# Is there a way to embed Clifford algebras into the corresponding tensor algebra?


There are simple and straightforward embeddings of the underlying vector space $V$ into its corresponding tensor algebra $\talg$ and any of its corresponding Clifford algebras $\clalg$ (where $q$ denotes the quadratic form defining the Clifford algebra).

This fact is what makes both tensor analysis and geometric (Clifford) algebra compatible with ordinary vector algebra or calculus.

However, even though any $\clalg$ can be formed as a quotient of $\talg$, or perhaps because of that fact, no $\clalg$ seems to be "compatible" with $\talg$ in a simple way, at least in the sense that there does not appear to exist any "simple" embedding of $\clalg$ into $\talg$, such that we could use all of the structure and geometric intuition afforded by the $\clalg$ framework while still working in the most general possible space, $\talg$.

This seems like a big problem to me, because even if one isn't interested in tensors of rank $>2$, rank 2 tensors (also known as matrices) are ubiquitous, and I have the impression that the greater compatibility of vector algebra with matrix algebra is the main obstacle to more widespread implementation of geometric algebra.

(I.e. because unlike vector algebra, geometric algebra doesn't "play nice" with matrix algebra or tensor algebra in general.)

This seems like a major pedagogical problem, since it seems like many subjects could be much more easily understood through the prism of geometric algebra, but the (seeming) incompatibility of any $\clalg$ with $\talg$ vastly (and arguably rightly) dampens any enthusiasm to pursue such an approach.

Even if there can not exist any simple embedding, what about a simple way to switch between the two systems? I am also doubtful in this regard, since the definition of exterior algebra and outer products in terms of tensor products is very unwieldy (a sum over the symmetric group). And the exterior algebra is the most degenerate type of Clifford algebra possible.

Here is a similar question on Math.SE, corroborating my claim that every multivector corresponds to a tensor but not vice versa: What is the relationship of tensor and multivector. These two documents explain how to represent some multivectors as tensors for the special case that $V=\mathbb{R}^3$ and $q=I$ i.e. just the identity, so that the inner product is just the dot product: (1) (2).

The best I could find so far is the following (see page 7 of this document):

Given $\qalg:= \{ \sum_k A_k \otimes (v\otimes v -q(v))\otimes B_k: v \in V, A_k, B_k \in \talg \}$

We have for $A,B \in \clalg$: $AB = A \otimes B + \qalg$, where $AB$ denotes the Clifford (geometric) product and $\otimes$ is of course the tensor product.

EDIT: Thinnking about Oscar Cunningham's comments below, I think we can write $$\talg = \clalg \oplus \qalg$$ (or something that's actually mathematically correct but similar "in spirit").

$\clalg$ are exactly those tensors which have a unique embedding as an element of the Clifford algebra, and $\qalg$ consists of exactly those tensors which can not be represented in the Clifford algebra, and hence are mapped to 0 by the quotient map.

Thus the problem reduces to:

Given any arbitrary tensor $t \in \talg$, determine its (unique since direct sum?) representation as $$t= C + \tau,$$ where $C \in \clalg$ is a member of the Clifford algebra, and $\tau \in \qalg$ is a member of the ideal $\qalg=\{ \sum_k A_k \otimes (v\otimes v -q(v))\otimes B_k: v \in V, A_k, B_k \in \talg \}$.

Since we have this explicit representation for $\qalg$, I think the problem might be much easier than I originally anticipated. For example, if $t$ is a rank 4 tensor in some four dimensional vector space $V$, we could write something like $$t=\sum_{\sigma \in S_4} \left[e_{\sigma(1)}\otimes(\langle e_{\sigma(2)},e_{\sigma(3)}\rangle_q)\otimes e_{\sigma(4)}\right] + \sum_{\sigma\in S_4} \left[e_{\sigma(1)}\otimes(e_{\sigma(2)}\otimes e_{\sigma(3)} - \langle e_{\sigma(2)}, e_{\sigma(3)} \rangle_q )\otimes e_{\sigma(4)}\right],$$ where $e_1, e_2, e_3, e_4$ are basis vectors and $\langle \cdot, \cdot\rangle_q$ is the inner product formed from the quadratic form $q$ via polarization.

