I'll start with a physical example. Let's say we have the angular velocity (in 3D euclidean space with orthonormal basis spanning it), which has the bivector representation $$\Omega = \omega_x \mathbf e_{yz} + \omega _y \mathbf e_{xz} + \omega_z \mathbf e_{xy}$$ for unit bivectors $\mathbf e_{yz}, \mathbf e_{xz}, \mathbf e_{xy}$,
and the skew symmetric matrix representation $$\Omega = \begin{bmatrix} 0 & -\omega_z & \omega_y \\ \omega_z & 0 & -\omega_x \\ -\omega_y & \omega_x & 0 \end{bmatrix}$$
The hodge dual of this bivector is the axial vector $$\star\Omega = \boldsymbol\omega = \omega_x\mathbf e_x + \omega _y \mathbf e_{y} + \omega_z \mathbf e_{z}$$
When we calculate the velocity of a particle with a position vector $\mathbf r = r_x\mathbf e_x + r_y \mathbf e_{y} + r_z \mathbf e_{z}$ and angular velocity $\Omega$, then the velocity of the particle is said to be given by $$\mathbf v = \Omega \mathbf r \iff \mathbf v = \boldsymbol\omega\times \mathbf r$$
What I don't understand is what the operation between $\Omega$ and $\mathbf r$ is.
I understand that the hodge dual of a wedge product is a cross product, which gives us the relationship that $$\text{vector}\wedge\text{vector} = \text{bivector} \iff \text{vector}\times \text{vector}=\text{axial vector}$$
However, there is also the relationship that $$\text{vector}\times \text{axial vector} = \text{vector}$$ which seems to be the relationship here. Now, if we replace the axial vector with the actual bivector (as seen in the angular velocity example above), what is the operation between the vector and the bivector??? $$ \text{bivector } ??? \text{ vector} = \text{vector}$$
Obviously, when we write the bivector as a skew symmetric matrix, we're just performing regular matrix multiplication. However, if we're writing it as just the sum of unit bivectors, what is this operation??
$$\langle\omega_x \mathbf e_{yz} + \omega _y \mathbf e_{xz} + \omega_z \mathbf e_{xy}\rangle \text{ ??? } \langle r_x\mathbf e_x + r_y \mathbf e_{y} + r_z \mathbf e_{z} \rangle$$
I initially thought it might be a wedge product, but quickly realized that it would just be wedging a bivector and a vector which returns a trivector, not a regular vector that I want. I can't think of any other operation that would return a vector between a bivector and a vector.
Any help?
Thanks!
It is 'almost' the interior product (not to be confused with the inner product, which will also play a role, hence the ‘almost’). In general, suppose $V$ is a real vector space, and you have integers $0\leq k\leq p$. Given a $p$-vector $v$ (i.e $v\in \bigwedge^pV$) and a $k$-covector $\alpha$ (i.e $\alpha\in\bigwedge^k(V^*)$), one can define a $(p-k)$-vector $\iota_{\alpha}v\equiv\alpha\,\lrcorner\, v\in \bigwedge^{p-k}V$, and the result is a bilinear function of $(\alpha,v)$. Usually, one flips the role of $V$ and $V^*$. Now if you have an inner product, then you can establish an isomorphism between $\bigwedge^k(V^*)$ and $\bigwedge^kV$, so rather than requiring $\alpha$ to be a $k$-covector, you can simply work with a $k$-vector.
I don’t feel like going full general, so I’ll just tell you how things work here. Let $(V,\langle\cdot,\cdot\rangle)$ be an inner product space, $\xi\in V$ a given vector (i.e we’re taking $k=1$) and suppose you have $p$ many vectors $v_1,\dots,v_p\in V$. Then, once you unwind the mumbo jumbo, we can define the interior product $\xi\,\lrcorner\,(v_1\wedge\cdots\wedge v_p)$, and it will equal \begin{align} \xi\,\lrcorner\,(v_1\wedge\cdots\wedge v_p)&=\sum_{j=1}^p(-1)^{j-1}\langle\xi,v_j\rangle\,v_1\wedge\cdots\wedge\widehat{v_j}\wedge \cdots\wedge v_p. \end{align} In other words, you run through $v_1\wedge\cdots\wedge v_p$, and starting with a positive sign, delete the $v_j$ and replace it instead with the number $\langle\xi,v_j\rangle$. So, from a $p$-vector, we have converted it into a $(p-1)$-vector. Btw, the ‘ordering’ isn’t really important. I could just as well define a new operation $\llcorner$ which acts like $(v_1\wedge\cdots\wedge v_p)\,\llcorner\,\xi:= \xi\,\lrcorner\,(v_1\wedge\cdots\wedge v_p) $. In particular, when $p=2$, we have for vectors $\xi,v,w\in V$, \begin{align} \xi\,\lrcorner \,(v\wedge w)\equiv (v\wedge w)\,\llcorner\,\xi =\langle\xi,v\rangle w-\langle\xi,w\rangle v.\label{1}\tag{$*$} \end{align}
So, anyway now with your angular velocity bivector (I’m pretty sure you have a typo in its definition) and position vector, you can play this game ($p=2,k=1$) to produce a vector $\Omega\,\llcorner\,\mathbf{r}$: \begin{align} \Omega\,\llcorner\,\mathbf{r} &= (\omega_x\,e_y\wedge e_z+\omega_y\,e_z\wedge e_x+\omega_z\,e_x\wedge e_y)\,\llcorner\,(r_xe_x+r_ye_y+r_ze_z). \end{align} Now, just open up the brackets (we can do this because as I mentioned above, the interior product is bilinear). You’ll get a total of $9$ interior products to compute (as you’d expect when opening up a bracket of 3$\times$ 3 terms). For each interior product of a bivector with a vector, we produce $2$ terms, as in \eqref{1}. Fortunately, many of these terms vanish due to orthonormality of $\{e_x,e_y,e_z\}$. We end up with the expected formula for the velocity.
But, note also that the cross product of two vectors in $\Bbb{R}^3$ (or more generally, an oriented 3-dimensional inner-product space) is $v\times w=\star(v\wedge w)$. So, in the case of angular velocity, you have $\omega\times r=\star(\omega\wedge r)=\star((\star\Omega)\wedge r)$.