How to compute if a multivector inverse exists in Clifford Algebra Suppose we have a 4 dimension positive signature clifford algebra. In Calculating the inverse of a multivector and Inverse of a general nonfactorizable multivector, the inverse of a multivector is presented as a solution when vectors/bivectors are present 
$B^{-1} = \frac{B^\dagger}{B B^\dagger}$
but the above is not true for any multivector. For example, how to know if  
$(1+e_{1234})^{-1}$ 
exists and how to compute it?
 A: Naively speaking, the existence of inverses will depend on the signature $(p,q)$ of the quadratic space $\mathbb{R}^{p,q}=(\mathbb{R}^{p+q},g)$, in which for an orthonormal basis $\{e_i\}_{i=1}^{n=p+q}$ and $v=\sum v^ie_i$ we have
$$g(v,v)=(v^1)^2+(v^2)^2+\cdots+(v^p)^2-(v^{p+1})^2-\cdots-(v^{p+q})^2.$$
For your example, notice that if $(e_{1234})^2=-1$, then
$$(1+e_{1234})\frac{1}{2}(1-e_{1234})=1,$$
which means that $(1+e_{1234})^{-1}=\frac{1}{2}(1-e_{1234})$. 
Now, if $(e_{1234})^2=1$, then there is no inverse for $(1+e_{1234})$, which is due to the fact that $x\overline{x}=0$. More specifically, one can derive conditions for which there are inverses for the cases where $p+q=n\leq 5$. The discovery of new faster methods for higher dimensions, which do not depend on the signature is a problem still under development as of today. As an example, we could cite this. 
A: This paper by D. S. Shirokov (2021) claims to give a basis-free formula using only basic operations and various involutions on multivectors. Preprint.
