# What is the relationship between Clifford algebra and differential forms?

Counting the basis vectors, Clifford algebra has 1 scalar, 4 vectors, 6 bivectors, 4 trivectors, 1 pseudoscalar

Differential forms have 1 scalar, 4 one-forms, 6 two-forms, 4 three-forms, 1 4-form (the Levi civita Tensor)

Product of a differential form with the Levi Civita tensor gives you the Hodge dual. Clifford algebra also has a similar idea about mapping bivectors to vectors, using the pseudoscalar.

But Clifford algebra has a metric while differential forms don't need one (But the Hodge duality does need a metric, in order to raise the indices of the Levi Civita tensor)

Also, Clifford algebra has the geometric product while differential forms have the wedge product. The products are different. But the geometric product has a wedge product in its definition. Is Clifford algebra sort of an extension of differential forms? What would be the motivation for this extension?

What is the precise relationship between Clifford algebra and differential forms?

• Short answer: differential forms are sections of the exterior powers of the cotangent bundle, so you really should be comparing the exterior algebra with the Clifford algebra, in which case there's is a specific relation: the exterior algebra is the associated graded algebra of the Clifford algebra (see en.m.wikipedia.org/wiki/Filtered_algebra). One can think of exterior algebra as the Clifford algebra with forgotten metric (all dot products equal to zero). Nov 11, 2022 at 6:06

$$\newcommand\Cl{\mathrm{Cl}} \newcommand\Ext{{\textstyle\bigwedge}} \newcommand\rev\widetilde \newcommand\rintr{\mathbin{{\llcorner}}} \newcommand\R{\mathbb R} \newcommand\form{\langle#1\rangle} \newcommand\dd{\mathrm{d}} \newcommand\lcontr{\mathbin{{\rfloor}}} \newcommand\rcontr{\mathbin{{\lfloor}}} \newcommand\doub{\mathfrak} \newcommand\doubnabla{\boldsymbol\nabla} \newcommand\adj{\overline}$$

Let $$Q$$ be a quadratic form over an $$n$$-dimensional vector space $$V$$ (which for this discussion we will assume is real though this isn't strictly necessary.) Quadratic forms $$Q$$ and symmetric bilinear forms $$B$$ are in one-to-one correspondence via $$B(v, w) = \frac12(Q(v + w) - Q(v) - Q(w)),\quad Q(v) = B(v, v).$$ In this context, metric is another word for symmetric bilinear form.

The Clifford algebra $$\Cl(V, Q)$$ is essentially the associative algebra generated by $$V$$ with the relations $$v^2 = Q(v)$$ for all $$v \in V$$. This can be formalized by defining $$\Cl(V, Q)$$ as the algebra with a certain universal property, or by quotienting the tensor algebra by the previously mentioned relations. It is not defined in terms of a wedge product; I would consider such a definition a bad definition.

But there are two important relationships between Clifford algebras and the exterior algebra $$\Ext V$$. First, $$\Cl(V, 0)$$ is exactly $$\Ext V$$, where $$0$$ is the trivial quadratic form. Second, there is a canonical linear isomorphism (not an algebra isomorphism) $$\Cl(V, Q) \cong \Ext V$$ for any $$Q$$. This means that we can view $$\Cl(V, Q)$$ as $$\Ext V$$ endowed with a second product, the Clifford product, and one way of constructing $$\Cl(V, Q)$$ is by defining such a product on $$\Ext V$$ (but I would never take this as a definition).

In light of this, we can view various operations on $$\Ext V$$ as operations on $$\Cl(V, Q)$$. This is very natural with e.g. the Hodge star $$\star$$. Denote by $$\Cl^k(V, Q)$$ the linear subspace of $$\Cl(V, Q)$$ corresponding to $$\Ext^k V$$. There is an antiautomorphism $$A \mapsto \rev A$$ of $$\Cl(V, Q)$$ which reverses all products and is the identity on $$V$$. We get the Hodge star when $$Q$$ is non-degenerate and when we choose $$I \in \Cl^n(V, Q)$$ such that $$I^2 = \pm1$$ (such elements always square to scalars); this is equivalent to choosing an orientation on $$V$$. Then the Hodge star is exactly $$\star A = \rev AI$$ for any $$A \in \Cl(V, Q) \cong \Ext V$$.

