For the 3D case, it is described in Doran & Lasenby (2003) Geometric Algebra for Physicists (and also in my book https://www.springer.com/gp/book/9783030017552).

The 4D case:

First, a four vector is a pseudoscalar in 4D, its reverse does not change a sign, therefore it cannot be a part of a rotor (it can be a part of a 4D spinor).

Second, given the vectors a and b, the element B = b/a (or a/b) will do the desired rotation in all dimensions (of course, we have to exclude degenerate signatures), but in the operator form (acting from the left). From Ba = (b/a)a = b, using the fact that Ba = -aB and the exponential form of B, we can get your rotor. Note that, apart from real scalar, B contains only bivectors, in any dimension.

I hope that this helps.

Josipovic’s solution looks correct, so I will just explain what he is doing In more detail. If my guess is correct, b/a means $\vec b \vec a^{-1}$, where $\vec a^{-1} = \vec a/|\vec a|^2$. This makes $B = \vec b /\vec a^{-1} = \vec b\vec a/|\vec a|^2$$. Thus, $B\vec a = \vec b(\vec a\vec a)/|\vec a|^2 = \vec b$.

Quirino M Sugon Jr.

Of course, a/b means ab^(-1).

Thank you Miroslav Josipovic for your answer. Also thanks to Quirino

Interesting!!! If I've well understood, you are not proposing a sandwich operator, but a left operator that applies to a in order to get b in any dimension.

At first it solves my question, so I need to elaborate more on this.

The final goal here is to find the rotor that allow the orthonormal canonical basis {e} (let's say in 4D) to become another orthonormal basis {f}. Both {e} and {f} are already known.

This way, any arbitrary vector a = v_1*e1+v_2*e1+v_3*e3+v_4*e4 can be viewed as a=v_a*f1+v_b*f2+v_c*f3+v_d*f4 in the new basis (so we rotate the basis but the vector is the same). Is there a general expression to find this "global" rotor?

Note: I have your book, and, yes, found on page 41 this demonstration for 3D

Francisco G. Montoya

I started with the operator form, but there is an additional comment. The element B spans a plane and this plane contains the vectors a and b, which means that B anti-commutes with a and b. Now, you can find the exp. form, B = exp(fi b), b anti-commutes with a and b, whence

Ba = exp(fi b)a = exp(fi b/2)exp(fi b/2)a = exp(fi b/2) a exp(-fi b/2).

Thus, you have the sandwich form (i.e., you have the rotor).

However, if you want to find a rotor to rotate a base to another base in higher dimensions, I think the result can be obtained similarly as for the 3D case, but this should be checked. For example, we can always find number of the base vectors that commute with a bivector, in any dimension.

Yes, dealing with vectors that lie on a plane, and perform rotations using this plane is the easy part :P. It remains to discover the general form in n-dimensions!!

Hello everyone,

I share with you the following paper which covers the subject fairly well to help answer this interesting topic:

• J. Denker, " How to Represent Multi-Dimensional Rotations, Including Boosts ", 2003 - https://www.av8n.com/physics/rotations.pdf

Both sandwich and left/right operators for modeling rotors are discussed. The rotors may be composed to form other rotors.

Regards

Thank you Stéphane Breton . I've already read this paper yesterday, specially page 15 and onwards. Unfortunately, I could not derive the general expression I'm looking for :(

Also, it contradicts Miroslav with:

"In four dimensions, a rotor has eight components, namely one scalar component, six bivector components, and one quadvector component"

Francisco G. Montoya

The formula from my book is valid generally, only the coefficients in the quantity X must be generalized (I just found the general formula). The solution for Euclidean v. spaces is easy (just count the number of bivectors and the number of the base unit vectors that commute with bivectors). For example, to get the solution in 4D, just remove 1 from X. Signatures (p, q) only require more attention in the calculations, nothing is fundamentally different.

Enjoy!

Impressive!!! You've provided the right solution. I already checked with a real example in 4D and works!!.

I will study the general formula based on your comments.

Thank you very much Miroslav Josipovic

Francisco G. Montoya

You are welcome!

Of course, I would like to see your solution, just to compare. I have no detailed solution for negative signatures (I have no interest to do that), so, you can calculate for that case.

For interested readers, the solution drawn by Miroslav Josipovic is fully understood by combining sections §1.9.5, §1.9.9 and §1.9.11 from his book Book Geometric Multiplication of Vectors: An Introduction to Geom...

Thank you Miroslav Josipovic.

Stéphane Breton

you are welcome!

I prepared this document to prove the provided solution by Miroslav Josipovic in 4 dimensions. I already checked with numerical values.

Francisco G. Montoya

That is it. Your footnote 1 is ok, now, just write the formula for the general case.

I'm interested in an Euclidean vector space, so for a Cl(p,0,0), the general formula is:

If p is odd --> R=\frac{ 1+\mu_p\sigma_p }{\| 1+\mu_p\sigma_p \| }

if p is even --> R=\frac{ \mu_p\sigma_p }{\| \mu_p\sigma_p \| }

Nota that Einstein notation has been used.

Francisco G. Montoya

Can you write it in pdf, I have problems to see the solution?

I think that the generalization in my previous answer is wrong...

I have also problems to compute $\sum e_k R^{\dagger} e_k$ for any arbitrary dimension.

ok, I think i have it... please, correct me if I'm wrong:

$$ R= \frac{ A }{ \| A \| } $$

where

$$ A = \sum_p \mu_p\sigma_p + (4-p) $$

Francisco G. Montoya

I cannot read your formulas, but if this is Sigma + n-4 (n is a dimension), then it is correct. Simple, each bivector from the Clifford basis have a form eiej, which means that it anti-commutes with exactly 2 vectors (ei and ej).

There are problems with negative signatures. First, there is no simple formula for the Sigma, because the result depends on the type of a basis bivector (check directly). Note also that "mixed" bivectors (with ei of positive and ej of negative signature) square to 1 and represent boosts, not ordinary rotations. Therefore, this approach will not give us the result. However, there is a possibility to write the "rotor" R as a product of a boost and a pure rotor. It is easy to find the reverse, try to implement commutativity rules, etc. However, I am not interested to do this. We can formulate all physics without strange and un-intuitive negative signatures.

I updated the document with correct results.

Francisco G. Montoya

In your formula (7) you should replace 4 - k by 4 - n. Of course, the relations (5) and (6) should also be corrected.

thank you Miroslav Josipovic , it was a typo that I've corrected. I've updated the pdf.

Francisco G. Montoya

You can now share the result with the group

https://www.linkedin.com/groups/8278281/

:)

I came across this paper by Leo Dorst by chance!!! It is more mathematically profound than mine, but essentially, it has the same goal.

Article Determining a versor in nD geometric algebra from the known ...

you all should consider joining the discord for GA. lot of experts there

the link can be found on this page

https://bivector.net/index.html