First I will play the old gauge symmetry game repeated everywhere for EM. I will then teach that old dog a new trick that ends up with the same bone. I will then struggle a little over the lingo to describe what I am interested in, the center of the quaternion group. I will then show how a new new trick, the new trick shifted to the right just enough to work, will lead to a gauge symmetry proof.

Let's review gauge symmetry in EM. Start with an arbitrary scalar function f. Take the conjugate of the 4-derivative operator and apply it to f:

$\left( \frac{\partial}{\partial t}, \vec{\nabla} \right)^* \times (f, \vec{0}) = \left( \frac{\partial f}{\partial t}, -\vec{\nabla} f \right)$

No one writes it this way. The contempt of Lord Kelvin toward quaternions is alive and well today. We can expect anyone replying to this blog to write things by components, playing their "same up to an isomorphism" card.

I would rather be tied into math ideas. A quaternion conjugate is an involution because q** = q. The conjugate is an antiautomorphism because (a b)* = b* a*. After I wrote the line which I was certain was true for complex numbers, I wondered if it was still true for quaternions that don't commute? Yup. A quaternion product has two parts: the part that commutes and the part that anti-commutes (the cross product). The only part that matters here is the cross product. The (a b)* term flips the sign of the cross product once. The b* a* flips the sign three times which is the same as flipping it once. Nice. The gauge field is using an involutive antiautomorphism which sounds fancy.

Add the gauge field to the 4-potential, and see how they effect the E and B fields:

\begin{align*} E_i &= - \frac{\partial A_i}{\partial t} - \frac{\partial \phi}{\partial x_i}\\ E'_i &= - \frac{\partial A_i}{\partial t} + \frac{\partial}{\partial t} \frac{\partial f}{\partial x_i} - \frac{\partial \phi}{\partial x_i} - \frac{\partial}{\partial x_i} \frac{\partial f}{\partial t} = E_i\\ \\ B_i &= \frac{\partial A_k}{\partial x_j} - \frac{\partial A_j}{\partial x_k}\\ B'_i &= \frac{\partial A_k}{\partial x_j} - \frac{\partial}{\partial x_j} \frac{\partial f}{\partial x_k} - \frac{\partial A_j}{\partial x_k} + \frac{\partial}{\partial x_k} \frac{\partial f}{\partial x_j} = B_i\\ \end{align*}

Adding the gauge field to the potential leaves the E and B fields unaltered, so this is a gauge symmetry. Plug these fields into an action, and where there is a symmetry, there is a conserved charge. In this case, gauge symmetry leads to electric charge conservation.

In physics as in math, there are sometimes multiple ways to demonstrate the same darn thing. As an example, consider demonstrating that a force is conservative. I know of at least 4 ways to demonstrate it: the curl is zero, there is a scalar potential function that can serve as the force, and two statements involving line integrals. There are at least 5 ways to show a complex-valued function is analytic. Let's see if we can find another way to demonstrate gauge symmetry for the E and B fields of EM.

In complex analysis, one can work on the manifold $\inline \mathbb{C}^1$using a complex number z and its conjugate z*. A similar effort comes up short with quaternion manifold since there are four degrees of freedom. In effect, one needs more conjugates for a quaternion manifold. The conjugate we are all familiar with keeps the sign of the real number the same but flips the imaginary terms (one for a complex number, three for quaternions the union of three complex numbers). I call them the first, second, and third conjugates those that flip all but that imaginary. Here is how the first conjugate can be defined:

\begin{align*} (i q i)^* &= (i (t, x, y, z) i)^* \\ &= i (t, -x, -y, -z) i \\ &= (x, t, z, -y) \\ &= (-t, x, -y, -z) \\ &\equiv q^{*1} \end{align*}

These are still involutions: do two in a row, and you get back to go. But are they antiautomorphisms, does (a b)* = b* a*? No. What is true is:

$(a \times b)^{*1} = -b^{*1} \times a^{*1}$

The right side has one pair of i's while the left has 4 i's, two of which meet and generate the extra factor of minus one. Let's call that extra factor of -1 a twist. The first, second, and third conjugates are twisted involutive antiautomorphisms.

