Dual Gauge Curvature for U(1) and SU(3)

[lark]

http://camoo.freeshell.org/25.8.pdf"

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}

 

[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.

 

[lark]

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.

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

 

[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.

 

[lbrits]

scary...