How does a Lie algebra encode a space? (Part 1)

A fundamental principle of derived deformation theory is that a formal “space” (i.e. formal moduli problem) can be encoded by a differential graded Lie algebra, or more generally, an $L_{\infty}$ -algebra. This is usually attributed to a number of famous mathematicians, such as Quillen, Deligne, Drinfeld, Goldman, Millson, Feigin, Manetti, and Kontsevich. More recently, a formalization of this principle was independently obtained by Lurie and by Pridham.

In this post, I want to discuss a simple example of this: how to encode the ‘derived’ vanishing-locus of a function using an $L_{\infty}$ -algebra. My goal is to show that, far from being a high-brow construction, this essentially boils down to the familiar Taylor series expansion. After explaining this fact, I will go through a number of concepts and constructions in the world of $L_{\infty}$ -algebras and show how they can be translated into the setting of a function and its zero locus. Hopefully, this will help to demystify the notion of derived manifolds for myself and maybe for others. In a future post, I will discuss the ‘stacky’ version of this story: how to describe a Lie algebra action, or a Lie algebroid, in terms of an $L_{\infty}$ -algebra. As usual, there might be some (hopefully minor) errors in what follows, so proceed with caution!

The vanishing locus of a function

Let $V$ and $E$ be finite-dimensional vector spaces of respective dimensions $n$ and $k$ , and let $f: V \to E$ be a polynomial function. If we choose a basis for $E$ , we can write $f$ as a $k$ -tuple

$f = (f_{1}, ..., f_{k})$ .

Our goal is to study the vanishing-locus $Z(f)$ of $f$ , defined as the following set:

$Z(f) = \{ v \in V \ | \ f(v) = 0 \} = f^{-1}(0)$ .

Let’s choose a point $a \in Z(f)$ and study the vanishing locus in a neighborhood of $a$ . Now if you’ve taken a differential topology class, your first instinct at this point might be to check whether $a$ is a regular point of $f$ . Recall that a point $a \in V$ is regular if the derivative $df_{a}: V \to E$ is surjective. In this case, the constant rank theorem tells us that we can find local coordinates on $V$ and $E$ such that the function has the form

$f(x_{1}, ..., x_{n}) = (x_{n-k +1}, ..., x_{n})$ .

Hence, in these coordinates, the vanishing locus is given by

$Z(f) = \{ (x_{1}, ..., x_{n-k}, 0, ..., 0) \}$ ,

and is (locally around $a$ ) a smooth submanifold of dimension $n-k$ .

The situation becomes more interesting if $a$ is not regular. This indicates some sort of ‘degeneracy’ of the function around the point $a$ where the components $f_{i}$ fail to be ‘independent’. In this case, the constant rank theorem fails and the vanishing locus might not be smooth!

For example, suppose our function is

$f = x^2 - y^3 : \mathbb{R}^2 \to \mathbb{R}$ ,

whose derivative $df = (2x, -3y^2)$ vanishes at the origin. In this case, $Z(f)$ has a cusp singularity at $0$ :

If we were classical geometers we would leave it at that: our manifold $Z(f)$ is singular and there’s nothing we can do about it. But as derived geometers, we know that $Z(f)$ is singular because we’ve thrown away something important: the ‘reason’ for the singularity, i.e. the way in which the components of $f$ fail to be independent. So to salvage the situation, we don’t throw out this information. Namely, we keep the function $f$ , or better yet, it’s Taylor coefficients! In the next section, I want to explain how to think of these coefficients as the brackets defining an $L_{\infty}$ -algebra. This will allow us to apply all the tools of homological algebra to the study of $Z(f)$ .

The $L_{\infty}$ -algebra of a function

Choose linear coordinates $(x_{1}, ..., x_{n})$ for $V$ . The $m^{th}$ Taylor coefficient of the function $f$ at the point $a$ is the symmetric $m$ -tensor

$D^{m}f_{a} \in Hom(Sym^{m}(V), E)$

with coefficients

$\displaystyle (D^{m}f_{a})_{i_{1}...i_{m}} = \frac{\partial^m f}{\partial x_{i_1} \partial x_{i_2} ... \partial x_{i_m}}(a)$ .

