# Space-time split in Geometric Algebra

I was performing some calculations with Geometric Algebra today and I've found myself stuck with a simple operation that I don't know how to answer.

Considering that you have the usual vector derivative $$\nabla = \gamma^\mu \partial_\mu = \gamma_0(\partial_0 + \vec{\nabla}) =(\partial_0 - \vec{\nabla})\gamma_0,\tag{1}$$ and performing the scalar multiplication by $$\gamma_0$$. One one side one would have

$$\gamma_0 \cdot \nabla = \partial_0\tag{2}$$

but on the other hand, if one performs a time-split before the multiplication

$$\gamma_0 \cdot \nabla = \gamma_0 \cdot \gamma_0(\partial_0 + \vec{\nabla}) = \partial_0 + \vec{\nabla}\tag{3}$$

Both results don't seem compatible to be, so I would like to ask. What I am missing?

• @NDewolf It comes from the anticommutativity of the $\gamma_\mu$. Because $\vec{\nabla} = \gamma_0 \wedge \gamma_i \partial_i$ Jan 25 at 17:03

In the first case, you are considering the direct application of the inner product of $$\gamma_0$$ and $$\nabla$$, $$\gamma_0\cdot\nabla=\gamma_0\cdot\gamma_\mu\partial_\mu=\partial_0.\tag{1}$$ However, in the second case, you are inserting $$\gamma_0^2=1$$ in between the two terms, then multiplying this to the left and to the right: \begin{align} \gamma_0\cdot\nabla &= \gamma_0\cdot\left(\gamma_0^2\right)\nabla \\ &= \left(\gamma_0\cdot\gamma_0\right)\left(\gamma_0\nabla\right) \tag{2}\\ &= \gamma_0\nabla \end{align} And this is different from Equation (1) here because you've changed the order of products between the first and second lines. Instead of dotting the frame with the gradient (Eq 1 here), you've dotted frame with itself and multiplied that to the spacetime split gradient (Eq 2).
So it seems to me that all that you have discovered is that, $$a\cdot bc\equiv\left(a\cdot b\right)c\neq a\cdot\left(bc\right)$$ which is fine, since product orders matter.