# How do you take this multivector derivative that has a nested variable?

I want to compute the multivector derivative $$\partial_\psi$$ of $$\Omega_B\cdot \mathcal{I}(\Omega_B)$$ where $$\mathcal{I}$$ is a function that is linear in bivectors, $$\mathcal{I}(\Omega_B) = \int d^3x \rho(x) x\wedge (x\cdot \Omega_B)$$. $$\psi$$ is an even multivector in 3D space (so it contains scalar and bivector terms). $$\Omega_B$$ is a function of $$\psi$$, $$\Omega_B\equiv -\psi^\dagger \dot{\psi} + \dot{\psi}^\dagger \psi$$. $$\Omega_B$$ is a bivector because the second term is the negative reverse of the first term. The overdot signifies time derivative and the dagger signifies reverse. This is for a Lagrangian method so we are treating $$\dot{\psi}$$ and $$\dot{\psi}^\dagger$$ as constants. I use $$\hat{\partial}$$ instead of $$\partial$$ when a specific term is being operated on. This question relates to the geometric algebra Lagrangian for rigid body motion.

What I have so far: \begin{align} \partial_\psi \Omega_B\cdot (\Omega_B) &= \partial_\psi \langle \Omega_B \mathcal{I}(\Omega_B) \rangle \\ &= \hat{\partial}_\psi\langle\hat{ \Omega}_B \mathcal{I}((\Omega_B)\rangle + \hat{\partial}_\psi\langle \Omega_B \hat{\mathcal{I}(}(\Omega_B) \rangle \\ \end{align}

The first term is: \begin{align} \hat{\partial}_\psi\langle\hat{ \Omega}_B \mathcal{I}(\Omega_B)\rangle &= \hat{\partial}_\psi \langle -\hat{\psi}^\dagger \dot{\psi}\mathcal{I}(\Omega_B) + \dot{\psi}^\dagger\hat{\psi} \mathcal{I}(\Omega_B) \rangle \\ &= \hat{\partial}_\psi \langle -\hat{\psi}^\dagger \dot{\psi}\mathcal{I}(\Omega_B) + \hat{\psi} \mathcal{I}(\Omega_B) \dot{\psi}^\dagger\rangle \\ &= -(\dot{\psi}\mathcal{I}(\Omega_B))^\dagger + \mathcal{I}(\Omega_B)\dot{\psi}^\dagger \\ &= 2\mathcal{I}(\Omega_B)\dot{\psi}^\dagger \end{align} where I used $$\Omega_B^\dagger = -\Omega_B$$.

How is the $$\hat{\partial}_\psi\langle \Omega_B \hat{\mathcal{I}}(\Omega_B) \rangle$$ term calculated? This is also supposed to end up as $$2\mathcal{I}(\Omega_B)\dot{\psi}^\dagger$$.

• I'm guessing that the use of an adjoint will lead to an elegant solution for the second term, but not sure how to formulate it Nov 27, 2021 at 3:20

The second term is:

\begin{align} \hat{\partial}_\psi\langle \Omega_B \hat{\mathcal{I}}(\Omega_B) \rangle &= \hat{\partial}_\psi \langle \hat{\mathcal{I}}(\Omega_B) \Omega_B \rangle \\ &= \int d^3x\rho \sum_J a^J \langle x\wedge (x\cdot ((a_J*\partial_\psi)(-\psi^\dagger\dot{\psi} + \dot{\psi}^\dagger\psi)))\Omega_B\rangle \\ &= \int d^3x\rho \sum_J a^J \langle x\wedge (x\cdot (-a_{J+}^\dagger\dot{\psi} + \dot{\psi}^\dagger a_{J+}))\Omega_B\rangle \\ &= 2\int d^3x\rho \sum_J a^J \langle x\wedge (x\cdot (\dot{\psi}^\dagger a_{J+})) \Omega_B\rangle \\ &= 2\int d^3x\rho \sum_J a^J \left\{\langle xx\cdot(\dot{\psi}^\dagger a_{J+})\Omega_B\rangle - \underbrace{\langle x\cdot(x\cdot(\dot{\psi}^\dagger a_{J+}))\Omega_B \rangle}_{=0}\right\} \\ &= 2\int d^3x\rho \sum_J a^J \frac{1}{2}\langle x\left(x\dot{\psi}^\dagger a_{J+} - \dot{\psi}^\dagger a_{J+}x\right)\Omega_B\rangle \\ &= 2\int d^3x\rho \sum_J a^J \frac{1}{2}\langle a_{J+}\Omega_B x^2\dot{\psi}^\dagger - a_{J+}x\Omega_Bx\dot{\psi}^\dagger \rangle \\ &= 2\int d^3x\rho \sum_J a^J \frac{1}{2}\langle a_{J+}x\left( x\Omega_B - \Omega_B x)\right) \dot{\psi}^\dagger\rangle \\ &= 2 \sum_J a^J a_{J+} * \left(\int d^3x \rho x\wedge (x\cdot \Omega_B)\dot{\psi}^\dagger \right)\\ &= 2\mathcal{I}(\Omega_B)\dot{\psi}^\dagger \end{align}

It follows from the fact that $$\mathcal I$$ is its own adjoint, whence $$\hat\partial_\psi\langle\Omega_B\mathcal I(\hat\Omega_B)\rangle = \hat\partial_\psi\langle\mathcal I(\Omega_B)\hat\Omega_B\rangle = \hat\partial_\psi\langle\hat\Omega_B\mathcal I(\Omega_B)\rangle,$$ exactly the expression you already computed.

It's straightforward to show that $$\mathcal I$$ is self-adjoint if you use the left and right contractions[1] $$X_r{\rfloor}Y_s = \langle X_rY_s\rangle_{s-r},\quad X_r{\lfloor}Y_s = \langle X_rY_s\rangle_{r-s},$$ for $$X_r$$ with grade $$r$$ and $$Y_r$$ with grade $$s$$ and with the convention that the projections are zero if the grade is negative. These can then be extend by linearity to arbitrary multivectors. Alternatively, they are defined by the identities $$\langle A{\wedge} B\,C\rangle = \langle A\,B{\rfloor}C\rangle,\quad \langle A\,B{\wedge} C\rangle = \langle A{\lfloor}B\,C\rangle, \tag{*}$$ for arbitrary multivectors $$A,B,C$$.

I adopt the convention of using $$X\cdot Y$$ only when $$X$$ and $$Y$$ have the same grade.

Now, for bivectors $$A, B$$ $$A\cdot(x\wedge(x{\rfloor}B)) = (A{\lfloor}x)\cdot(x{\rfloor}B) = -(x{\rfloor}A)\cdot(x{\rfloor}B),$$ with the first equality from applying ($$*$$). It's clear from this form that $$A\cdot\mathcal I(B) = \mathcal I(A)\cdot B$$.

[1] L. Dorst, The Inner Products of Geometric Algebra, in: L. Dorst, C. Doran, J. Lasenby (Eds.), Applications of Geometric Algebra in Computer Science and Engineering, Birkhäuser, Boston, 2002: pp. 35–46.