This means that for vectors $v^{(i)} \in V$ , written in coordinates as $v^{(i)} = (v^{(i)}_{1}, ..., v^{(i)}_{n})$ , we have the expression

$\displaystyle D^{m}f_{a}(v^{(1)}, ..., v^{(m)}) = \sum_{i_{1}, ..., i_{m} = 1}^{n} \frac{\partial^m f}{\partial x_{i_1} \partial x_{i_2} ... \partial x_{i_m}}(a) v^{(1)}_{i_{1}} ... v^{(m)}_{i_{m}} \in E$ ,

which is symmetric in the $v^{(i)} \in V$ because of the equality of mixed partials. Taylor’s theorem then states that we have the series expansion

$\displaystyle f(x) = \sum_{m \geq 0} \frac{1}{m!} D^{m}f_{a}(x-a, ..., x-a).$

In our case, this series is finite because $f$ is a polynomial.

To define an $L_{\infty}$ -algebra we let

$L = V[-1] \oplus E[-2]$

be the graded vector space with $V$ in degree $1$ and $E$ in degree $2$ . Let $l_{m} = D^{m}f_{a}$ , but now we will view it as a graded-antisymmetric map

$\displaystyle l_{m} : L^{\otimes m} \to L$

which has degree $(2-m)$ . We denote this $L_{\infty}$ -algebra by $L(f,a)$ since it depends on the function $f$ as well as the basepoint $a$ .

Okay, I anticipate a few objections at this stage. This construction might seem to come out of the blue and is somewhat unmotivated. For this I’m just going to ask you to suspend your disbelief, at least for now. The point is that a function can be viewed as a special case of a much more general structure. More importantly, I haven’t yet even told you what an $L_{\infty}$ -algebra is. This is what I’m about to do, but I’m going to build up to it, starting with the more digestible concept of a differential graded Lie algebra (dgla for short), followed by some simple examples.

Definition. A dgla is a $\mathbb{Z}$ -graded vector space

$L = \bigoplus_{n \in \mathbb{Z}} L_{n}$

which is equipped with a degree 1 linear map $d: L_{n} \to L_{n+1}$ that squares to $0$ and a degree 0 bilinear bracket $[ -, - ] : L_{n} \times L_{m} \to L_{n + m}$ , satisfying the following list of axioms:

- skew symmetry: $[x,y] = -(-1)^{ij}[y,x]$
- Jacobi identity: $[x, [y, z]] = [[x,y], z] + (-1)^{ij}[y, [x,z]]$
- Leibniz identity: $d[x,y]=[dx,y]+(-1)^{i}[x,dy]$ ,

where $x \in L_{i}$ and $y \in L_{j}$ are elements with respective degrees $i$ and $j$ .

There are two basic cases of the definition that you should keep in mind. First, an abelian dgla, namely one for which the bracket is $0$ , is nothing other than a chain complex. Second, if the vector space $L = L_{0}$ consists of a single vector space in degree $0$ , then the differential $d$ is forced to vanish and we recover the definition of an ordinary Lie algebra. More generally, if our vector space has several graded components, but the differential is still $0$ , then we have a graded Lie algebra. By the way, looking back at the construction of $L(f,a)$ , we see that our dglas can only live in degrees $1$ and $2$ . In particular, we won’t actually be seeing any ordinary Lie algebras. Those are for next time!

Examples:

1. Suppose first that the function $f = \lambda: V \to E$ is linear, and let $a = 0 \in V$ . Then

$D^{0}f_{0} = f(0) = 0, \ \ D^{1}f_{0} = \lambda, \ \ D^{m}f_{0} = 0$ ,

for $m \geq 2$ . Then we just get the $2$ -term chain complex

$\lambda: V \to E$ ,

with $V$ in degree $1$ and $E$ in degree $2$ .

