Contents

### Context

#### Differential geometry

synthetic differential geometry

Introductions

from point-set topology to differentiable manifolds

Differentials

V-manifolds

smooth space

Tangency

The magic algebraic facts

Theorems

Axiomatics

cohesion

tangent cohesion

differential cohesion

singular cohesion

$\array{ && id &\dashv& id \\ && \vee && \vee \\ &\stackrel{fermionic}{}& \rightrightarrows &\dashv& \rightsquigarrow & \stackrel{bosonic}{} \\ && \bot && \bot \\ &\stackrel{bosonic}{} & \rightsquigarrow &\dashv& \mathrm{R}\!\!\mathrm{h} & \stackrel{rheonomic}{} \\ && \vee && \vee \\ &\stackrel{reduced}{} & \Re &\dashv& \Im & \stackrel{infinitesimal}{} \\ && \bot && \bot \\ &\stackrel{infinitesimal}{}& \Im &\dashv& \& & \stackrel{\text{étale}}{} \\ && \vee && \vee \\ &\stackrel{cohesive}{}& ʃ &\dashv& \flat & \stackrel{discrete}{} \\ && \bot && \bot \\ &\stackrel{discrete}{}& \flat &\dashv& \sharp & \stackrel{continuous}{} \\ && \vee && \vee \\ && \emptyset &\dashv& \ast }$

Models

Lie theory, ∞-Lie theory

differential equations, variational calculus

Chern-Weil theory, ∞-Chern-Weil theory

Cartan geometry (super, higher)

# Contents

## Idea

The Hadamard lemma says that the Taylor series of a smooth function on the real line around the origin has a remainder at order $n$ which is the product of $x^{n+1}$ (for $x$ the canonical coordinate function) with another smooth function.

Simple as this may sound, it has profound consequences, as it means that smooth functions behave more like polynomials than the classical definition might suggest. For instance the Hadamard lemma directly implies that:

1. derivations of smooth functions are vector fields: For $X$ a smooth manifold and $C^\infty(X)$ its $\mathbb{R}$-algebra of smooth functions (under pointwise multiplication), then there is a natural bijection between the smooth tangent vector fields on $X$ and the purely algebraic derivations of the algebra $C^\infty(X)$;

2. the smooth infinitesimally thickened point $\mathbb{D}$ is the same as in algebraic geometry: The quotient of the algebra of smooth functions $C^\infty(\mathbb{R}^1)$ on the real line by the ideal generated by the square $x^2$ of the canonical coordinate function $x$ is the ring of dual numbers:

$C^\infty(\mathbb{R})/(x^2) \simeq (\mathbb{R} \oplus \epsilon \mathbb{R})/(\epsilon^2) \,.$
3. together this implies that a tangent vector in a smooth manifold $X$ is equivalently a morphism of the form

$\mathbb{D} \longrightarrow X$

of formal duals of $\mathbb{R}$-alghebras, from the infinitesimally thickened point $\mathbb{D}$.

This means that differential geometry has more in common with algebraic geometry than is manifest from the traditional definitions. In synthetic differential geometry one makes use of these facts to find a useful unified perspective. For exposition of this point see at geometry of physics – supergeometry.

More generally one may ask for other types of function algebras which satisfy the conclusion of the Hadamard lemma. These turn out to be the algebras over those algebraic theories which are called Fermat theories. These are hence a crucial ingredient for well-adapted models of synthetic differential geometry.

## Statement

###### Proposition

For every smooth function $f \in C^\infty(\mathbb{R})$ on the real line there is a smooth function $g$ such that

$f(x) = f(0) + x \cdot g(x) \,.$

This function $g$ is also called a Hadamard quotient.

###### Corollary

It follows that $g(0) = f'(0)$ is the derivative of $f$ at 0. By applying this repeatedly the lemma says that $f$ has a partial Taylor series expansion whose remainder $h$ is a smooth function:

$f(x) = f(0) + x f'(0) + \frac1{2} x^2 f''(0) + \cdots + \frac1{n!} x^n h(x) \,.$

More generally, for smooth functions on any Cartesian space $\mathbb{R}^n$ the lemma says that there are for each $f \in C^\infty(X)$ $n$ smooth functions $g_i$ such that

$f(\vec x) = f(0) + \sum_i x_i g_i(x) \,.$

So at the origin these smooth functions compute the partial derivatives of $f$

$g_i(0) = \frac{\partial f}{\partial x_i}(0) \,.$
###### Proof

Holding $x$ fixed, put $h(t) = f(t x)$. Then

$f(x) - f(0) = \int_{0}^{1} h'(t) d t = \int_{0}^{1} \sum_{i=1}^n \frac{\partial f}{\partial x_i} (t x) \cdot x_i d t$

where the second equality uses the chain rule. The lemma follows by putting

$g_i(x) = \int_{0}^{1} \frac{\partial f}{\partial x_i}(t x) d t \,.$

## Applications

The Hadamard lemma implies in particular that

## Fermat theory

The notion of a Fermat theory makes Hadamard’s lemma into an axiom. See there for more information.