Geometric calculus

In, geometric calculus extends the to include  and. The formalism is powerful and can be shown to encompass other mathematical theories including and s.

Differentiation
With a geometric algebra given, let $$a$$ and $$b$$ be and let $$F(a)$$ be a -valued function. The of $$F(a)$$ along $$b$$ is defined as


 * $$\nabla_b F(a) = \lim_{\epsilon \rightarrow 0}{\frac{F(a + \epsilon b) - F(a)}{\epsilon}},$$

provided that the limit exists, 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 s $$\{e_i\}$$ 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, consider the operator:
 * $$e^i\partial_i,$$

which means
 * $$F \mapsto e^i\partial_i F,$$

or, more verbosely:
 * $$F \mapsto (x\mapsto e^i\nabla_{e_i} F(x)).$$

It can be shown that this operator is independent of the choice of frame, and can thus be used to define the geometric derivative:
 * $$\nabla = e^i\partial_i.$$

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

It can be shown that 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 geometric 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 for the geometric 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, the geometric derivative only partially inherits this property. Consider two functions $$F$$ and $$G$$:


 * $$\begin{align}\nabla(FG) &= e^i\partial_i(FG) \\

&= e^i((\partial_iF)G+F(\partial_iG)) \\ &= e^i(\partial_iF)G+e^iF(\partial_iG). \end{align}$$

Since the geometric product is not with $$e^iF \ne Fe^i$$ in general, we cannot proceed further without new notation. A solution is to adopt the  notation, in which the scope of a geometric derivative with an overdot is the multivector-valued function sharing the same overdot. In this case, if we define


 * $$\dot{\nabla}F\dot{G}=e^iF(\partial_iG),$$

then the product rule for the geometric 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 and  as


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

Note, however, that these two operators are considerably weaker than the geometric derivative counterpart for several reasons. Neither the interior derivative operator nor the exterior derivative operator is.

Integration
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 $$e_1 \wedge e_2 \wedge\cdots\wedge e_n$$ to be the  of the $$n$$- subtended by these basis vectors. If the basis vectors are, then this is the unit pseudoscalar.

More generally, we may restrict ourselves to a subset of $$k$$ of the basis vectors, where $$1 \le k \le n$$, 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 is


 * $$\begin{align}d^kX &= \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{align}$$

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


 * $$\int_V F(x)\,d^kX = \int_V F(x) \left( e_{i_1}\wedge e_{i_2}\wedge\cdots\wedge e_{i_k} \right) dx^{i_1} dx^{i_2} \cdots dx^{i_k}.$$

More formally, consider some directed volume $$V$$ of the subspace. We may divide this volume into a sum of. 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:


 * $$\int_V F\,dU = \lim_{n \rightarrow \infty} \sum_{i=1}^n F(x_i)\,\Delta U_i(x).$$

Fundamental theorem of geometric calculus
The reason for defining the geometric derivative and integral as above is that they allow a strong generalization of. 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:

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


 * $$\begin{align}\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{align}$$

Likewise,


 * $$\begin{align}\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{align}$$

Thus we recover the ,


 * $$\int_V \nabla \cdot F(x)\,|dX| = \oint_{\partial V} F(x) \cdot \hat{n}\,|dS|.$$

Covariant derivative
A sufficiently smooth $$k$$-surface in an $$n$$-dimensional space is deemed a. To each point on the manifold, we may attach a $$k$$-blade $$B$$ that is tangent to the manifold. Locally, $$B$$ acts as a pseudoscalar of the $$k$$-dimensional space. This blade defines a of vectors onto the manifold:


 * $$\mathcal{P}_B (A) = (A \cdot B^{-1}) B.$$

Just as the geometric derivative $$\nabla$$ is defined over the entire $$n$$-dimensional space, we may wish to define an intrinsic derivative $$\partial$$, locally defined on the manifold:


 * $$\partial F = \mathcal{P}_B (\nabla )F.$$

(Note: The right hand side of the above may not lie in the tangent space to the manifold. Therefore, it is not the same as $$\mathcal{P}_B (\nabla F)$$, which necessarily does lie in the tangent space.)