2. Suppose next that $f = Q$ is a quadratic form. Then

$D^{0}f_{0} = f(0) = 0, \ \ D^{1}f_{0} = 0, \ \ D^{2}f_{0} = 2Q, \ \ D^mf_{0} = 0$ ,

for $m \geq 3$ . Then we get a graded Lie algebra $L = V[-1] \oplus E[-2]$ with bracket

$[(v_1, e_1), (v_2, e_2)] = (0, 2Q(v_1, v_2)).$

If instead, we take $f = \lambda + Q$ , where $\lambda: V \to E$ is a linear function, then we get a dgla with differential $\lambda$ and bracket $2 Q$ . In all these cases, checking the dgla axioms follows automatically because of the degrees involved.

3. Let’s take a look at our earlier example $f = x^2 - y^3 : \mathbb{R}^2 \to \mathbb{R}$ . At the origin, we have $D^{0}f_{0} = 0$ and $D^{1}f_{0} = 0$ , but both $D^{2}f_{0}$ and $D^{3}f_{0}$ are non-zero. As a result, we get non-zero brackets $l_{2}$ and $l_{3}$ on $L = \mathbb{R}^2[-1] \oplus \mathbb{R}[-2]$ . Let’s use coordinates $(a,b,c) \in L$ . Then the brackets are given by

$l_{2}( (a^{(1)}, b^{(1)}, c^{(1)}), (a^{(2)}, b^{(2)}, c^{(2)})) = (0, 0, 2a^{(1)}a^{(2)})$

and

$l_{3}( (a^{(1)}, b^{(1)}, c^{(1)}), (a^{(2)}, b^{(2)}, c^{(2)}), (a^{(3)}, b^{(3)}, c^{(3)})) = (0, 0, -6b^{(1)}b^{(2)}b^{(3)})$ .

This last example is the $L_{\infty}$ -algebra of the cusp singularity. It’s more general than a dgla because of the bracket $l_{3}$ . So I’ll finally, sort of, give you the definition of an $L_{\infty}$ -algebra:

Definition. An $L_{\infty}$ -algebra is a $\mathbb{Z}$ -graded vector space

$L = \bigoplus_{n \in \mathbb{Z}} L_{n}$

which is equipped with a sequence of graded skew-symmetric multi-linear maps

$l_{n}: L^{\otimes n} \to L$

of degrees $2-n$ , for $n \geq 1$ . These maps satisfy a sequence of identities, called generalized Jacobi identities: for all $n \ge1$ and all $n$ -tuples $(v_{1}, ..., v_{n})$ of homogeneous vectors in $L$ , we have

$\displaystyle \sum_{i + j = n + 1} \sum_{\sigma \in \text{UnShuff}(i, n-i)} \pm l_{j}(l_{i}(v_{\sigma(1)}, ..., v_{\sigma(i)}), v_{\sigma(i+1)}, ..., v_{\sigma(n)}) = 0.$

The $\sigma$ are unshuffles, and the $\pm$ are signs, which I’m omitting.

A few mandatory comments about the definition. Looking at the generalized Jacobi identity for $n = 1$ , we see that $l_{1}^2 = 0$ , so $l_{1}$ is a differential. For $n = 2$ , we get the Leibniz identity from dglas for $l_{1}$ and $l_{2}$ . Finally, for $n = 3$ , we almost get the Jacobi identity, but there is a correction term involving $l_{1}$ and $l_{3}$ . Hence, this reduces to the ordinary Jacobi identity if $l_{3} = 0$ and therefore, we recover the definition of a dgla when $l_{n} = 0$ for $n \geq 3$ . When $l_{3} \neq 0$ , we say that the Jacobi identity holds ‘up to homotopy’.

By the way, it’s possible to extend the definition of an $L_{\infty}$ -algebra to include a ‘ $0$ -ary bracket’ $l_{0} \in L_{2}$ , called the curvature. This actually shows up in our examples when we choose a point $a \in V$ with non-zero $f(a) \in E$ . Indeed, the value $f(a)$ is the curvature.

Looking back at the $L_{\infty}$ -algebra associated to a function $f$ , it is easy to check that all the generalized Jacobi identities are satisfied: indeed, all brackets are valued in $E$ , but the brackets are only non-zero when all entries come from $V$ . Therefore $l_{i} \circ l_{j} = 0$ .

