilon^2)\ :$$ $$=[x_1,x_2]+\varepsilon d^1 b_1^1(x_1,x_2)+O(\varepsilon^2)$$ One sees that if $\phi_\varepsilon$ is required to be a Lie algebra homomorphism, $b_1^1\in Z^1(\mathfrak{g},\mathfrak{g})\ .$ This can be trivially done by letting $b_1^1\in B^1(\mathfrak{g},\mathfrak{g})\ ,$ that is, $$b_1^1(x)=d^0 b_1^0 (x)=-[b_1^0,x]\ .$$ In this case $b_1^0\in\mathfrak{g}$ is called the infinitesimal generator of $\phi_\varepsilon\ .$ This is the usual situation in normal form theory.

In the case of formal deformations one sees that a trivial deformation $a_1^2=d^1 b_1^1$ can be realized by a formal coordinate transformation $x\mapsto x+\varepsilon b_1^1(x) +O(\varepsilon^2)\ .$ This implies that the existence of a nontrivial formal deformation implies the existence of a nontrivial element $[a_1^2]\in H_\wedge^2(\mathfrak{g},\mathfrak{g})\ .$

The question now is whether this is enough, in other words, is this the only obstruction? Consider (2) at order two. One obtains

$$0=a_0^2(a_2^2(x_1,x_2),x_3)-a_0^2(x_1,a_2^2(x_2,x_3))+a_0^2(x_2,a_2^2(x_1,x_3))\ :$$ $$+a_1^2(a_1^2(x_1,x_2),x_3)-a_1^2(x_1,a_1^2(x_2,x_3))+a_1^2(x_2,a_1^2(x_1,x_3))\ :$$ $$+a_2^2(a_0^2(x_1,x_2),x_3)-a_2^2(x_1,a_0^2(x_2,x_3))+a_2^2(x_2,a_0^2(x_1,x_3))$$ $$=[a_2^2(x_1,x_2),x_3]-[x_1,a_2^2(x_2,x_3)]+[x_2,a_2^2(x_1,x_3)]\ :$$ $$+a_1^2(a_1^2(x_1,x_2),x_3)-a_1^2(x_1,a_1^2(x_2,x_3))+a_1^2(x_2,a_1^2(x_1,x_3))\ :$$ $$+a_2^2([x_1,x_2],x_3)-a_2^2(x_1,[x_2,x_3])+a_2^2(x_2,[x_1,x_3])$$ $$= -d^2a_2^2 (x_1,x_2,x_3)+\frac{1}{2}[a_1^2,a_1^2]_{\hat{\mathcal{G}}}(x_1,x_2,x_3)$$ One needs the equivalence class of $[a_1^2,a_1^2]_{\hat{\mathcal{G}}}$ to be trivial. It is already shown that $[a_1^2,a_1^2]_{\hat{\mathcal{G}}}\in Z_\wedge^3(\mathfrak{g},\mathfrak{g})\ ,$ so this is not an impossibility, but is is a possible obstruction if $H_\wedge^3(\mathfrak{g},\mathfrak{g})\neq 0\ .$

Consider (2) at order three. One obtains

$$d^2a_3^2 =[a_1^2,a_2^2]_{\hat{\mathcal{G}}}$$ One computes $$d^3[a_1^2,a_2^2]_{\hat{\mathcal{G}}}=[a_1^2,d^2 a_2^2]_{\hat{\mathcal{G}}}=\frac{1}{2}[a_1^2,[a_1^2,a_1^2]_{\hat{\mathcal{G}}}]_{\hat{\mathcal{G}}}$$ and so one needs $[a_1^2,[a_1^2,a_1^2]_{\hat{\mathcal{G}}}]_{\hat{\mathcal{G}}}$ to be zero.

