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 Apr 14 comment A Pure Maths Approach to Thinking About Vectors I'm curious: you've tried to develop vectors and their algebra through a semi-intuitive process. What are your thoughts on this compared to, say, the traditional axiomatic development of linear algebra? Apr 12 comment Coordinate free Geometric Algebra vs. Linear Algebra I mean, since the clifford algebra can be described as a quotient of the tensor algebra, any products can be written in terms of the tensor product, so I'm not sure what you're asking for exists. Apr 7 comment Scalar curvature of metric? Is there a factor of $f$ or $\sqrt{f}$ missing? It does not seem clear to me that $e^0$ is unit in this way. Mar 31 answered Show that every rotation in $\mathbb{R^3}$ can be written as the product of two rotations of order 2. Mar 26 answered Why is the pseudoscalar called pseudoscalar in Geometric Algebra Feb 16 answered Question about cross product of images of linear transformation Feb 6 comment How does $\frac{\sin\theta}{\cos\theta}$ become $\frac{y}{x}$ @Michael You just nest them. e.g. \frac{ \frac{a}{b} }{c} gives $\frac{\frac{a}{b} }{c}$. Feb 6 comment How does $\frac{\sin\theta}{\cos\theta}$ become $\frac{y}{x}$ You might want to try \frac{\sin \theta}{\cos \theta} for fractions, and similarly elsewhere. Also, \sqrt{15} will look better than \sqrt 15. Feb 6 comment Show if $A^TA = I$ and $\det A = 1$ , then $A$ is a rotational matrix What happens if you substitute $\phi = -\theta$? Jan 30 comment Signature of the inner product. @user48672 Ah, well, I'm going to treat quaternions as even graded elements of the the 3d clifford algebra anyway, not as vectors in their own right, so any problem there, if it does exist, is avoided. Jan 30 comment Signature of the inner product. @user48672 I think the Clifford conjugate may be necessary when dimensions with negative signature are involved. For instance, you might notice that, for odd $k$, $k$-blades are not assigned positive norms when the metric signature is $(-1, -1, \ldots)$. I suspect using grade involution (which together with reversion is conjugation) fixes this problem. Jan 23 comment Metric tensor for just one index $Z^{ii} Z_i$ is not a tensor. It's gibberish. Jan 11 comment What good is the Commutator product? @Rodrigo The wedge product of a bivector and a $k$-vector is grade $k+2$. But unlike the commutator, this wedge is symmetric for a bivector. Jan 10 comment How many components of an antisymmetric rank five tensor on $\mathbb{R}^5$ are independent? The single independent component is multiplied by $\det R$ (does not need to be orthogonal). Jan 8 comment Contravariant vector example with polar coordinates No, you're still conflating positions with vectors. As an example, it would be entirely well-formed for me to say I want to consider the vector $(a^1, a^2) = (1, 1)$ at the point $(u^1, u^2) = (3, 5)$. That's because the vector need not have any relationship to the coordinates of position. Jan 7 comment Contravariant vector example with polar coordinates You haven't provided enough information to completely determine the answer. You need (1) a tuple $(u^1, u^2)$ that describes a point in Cartesian coordinates, (2) a tuple $(a^1, a^2)$ that describes a direction as a linear combination of the Cartesian basis vectors, and (3) a function that transforms the Cartesian coordinates $(u^1, u^2)$ to the polar coordinates $(v^1, v^2)$, which you already have. Once you have all these pieces of information, you can find the tuple $(b^1, b^2)$. Jan 4 comment Rotors/Quaternions: double reflection question I mean, your understanding of rotors seems to be correct, but I do not think that is getting you any closer to understanding the construction of building a rotation from two reflections (which is not unique to clifford algebra, but it is crucial to deriving the algebraic form of a rotor from that of a single reflection). Dec 13 comment Projection on a Non-Orthonormal Subspace Naturally the subspace doesn't change whether you use an orthonormal basis or not. The computation of the orthogonal projection may be more complicated, however, than the convenient case of using an orthonormal basis. Dec 9 revised Closed formula to transform roll-pitch-yaw angles into Axis-angle representation added cyclic qualifier Dec 9 comment Closed formula to transform roll-pitch-yaw angles into Axis-angle representation Sorry, these are cyclic permutations, and so there are only three cases to consider (one for each component of the axis vector).