Conversely, if $L = V[-1] \oplus E[-2]$ is an $L_{\infty}$ -algebra which is concentrated in degrees $1$ and $2$ , then, purely from degree considerations, each map $l_{n}$ is valued in $E$ and vanishes whenever any input comes from $E$ . In other words, there is a correspondence between functions and $L_{\infty}$ -algebras concentrated in degrees $1$ and $2$ . Let’s formalise this as the following theorem.

Theorem. There is a bijection between polynomial functions $f: V \to E$ and curved $L_{\infty}$ -algebras $L = V[-1] \oplus E[-2]$ which are concentrated in degrees $1$ and $2$ and such that only finitely many brackets $l_{n}$ are non-zero.

Using this correspondence, we can translate definitions from the theory of $L_{\infty}$ -algebras to more down-to-earth definitions involving polynomial functions. Let’s do this in a few important cases.

The Maurer-Cartan equation

Given an $L_{\infty}$ -algebra $L$ , the Maurer-Cartan function (or curvature) is the map $F: L_{1} \to L_{2}$ defined as follows

$\displaystyle F(x) = \sum_{m \geq 0} \frac{1}{m !} l_{m}(x,..., x)$ .

The Maurer-Cartan equation is obtained by setting $F$ equal to $0$ and it gives rise to the Maurer-Cartan locus:

$MC(L) = \{ x \in L^1 \ | \ F(x) = 0 \}.$

This locus is of fundamental importance. For the $L_{\infty}$ -algebra $L(f,a)$ , we see that the Maurer-Cartan function just recovers $f$ , via Taylor’s theorem, and hence the Maurer-Cartan locus is just the vanishing locus $Z(f)$ .

The twisting procedure

Consider a function $f : V \to E$ and two basepoints $a, b \in V$ giving rise to two different $L_{\infty}$ -algebras $L(f,a)$ and $L(f,b)$ with the same underlying graded vector space. Let’s try to see how these two $L_{\infty}$ -algebras are related. For simplicity, we will assume that $a = 0$ . Let’s label the brackets of $L(f,0)$ by $l_{n}$ . Then, as we’ve seen above, the function $f$ is given by

$\displaystyle f(x) = \sum_{m \geq 0} \frac{1}{m !} l_{m}(x,..., x)$ .

Label the brackets of $L(f,b)$ by $l^b_{n}$ . In order to obtain these, we simple expand $f(b + x)$ :

$\displaystyle f(b + x) = \sum_{m \geq 0} \frac{1}{m !} l_{m}((b+x)^m) = \sum_{m \geq 0} \frac{1}{m !} \sum_{k + r = m} \frac{m!}{k! r!}l_{m}(b^r, x^k) = \sum_{k \geq 0} \frac{1}{k!} \big( \sum_{r \geq 0} \frac{1}{r!} l_{k + r}(b^r, x^{k})\big)$ .

For the second equality we’ve used the binomial theorem. As a result, we obtain the following expression for the brackets $l^b_{n}$ in terms of $l_{n}$ :

$\displaystyle l_{m}^b(v^{(1)}, ..., v^{(m)}) = \sum_{r \geq 0} \frac{1}{r!} l_{m + r}(b^{r}, v^{(1)}, ..., v^{(m)}). \displaystyle$

In fact, this formula works in general. For any $L_{\infty}$ -algebra $L$ and degree $1$ element $b \in L_{1}$ , then, assuming that there are no issues with convergence, the above formula defines a new collection of brackets which satisfy the generalized Jacobi identities!

Cohomology

An $L_{\infty}$ -algebra $L$ has an underlying chain complex $(L, l_{1})$ and we can take its cohomology:

$\displaystyle H^{i}(L) = \frac{\text{Ker}(l_{1} : L_{i} \to L_{i+1})}{\text{Im}(l_{1}: L_{i-1} \to L_{i})}$ .

For our example $L(f,a)$ , where $f(a) = 0$ , we have $l_{1} = df_{a} : V \to E$ , so that

$\displaystyle H^{0}(L(f,a)) = \text{Ker}(df_{a}) = T_{a}(Z(f)) \subset V$ ,

the (Zariski) tangent space of $Z(f)$ at $a$ . In the next degree, we have

$\displaystyle H^{1}(L(f,a)) = \frac{E}{df_{a}(V)}$ .

