## two definitions for division in quaternion calculus?greenspun.com : LUSENET : quaternions : One Thread |

Hello Doug,Quaternion calculus is a division algebra and every quaternion Q has an inverse 1/Q. Q1/Q2 can be Q1(1/Q2) and also (1/Q2)Q1, and Q1(1/Q2) <> (1/Q2)Q1 (not equal)

Division does not seem to be unambiguous. Or is this reasoning flawed?

Doug, I am grateful for the quality information that you published on your website. Thanks!

-- Koen van Vlaenderen (nl88494@nl.ibm.com), October 16, 2002

Hello Tom:Naturally I agree with your perspective :-) To be more technical, there are two places in physics were non-commuting come up. One is angular momentum. The cross product is key to understanding angular mo, and it is the anticommuting part of the quaternion product (the other terms are all commuting, so together, the sum in non-commuting). The commutator in quantum mechanics is another critical area. It takes a little work to do this using only complex numbers, but it happens straight out of the box with quaternions.

doug

-- Doug Sweetser (sweetser@theworld.com), November 06, 2002.

Hello Koen:Quaternion multiplication usually does not commute, so your observation is consistent with that. Order does matter. So Q1/Q2 equals Q1 * (Q2^-1), and it is not proper to suggest to rearrange the order of multiplication in the definition of division.

Unfortunately,most folks don't understand how to work with numbers that don't commute. There are a few ways to cope with the situation. First, quaternions do commute if one of the quaternions is a scalar. This was key to my (so far unsupported by a professional mathematician) definition of a derivative with respect to a quaternion variable. In the limit definition, the infinitesimal could be written on the right or left, and they are not the same. What I did was use a 2 limit process, where the 3-vector goes to zero first, then the scalar. After the 3-vector goes to zero, the infinitesimal will commute, so at that point it does not matter which side it is written on. Quite cute, and possibility very important, but I need to seduce a mathematician to that project.

Second, two quaternions that happen to point in the same direction up to a scalar multiple will commute. An important class like this are polynomials, where q q^2 = q^2 q. This is vital for the series definition of sin(q).

There is a third class one should be aware of, that flip signs if the order is reverse, Q1 Q2 = - Q2 Q1. This is not abstract at all, for example (0, 4, 0, 0) (0, 0, 5, 0) have this property because ij = -ji. This is an important step to the big picture: all quaternion products can be broken into two pairs of numbers, one pair that commutes, the other pair that anti-commutes. The pair that commutes has every property found for complex numbers - which covers a huge amount of territory. The anti-commuting set are also well understood, usually under the name of outer product.

I cannot overemphasis how important this perspective is. When mathematician say, quaternions don't commute, it is usually delivered with a somber tone, because that indicates quaternions not too useful. As the union of inner and outer products, it has the formal possibility of being more useful that either in isolation, which appears to be how the professionals deal with inner and outer products.

Thanks for the compliment on quality. The site gets 2-7,000 hits a day. I do not want to waste anyone's time, so I am always eager to address issues.

doug

-- Doug Sweetser (sweetser@alum.mit.edu), October 16, 2002.

Ok, it is possible to define subclasses, but this is equaly true for complex quaternions, such that a subclass of complex quaternions yield an unambiguous division calculus.Your example of derivatives is not quite understood, and I understand the importance of your work here. I tried to express for instance coordinate transformations in quaternions. Let coordinate quaternion X = t +ix +jy +kz be a function of X' = t' +ix' + jy' + kz' then we have X = X(X'). Any function of X, for instance A(X), where quaternion A is the electrodynamic potential, can be transformed into a function A(X(X') and I was particularly interested in dA/dt' (a total time derivative of the electrodynamic potentials).

But there isnīt an unambiquous way to calculate dA/dt'. There are two different expressions: (dX/dt')(dA/dX) or (dA/dX)/(dX/dt') so a choice has to be made.

Also I assume that dA/dX = Nabla A where Nabla = 1/dX = d/dt + id/dx + jd/dy + kd/dz By associating 1/dX with Nabla, the choise for division is X/Y = (1/Y)X

Then dA/dt' = (1/dt')dA = (1/dt')(dX/dX)dA = (1/dt')(1/dX) dX dA = * (1/dt')dX(1/dX)dA = (* division by itself commutes, because it yields 1) (1/dt')dX(1/dX)dA = dX/dt' dA/dX

The applied chain rule, or the association Nabla A = dA/dX might be wrong. The Jacobi matrix comes to mind, and Nabla A looks like a particular choice of dA/dX

Regards, Koen

-- Koen van Vlaenderen (nl88494@nl.ibm.com), October 17, 2002.

Hello Koen:Quaternions that have complex-values are not a division algebra, and I never work with them. There are a number of technical reasons for this choice, but it is a choice I have been consistent with throughout the site. We all have noticed how hugely important real and complex numbers are in math and physics. A question is why these two, what are their key properties? Briefly, they are topological algebraic fields. This means they have neighborhoods, which is key to defining calculus, and addition, subtraction, multiplication and division. The only other finite field with these properties is the quaternions. If one drops the need to be associative, one could add the octonions, but I have yet to work with them becasue I have yet to hear were they arise in physics applications.

There is also a reason in the logical foundations of quantum mechanics to require one work with a division algebra. The reason roughly is that if one has a field equation, that equation must be invertible to have an equation for the propagators. One can only do that over a division algebra such as the real, complex, and quaternion numbers. Quantum mechanics over the real numbers is boring because all the fun interference stuff does not work. Complex numbers is virtually all that is done. A fellow named Adler does do some quantum mechanics, but since there is no quaternion analysis, his attempts fall short of using quaternions consistently instead of in a restricted fashion.