Of course, I am going somewhat out on a limb here in assuming, i.e. I do not know how to prove that $$\sum_{\sigma \in S_4} \left[e_{\sigma(1)}\otimes(\langle e_{\sigma(2)},e_{\sigma(3)}\rangle_q)\otimes e_{\sigma(4)}\right]=C \in \clalg$$ or that $$\sum_{\sigma\in S_4} \left[e_{\sigma(1)}\otimes(e_{\sigma(2)}\otimes e_{\sigma(3)} - \langle e_{\sigma(2)}, e_{\sigma(3)} \rangle_q )\otimes e_{\sigma(4)}\right]=\tau\in \qalg.$$

The other strategy I was thinking of was to represent tensors of rank 2 or greater only via their isomorphisms with multilinear maps, since then that would come down to representing them as functions of (contravariant) vectors, a function of $k$ vectors, a $k$-linear map, for a $k$ tensor. This would work insofar as all contravariant vectors have canonical embeddings in both the Clifford and tensor algebras, but I think we still have as unresolved the problem of how to represent covariant vectors (row vectors) as well as general covariant/mixed variance tensors.

Some possible problems: The 0 element of a Clifford/geometric algebra has all grades (i.e. for any $k$ it is a $k$-blade/multivector), whereas I don't think this is true in the tensor algebra (at least thinking in terms of representing tensors via k-dimensional arrays -- you can have the scalar 0, but also 0 column vectors, 0 row vectors, 0 matrices, etc.).

Intuitively, one wants to think of a geometric/Clifford algebra as "symmetric algebra + exterior algebra" (commutative inner product $\frac{1}{2}(v\otimes w + w \otimes v)$ and anticommutative outer product $\frac{1}{2}(v\otimes w - w \otimes v)$), especially since (at least for vector spaces over fields of characteristic zero, which is what I personally am interested in) both the symmetric and exterior algebras can be identified with certain subclasses of tensors (see: Wikipedia - symmetric algebra vs. symmetric tensors). But how do we do this for a general Clifford algebra? Maybe $$vw = v \otimes w + \qalg \\= \frac{1}{2}(\langle v, w \rangle_q + v \wedge w)= \frac{1}{2}(v\otimes w + w \otimes v + \qalg)+\frac{1}{2}(v \otimes w - w \otimes v)?$$ It's clear that $v \wedge w$ can always be identified as a subalgebra of the Clifford algebra, and this sort of definition somewhat explains that, but then is it true that the inner product generated by polarization of $q$ must equal $\frac{1}{2}(v\otimes w + w \otimes v + \qalg)$? And even if it does, does that mean that the inner product of a Clifford algebra is commutative if and only if $q$ is a symmetric quadratic form? Since in general the product derived from a quadratic form via polarization need only be bilinear, but not necessarily symmetric (i.e. commutative) unless $q$ is symmetric too.

Related: Which concepts in differential geometry cannot be represented using geometric algebra? (The answer is essentially: tensors.)

How would one express the result of a tensor product (of two vectors) in the geometric algebra?

Is geometric algebra isomorphic to tensor algebra? (No. Most tensors do not admit of a unique representation in geometric algebra due to the quotient structure.)

What is the hierarchy of algebraic objects meant to capture geometric intuition? (Tensor algebras are essentially the most general possible, even though geometric algebras are in general less unwieldy.)

• If $q$ is an inner product, do $\mathcal T(V)$ and $\mathcal{Cl}_q(V)$ get a canonical inner product? If so we could take the adjoint of the quotient map... – Oscar Cunningham Jul 1 '16 at 20:16
• @OscarCunningham I suppose we get a canonical inner product by taking the polarization of the quadratic form $q$. Could you elaborate on "taking the adjoint of the quotient map"? I don't understand. – Chill2Macht Jul 1 '16 at 20:20
• There's a canonical quotient map $\mathcal T(V)\rightarrow \mathcal{Cl}_q(V)$. Since $\mathcal T(V)$ is an inner product space, I think that $\mathcal{Cl}_q(V)$ has to be isomorphic to the space orthogonal to the kernel of this map. – Oscar Cunningham Jul 1 '16 at 20:29
• I don't think ${\cal T}(V)$ has any interesting finite-dimensional subalgebras. – მამუკა ჯიბლაძე Jul 2 '16 at 5:44
• Might the embedding of a Clifford algebra into the endomorphisms of an exterior algebra, as described in the accepted answer for mathoverflow.net/questions/68378/clifford-algebra-non-zero, be helpful for you? – KConrad Jul 2 '16 at 9:55

$\newcommand{\qalg}{\mathcal{I}_q(V)}$As K. Conrad points out, this question has actually been answered already on MathOverflow by user MTS: see these answers here and here.

The essential idea is this: despite the fact that Clifford algebras have non-zero quadratic form in general, we can still "piggyback" on the embedding of the exterior algebra in the tensor algebra to represent the Clifford algebra, regardless of the choice of quadratic form.

Note: The embedding of the exterior algebra into the tensor algebra can be given by defining the exterior=wedge product in terms of the tensor product. Discussion of two canonical ways of doing this are given here on MathOverflow (the top answer being again by MTS) as well as on Math.SE.