In general one finds

$$d^2a_n^2 =\frac{1}{2}\sum_{i=1}^{n-1} [a_i^2,a_{n-i}^2]_{\hat{\mathcal{G}}}$$ This implies that a necessary condition is that $[a_1^2,\dots,[a_1^2,a_1^2]_{\hat{\mathcal{G}}}\dots]_{\hat{\mathcal{G}}}$ be zero (The Massey powers of the class $[a_1^2]\in H^2(\mathfrak{g},\mathfrak{g})$ vanish). It is sufficient for the existence of a formal deformation to require that $[a_i^2,a_j^2]$ be trivial for $i,j>0\ .$

exercise

Assume $a_1^2$ to be trivial. What does this imply for all the obstructions?

example

Let $\mathfrak{g}=P[x,y]\ .$ Define

$$a_0^2(f,g)=[f,g]=\frac{\partial f}{\partial x}\frac{\partial g}{\partial y} -\frac{\partial g}{\partial x}\frac{\partial f}{\partial x}$$ One verfies that $[f,gh]=[f,g]h+g[f,h]$ Define $$a_1^2(f,g)=f_{xx}g_{yy}-f_{yy}g_{xx}$$ One computes $$d^2 a_1^2(f,g,h)=[f,a_1^2(g,h)]-[g,a_1^2(f,h)]+[h,a_1^2(f,g)]-a_1^2([f,g],h)-a_1^2(g,[f,h])+a_1^2(f,[g,h])\ :$$ $$=[f,g_{xx}h_{yy}]-[f,g_{yy}h_{xx}]-[g,f_{xx}h_{yy}]+[g,f_{yy}h_{xx}]+[h,f_{xx}g_{yy}]-[h,f_{yy}g_{xx}]\ :$$

$$-(f_xg_y-f_yg_x)_{xx}h_{yy}+(f_xg_y-f_yg_x)_{yy}h_{xx}-g_{xx}(f_xh_y-f_yh_x)_{yy}+g_{yy}(f_xh_y-f_yh_x)_{xx}+f_{xx}(g_xh_y-g_yh_x)_{yy} -f_{yy}(g_xh_y-g_yh_x)_{xx}\ :$$

$$=[f,g_{xx}]h_{yy}-[f,g_{yy}]h_{xx}-[g,f_{xx}]h_{yy}+[g,f_{yy}]h_{xx}+[h,f_{xx}]g_{yy}-[h,f_{yy}]g_{xx}\ :$$

$$+g_{xx}[f,h_{yy}]-g_{yy}[f,h_{xx}]-f_{xx}[g,h_{yy}]+f_{yy}[g,h_{xx}]+f_{xx}[h,g_{yy}]-f_{yy}[h,g_{xx}]\ :$$

$$-(f_xg_y-f_yg_x)_{xx}h_{yy}+(f_xg_y-f_yg_x)_{yy}h_{xx}-g_{xx}(f_xh_y-f_yh_x)_{yy}+g_{yy}(f_xh_y-f_yh_x)_{xx}+f_{xx}(g_xh_y-g_yh_x)_{yy}\ :$$

$$-f_{yy}(g_xh_y-g_yh_x)_{xx}\ :$$

$$=([f,g_{xx}]-(f_xg_y-f_yg_x)_{xx}-[g,f_{xx}])h_{yy} +(-[f,g_{yy}]+[g,f_{yy}]+(f_xg_y-f_yg_x)_{yy})h_{xx} +([h,f_{xx}]-[f,h_{xx}]+(f_xh_y-f_yh_x)_{xx})g_{yy} \ :$$

$$+(-[h,f_{yy}]+[f,h_{yy}]-(f_xh_y-f_yh_x)_{yy})g_{xx} +(-[g,h_{yy}]+[h,g_{yy}]+(g_xh_y-g_yh_x)_{yy})f_{xx} +([g,h_{xx}]-[h,g_{xx}] -(g_xh_y-g_yh_x)_{xx})f_{yy}\ :$$

$$=2(f_{xy}g_{xx}-f_{xx}g_{xy})h_{yy} -2(f_{yy}g_{xy}-f_{xy}g_{yy})h_{xx} -2(f_{xy}h_{xx}-f_{xx}h_{xy})g_{yy} \ :$$

$$+2(f_{yy}h_{xy}-f_{xy}h_{yy})g_{xx} -2(g_{yy}h_{xy}-g_{xy}h_{yy})f_{xx} +2(g_{xy}h_{xx}-g_{xx}h_{xy})f_{yy}\ :$$