This is usually called the obstruction space. To see why, consider the following problem: given a tangent vector $v \in T_{a}Z(f)$ , we would like to extend it to a path

$\gamma(t): (-\epsilon, \epsilon) \to Z(f)$ ,

such that $\gamma(0) = a$ and $\gamma'(0) = v$ . Let’s start with the easiest guess $\gamma(t) = a + t v$ . Plugging this into $f$ we get

$\displaystyle f(\gamma(t)) = \sum_{m \geq 1} \frac{1}{m !} l_{m}(tv,..., tv) = \frac{t^2}{2}l_{2}(v,v) + \mathcal{O}(t^3)$ .

Hence, $l_{2}(v,v) \in E$ is obstructing the function from being $0$ . So let’s try to fix the problem by modifying our guess: $\gamma(t) = a + t v + \frac{t^2}{2} u$ , for some $u \in V$ . Now we get

$\displaystyle f(\gamma(t)) = \frac{t^2}{2}\big(l_{1}(u) + l_{2}(v,v)\big) + \mathcal{O}(t^3)$ .

We can solve the equation

$\displaystyle l_{1}(u) + l_{2}(v,v) = 0$

if and only if the projection of $l_{2}(v,v)$ to $H^{1}(L(f,a))$ is $0$ . In other words, the cohomology class $[l_{2}(v,v)] \in H^{1}(L(f,a))$ is an obstruction to constructing the path $\gamma(t)$ .

Chevalley-Eilenberg = Koszul

An important construction is the Chevalley-Eilenberg algebra of a dgla (or $L_{\infty}$ -algebra). This construction takes in an $L_{\infty}$ -algebra and outputs a commutative differential graded algebra (cdga for short). In fact, this is sometimes used to give the definition of an $L_{\infty}$ -algebra, since it encodes the infinite list of generalized Jacobi identities into the single equation $d^2 = 0$ . I’m going to be a bit sparse on details, but here’s the basic idea. Starting from the graded vector space $L = \oplus_{n} L_{n}$ , we first define the following graded commutative algebra

$CE(L) = Sym^{\bullet}(L^*[-1])$ .

Namely, we take the dual space of $L$ , shift all the degrees up by $+1$ , and then take the free graded symmetric algebra. By the property of symmetric algebras, a degree $+1$ derivation $d$ of $CE(L)$ is uniquely specified by a linear map

$d : L^*[-1] \to Sym^{\bullet}(L^*[-1])$ ,

and this decomposes into components $d_{m} : L^*[-1] \to Sym^{m}(L^*[-1])$ . We take these to be the duals of the brackets $l_{m}$ . As I said above, the generalized Jacobi identities for $l_{m}$ are equivalent to the property that $d^2 = 0$ .

I encourage you to try working out the Chevalley-Eilenberg algebra of $L(f,a)$ , but here’s the answer. The underlying graded commutative algebra turns out to be isomorphic to

$CE(L(f,a)) = \mathbb{R}[V] \otimes \wedge^{\bullet}(E^*)$ ,

with the polynomial functions on $V$ in degree $0$ and the elements of $\wedge^{r}(E^*)$ in degree $-r$ . And the derivation is given by the interior product with the function $f$ :

$\iota_{f} : \mathbb{R}[V] \otimes \wedge^{\bullet}(E^*) \to \mathbb{R}[V] \otimes \wedge^{\bullet}(E^*)$ .

Here, we view $f \in E \otimes \mathbb{R}[V]$ .

This cdga is actually well-known under another name: it is the Koszul complex associated to the function $f$ . There are a few important things to note. As a graded algebra, the degrees are all non-positive, but the differential has degree $+1$ . Furthermore, the degree $0$ cohomology is given by

$H^{0}(CE(L(f,a))) = \mathbb{R}[V]/(f_{1}, ..., f_{k}) = \mathbb{R}[Z(f)]$ ,

which is the ring of functions on the vanishing locus! This should encourage us to think of the Chevalley-Eilenberg algebra as defining a space in its own right. Indeed, just as the spectrum of a commutative ring is an affine scheme, we should think of the “spectrum” of a cdga which is concentrated in non-positive degrees as a derived affine scheme. In this case, we see that $\text{Spec}(CE(L(f,a)))$ is some sort of enhancement of the vanishing locus $Z(f)$ , which we call the derived vanishing locus. This is what we get when we don’t throw away the function.