This can be generalized using basis vectors:

$(e_i \times q \times e_i)^* = (-t, x, -y, -z)$

The first method used to show gauge symmetry can be rewritten like so:

$\left( e_0 \left( \frac{\partial}{\partial t}, \vec{\nabla} \right) e_0 \right)^* \times (f, \vec{0}) = \left( \frac{\partial f}{\partial t}, -\vec{\nabla} f \right)$

This is not much of a rewrite since the e0's are just 1's. It does provide the structure for a variation:

$\left( e_i \left( \frac{\partial}{\partial t}, \vec{\nabla} \right) e_i \right)^* \times (f, \vec{0}) = \left( -\frac{\partial f}{\partial t}, +\frac{\partial f}{\partial x_i}, -\frac{\partial f}{\partial x_j}, -\frac{\partial f}{\partial x_k} \right)$

The terms that will go into the B field are exactly the same. The two terms that go into the E field both flip signs, but that will not alter the cancellation. This gauge field can be used to prove gauge symmetry of the E and B fields of EM.

This alternative has probably been done before, but it is super obscure. The overall signs don't matter since they will cancel anyway. The gauge terms going into the E field have to have opposite signs, while those in the B field need the same. One is free to pick and choose after that.

What is important in this exercise is that the way to create this funky alternative gauge field is precisely defined.

Now it is time for a math interlude. I was reading up on the quaternion group at a site dedicated to the details of group theory. The quaternion group has eight elements

(1, 0, 0, 0)

(-1, 0, 0, 0)

(0, 1, 0, 0)

(0, -1, 0, 0)

(0, 0, 1, 0)

(0, 0, -1, 0)

(0, 0, 0, 1)

(0, 0, 0, -1)

The center of the quaternion group is the Klein 4-group.  Why is this so? The center of the group - the parts that commute with all other members of the group - are the first two:

(1, 0, 0, 0)

(-1, 0, 0, 0)

This pair is known as the cyclic group Z2. There are four basis coordinates, so one needs the Klein 4-group the direct product of Z2xZ2.

Using the center of the quaternion group sounds more solid than I just decided to use this rule for forming a product. But here is the magic of trying to do things with more math jargon. I was doing my random reading about the Klein 4-Group, so popular in some circles it gets the name V. Down at the bottom of the page under "Other Properties", they said the following:

$V$ is the symmetry group of the Riemannian curvature tensor.

That jumped off the proverbial screen. I am not going to make too much about it, but it was a surprising link.

I have used quaternion products to derive the Maxwell equations. Once the action is written, everything else that follows is standard. What happens if instead of the quaternion product, one uses the center of the quaternion group to form the field? That is the issue I have explored.

As readers of this blog know, there is a technical problem with using the Klein 4-group to form the product: the field is not invariant under the gauge transformations discussed so far.

Calculate what happens if one replaces the quaternion product with the Klein 4-group product:

$vec(\nabla \boxtimes A) = (0, \frac{\partial A}{\partial t} + \nabla \phi + \nabla \otimes A)$

The otimes is a symbol for the symmetric curl, one with all positive signs. This has the standard electromagnetic field -E and a term with a symmetric curl. A symmetric curl is not going to be invariant under a rotation. I recall one commenter claiming that all laws of physics had to be invariant under rotation. If that was true then everything would need to be shaped like a sphere. Tell that to a skipping stone. In this blog, I want to keep a focus on gauge symmetry.

Since all the signs are the same, the gauge field must have the pattern that every other term has the opposite sign. Instead of three minus signs that appear in either of EM gauge fields, two and only two sign flips are required here. The way to get the right number of minus signs is to multiply on both sides by the basis vector, but skip taking the conjugate:

$(e_i \times q \times e_i) &= (-t, -x, y, z)$

This has the right count of minus signs, but not the right distribution: the x needs to be positive while either the y or z needs to be negative for the first gauge field. Just shift the basis vector over by one:

$(e_{i + 1 \% 3} \times q \times e_{i + 1 \% 3}) &= (-t, x, -y, z)$

[technical detail: let the indexes for the basis vectors be i=0, j=1, k=2. The % is a symbol for the modulo operator, the amount left over after dividing by 3 in this case. For ek, k+1 % 3 would result in ei.] This operation is still an involution which I choose to symbolize as *2* (the conjugate of the second conjugate):

$(q^{*2*})^{*2*} = q$

It is not an antiautomorphism, nor is it a twisted antiautomorphism. It is a third animal, a twisted automorphism:

$(a \times b)^{*2*} = -a^{*2*} \times b^{*2*}$

I prove these to myself numerically. Apply this twisted autormorphism to generate a gauge field:

$\left( e_{i+1 \% 3} \left( \frac{\partial}{\partial t}, \vec{\nabla} \right) e_{i+1 \% 3} \right) \times (f, \vec{0}) = \left( -\frac{\partial f}{\partial t}, +\frac{\partial f}{\partial x_i}, -\frac{\partial f}{\partial x_j}, \frac{\partial f}{\partial x_k} \right)$

It is simple enough to show neither the field Ei nor bi would be altered by adding in this gauge field.

The field I worked with more extensively was not quite this one. Maxwell's equations have stood the test of time in physics. The equations feel complete to me. I was interested in fields that used the Klein 4-group - the center of the quaternion group - for the product but were distinct from the E and B fields of EM. That requires a conjugate operator like so:

$vec(\nabla^* \boxtimes A) = (0, \frac{\partial A}{\partial t} - \nabla \phi - \nabla \otimes A)$

Being complete is an important driving factor in physics. The Maxwell equations, good and grand and successful as they have been, only has the sum of the time derivative of A and the gradient of phi. There is only the difference of two terms going into a curl. With this field, Nature may have access to two different players, namely the difference between the time derivative of A and the gradient of phi, as well as the sum of two terms going into a curl.

In early November of 2011, I retracted a proposal I had for gravity because I could see no way to prove that the field was invariant under a gauge transformation. The first two terms, the time derivative of A minus gradient of phi need terms that have the same sign, while the symmetric curl needs two different signs. A simple conjugate does not do the trick. One needs three minus signs again. The shift trick is required to get a signs in the right spots:

$(e_{i + 1 \% 3} \times q \times e_{i + 1 \% 3})^* &= (-t, -x, y, -z)$

This is a twisted involutive antiautomorphism like before, the only difference being the shift. Use it to create a gauge field:

$\left( e_{i+1\%3} \left( \frac{\partial}{\partial t}, \vec{\nabla} \right) e_{i+1\%3} \right)^* \times (f, \vec{0}) = \left( -\frac{\partial f}{\partial t}, -\frac{\partial f}{\partial x_i}, +\frac{\partial f}{\partial x_j}, -\frac{\partial f}{\partial x_k} \right)$

Add this gauge field to the symmetric fields:

\begin{align*} e_i &= \frac{\partial A_i}{\partial t} - \frac{\partial \phi}{\partial x_i}\\ e'_i &= \frac{\partial A_i}{\partial t} - \frac{\partial}{\partial t} \frac{\partial f}{\partial x_i} - \frac{\partial \phi}{\partial x_i} + \frac{\partial}{\partial x_i} \frac{\partial f}{\partial t} = e_i\\ \\ b_i &= \frac{\partial A_k}{\partial x_j} + \frac{\partial A_j}{\partial x_k}\\ b'_i &= \frac{\partial A_k}{\partial x_j} + \frac{\partial}{\partial x_j} \frac{\partial f}{\partial x_k} + \frac{\partial A_j}{\partial x_k} - \frac{\partial}{\partial x_k} \frac{\partial f}{\partial x_j} = b_i \end{align*}

[technical note: there is a way to write the gauge field used here with the Klein 4-group product which I left as a snarky puzzle.]

I now know how to prove the fields e and b are invariant under a gauge transformation. The proof has the same structure as for EM, but a different gauge field.

I am comfortable with the math done here. The games are not hard, using different combinations of basis vectors and conjugate operators. The proofs are all similar: take this gauge field and add it to that field, the field is not altered, therefore it is has gauge invariance. Where there is gauge invariance, there will be a conservation law if the field is put into an action. For the symmetric e and b fields, the conserved charge is necessarily different from the one for EM because the gauge field is different. Nice.