$$=0$$ But is this a nontrivial cocycle, as claimed in Fuchs, page 38? The first candidate for a coboundary that comes to mind is $$a_1^1(f)=f_{xy}$$ One computes $$d^1 a _1^1(f,g)=[f,a_1^1(g)]-[g,a_1^1(f)]-a_1^1([f,g])\ :$$ $$=[f,g_{xy}]-[g,f_{xy}]-(f_xg_y-f_yg_x)_{xy}\ :$$ $$=f_xg_{xyy}-f_yg_{xxy}-g_xf_{xyy}+g_yf_{xxy}-(f_{xx}g_y +f_xg_{xy}-f_{xy}g_x-f_yg_{xx})_y\ :$$ $$=f_xg_{xyy}-f_yg_{xxy}-g_xf_{xyy}+g_yf_{xxy}-f_{xxy}g_y -f_{xy}g_{xy}+f_{xyy}g_x+f_{yy}g_{xx}-f_{xx}g_{yy} -f_xg_{xyy}+f_{xy}g_{xy}+f_yg_{xxy}\ :$$ $$=f_{yy}g_{xx}-f_{xx}g_{yy} \ :$$ $$=-a_1^2(f,g)$$ One can identify $P[x,y]$ with the space polynomial differential operators $P[x,\frac{d}{dx}]$ on the line by putting $$\widehat{x^m\frac{d^k}{dx^k}}=x^m y^k$$ One observes that $$[x^k \frac{d^l}{dx^l},x^m \frac{d^n}{dx^n}]=x^k \frac{d^l}{dx^l}x^m \frac{d^n}{dx^n}-x^m \frac{d^n}{dx^n}x^k \frac{d^l}{dx^l}$$ Since $$\frac{d^k}{dx^k}f=\sum_{i=0}^{k} {k\choose i} f_i \frac{d^{k-i}}{dx^{k-i}}$$ and, with $f=x^m\ ,$ $$f_i={m\choose i} x^{m-i}$$ one obtains $$[x^k \frac{d^l}{dx^l},x^m \frac{d^n}{dx^n}]=\sum_{i=0}^{\min(l,m)} { {l}\choose{i}} { {m}\choose{i}} x^{k+m-i} \frac{d^{n+l-i} }{dx^{n+l-i} } -\sum_{i=0}^{\min(n,k)} { {k}\choose{i}}{ {n}\choose{i}} x^{m+k-i} \frac{d^{l+n-i} }{dx^{l+n-i} }\ :$$ $$=\sum_{i=1}^{\max(\min(l,m),\min(n,k))} \left( { {l}\choose{i}} {{m}\choose{i} } -{ {k}\choose{i}}{ {n}\choose{i}}\right) x^{m+k-i} \frac{d^{l+n-i} }{dx^{l+n-i} }$$ This can be written, with $f=x^k \frac{d^l}{dx^l}=\widehat{x^ky^l}$ and $g=x^m \frac{d^n}{dx^n}=\widehat{ x^m y^n}\ ,$ as $$\widehat{[f,g]}= \sum_{i=1}^\infty \left( \frac{\partial^i \hat{f}}{\partial x^i}\frac{\partial^i \hat{g}}{\partial y^i}-\frac{\partial^i \hat{f}}{\partial y^i}\frac{\partial^i \hat{g}}{\partial x^i}\right)= \sum_{i=0}^\infty a_i^2(\hat{f},\hat{g})$$ This is, by the way, another proof of the exactness of $a_1^2\ ,$ one that illustrates the power of looking at the same object from different angles.

deformation of an associative structure

Observe that similiar arguments hold for deformations of associative strucures.

References

• Fuks, D. B. Cohomology of infinite-dimensional Lie algebras. Translated from the Russian by A. B. Sosinski\u\i. Contemporary Soviet Mathematics. Consultants Bureau, New York, 1986. xii+339 pp. ISBN: 0-306-10990-5
• Gerstenhaber, Murray . The cohomology structure of an associative ring. Ann. of Math. (2) 78 1963 267--288. [2]