# Geometric calculus

## Infinitesimal calculus on functions defined on a geometric algebra / From Wikipedia, the free encyclopedia

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In mathematics, **geometric calculus** extends the geometric algebra to include differentiation and integration. The formalism is powerful and can be shown to encompass other mathematical theories including vector calculus, differential geometry, and differential forms.^{[1]}

With a geometric algebra given, let $a$ and $b$ be vectors and let $F$ be a multivector-valued function of a vector. The directional derivative of $F$ along $b$ at $a$ is defined as

- $(\nabla _{b}F)(a)=\lim _{\epsilon \rightarrow 0}{\frac {F(a+\epsilon b)-F(a)}{\epsilon }},$

provided that the limit exists for all $b$, where the limit is taken for scalar $\epsilon$. This is similar to the usual definition of a directional derivative but extends it to functions that are not necessarily scalar-valued.

Next, choose a set of basis vectors $\{e_{i}\}$ and consider the operators, denoted $\partial _{i}$, that perform directional derivatives in the directions of $e_{i}$:

- $\partial _{i}:F\mapsto (x\mapsto (\nabla _{e_{i}}F)(x)).$

Then, using the Einstein summation notation, consider the operator:

- $e^{i}\partial _{i},$

which means

- $F\mapsto e^{i}\partial _{i}F,$

where the geometric product is applied after the directional derivative. More verbosely:

- $F\mapsto (x\mapsto e^{i}(\nabla _{e_{i}}F)(x)).$

This operator is independent of the choice of frame, and can thus be used to define what in geometric calculus is called the *vector derivative*:

- $\nabla =e^{i}\partial _{i}.$

This is similar to the usual definition of the gradient, but it, too, extends to functions that are not necessarily scalar-valued.

The directional derivative is linear regarding its direction, that is:

- $\nabla _{\alpha a+\beta b}=\alpha \nabla _{a}+\beta \nabla _{b}.$

From this follows that the directional derivative is the inner product of its direction by the vector derivative. All needs to be observed is that the direction $a$ can be written $a=(a\cdot e^{i})e_{i}$, so that:

- $\nabla _{a}=\nabla _{(a\cdot e^{i})e_{i}}=(a\cdot e^{i})\nabla _{e_{i}}=a\cdot (e^{i}\nabla _{e_{i}})=a\cdot \nabla .$

For this reason, $\nabla _{a}F(x)$ is often noted $a\cdot \nabla F(x)$.

The standard order of operations for the vector derivative is that it acts only on the function closest to its immediate right. Given two functions $F$ and $G$, then for example we have

- $\nabla FG=(\nabla F)G.$

### Product rule

Although the partial derivative exhibits a product rule, the vector derivative only partially inherits this property. Consider two functions $F$ and $G$:

- ${\begin{aligned}\nabla (FG)&=e^{i}\partial _{i}(FG)\\&=e^{i}((\partial _{i}F)G+F(\partial _{i}G))\\&=e^{i}(\partial _{i}F)G+e^{i}F(\partial _{i}G).\end{aligned}}$

Since the geometric product is not commutative with $e^{i}F\neq Fe^{i}$ in general, we need a new notation to proceed. A solution is to adopt the *overdot notation*, in which the scope of a vector derivative with an overdot is the multivector-valued function sharing the same overdot. In this case, if we define

- ${\dot {\nabla }}F{\dot {G}}=e^{i}F(\partial _{i}G),$

then the product rule for the vector derivative is

- $\nabla (FG)=\nabla FG+{\dot {\nabla }}F{\dot {G}}.$

### Interior and exterior derivative

Let $F$ be an $r$-grade multivector. Then we can define an additional pair of operators, the interior and exterior derivatives,

- $\nabla \cdot F=\langle \nabla F\rangle _{r-1}=e^{i}\cdot \partial _{i}F,$
- $\nabla \wedge F=\langle \nabla F\rangle _{r+1}=e^{i}\wedge \partial _{i}F.$

In particular, if $F$ is grade 1 (vector-valued function), then we can write

- $\nabla F=\nabla \cdot F+\nabla \wedge F$

and identify the divergence and curl as

- $\nabla \cdot F=\operatorname {div} F,$
- $\nabla \wedge F=I\,\operatorname {curl} F.$

Unlike the vector derivative, neither the interior derivative operator nor the exterior derivative operator is invertible.

### Multivector derivative

The derivative with respect to a vector as discussed above can be generalized to a derivative with respect to a general multivector, called the **multivector derivative**.

Let $F$ be a multivector-valued function of a multivector. The directional derivative of $F$ with respect to $X$ in the direction $A$, where $X$ and $A$ are multivectors, is defined as

- $A*\partial _{X}F(X)=\lim _{\epsilon \to 0}{\frac {F(X+\epsilon A)-F(X)}{\epsilon }}\ ,$

where $A*B=\langle AB\rangle$ is the scalar product. With $\{e_{i}\}$ a vector basis and $\{e^{i}\}$ the corresponding dual basis, the multivector derivative is defined in terms of the directional derivative as^{[2]}

- ${\frac {\partial }{\partial X}}=\partial _{X}=\sum _{i<\dots <j}e^{i}\wedge \cdots \wedge e^{j}(e_{j}\wedge \cdots \wedge e_{i})*\partial _{X}\ .$

This equation is just expressing $\partial _{X}$ in terms of components in a reciprocal basis of blades, as discussed in the article section Geometric algebra#Dual basis.

