In what follows, let $V=\mathbb{R}^n$ (although the following probably applies also to a larger number of finite-dimensional spaces).
We assume throughout that we have made a choice for an inner product $\langle \cdot, \cdot \rangle$ on $V$.
Geometric/Clifford Product: The geometric/Clifford product of any two vectors $v,w \in V$ decomposes as: $$v w = \langle v, w\rangle + v \wedge w\,, $$ where the first term on the right-hand side is the inner product of $v,w$, and the second term corresponds to the bivector spanned by $v$ and $w$ in that orientation.
Trace: Our choice of inner product $\langle \cdot, \cdot \rangle$ on $V$ induces a canonical (w.r.t. $\langle \cdot, \cdot \rangle$) isomorphism between $V$ and $V^*$. This allows us to identify $V \otimes V$ and $V^* \otimes V$, and $V^* \otimes V$ can always be identified with $\operatorname{End}(V)=\operatorname{Hom}(V,V)=\mathscr{L}(V)$ the space of linear operators on $V$ (I think).
Then the bilinear function $\langle \cdot, \cdot \rangle : V \times V \to \mathbb{R}$ induces under the identification $V^* \cong V$ a bilinear function $V^* \times V \to \mathbb{R}$ which via the universal property of tensor products induces a linear function $V^* \otimes V \cong \operatorname{End}(V) \to \mathbb{R}$ which turns out to be the trace. (See my two previous questions (1)(2) of which this question is a follow-up.)
Now note that the kernel of the trace $\mathfrak{sl}(n,\mathbb{R})$, and also note that an important subspace of $\mathfrak{sl}(n,\mathbb{R})$ is $\mathfrak{o}(n,\mathbb{R})=\mathfrak{so}(n,\mathbb{R})$, which can be identified with the space of all skew-symmetric matrices.
The latter space, $\mathfrak{so}(n,\mathbb{R})$, can be identified with the space of all bivectors (see here or here).
Thus, by factoring through $V^* \otimes V$, the Clifford/geometric product decomposition above assumes the form $$(v,w) \mapsto (v^*,w) \mapsto v^* \otimes w \mapsto \operatorname{trace}(v^* \otimes w) + v \wedge w = \operatorname{trace}(v^* \otimes w) + (v^* \otimes w - w^* \otimes v)\,, $$ with the first term $\operatorname{trace}(v^* \otimes w)$ being in the image of the trace and the second term $v \wedge w$ being in the kernel. Thus the decomposition of the Clifford product $vw$ seems to (implicitly) have the form of an image-kernel decomposition with respect to the linear map trace.
Question: Is this correct? Can the Clifford product really be interpreted as being related to or derived from the trace?
Specifically, it seems like in the above we are disregarding the elements $\mathfrak{sl}(n,\mathbb{R}) \setminus \mathfrak{so}(n,\mathbb{R})$.
Thus (assuming that there is a natural splitting of $\mathfrak{sl}(n,\mathbb{R})= S \oplus \mathfrak{so}(n,\mathbb{R}) $, and I don't know whether there is or isn't), can the quotienting process which gives us the Clifford algebra from the tensor algebra be considered as equivalent to "quotienting out" those elements of the kernel of the trace which are not also skew-symmetric operators? Or am I over-simplifying?
Also for some reason I thought there might be some relationship between this and semidirect products of Lie groups, but I don't know that topic anywhere near well enough for that to be considered anything besides wild speculation based on visual similarities of the formulas.
