$\newcommand{\talg}{\mathcal{T}(V)}$$\newcommand{\clalg}{\mathcal{Cl}_q(V)}$$\newcommand{\qalg}{\mathcal{I}_q(V)}$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.

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 $$\sum_{\sigma\in S_4} \left[e_{\sigma(1)}\otimes(e_{\sigma(2)}\otimes e_{\sigma(3)} - \langle e_{\sigma(2)}, e_{\sigma(3)} \rangle_q )\otimes e_{\sigma(4)}\right]=\tau\in \qalg.$$

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.)