# Graded Algebras and Classical Mechanics

Certain algebras possess an important classifying property known as a grading.  This essentially amounts to being able to express these algebras as a special kind of direct sum that respects multiplication.

Definition[Graded Algebra]:  An algebra $A$ over a field $\kappa$ is said to be graded if, as a vector space over $\kappa$, $A$ can be written as the direct sum of a family of subspaces $(A_{i})_{i\in \mathbb{N}}$ – i.e.

$\displaystyle A = \bigoplus_{i\in \mathbb{N}}A_{i}$

and such that multiplication behaves according to

$A_{i}\cdot A_{j} \subseteq A_{i+j}$, for all $i,j\in \mathbb{N}$

In classical mechanics, one deals with smooth manifolds. To understand what these are we need some definitions.

Definition[Chart]:  Let $M$ be a manifold of dimension $n$.  Then a chart is a pair $(U, \psi)$, where $U\subset M$ is open and $\psi: U\to V\subset \mathbb{R}^{n}$ is a homeomorphism to some open $V$.

The collection of all charts such that each $x\in M$ belongs to some chart is called an atlas.  An atlas characterizes a manifold.  More formally,

Definition[Atlas]:  An atlas on a topological space $T$ is a collection of charts $\{(U_{\alpha},\varphi_{\alpha})\}_{\alpha\in I}$ where the $U_{\alpha}$ are open sets that cover $T$, and for each index $\alpha$

$\varphi_{\alpha}: U_{\alpha}\to \mathbb{R}^n$

is a homeomorphism of $U_{\alpha}$ onto an open subset of $n$-dimensional Euclidean space.

Definition[Smooth Atlas]: An atlas $\mathcal{A}$ is called smooth if for all $\varphi_{i}, \varphi_{j}$ we have that

$\varphi_{j}\circ \varphi_{i}^{-1} : \varphi_{i}(U_{i}\cap U_{j})\to \varphi_{j}(U_{i}\cap U_{j})$

is a diffeomorphism.

Definition[Diffeomorphism]: Given two manifolds $M$ and $N$, a differentiable map $f: M\to N$ is called a diffeomorphism if it is a bijection, and its inverse $f^{-1}: N\to M$ is also differentiable.

Now, a manifold may not be generated by a unique atlas.  To account for this, we seek a preferred atlas known as a maximal atlas.  In defining such a thing, we need to know what it means for atlases to be compatible.

Definition[Compatibility]: Let $\mathcal{A}$ and $\mathcal{A}'$ be two smooth atlases.  Then $\mathcal{A}$ and $\mathcal{A}'$ are called compatible if and only if $\mathcal{A}\cup \mathcal{A}'$ is again a smooth atlas.

This notion of compatibility is actually an equivalence relation.  The union over an equivalence class of atlases gives us our maximal atlas.  If we denote an equivalence class by $[\mathcal{A}]$, then we have

Definition[Maximal Atlas]:  A maximal atlas for a manifold $M$ is the union of all smooth atlases in an equivalence class – i.e.

$\mathcal{A}_{max} = \bigcup\{\mathcal{B} : \mathcal{B}\in [\mathcal{A}]\}$

This maximal atlas is said to be the differentiable structure of the manifold.  We are now ready to understand the idea of a smooth manifold.

Definition[Smooth Manifold]:  A smooth manifold is a pair $(M, \mathcal{A}_{max})$ where $M$ is a manifold and $\mathcal{A}_{max}$ is a differentiable structure of $M$.

Smooth manifolds are important in classical mechanics because they allow us to do calculus on them.  Now, to return to the original topic, we are also interested in the graded algebra of smooth differentiable forms on a smooth manifold $M$ with respect to the wedge product, which is denoted by

$\displaystyle \mathcal{A}^{\bullet}(M) = \bigoplus_{k=0}^{n}\mathcal{A}^{k}(M)$

# – Differential Forms –

For now, let’s restrict our focus to what are called differential $1$-forms, or just $1$-forms.  According to one source, “Informally, a differential form is what can be integrated along a path.”

