# Dual gauge curvature

1. Apr 27, 2008

### lark

http://camoo.freeshell.org/25.8.pdf" [Broken]

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}

Last edited by a moderator: May 3, 2017
2. Apr 28, 2008

### lbrits

Confusing to say the least. But I believe that $$D_{[a} D_{b]}$$ where $$D_a = \nabla_a +A_a$$ should behave multiplicatively. Perhaps that is not the best definition of $$D$$. In any event, the field-strength $$F$$ doesn't reflect the curvature of the manifold since it is determined as an exterior derivative. But $$*F$$ will because the metric is needed to construct the Hodge dual.

So normally one would write instead $$D = dx^\mu \partial_\mu \wedge + \mathbf{A}_\mu dx^\mu \wedge = dx^\mu (\partial_\mu + \mathbf{A}_\mu) \wedge = dx^\mu D_\mu \wedge$$ so that the covariant derivative doesn't carry a Christoffel symbol. The christoffel symbol part does start popping up when you have expressions like $$D_\mu D^\mu \phi = 0$$ from equations of motion, say, but for the definition of $$F$$, it is absent.

3. Apr 30, 2008

### lark

I adapted the language of Christoffel symbols to the color transformation - so my "Christoffel" symbols have two color indices. I adapted the derivation of the Riemann tensor to the gauge curvature. I was doing it in flat spacetime - I made a comment in there that I'd tried to get a "curvature tensor" when there was spacetime curvature as well as the gauge curvature - but I wasn't getting a tensor out of it, i.e.
$$(\nabla_b\nabla_a - \nabla_a\nabla_b)\psi$$ wasn't a function just of $$\psi$$. So the curvature tensor $$S_{ab}$$ only refers to the gauge curvature. If you had curved spacetime, the gauge transformation matrices $$C_{a}$$ have one spacetime index, so $$\nabla_aC_{a}\neq\partial C_{a}/\partial x^a$$. When you were deriving $$\nabla_b\nabla_a - \nabla_a\nabla_b)$$ you'd have spacetime Christoffel symbols applied to the partial derivatives of the $$C_{a}$$'s. A mess! Maybe it would be a tensor if I did it right, but it's be a complicated thing!!

The Christoffel symbols are essentially matrices in various directions. If you're approximating something on a small scale you use a linear (first order) approximation. So the Christoffel symbols are the general linear approximation of some alternative definition for the "change" along a curve. For Riemannian curvature, it's defining what is parallel transport. For the gauge curvature, it's defining what it means for the wavefunction to "not change" along a path.

So what I called $${C_{ab}}^c$$, in the usual language of quantum mechanics are expressed in terms of sums $$d_a^bG_b$$ where the $$G_b's$$ are the 8 Gell-Mann matrices (I don't remember which exact letters they use but that's more or less how they'd write it). The Lie algebra of SU(3) has diimension 8 so there are 8 GM matrices.
Laura

4. Apr 30, 2008

### lark

The covariant derivative $$\nabla_a\psi = \partial\psi/\partial x^a - ieA_a\psi$$ is kind of strange, because it doesn't follow the Leibnitz rule if you applied it to 2 wavefunctions multiplied together. A covariant derivative is supposed to be a linear operator and to follow the Leibnitz rule. At least before you start including spin states etc., the wavefunction is simply a scalar field on spacetime and $$\nabla_a(\psi\phi)$$ should $$=\psi\nabla_a\phi+\phi \nabla_a\psi$$. So I guess you have to make special rules as to what kind of scalar field the Leibnitz rule applies to, or in what situations exactly you're using the covariant derivative.

5. Apr 30, 2008

### lbrits

scary...

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