Parallel transport of a 1-form aound a closed loop

[Phinrich]

TL;DR Summary

If we parrallell transport a gradient vector of a scalar around a closed loop do we get a none-zero result?

Good day all.

Since the gradient theorem, also known as the fundamental theorem of calculus for line integrals, says that a line integral through a gradient field can be evaluated by evaluating the original scalar field at the endpoints of the curve. Then If we form the Gradient vector field of the Scalar (a 1-form) and if we parrallell transport it around a Closed Curve would we not get a result of zero since we would evaluate the scalar at the start and end (the same point) and subtract. If this is yes then since we define the Riemann Curvature Tensor by parallell transporting a vector around a closed loop, If we tried to similarly define a Riemann Curvature Tensor by parrallell transport of the Gradient of a scalar around a closed point and get zero that suggests we cannot define such a Riemann Curvature Tensor for the Gradient Field?
Or am I TOTALLY wrong here ?

Thanks

 

[Orodruin]

A general 1-form is not necessarily a gradient field.

 

[Phinrich]

Thank you. I agree. But of we agree that a Gradient is a 1-form then do we have the issue I mentioned?

 

[Phinrich]

A general 1-form is not necessarily a gradient field.

Interesting reply. What does the relativity of simultaneity to do with it ? Are we suggesting that not all observers would agree the two points are the same ?

 

[Orodruin]

Thank you. I agree. But of we agree that a Gradient is a 1-form then do we have the issue I mentioned?

Your issue is non-existent. The parallel transport equation for 1-forms would still be valid.

Interesting reply. What does the relativity of simultaneity to do with it ? Are we suggesting that not all observers would agree the two points are the same ?

Nothing. You are reading my signature, which is in all my posts.

 

[Phinrich]

That explains my confusion about where that answer came from. And I hear what you say about the non-existence of my issue. I will think that one over carefully.

Thanks

 

[Phinrich]

I agree that the formula remains valid. My only question is will the answer always be zero for the transport of a Gradient vector of a scaar? OK I am here associating parallel transport of a vector/1-form along a line with the line integral of the vector or 1-form. If that's correct?

 

[Orodruin]

I agree that the formula remains valid. My only question is will the answer always be zero for the transport of a Gradient vector of a scaar? OK I am here associating parallel transport of a vector/1-form along a line with the line integral of the vector or 1-form. If that's correct?

The exterior derivative has no a priori dependence on the connection. As such, the parallel transport (which is given by the connection) is not the same concept. I am sorry, but I do not see how you make this connection. For an exact 1-form $df$, you have $\int_\gamma df = f(b) - f(a)$, but there is nothing here telling you that the one-form is parallel along $\gamma$. For this to be the case you would need $\nabla_{\dot\gamma} df = 0$, which in coordinate form would be $\dot x^\nu [\partial_\nu \partial_\mu f - \Gamma_{\nu\mu}^\rho \partial_\rho f] = 0$. For example, if you take $f(x,y) = x^2 + y^2$ in $\mathbb R^2$, you would have $\nabla f = 2 \vec x$. This is certainly not parallel along any curve with the usual connection on $\mathbb R^2$.

 

[Phinrich]

The exterior derivative has no a priori dependence on the connection. As such, the parallel transport (which is given by the connection) is not the same concept. I am sorry, but I do not see how you make this connection. For an exact 1-form $df$, you have $\int_\gamma df = f(b) - f(a)$, but there is nothing here telling you that the one-form is parallel along $\gamma$. For this to be the case you would need $\nabla_{\dot\gamma} df = 0$, which in coordinate form would be $\dot x^\nu [\partial_\nu \partial_\mu f - \Gamma_{\nu\mu}^\rho \partial_\rho f] = 0$. For example, if you take $f(x,y) = x^2 + y^2$ in $\mathbb R^2$, you would have $\nabla f = 2 \vec x$. This is certainly not parallel along any curve with the usual connection on $\mathbb R^2$.

Thank you for your response. It is obvious that I do not fully understand the concept of the 1-form. I will study your response in detail ad hope that it will improve my understanding.