Beautiful and rigorous proofs of this fact are given in the two answers (1) (2) linked to above by user MTS; in what follows I will merely try to motivate this result for novices like me.

1. First, we might expect this result intuitively by considering the decomposition of the Clifford product for vectors into an "inner" and "outer" product. The former is the inner product generated by polarization of the quadratic form $q$ and the latter is "equivalent" in some sense to the exterior/wedge product from the exterior algebra. Since the inner product is grade-reducing, while the outer product is grade-increasing, we might expect that the inner product would be "incapable of producing new basis elements", hence the only basis elements of the Clifford algebra we should expect are those created by the corresponding exterior algebra, i.e. there is some reason to expect a priori that a given Clifford algebra is "no more rich than" making the corresponding exterior algebra an inner product space, i.e. by defining a quadratic form on the exterior algebra.

2. MTS's answers make this suspicion rigorous by showing that the ideals for the Clifford algebra and the exterior algebra are sufficiently similar so that the quotients of the tensor algebra by them can be related in a straightforward manner.

Let us motivate this fact by considering $\mathcal{I}_q(V)$ explicitly. Remember that $$\mathcal{I}_q(V):= \left\{ \sum_k A_k \otimes (v \otimes v - q(v)) \otimes B_k : v \in V, A_k, B_k \in \mathcal{T}(V) \right\}$$ Using the distributivity of the tensor product (e.g. here) I rewrite this as $$\qalg = \left\{ \sum_k \left[A_k \otimes v \otimes v \otimes B_k - A_k \otimes q(V) \otimes B_k \right] : v \in V, A_k, B_k \in \mathcal{T}(V)\right\} \\ \overset{?}{=} \left\{ \sum_k \left[A_k \otimes v \otimes v \otimes B_k - q(V) \cdot( A_k \otimes B_k) \right] : v \in V, A_k, B_k \in \mathcal{T}(V)\right\}$$ In any case, the motivating idea here is that the right term arising from the quadratic form, always being of strictly lower grade than the left term, should in some sense be "negligible" compared to the left term (MTS's second answer makes this idea explicit and rigorous). In other words: $$\qalg \approx \left\{ \sum_k A_k \otimes v \otimes v \otimes B_k: v \in V, A_k, B_k \in \mathcal{T}(V) \right\}$$ where the right hand side is obviously $\mathcal{I}_0(V)$, the ideal which forms the exterior algebra.

3. In the very simple case of geometric algebra over $\mathbb{R}^n$, this fact is used all the time in creating a basis for the geometric algebra. Namely, it is stated that $\mathbb{G}^n \cong \mathbb{R}^{2^n}$ (vector space isomorphism), with the basis being given by: $$\{e_1, \dots, e_n, e_{\sigma_2(1)}\wedge e_{\sigma_2(2)}, e_{\sigma_3(1)}\wedge e_{\sigma_3(2)} \wedge e_{\sigma_3(3)}, \dots, e_1 \wedge \dots \wedge e_n : \sigma_i \in A_i \}$$ where $A_i$ denotes the set of all even permutations of $i$ elements selected from $\{1, \dots, n\}$ and $\{e_1,\dots,e_n\}$ is an orthonormal basis of $\mathbb{R}^n$, orthogonal with respect to the inner product of $\mathbb{G}^n$.

It should be evident that this basis is isomorphic to the basis of the exterior algebra over $\mathbb{R}^n$.

Since in any Clifford algebra $vw = v \wedge w$ ($vw$ denoting the Clifford/geometric product) if and only if $\langle v, w \rangle_q=0$ i.e. if and only if $v$ and $w$ are orthogonal with respect to the inner product induced by $q$ via polarization, the possible choices of orthonormal basis will in general depend on the choice of quadratic form for the geometric algebra. Nevertheless, the one-to-one correspondence of the bases with the basis for the exterior algebra $\mathbb{R}^n$ will still remain and will be sufficient to generate a vector space isomorphism, since vector space isomorphisms only depend on the linear independence and dimensions of the bases, and not their orthogonality under an inner product.

In other words, we can always use the vector space basis of the exterior algebra over $\mathbb{R}^n$ as a vector space basis for the geometric algebra $\mathbb{G}^n$ regardless of our choice of quadratic form, although this basis will not always be orthogonal. Hence by identifying the geometric algebra with the exterior algebra + inner product structure, we get a linear embedding of the geometric algebra into the tensor algebra "for free" by using the exterior algebra's embedding.

(See p.4 of this document for discussion of the canonical basis of a geometric algebra generated by an orthonormal basis of $\mathbb{R}^n$.)