Complex-valued quaternions are just another Clifford algebra. There are lots of folks who do fine work with such algebras, and I wish them luck. However, I have never heard a good reason why one should work with a particular Clifford algebra, and have not have a heard a reply to the basic requirement in quantum mechanics pointed out by Birkhoff and von Neumann in the 1930s. Until I get those two questions addressed, personally I will stick with real-valued quaternions, the division algebra.

The derivative of a quaternion with respect to a quaternion variable is entirely separate from what you are discussing. The reason is that I was talking about working on the manifold H^1, where only quaternions live, not R^4, which has four real parts. On H^1, with a function f(q, q^*, q^*1, q^*2), one can define df/dq, df/dq^*, df/dq^*1, or df/dq^*2 with my definition. Over R^4, one can define g(t, x, y, z) and dg/dt using a standard approach and respecting the order of things. There is no problem with doing quaternion calculus over R^4. The problem is with doing calculus on H^1.

The way I write quaternion operators and multiplication is with what I call the ASCII convention. A quaternion can be a letter, but inside of a (,), a small letter is a scalar, a capital letter is a 3-vector. I thought Nabla was an upside-down triangle, so only dealt with x, y, z, and not the scalar. Anyway, I call the upside-down 3-vector operator Del, and (d/dt, Del) I call Box. So...

Box A = (d/dt, Del) (a, A) = (da/dt - Del.A, dA/dt + Dela + DelxA)

One might need to work with A Box, which flips the sign of the curl:

A Box = (d/dt, Del) (a, A) = (da/dt - Del.A, dA/dt + Dela - DelxA)

Fortunately, this does not come up that often.

Let's see how the chair rule applies to a function that is the product of two other functions: A = F G. One varies F, keeping G constant, then varies G keeping F constant. Since F and G do not commute, one must leave the order of F first, G second, in tact during the process:

Box A = Box (F G) = (Box F) G + F (Box G)

= ((d/dt, Del) (f, F)) G + F ((d/dt, Del) (g, G))

= (df/dt - Del.F, dF/dt + Delf + DelxF) (g, G) + (f, F) (dg/dt - Del.G, dG/dt + Delg + DelxG)

= (df/dt g - Del.F g - dF/dt.G - Delf.G - DelxF.G, df/dt G - Del.F G + dF/dt g + Delf g + DelxF g + dF/dtxG + DelfxG + DelxFxG)

+ (f dg/dt - f Del.G - F.dG/dt - F.Delg - F.DelxG, F dg/dt - F Del.G + f dG/dt + f Delg + f DelxG + FxdG/dt + FxDelg + FxDelxG)

= (df/dt g - Del.F g - dF/dt.G - Delf.G - DelxF.G + f dg/dt - f Del.G - F.dG/dt - F.Delg - F.DelxG,

df/dt G - Del.F G + dF/dt g + Delf g + DelxF g + dF/dtxG + DelfxG + DelxFxG + F dg/dt - F Del.G + f dG/dt + f Delg + f DelxG + FxdG/dt + FxDelg + FxDelxG)

The chain rule applied to a quaternion product has 4^3 = 64 parts to it, which is why it looks so complicated. Fortunately I didn't write it out in individual components!

If I had to work with coordinate transformations, I would probably try and figure out covariant derivatives and the Christoffel symbols. Or I would work in H^1, where no coordinate choice is made, then map those results to a particular coordinate system in R^4. Any way it is done gets very confusing.

doug

-- Doug Sweetser (Sweetser@alum.mit.edu), October 17, 2002.

The first answer raises some perfectly correct mathematical points to do with quaterions not commuting. However, it is incorrect to assume that not commuting dooms them to the waste-basket of mathematical oddities. In my opinion, operations that commute are the quirks. A nice quirk, that can be useful, but still a quirk. It only seems odd beacause you spent at least the first 17 years of your life not meeting no-commuting systems, and so it seems odd. (matrices also don't commute, and trying doing computer graphics without them). Whats really, really important is that quaternions associate under multiplication (ie a(bc)=(ab)c).

-- Tom Willis (thomas.willis@durham.ac.uk), November 06, 2002.

Alfred North Whitehead wrote that a good notation does half your thinking for you. Fortunately, the standard ASCII keyboard supplies a way to denote two distinct ratios arising from the non-commutativity of division of quaternions. Just use the slash (/) and backslash (\). Thus Q1 /Q2 would be distinct from Q2 \Q1.Incidentally, commutativity does not require that Q1 and Q2 both be scalar. If the vector parts of Q1 and Q2 are co-directional, their divisions will commute.

Best regards,

Matthew

-- Matthew McCann PE (mmccann@franciscan.edu), March 06, 2004.

Hello Matthew:I will continue to write Q2\Q1 as Q1/Q2 to keep with the tradition of left to right information flow in my native language of English. An ASCII notation issue for quaternions that I use is for scalars to be lower case, 3-vectors as upper case, Q1 = (a, B), Q2 = (c, D), so Q1/Q2 = (ac + B.D, -aD + Bc - BxD)/(cc + D.D).

The point on co-directionality is HUGE. Think of everything proven for complex numbers. All that stuff applies to co-directional quaternions. The riddle that remains is how to include the stuff that anticommutes. I have used this trick many times: understand the complex number case, figure out how to add in the antisymmetric product.

doug

-- Douglas Sweetser (sweetser@alum.mit.edu), March 07, 2004.

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