A final thing to note is that, even though $Z(f)$ may be singular, $CE(L(f,a))$ looks very much like a polynomial algebra. Hence, in some sense, $\text{Spec}(CE(L(f,a)))$ is always smooth. I believe this phenomenon is called “hidden smoothness” and is one of the advantages of going to the world of derived geometry.

Quasi-isomorphisms

So far, everything we’ve been doing with the $L_{\infty}$ -algebra $L(f,a)$ is equivalent to just working with the function. We did this because we didn’t want to ‘throw away’ the data of $f$ when the variety is singular. But we don’t really want to remember everything: for example, if $f$ is a submersion, then all we care about is the vanishing locus $Z(f)$ ; the rest is redundant. In other words, we would like to think of $L(f,a)$ in a more intrinsic way. This can be acheived using the notion of an $L_{\infty}$ -quasi-isomorphism.

Definition. Let $L$ and $M$ be $L_{\infty}$ -algebras. An $L_{\infty}$ -morphism $\varphi: L \to M$ consists of a sequence of multilinear graded skew-symmetric maps

$\varphi_{n}: L^{\otimes n} \to M$

of degree $1- n$ , for $n \geq 1$ . These maps must satisfy the following sequence of identities: for all $n \ge1$ and all $n$ -tuples $(v_{1}, ..., v_{n})$ of homogeneous vectors in $L$ , we have

$\displaystyle \sum_{i + j = n + 1} \sum_{\sigma \in \text{UnShuff}(i, n-i)} \pm \varphi_{j}(l_{i}(v_{\sigma_{1}},...,v_{\sigma_{i}}), v_{\sigma_{i+1}}, ..., v_{\sigma_{n}}) = \sum_{j \geq 1} \sum_{i_{1} + ... + i_{j} = n} \sum_{\sigma \in \text{UnShuff}(i_{1}, ..., i_{j})} \pm l_{j}(\varphi_{i_{1}}(v_{\sigma_{1}}, ...), ... \varphi_{i_{j}}(..., v_{\sigma_{n}})).$

In particular, $\varphi_{1}: L \to M$ is a chain map. We say that $\varphi$ is a quasi-isomorphism if it induces an isomorphism on cohomology.

We will think of two $L_{\infty}$ -algebras as being “the same” if they are related by an $L_{\infty}$ -quasi-ismorphism. The importance of this notion stems from the following two facts:

Quasi-isomorphic $L_{\infty}$ -algebras have isomorphic cohomology groups.
Quasi-isomorphic $L_{\infty}$ -algebras have equivalent ‘local’ Maurer-Cartan loci. More precisely, if $L$ and $M$ are $L_{\infty}$ -algebras concentrated in strictly positive degrees, then an $L_{\infty}$ -quasi-ismorphism $\varphi: L \to M$ induces a bijection between neighborhoods of the basepoints in $MC(L)$ and $MC(M)$ .

Hence, quasi-isomorphic $L_{\infty}$ -algebras represent the same (local) derived varieties!

Let’s look at a quasi-isomorphism between two algebras $L(f_{1},0)$ and $L(f_{2},0)$ , where $f_{i} : V_{i} \to E_{i}$ are polynomials which vanish at the origin. By degree considerations, the component $\varphi_{n}: L(f_{1},0)^{\otimes n} \to L(f_{2}, 0)$ breaks up into two pieces:

$g_{n}: V_{1}^{\otimes n} \to V_{2}, \qquad h_{n} : V_{1}^{\otimes(n-1)}\otimes E_{1} \to E_{2}.$