I am less comfortable with the words to describe the math. People are so familiar with gauge symmetry as it appears in EM. I suspect there will be people who impose wiki-law which says if it is not on wikipedia, it does not exist. If I bring up automorphisms of any type, such lingo does not connect with a component approach. Let me save the "do it by components" crowd some effort and write the three gauge fields used in the last example out explicitly:

\begin{align*} f_i = \left( -\frac{\partial f}{\partial t}, -\frac{\partial f}{\partial x_i}, +\frac{\partial f}{\partial x_j}, -\frac{\partial f}{\partial x_k} \right) \\ \\ f_j = \left( -\frac{\partial f}{\partial t}, -\frac{\partial f}{\partial x_i}, -\frac{\partial f}{\partial x_j}, +\frac{\partial f}{\partial x_k} \right) \\ \\ f_k = \left( -\frac{\partial f}{\partial t}, +\frac{\partial f}{\partial x_i}, -\frac{\partial f}{\partial x_j}, -\frac{\partial f}{\partial x_k} \right) \end{align*}

I can imagine someone complaining that I chose the signs just to show the e and b fields were invariant under a gauge transformation. I am guilty as charged. The same "crime" happens in EM. There are the same number of pluses and minuses in this gauge field as the one for EM, but they line up differently.

Now I can see why I couldn't spot the right gauge field: it was not obvious. I can appreciate that someone toying with the difference of the time derivative of A and the gradient of phi would give up once they spotted the gauge symmetry issue. I wrote a retraction :-) We will see if the skeptical critics think this is a workable solution to the gauge issue (and the gauge issue only).

One good thing about writing blogs is it points out the obvious: there is a huge world of automorphisms out there.

Doug

Snarky puzzle:

Three odd gauge fields used in this blog used a quaternion product.

\begin{align*} \left( e_i \left( \frac{\partial}{\partial t}, \vec{\nabla} \right) e_i \right)^* \times (f, \vec{0}) &= \left( -\frac{\partial f}{\partial t}, +\frac{\partial f}{\partial x_i}, -\frac{\partial f}{\partial x_j}, -\frac{\partial f}{\partial x_k} \right) \\ \\ \left( e_{i+1 \% 3} \left( \frac{\partial}{\partial t}, \vec{\nabla} \right) e_{i+1 \% 3} \right) \times (f, \vec{0}) &= \left( -\frac{\partial f}{\partial t}, +\frac{\partial f}{\partial x_i}, -\frac{\partial f}{\partial x_j}, \frac{\partial f}{\partial x_k} \right) \\ \\ \left( e_{i+1\%3} \left( \frac{\partial}{\partial t}, \vec{\nabla} \right) e_{i+1\%3} \right)^* \times (f, \vec{0}) &= \left( -\frac{\partial f}{\partial t}, -\frac{\partial f}{\partial x_i}, +\frac{\partial f}{\partial x_j}, -\frac{\partial f}{\partial x_k} \right) \end{align*}

Redo them with the Klein 4-group product rule for multiplication.

Next week: Gravity breaking spherical symmetry

Golden Rule Policy: If a comment is not consistent with the golden rule, it will be moved to its own blog, a collection of Golden Rule Breakers.

The Golden Rule or ethic of reciprocity is a maxim,[2] ethical code, or morality[3] that essentially states either of the following:
• (Positive form): One should treat others as one would like others to treat oneself.[2]
• (Negative/prohibitive form, also called The Silver Rule): One should not treat others in ways that one would not like to be treated.

This concept describes a "reciprocal" or "two-way" relationship between one's self and others that involves both sides equally and in a mutual fashion

I will post a blog that once I have five. This includes comments like "I will read your work no more" which is a perfectly fine conclusion to make, but will not facilitate the discussion at hand. In a tit-for-tat style, I will reply back, and keep the comments frozen. To quote the motto of "The Shame Report", "I didn't put you on this list, you put you on this list." Not surprisingly, one AnnoyedRead inspired this idea.