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Quick question: Curvature form of a connection on the trivial bundle

+ 5 like - 0 dislike
176 views

Let $L=\mathbb{R}^2\times U(1)$ be the trivial $U(1)$-bundle over $\mathbb{R}^2$. Define a connection $\nabla=d+A$ where $A=fdx+gdy$ is an $\mathbb{R}$ valued $1$-form on $L$. That is, $\nabla$ gives a distribution $\mathcal{H}$ on $L$ - the horizontal distribution.

The distribution $\mathcal{H}$ is obtained as the graph of $-A$ as a linear map from $\mathbb{R}^2\rightarrow\mathbb{R}$. A horizontal lift $\tilde{X}$ of a vector field $X$ on $\mathbb{R}^2$ is given by $\tilde{X}=(X,-A(X))$.

Let $\alpha$ be the projection onto the vertical direction on $L$ i.e. $\ker(\alpha)=\mathcal{H}$, and define the curvature $2$-form $\Omega_{\nabla}$ of the connection $\nabla$ by

$$\Omega_{\nabla}(X,Y)=\alpha([\tilde{X},\tilde{Y}])$$

The following is expected to be true $$\Omega_{\nabla}=-dA?$$

Here is my confusion:

Let $z$ be the local vertical coordinate, $X=\partial_x$ and $Y=\partial_y$. Then $\tilde{X}=-f\partial_z+\partial_x$ and $\tilde{Y}=-g\partial_z+\partial_y$. And $$-dA(X,Y)=-X(A(Y)+Y(A(X))+A([X,Y])=-\partial_xg+\partial_yf$$ while $$[\tilde{X},\tilde{Y}]=(f(\partial_zg)-g(\partial_zf)-\partial_xg+\partial_yf)\partial_z$$ therefore $$\Omega_{\nabla}(X,Y)=\alpha([\tilde{X},\tilde{Y}])=f(\partial_zg)-g(\partial_zf)-\partial_xg+\partial_yf.$$ Where is the mistake? Thank you.

This post imported from StackExchange Physics at 2016-07-29 20:36 (UTC), posted by SE-user Student
asked Jul 29, 2016 in Mathematics by student (85 points) [ no revision ]
I don't have time to go through this right now, so I dunno if this causes any problems or not, but $A$ should be imaginary valued. The Lie algebra of $\mathrm{U}(1)$ is $i\mathbb{R}$, the reason why the vector potential is real in QED is because you take $A=iq\mathcal{A}$, where $A$ is the connection and $\mathcal{A}$ is the vector potential.

This post imported from StackExchange Physics at 2016-07-29 20:36 (UTC), posted by SE-user Uldreth

1 Answer

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Your computation is correct but a final simplification is possible.  Remember that $g$ and $f$ are functions only of $x$ and $y$.  Then $\partial_zg =0$  and $ \partial_zf=0$.  Applying this we obtain

$$\Omega_{\nabla}(X,Y)=f(\partial_zg)-g(\partial_zf)-\partial_xg+\partial_yf=-\partial_xg+\partial_yf=-dA(X,Y)$$

Do you agree?  All the best.

answered Aug 2, 2016 by juancho (860 points) [ revision history ]

Isn't $z$ the coordinate parameterizing the $U(1)$? Why should $f,g$ be independent of it? Then it would be a 1-form on $R^2$, which is much more restrictive than one of the given form on $L$.

@ArnoldNeumaier  : Your question is very interesting.  The user Student writes:

"$A=fdx+gdy$ is an $\mathbb{R}$ valued $1$-form on $L$"

according with that $A$ is a 1-form on  $\mathbb{R}^2$.  The truly 1-form on $\mathbb{R}^2\times U(1)$ must be written as $A=fdx+gdy+ dz$ where $f$ and $g$ are functions on $\mathbb{R}^2$, it is to say, $f$ and $g$ are functions only of $x$ and $y$.  Do you agree?

I think student meant indeed to say that $A$ is a 1-form on $R^2$, and his mistake consisted in making it a 1-form on $L$. Then everything is consistent, as you explain. The most general 1-form on $L$ would have the form $fdx+gdy+hdz$ with functions $f,g,h$ depending on $x,y,z$, and nothing could be concluded. Do you agree?

@ArnoldNeumaier : You are right, "the most general 1-form on $L$ would have the form $fdx+gdy+hdz$ with functions $f,g,h$ depending on $x,y,z$".  In the case proposed by Student, a trivial bundle is considered and the 1-form $A$ is a connection.  In such case, $h =1$; and $f$ and $g$ are functions only of $x$ and $y$. Do you agree?  All the best.

Yes, I agree.

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