[$1$-Forms]

Starting in $\mathbb{R}^{2}$, let $U$ be some open subset of $\mathbb{R}^{2}$ and let $F, G$ be two real-valued functions defined on $U$.  Then an expression of the form

$F(x,y)dx+G(x,y)dy$

is called a differential $1$-form on $U$.  A particularly important example of a differential $1$-form is the total differential of a $C^{1}$ real-valued function $f$ defined on some open subset $U$.

Definition[Total Differential]:  Let $f$ by a $C^{1}$ real-valued function defined on an open subset $U$ of $\mathbb{R}^{2}$.  Then its total differential $df$ is defined by

$\displaystyle df = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy$

This can also be expressed as a kind of dot product of vectors, namely $\nabla f\cdot (dx,dy)$, where $\nabla f$ is the gradient of $f$.  Note that in this case we have $\displaystyle F(x,y) = \frac{\partial f}{\partial x}$ and $\displaystyle G(x,y) = \frac{\partial f}{\partial y}$.

$1$-forms can be defined similarly for higher dimensions.  For instance, a differential $1$-form on an open subset $U$ of $\mathbb{R}^{3}$ is an expression of the form

$F(x,y,z)dx+G(x,y,z)dy+H(x,y,z)dz$

where $F, G$ and $H$ are real-valued functions on $U$.  Again, if $f$ is a $C^{1}$ function defined on $U$, then the total differential

$\displaystyle df = \frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy+\frac{\partial f}{\partial z}dz$

is an example.

In general, a $1$-form on an open subset of $\mathbb{R}^n$ is an expression of the form

$F_{1}(x_{1},x_{2},...,x_{n})dx_{1}+F_{2}(x_{1},x_{2},...,x_{n})dx_{2}+\ldots + F_{n}(x_{1},x_{2},...,x_{n})dx_{n}$

or simply $\displaystyle \sum_{i = 1}^{n}F_{i}(x_{1},x_{2},...,x_{n})dx_{i}$.  If we let $\boldsymbol{x} = (x_{1}, x_{2},..., x_{n})$, then we can write a $1$-form even more succinctly as

$\displaystyle \varphi = \sum_{i=1}^{n}F_{i}(\boldsymbol{x})dx_{i}$

So, if $f$ is a $C^{1}$ real-valued function defined on $U$, the the general total differential is

$\displaystyle df = \sum_{i=1}^{n}\frac{\partial f}{\partial x_{i}}dx_{i}$

Note that each $dx_i$ is a $1$-form.  These are defined by the property that for each vector $\boldsymbol{v} = (v_{1},v_{2},...,v_{n})\in T_{x}\mathbb{R}^{n}$

$dx_{i}(\boldsymbol{v}) = v_{i}$

Clearly, these form a basis for the $1$-forms on $\mathbb{R}^{n}$.

Definition[Smooth $1$-Form]:  A smooth $1$-form $\varphi$ on $\mathbb{R}^{n}$ is a real-valued function on the set of all tangent vectors to $\mathbb{R}^{n}$, i.e. –

$\varphi : T\mathbb{R}^{n}\to \mathbb{R}$

with the properties that

1. $\varphi$ is linear on the tangent space $T_{x}\mathbb{R}^{n}$ for each $x\in \mathbb{R}^{n}$ – i.e. if $\boldsymbol{v}_{1}, \boldsymbol{v}_{2}\in T_{x}\mathbb{R}^{n}$ and $c_{1},c_{2}\in \mathbb{R}$,then $\varphi(c_{1}\boldsymbol{v}_{1}+c_{2}\boldsymbol{v}_{2}) = c_{1}\varphi(\boldsymbol{v}_{1})+c_{2}\varphi(\boldsymbol{v}_{2})$.
2. For any smooth vector field $v = v(x)$, the function $\varphi(v):\mathbb{R}^{n}\to \mathbb{R}$ is smooth.

A smooth $1$-form on an $n$-dimensional manifold $M$ is defined similarly.

Note:  Another source simply says that a smooth $1$-form on an open subset $U$ of $\mathbb{R}^{n}$ is given by an expression

$\displaystyle \varphi = \sum_{i=1}^{n}f_{i}dx_{i}$

where $f_{i}\in C^{\infty}(U)$.

