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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).
Dec
9
answered Closed formula to transform roll-pitch-yaw angles into Axis-angle representation
Dec
9
comment Closed formula to transform roll-pitch-yaw angles into Axis-angle representation
Your question is unclear; you used the angle $\phi$ twice in breaking down the rotation into three separate rotations. Is $\psi$ associated with the $x$ rotation or the $z$ rotation?
Dec
3
comment The “inverse” of $\nabla\times$ operator
A note: one can use clifford algebra to define $\nabla A$ even for $A$ a vector, and in turn, there is a Green's function for this "vector gradient". The curl forms part, but not all, of the components of this vector gradient, but just as expected, the missing information is the divergence.
Dec
2
comment Why is this a fake proof?
Differential forms would typically be introduced in the context of differential geometry--the study of curves, surfaces, and the like, that are smooth enough to be described by differentiable functions. Differential forms are very convenient for integration on such regions. - My point of view on "small changes in $x$" is that you'll stop thinking about small intervals and start thinking about weighted directions. Here, that direction is $C'$, the tangent direction to the level curve. Anything you can do with a differential could be done with tangent directions instead, using vectors.
Dec
2
answered Why is this a fake proof?
Dec
1
comment How to rotate the origin of rotation of a quaternion
When you say origin that makes it sound like a translation is happening--that the central point of the object is moved. Is that what you mean? Or so you say origin and mean a set of axes?
Dec
1
comment How to rotate the origin of rotation of a quaternion
It sounds like you want to rotate with respect to the axes of some previously-rotated body. Is that the case?
Nov
30
comment How to Prove $(A \times B) \otimes C + (B \times C) \otimes A + (C \times A) \otimes B = [(A \times B) \cdot C] \bf{I}$?
A clifford algebra uses a vector multiply operation--the "clifford product" that I have used here. This operation obeys $mm = |m|^2$ and $ijk = (ij)k = i(jk)$. The resulting algebra has as its elements multivectors, some of which correspond to planes or volumes instead of lines (as vectors do). - The symbol $\wedge$ here should be read as "wedge". The wedge product of two vectors is a 2-vector or bivector and corresponds to a plane. The wedge product of three vectors corresponds to a volume. These can be written in terms of the clifford product. - Note that $i^a m^b \delta_{ab} = i \cdot m$.