http://camoo.freeshell.org/25.8.pdf" [Broken](adsbygoogle = window.adsbygoogle || []).push({});

Laura

Latex source below. I won't be changing this if I edit the file, it's just for convenience if you want to grab latex code.

n sec. 25.8, he

says "recall the dual $^\ast F$ of the Maxwell tensor F. We could imagine

a 'dual' U(1) gauge connection that has $^\ast F$ as its bundle curvature"

and then he says there's a problem with using the dual curvature of a

nonabelian gauge. First, though, what happens if you try to dualize the

abelian electromagnetic gauge?

If you have the Maxwell tensor $F_{ab}$, you can recover a vector

potential $A$ from it: $$A_b(\vec{x}) = \int_0^1 {uF_{ab}x^adu}$$

Similarly, if you take the Hodge dual $^\ast F_{ab}$ of $F_{ab}$, you can

define a gauge potential from it using a potential $Z$ derived from $^\ast

F_{ab}$ $$Z_b(\vec{x}) = \int_0^1 {u^\ast F_{ab}x^adu}$$

This only works if $d^\ast F_{ab} = 0$, i.e. the charge-current vector

$J=0$. It's an application of the Poincare' lemma, which says that in a

contractible (small, topologically simple) region, a form $F$ with $dF=0$

is the exterior derivative of another form.

I totally wracked my brains about it and I couldn't

see how you could use $^\ast F$ as a gauge curvature \emph{unless} $d^\ast

F=0$.

And after I thought about it some \underline{more} I figured that in the

application of a dual gauge connection, the field probably \emph{would} be

source-free, because the gauge connection's applied to quantum

wavefunctions and when you're at the quantum level, you wouldn't have a

charge-current vector. Any charges and currents would be explicit as

particle wavefunctions, not as the field.

You could add any gradient $d\phi$ to $Z_b$: $Z^\prime_b = Z_b +

\partial\phi/\partial x^b$ gives the same $^\ast F_{ab}$.

From Z you can define a covariant derivative $\nabla_a\psi =

\partial\psi/\partial x^a - ieZ_a\psi$. I guess this connection would be

applied to wavefunctions.

If you have a nonabelian gauge group SU(3), then you'd have a gauge

connection $$\nabla_a\psi= \partial\psi/\partial x^a - C_a\psi$$. Here the

$C_a$'s are matrices in the Lie group algebra of SU(3), operating on a

wavefunction that has a color index. So $\psi = y_1 |red> + y_2 |green> +

y_3 |blue>$ and $|y_1|^2 + |y_2|^2 + |y_3|^2 = 1$, so that the gauge group

SU(3) is acting as transformations on $S^6$. The dimension of the unitary

group U(3) is $3^2=9$ (see sec. 13.10), so the dimension of the Lie

algebra of SU(3), the unitary matrices of determinant 1, is 8. I read

later that there are basis elements for the Lie algebra, trace-free $3

\times 3$ Hermitian matrices called Gell-Mann matrices, for the inventor

of the color theory.

The $C_a$'s are $i\times$ a Hermitian matrix. Since $e^{iH}$ is unitary

if H is Hermitian, this gives you a unitary transform if you're

integrating $\nabla_a$; taking a path integral, with the Lie algebra

elements varying over space should (though I haven't shown it rigorously)

integrate to a matrix in SU(3). The gauge transformation has to be

unitary because it should preserve the inner product $<\psi|\phi>$ of two

wavefunctions. And the gauge transformation should not change the

wavefunction of a 3-quark combination that's been antisymmetrized with

respect to color, because such a particle is a free particle, so the

covariant gauge derivative shouldn't affect it. That implies it has

determinant 1.

The curvature of the connection $\nabla_a\psi= \partial\psi/\partial x^a -

C_a\psi$ is $$\nabla_a\nabla_b - \nabla_b\nabla_a =

\frac{\partial C_a}{\partial x^b} - \frac{\partial C_b}{\partial x^a} +

C_aC_b - C_bC_a$$

This is a 2-form $S_{ab}$ with hidden color indices. It satisfies a

Bianchi identity $\partial S_{[ab}/\partial x^{c]} =0$, I checked.

I tried to find a curvature tensor for a connection with \emph{both} a

spacetime curvature and curvature on the color indices (the gauge

curvature), but it didn't work, that is the commutator $(\nabla_a\nabla_b

- \nabla_b\nabla_a)\psi$ didn't work out to something multiplied by just

$\psi$. Trying to quantize gravity!

You can find the Hodge dual $^\ast S_{ab}$ and try to interpret it as a

curvature tensor. But, with a nonabelian gauge the commutator $C_aC_b -

C_bC_a$ doesn't disappear, so the gauge curvature doesn't look like the

exterior derivative of a form. So the Poincare' lemma might not apply.

If you could show that $^\ast S_{ab}$ doesn't

satisfy the Bianchi identity $\partial S_{[ab}/\partial x^{c]} =0$, that

would show that $^\ast S_{ab}$ isn't a curvature tensor, at least

for a connection of the form $\nabla_a\psi= \partial\psi/\partial x^a -

C_a\psi$ - since I checked that $S_{ab}$ does satisfy this Bianchi

identity! The terms in the Bianchi identity for $^\ast S_{ab}$ are a lot

of complicated stuff that doesn't look like it would have a habit of

summing to 0.

If $^\ast S_{ab}$ \emph{did} satisfy the Bianchi identity $\partial

S_{[ab}/\partial x^{c]} =0$, maybe that would mean it's a curvature tensor

for a connection of the form $\nabla_a\psi= \partial\psi/\partial x^a -

C_a\psi$. I don't know, since the

the Poincare' lemma doesn't necessarily apply.

So that is my best take on a confusing exercise!

\end{document}

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