To better understand the graded algebra of smooth differential forms on $M$ with respect to the wedge product we need to understand that the $1$-forms on $\mathbb{R}^{n}$ are part of an algebra called the algebra of differential forms on $\mathbb{R}^{n}$.  Multiplication in this algebra is known as the wedge product and is denoted by the symbol “$\wedge$“.  It has the property of being skew-symmetric or anti-commutative:

$dx_{i}\wedge dx_{j} = -dx_{j}\wedge dx_{i}$

From this it follows that $dx_{i}\wedge dx_{i} = -dx_{i}\wedge dx_{i}$, which implies that $dx_{i}\wedge dx_{i} = 0$ (provided we are not working over a field with characteristic 2).  This means that the wedge product is alternating.

From here we can build up $k$-forms on $\mathbb{R}^{n}$.  If each summand of a differential form $\varphi$ contains $k$ $dx_{i}$‘s, then $\varphi$ is called a $k$-form.

Note: Functions are considered to be $0$-forms.

The set

$\{dx_{i_{1}}\wedge \ldots \wedge dx_{i_{k}} : 1\leq i_{1} < i_{2} < \ldots < i_{k} \leq n\}$

is a basis for the $k$-forms on $\mathbb{R}^{n}$.  Thus, every $k$-form can be expressed in the form

$\displaystyle \varphi = \sum_{|I| = k}f_{I}dx_{i_{1}}\wedge \ldots \wedge dx_{i_{k}}$

where $I$ ranges over all multi-indecies $I = (i_{1}, i_{2},...,i_{k})$ of length $k$.

## The Exterior Derivative

The exterior derivative is an operation that sends a $k$-form to a $(k+1)$-form.

Definition[Exterior Derivative ($1$-Form)]:  Let $f$ be a $0$-form (function).  Then its exterior derivative $df$ is the $1$-form

$\displaystyle df = \sum_{i=1}^{n}\frac{\partial f}{\partial x_{i}}dx_{i}$

Note that this corresponds to the total differential of $f$.

Definition[Exterior Derivative ($k$-form)]:  Let $\varphi$ be a $k$-form.  Then its exterior derivative $d\varphi$ is the $(k+1)$-form obtained from $\varphi$ by applying $lated d$ to each of the functions involved in $\varphi$.

Definition[Directional Derivative]:  Let $f$ be a differentiable function defined on a region $R$ in $\mathbb{R}^{n}$.  Let $\boldsymbol{v}$ be a vector based at the point $p_{0}\in R$.  Then the derivative of $f$ along $\boldsymbol{v}$, denoted $D_{\boldsymbol{v}}f$, is defined as follows.  Let $F(t) = f(p_{0}+t\boldsymbol{v})$.  Then

$D_{\boldsymbol{v}}f = F'(0)$

Theorem:  Let $f$ be a differentiable function defined on a region $R$ of $\mathbb{R}^{n}$.  Let $\varphi$ be defined on tangent vectors to points of $R$ by

$\varphi(\boldsymbol{v}) = D_{\boldsymbol{v}}f$

Then

$\varphi = df$

Proof:

Let $\boldsymbol{v}$ be a vector based at a point $p$ of $R$ with coordinates $(p_{1}, p_{2},..., p_{n})$. Let $\boldsymbol{v}$ have coordinates $(v_{1}, v_{2}, ..., v_{n})$.  Let $\textbf{u}_{i}$ be the $i$th unit vector.  Then

$\displaystyle df(\boldsymbol{v}) = \left(\sum_{i=1}^{n}\frac{\partial f}{\partial x_{i}}dx_{i}\right)\left(\sum_{i=1}^{n}v_{i}\textbf{u}_{i}\right)$