If $$a$$ is a vector tangent to the manifold, then indeed both the geometric derivative and intrinsic derivative give the same directional derivative:


 * $$a \cdot \partial F = a \cdot \nabla F.$$

Although this operation is perfectly valid, it is not always useful because $$\partial F$$ itself is not necessarily on the manifold. Therefore, we define the covariant derivative to be the forced projection of the intrinsic derivative back onto the manifold:


 * $$a \cdot DF = \mathcal{P}_B (a \cdot \partial F) = \mathcal{P}_B (a \cdot \mathcal{P}_B (\nabla F)).$$

Since any general multivector can be expressed as a sum of a projection and a rejection, in this case


 * $$a \cdot \partial F = \mathcal{P}_B (a \cdot \partial F) + \mathcal{P}_B^{\perp} (a \cdot \partial F),$$

we introduce a new function, the $$\mathsf{S}(a)$$, which satisfies


 * $$F \times \mathsf{S}(a) = \mathcal{P}_B^{\perp} (a \cdot \partial F),$$

where $$\times$$ is the commutator product. In a local coordinate basis $$\{e_i\}$$ spanning the tangent surface, the shape tensor is given by


 * $$\mathsf{S}(a) = e^i \wedge \mathcal{P}_B^{\perp} (a \cdot \partial e_i).$$

Importantly, on a general manifold, the covariant derivative does not commute. In particular, the is related to the shape tensor by


 * $$[a \cdot D, \, b \cdot D]F=-(\mathsf{S}(a) \times \mathsf{S}(b)) \times F.$$

Clearly the term $$\mathsf{S}(a) \times \mathsf{S}(b)$$ is of interest. However it, like the intrinsic derivative, is not necessarily on the manifold. Therefore, we can define the to be the projection back onto the manifold:


 * $$\mathsf{R}(a \wedge b)=-\mathcal{P}_B (\mathsf{S}(a) \times \mathsf{S}(b)).$$

Lastly, if $$F$$ is of grade $$r$$, then we can define interior and exterior covariant derivatives as


 * $$D \cdot F = \langle DF \rangle_{r-1},$$
 * $$D \wedge F = \langle D F \rangle_{r+1},$$

and likewise for the intrinsic derivative.

Relation to differential geometry
On a manifold, locally we may assign a tangent surface spanned by a set of basis vectors $$\{e_i\}$$. We can associate the components of a, the , and the as follows:


 * $$g_{ij}=e_i \cdot e_j,$$
 * $$\Gamma^k_{ij}=(e_i \cdot De_j) \cdot e^k,$$
 * $$R_{ijkl}=(\mathsf{R}(e_i \wedge e_j) \cdot e_k) \cdot e_l.$$

These relations embed the theory of differential geometry within geometric calculus.

Relation to differential forms
In a ($$x^1, \ldots, x^n$$), the coordinate differentials $$dx^1$$, ..., $$dx^n$$ form a basic set of one-forms within the. Given a $$i_1, \ldots, i_k$$ with $$1 \le i_p \le n$$ for $$1 \le p \le k$$, we can define a $$k$$-form
 * $$\omega = f_I\,dx^I=f_{i_1,i_2\cdots i_k}\,dx^{i_1}\wedge dx^{i_2}\wedge\cdots\wedge dx^{i_k}.$$

We can alternatively introduce a $$k$$-grade multivector $$A$$ as


 * $$A = f_{i_1,i_2\cdots i_k}e^{i_1}\wedge e^{i_2}\wedge\cdots\wedge e^{i_k}$$

and a measure


 * $$\begin{align}d^kX &= \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{align}$$

Apart from a subtle difference in meaning for the exterior product with respect to differential forms versus the exterior product with respect to vectors (in the former the increments are covectors, whereas in the latter they represent scalars), we see the correspondences of the differential form


 * $$\omega \cong A^{\dagger} \cdot d^kX = A \cdot \left(d^kX \right)^{\dagger},$$

its derivative


 * $$d\omega \cong (D \wedge A)^{\dagger} \cdot d^{k+1}X = (D \wedge A) \cdot \left(d^{k+1}X \right)^{\dagger},$$

and its


 * $$\star\omega \cong (I^{-1} A)^{\dagger} \cdot d^kX,$$

embed the theory of differential forms within geometric calculus.