A differential form assigns an element of $$(\Ext V)^*$$ to each tangent space of a manifold (where we assume the tangent spaces are isomorphic to $$V$$). There is a canonical isomorphism $$(\Ext V)^* \cong \Ext V^*$$ induced by a natural bilinear pairing $$\form{{-}, {-}} : \Ext V^* \times \Ext V \to \R$$, so we usually identify a differential form with an element of $$\Ext V^*$$. Then the action of a differential form $$A^* \in \Ext V^*$$ on $$B \in \Ext V$$ is given by $$\form{A^*, B}$$, or equivalently the action on vectors $$v_1,\dotsc, v_k \in V$$ is given by $$\form{A^*, v_1\wedge\cdots\wedge v_k}.$$ But in the presence of a non-degenerate metric there is an isomorphism $$V \cong V^*$$, and this extends naturally to algebra isomorphisms $$\Ext V \cong \Ext V^*$$ and $$\Cl(V, Q) \cong \Cl(V^*, Q^*)$$ (where $$Q^*$$ is $$Q$$ applied to $$V^*$$ via the isomorphism). This means we can identify differential forms with elements of $$\Ext V$$. The pairing $$\form{{-}, {-}}$$ becomes a bilinear form on $$\Ext V$$, and is (almost) exactly the scalar part of the Clifford product: $$\form{A, B} = \form{\rev AB}_0,\quad A, B \in \Cl(V, Q) \cong \Ext V.$$ This scalar part is defined either by taking advantage of the grading of $$\Ext V$$, or as intrinsic to the Clifford algebra via the normalized trace: $$\form{A}_0 = \frac1{2^n}\mathrm{Tr}(B \mapsto AB),\quad B \in \Cl(V, Q).$$ The reversal $$\rev A$$ in the above and in the Hodge star is an artifact of how $$\form{{-},{-}}$$ is defined, and can be done away with if desired. For simplicity, we will use this convention and define $$\form{A, B} = \form{AB}_0,\quad \star A = AI.$$

In this way, we can represent differential forms as elements of $$\Cl(V, Q)$$ and use all the tools that comes with that.

As a particular example, we get the exterior derivative as $$[\dd\form{A_x, {-}}](B) = \form{(\nabla\wedge A_x)B}_0$$ and so the codifferential as $$[{\star^{-1}}\dd{\star}\form{A_x, {-}}](B) = \form{(\nabla\wedge(A_xI))I^{-1}B}_0 = \form{(\nabla\lcontr A_x)B}_0$$ where $$\lcontr$$ is the left contraction $$A\lcontr B = \form{AB}_{k-j},\quad A \in \Cl^j(V, Q),\quad B \in \Cl^k(V, Q),$$ i.e. the projection onto the exterior algebra grade $${k-j}$$ component, which we define to be zero if $$j > k$$. This is the manifestation of the interior product in the presence of a metric, as is the closely related right contraction $$\rcontr$$. To summarize, the exterior derivative and codifferentials are expressed in $$\Cl(V, Q)$$ as $$\nabla\wedge A_x,\quad \nabla\lcontr A_x.$$ But for any $$v \in V$$ and $$A \in \Cl(V, Q)$$ $$vA = v\wedge A + v\lcontr A$$ so we see that the sum of the exterior derivative and codifferential is the geometric derivative $$\nabla A_x = \nabla\wedge A + \nabla\lcontr A$$ where on the left we are using the Clifford product. This is extremely useful; for example, Maxwell's equations become exactly one equation $$\nabla F = J$$, and (at least in flat space) $$\nabla$$ has a Green's function which allows us to directly solve for $$F$$ in terms of $$J$$.

This is the beginning of geometric calculus; I would recommend chapter 6 of Doran and Lasenby's Geometric Algebra for Physicists (2003) for more. (Really, the whole book is worth a read.) In brief, this approach to differential geometry (of course with a focus on Riemannian geometry) involves embedding our manifold in flat space, where then one of the major objects of study becomes the bivector-valued shape tensor $$S_x$$ which encodes much of the extrinsic and intrinsic geoemtry of the manifold. For instance, the covariant derivative (i.e. Levi-Civita connection) can be expressed as $$DA_x = \partial A_x + \partial_a(S_x(a)\times A_x)$$ where $$X\times Y = \tfrac12(XY - YX)$$, the tangential derivative $$\dot\partial = P_x(\dot\nabla)$$ is the "projection" of $$\nabla$$ onto the tangent space of the manifold, and $$\partial_a$$ is the tangential derivative with respect to $$a$$ (actually representing a kind of tensor contraction since $$S_x$$ is linear).

The introduction of a metric obviously takes us away from pure differential forms. We can actually try to do away with the metric: the space $$\doub W = V^*\oplus V$$ has a natural bilinear form $$\form{v^* + v, w^* + w} = v^*(w) + w^*(v)$$ with quadratic form $$\form{v^* + v, v^* + v} = 2v^*(v).$$ We then consider the Clifford algebra $$\Cl(\doub W)$$. Since $$\form{v^*, v^*} = \form{v, v} = 0$$, the subalgebras generated by $$V^*$$ and $$V$$ are exactly $$\Ext V^*$$ and $$\Ext V$$. Not only that, but if $$M$$ is a manifold such that $$T_xM \cong V$$, then the tangent spaces of the cotangent bundle are naturally isomorphic to $$\doub W$$. This makes $$\Cl(\doub W)$$ a potentially very natural setting in which to study differential forms.

However, I have yet to study this construct to my satisfaction, so I will stop here.