# Intrinsic formality versus rigidity of a differential graded Lie algebra

Let $\mathfrak g:=(V,d,[\cdot,\cdot])$ be a differential graded Lie algebra (DGLA) where $d$ is the zero differential.

Intrinsic formality: The DGLA $\mathfrak g$ will be said intrinsically formal if any $L_\infty$-algebra $l$ admitting $\mathfrak g$ in cohomology (i.e. $H(l)=\mathfrak g$) is formal (i.e. there exists a $L_\infty$-quasi isomorphism $\mathfrak g\to l$).

Rigidity: The DGLA $\mathfrak g$ will be said rigid if any $L_\infty$-algebra $(V,l')$ such that $l'_1=0$, $l'_2=[\cdot,\cdot]$ is isomorphic to $\mathfrak g$. In other words, $\mathfrak g$ does not admit non-trivial deformations as a $L_\infty$-algebra.

Question 1: What is the relation (equivalence, implication) between these two notions?

Let me denote $NR(\mathfrak g)$ the DGLA controlling the deformations of $\mathfrak g$ as a $L_\infty$-algebra and let $H^1(NR(\mathfrak g))$ be the associated first cohomology group.

Question 2: Does any of the previous notions entails the vanishing of $H^1(NR(\mathfrak g))$, or conversely ?

• There is something garbled in your definition of rigidity. Feb 11 '17 at 20:29
• Could you elaborate on that? Feb 12 '17 at 19:34
• Something like "admitting $\mathfrak{g}$ in homology" or "linearly isomorphic to $\mathfrak{g}$ (by some fixed isomorphism)" is missing. As written the definition of $l'_2$ doesn't make sense because the bracket is only defined on $\mathfrak{g}$. Feb 13 '17 at 9:47
• You should also clarify what you mean by $H(l)=\mathfrak{g}$ for intrinsic formality. Does $H(l)$ denote the transferred $L_\infty$ structure quasi-isomorphic to $l$ or rather the truncated Lie algebra structure obtained by forgetting everything except $(l_2)_*$? Feb 13 '17 at 9:54
• As long as you don't mean something really weird, though, the answer to Question 1 will be that the two notions are equivalent (at least in characteristic zero). Rigidity will be a special case of intrinsic formality (using the fact that quasi-isomorphisms between structures with $0$ differential are isomorphisms) and the homological perturbation lemma can be combined with rigidity to give intrinsic formality. Feb 13 '17 at 9:57

Assume $\mathfrak{g}=(V,0,[\cdot,\cdot])$ is intrinsically formal. Now let $(V,l')=(V,0,l'_2=[\cdot,\cdot],l'_3,\ldots)$ be an $L_\infty$ algebra as in the definition of rigidity. Then the homology of $(V,0)$ is $(V,0)$ and so the induced truncated Lie algebra on $H(V,l')$ is precisely $\mathfrak{g}$. By intrinsic formality, there exists an $L_\infty$ quasi-isomorphism from $\mathfrak{g}$ to $(V,l')$. Any $L_\infty$ quasi-isomorphism between two $L_\infty$ algebras with zero differential is in fact an $L_\infty$ isomorphism, so $(V,l')$ is isomorphic to $\mathfrak{g}$. Thus $\mathfrak{g}$ is rigid.
On the other hand, assume $\mathfrak{g}$ is rigid and let $(W,l)$ be an $L_\infty$ algebra with $$H(W)^{\text{trunc}}:=(H(W),0,(l_2)_*)$$ isomorphic to $\mathfrak{g}$ as in the definition of intrinsic formality. The homological perturbation lemma says that there exists an $L_\infty$ algebra $$H(W)^{\text{trans}}:=(H(W),0,(l_2)_*,l_3^{\text{trans}},\ldots)$$ equipped with a quasi-isomorphism to $(W,l)$. Note that the truncation of $H(W)^{\text{trans}}$ is $H(W)^{\text{trunc}}\cong \mathfrak{g}$. Then by rigidity, $H(W)^{\text{trans}}$ is isomorphic to $\mathfrak{g}$. Composing this with the quasi-isomorphism $H(W)^{\text{trans}}\to (W,l)$ gives a quasi-isomorphism so that $\mathfrak{g}$ is intrinsically formal.
• If you give me a reference for your conventions for the definition of $NR(\mathfrak{g})$ I'll answer that part too. Feb 14 '17 at 1:43