$=\displaystyle \sum_{i=1}^{n}\frac{\partial}{\partial x_{i}}f(p)v_{i}$

Next, let’s evaluate $\varphi(\boldsymbol{v}) = D_{\boldsymbol{v}}f = F'(0)$ where we recall that $F(t) = f(p+t\boldsymbol{v})$.  So, really, we should write $\displaystyle F'(0) = \frac{\partial}{\partial t}f(0)$.  Now, $F$ is actually a composition of two functions, $F(t) = f(g(t))$, where $g(t) = p+t\boldsymbol{v} = (p_{1}+tv_{1}, p_{2}+tv_{2},...,p_{n}+tv_{n})$.  If we now apply the chain rule, then we get

$\displaystyle \frac{\partial}{\partial t}f(0) = \sum_{i=1}^{n}\frac{\partial}{\partial x_{i}}f(g(0))\frac{d}{dt}x_{i}(0)$

But $x_{i} = p_{i}+tv_{i}$, so $\displaystyle \frac{d}{dt}x_{i} = v_{i}$.  Therefore,

$\displaystyle \frac{\partial}{\partial t}f(0) = \sum_{i=1}^{n}\frac{\partial}{\partial x_{i}}f(p)v_{i}$

This finishes the proof.

The exterior derivative obeys both the Leibniz rule and the chain rule:

• $d(fg) = gdf+fdg$         Leibniz Rule
• $d(h(f)) = h'(f)df$        Chain Rule

If $\phi$ is a $p$-form and $\psi$ is  a $q$-form, then the Leibniz rule takes the form

$d(\phi \wedge \psi) = d\phi \wedge \psi+(-1)^{p}\phi\wedge d\psi$

Theorem:  For any differential form $\phi$,

$d(d\phi) = 0$

(Or more succinctly, $d^2 = 0$).

Proof:

First, let $f$ be a function (i.e. $0$-form).  Then

$\displaystyle d(df) = d\left(\sum_{i=1}^{n}\frac{\partial f}{\partial x_{i}}dx_{i}\right)$

$\displaystyle = \sum_{i=1}^{n}d\left(\frac{\partial f}{\partial x_{i}}\right)dx_{i}$

$\displaystyle = \sum_{i=1}^{n}\left(\sum_{j=1}^{n}\frac{\partial f_{x_{i}}}{\partial x_{j}}dx_{j}\right)dx_{i}$

[where $\displaystyle f_{x_{i}} : = \frac{\partial f}{\partial x_{i}}$]

$\displaystyle =\sum_{i=1}^{n}\left(\sum_{j=1}^{n}\frac{\partial^{2} f}{\partial x_{i}\partial x_{j}}dx_{j}\right)dx_{i}$

$\displaystyle =\sum_{i,j}\frac{\partial^{2} f}{\partial x_{i}\partial x_{j}}dx_{j}\wedge dx_{i}$

$\displaystyle =\sum_{i

since mixed partials commute.  Now, since $dx_{i}$ actually means $d(x_{i})$, where $x_{i}$ is the $i$-th coordinate function, we have that $d(dx_{i}) = 0$.  Let

$\displaystyle \phi = \sum_{|I| = k}f_{I}dx_{i_{1}}\wedge \ldots \wedge dx_{i_{k}}$

Then

$\displaystyle d(d\phi) = \left(\sum_{|I|=k}df_{I}\wedge dx_{i_{1}}\wedge \ldots \wedge dx_{i_{k}}\right)$

$\displaystyle = \sum_{|I|=k}\Big(d(df_{I})\wedge dx_{i_{1}}\wedge \ldots \wedge dx_{i_{k}}-df_{I}\wedge d(dx_{i_{1}})\wedge \ldots \wedge dx_{i_{k}}+\ldots\Big)$

[By the Leibniz Rule]

Per the reasoning above with functions, this is clearly $0$.

Much of the above has been described in terms of differential forms on $\mathbb{R}^{n}$.  Most of it transfers fairly seamlessly to general manifolds (i.e. the concepts are defined similarly).  So, to tie this all back to the graded algebra given above, we should understand $\mathcal{A}^{0}(M)$ to be $C^{\infty}(M)$ (i.e. the $0$-forms on $M$) and $\mathcal{A}^{k}(M)$ to be the algebra of smooth differential $k$-forms on $M$. We’ll look more at graded algebras in the next post.