A key property of the multivector derivative is that

- $\partial _{X}\langle XA\rangle =P_{X}(A)\ ,$

where $P_{X}(A)$ is the projection of $A$ onto the grades contained in $X$.

The multivector derivative finds applications in Lagrangian field theory.

Let $\{e_{1},\ldots ,e_{n}\}$ be a set of basis vectors that span an $n$-dimensional vector space. From geometric algebra, we interpret the pseudoscalar $e_{1}\wedge e_{2}\wedge \cdots \wedge e_{n}$ to be the signed volume of the $n$-parallelotope subtended by these basis vectors. If the basis vectors are orthonormal, then this is the unit pseudoscalar.

More generally, we may restrict ourselves to a subset of $k$ of the basis vectors, where $1\leq k\leq n$, to treat the length, area, or other general $k$-volume of a subspace in the overall $n$-dimensional vector space. We denote these selected basis vectors by $\{e_{i_{1}},\ldots ,e_{i_{k}}\}$. A general $k$-volume of the $k$-parallelotope subtended by these basis vectors is the grade $k$ multivector $e_{i_{1}}\wedge e_{i_{2}}\wedge \cdots \wedge e_{i_{k}}$.

Even more generally, we may consider a new set of vectors $\{x^{i_{1}}e_{i_{1}},\ldots ,x^{i_{k}}e_{i_{k}}\}$ proportional to the $k$ basis vectors, where each of the $\{x^{i_{j}}\}$ is a component that scales one of the basis vectors. We are free to choose components as infinitesimally small as we wish as long as they remain nonzero. Since the outer product of these terms can be interpreted as a $k$-volume, a natural way to define a measure is

- ${\begin{aligned}d^{k}X&=\left(dx^{i_{1}}e_{i_{1}}\right)\wedge \left(dx^{i_{2}}e_{i_{2}}\right)\wedge \cdots \wedge \left(dx^{i_{k}}e_{i_{k}}\right)\\&=\left(e_{i_{1}}\wedge e_{i_{2}}\wedge \cdots \wedge e_{i_{k}}\right)dx^{i_{1}}dx^{i_{2}}\cdots dx^{i_{k}}.\end{aligned}}$

The measure is therefore always proportional to the unit pseudoscalar of a $k$-dimensional subspace of the vector space. Compare the Riemannian volume form in the theory of differential forms. The integral is taken with respect to this measure:

- $\int _{V}F(x)\,d^{k}X=\int _{V}F(x)\left(e_{i_{1}}\wedge e_{i_{2}}\wedge \cdots \wedge e_{i_{k}}\right)dx^{i_{1}}dx^{i_{2}}\cdots dx^{i_{k}}.$

More formally, consider some directed volume $V$ of the subspace. We may divide this volume into a sum of simplices. Let $\{x_{i}\}$ be the coordinates of the vertices. At each vertex we assign a measure $\Delta U_{i}(x)$ as the average measure of the simplices sharing the vertex. Then the integral of $F(x)$ with respect to $U(x)$ over this volume is obtained in the limit of finer partitioning of the volume into smaller simplices:

- $\int _{V}F\,dU=\lim _{n\rightarrow \infty }\sum _{i=1}^{n}F(x_{i})\,\Delta U_{i}(x_{i}).$

### Fundamental theorem of geometric calculus

The reason for defining the vector derivative and integral as above is that they allow a strong generalization of Stokes' theorem. Let ${\mathsf {L}}(A;x)$ be a multivector-valued function of $r$-grade input $A$ and general position $x$, linear in its first argument. Then the fundamental theorem of geometric calculus relates the integral of a derivative over the volume $V$ to the integral over its boundary:

$\int _{V}{\dot {\mathsf {L}}}\left({\dot {\nabla }}dX;x\right)=\oint _{\partial V}{\mathsf {L}}(dS;x).$

As an example, let ${\mathsf {L}}(A;x)=\langle F(x)AI^{-1}\rangle$ for a vector-valued function $F(x)$ and a ($n-1$)-grade multivector $A$. We find that

- ${\begin{aligned}\int _{V}{\dot {\mathsf {L}}}\left({\dot {\nabla }}dX;x\right)&=\int _{V}\langle {\dot {F}}(x){\dot {\nabla }}\,dX\,I^{-1}\rangle \\&=\int _{V}\langle {\dot {F}}(x){\dot {\nabla }}\,|dX|\rangle \\&=\int _{V}\nabla \cdot F(x)\,|dX|.\end{aligned}}$

Likewise,

- ${\begin{aligned}\oint _{\partial V}{\mathsf {L}}(dS;x)&=\oint _{\partial V}\langle F(x)\,dS\,I^{-1}\rangle \\&=\oint _{\partial V}\langle F(x){\hat {n}}\,|dS|\rangle \\&=\oint _{\partial V}F(x)\cdot {\hat {n}}\,|dS|.\end{aligned}}$

Thus we recover the divergence theorem,

- $\int _{V}\nabla \cdot F(x)\,|dX|=\oint _{\partial V}F(x)\cdot {\hat {n}}\,|dS|.$