Timeline for Physicist's request for intuition on covariant derivatives and Lie derivatives
Current License: CC BY-SA 3.0
16 events
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| Jul 25 at 5:23 | comment | added | Liviu Nicolaescu | @BastamTajik Indeed, it is not a connection on the tangent bundle. If it were It would satisfy the property $L_{fX}Y= fL_X Y$, for any smooth function $f$ and any smooth vectors fields $X,Y$. Clearly $L_{fX}Y= [fX,Y]= f[X,Y]-(Yf)X=fL_Xy-(Yf)X.$ | |
| Jun 26 at 11:29 | comment | added | Bastam Tajik | I have got this feeling you are implicitly saying that the Lie derivative does not define a connection and parallel transport. And I guess there are examples to certify this. Right? | |
| Jun 19 at 20:39 | comment | added | A. Chen | @Liviu Nicolaescu May I know where the formal definition of "$T_{\gamma}$ should depend smoothly on $\gamma$" can be found? I see a few textbooks mentioned this but they all skipped the definition. | |
| Jan 20, 2022 at 11:44 | comment | added | Liviu Nicolaescu | The one-form you talk about defines the so-called horizontal distribution. It can be used to defined the covariant derivative. It can be described in terms of the $A$ above but it is not $A$. And no, neither $A$ nor the vector valued $1$-form are tensors since they are not sectors of tensor bundles. | |
| Jan 20, 2022 at 11:37 | comment | added | Alex | I guess the parts that confuses me is this: on a principal bundle, we have something like $D\psi = \partial\psi + A\psi$, where $A$ is a $\mathcal{g}$-valued one-form. Yet, 1. is $A$ here analogous to the Christoffel symbols which are not tensors? 2. Shouldn't $D$ here be analogous to $\nabla$ which is the "connection" which should be a vector-valued one-form? Why do we call $A$ $\mathcal{g}$-valued something? | |
| Jan 20, 2022 at 11:30 | comment | added | Liviu Nicolaescu | Here is how my adviser taught me. Regular one-forms are beasts that eat a vector and return a number. An operator valued 1-form eats a vector and returns an operator. If you feed a connection $\nabla$ you get an operator $\nabla_X$. This operator takes a vector $Y$ and returns the vector $\nabla_XY$. | |
| Jan 20, 2022 at 4:21 | comment | added | Alex | Thank you! A follow-up question is how do I view the connection as a vector-valued one-form? | |
| Jan 19, 2022 at 18:08 | comment | added | Liviu Nicolaescu | Yes, you got it right. | |
| Jan 19, 2022 at 13:54 | comment | added | Alex | I think I got it now - at every point on $M$, $f$ is just a scaling number. The notion of parallel transport (or to say connection) is defined on $M$ independent of $X$, hence $f$ can be pulled to the front at each point independently. Is my understanding in the 2nd comment above correct? | |
| Jan 19, 2022 at 13:18 | comment | added | Liviu Nicolaescu | The $C^\infty$-linearity is very intuitive. Think you are traveling by car with velocity $v$ and you measure the rate of change in temperature per unit of time. This rate of change is the rate of change per unit of distance $\times$ velocity. Different cars will report different temporal rates of change at the same location $x$ if they have different velocities at that location. The proportionality factor depends on $x$, and is a function. | |
| Jan 19, 2022 at 12:53 | comment | added | Alex | Or did you mean that with a vector field $X$ we can generate its flow - a bunch of curves. Then, with the extra structure of "parallel transport" defined over the entire manifold we can take the covariant derivative along these curves. But the covariant derivative practically has nothing to do with $X$ (other than that $\nabla_X Y$ is just on a curve defined by $X$), rather, covariant derivative is all specified by the parallel transport. Unlike Lie derivative which is about $X$ just as much it is about $Y$? | |
| Jan 19, 2022 at 12:47 | comment | added | Alex | This is the best answer I've found on the internet. I am trained as a physicist and I follow your intuition very well. I am with you till the end part - you suddenly say this construction allows us to define the covariant derivative $\nabla_X u$ over a vector field, which is $C^\infty$-linear. I was lost here. Up till here you were talking about derivative along a curve, so 1. where is a vector field $X$ from 2. why is it (naturally or can be made) $C^\infty$-linear? | |
| Mar 31, 2017 at 10:02 | history | edited | Liviu Nicolaescu | CC BY-SA 3.0 |
added 67 characters in body
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| Mar 31, 2017 at 9:51 | history | edited | Liviu Nicolaescu | CC BY-SA 3.0 |
added 67 characters in body
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| Jan 3, 2012 at 19:59 | history | edited | Liviu Nicolaescu | CC BY-SA 3.0 |
corrected spelling, included missing terminology
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| Jan 3, 2012 at 18:33 | history | answered | Liviu Nicolaescu | CC BY-SA 3.0 |