These are the Taylor coefficients of two maps: a map $g: V_{1} \to V_{2}$ sending the origin to the origin and a map $h: V_{1} \to Hom(E_{1}, E_{2})$ . It is useful to think of these maps in the following way: $W_{i} = V_{i} \times E_{i}$ is a trivial vector bundle over $V_{i}$ . Then $h$ defines a linear vector bundle map $\Phi : W_{1} \to W_{2}$ covering the map $g$ . Now to understand the generalized homomorphism condition for $\varphi$ , note that we may view $f_{i}$ as a section of the bundle $W_{i}$ . Then $\varphi$ is an $L_{\infty}$ -morphism if and only if the bundle map $\Phi$ sends the section $f_{1}$ to $f_{2}$ :

$\Phi \circ f_{1} = f_{2} \circ g.$

Note that the linear component of $\varphi$ , which we use to determine whether it is a quasi-isomorphism, is given by the two linear maps

$dg_{0}: T_{0}V_{1} \to T_{0}V_{2}, \qquad \Phi(0) : W_{1}|_{0} \to W_{2}|_{0}.$

Minimal model and homotopy transfer

Consider again a function $f: V \to E$ with its associated $L_{\infty}$ -algebra $L(f,0)$ . For simplicity we assume that $f$ vanishes at the origin. We will now use the homotopy transfer theorem to produce the smallest possible space containing the vanishing locus.

First, let $l_{1} = df_{0} : V \to E$ and let $K = \text{Ker}(df_{0}) \subseteq V$ be the kernel. Second, let $C \subset E$ be a subspace which is complementary to the image of $l_{1}$ , so that

$E = C \oplus l_{1}(V).$

Then $(L' = K[-1] \oplus C[-2], 0)$ , with zero differential, is a quasi-isomorphic subcomplex of $(L(f,0), l_{1})$ . Let $i : L' \to L(f,0)$ denote the inclusion map. Now choose a linear map $p_{1} : V \to K$ which is a left inverse to the inclusion. Then by this lemma, this determines a chain map $p : L(f,0) \to L'$ and a degree $-1$ map $h: E \to V$ such that

$\displaystyle p \circ i = id_{L'}, \ \ id_{L(f,0)} - i \circ p = l_{1} \circ h + h \circ l_{1}, \ \ h \circ i = 0, \ \ p \circ h = 0.$

This data is called a special deformation retract. Finally, by the homotopy transfer theorem we can use this data to construct

An $L_{\infty}$ -algebra structure on $L'$ with zero differential.
An $L_{\infty}$ -quasi-isomorphism $\phi: L' \to L(f,0)$ whose linear part is $i$ .

The resulting $L_{\infty}$ -algebra $(L', l'_{m})$ is called the minimal model of $L(f,0)$ .

Translating back to the world of functions, we have acheived the following: a function $f': K \to C$ which has zero linear part and a map $g: K \to V$ ‘extending’ the inclusion $K \subseteq T_{0}V$ and inducing a local isomorphism

$Z(f') \to Z(f)$

in a neighbourhood of the origin.

It’s instructive to consider the construction of the minimal model in the special case of a map $f: V \to E$ which is a submersion at $0$ . Then $C = 0$ , and this forces $f' = 0$ and $Z(f') = K$ . The induced map $g: K \to V$ , therefore, maps $K$ isomorphically onto $Z(f)$ in a neighborhood of the origin. Recall from above that the only data we chose in order to produce the map $g$ was the projection $p_{1} : V \to K$ . Indeed, in the present setting, this map restricts to an isomorphism from $Z(f) \to K$ and hence $g$ is its inverse.

How does a Lie algebra encode a space? (Part 1)

The vanishing locus of a function

The $L_{\infty}$ -algebra of a function

Examples:

The Maurer-Cartan equation

The twisting procedure

Cohomology

Chevalley-Eilenberg = Koszul

Quasi-isomorphisms

Minimal model and homotopy transfer

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The vanishing locus of a function

The -algebra of a function

Examples:

The Maurer-Cartan equation

The twisting procedure

Cohomology

Chevalley-Eilenberg = Koszul

Quasi-isomorphisms

Minimal model and homotopy transfer

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The $L_{\infty}$ -algebra of a function