# A Diffeomorphisms & the Lie derivative

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1. Jan 23, 2017

### Frank Castle

I've been studying a bit of differential geometry in order to try and gain a deeper understanding of the mathematics of general relativity (GR). As you may guess from this, I am approaching this subject from a physicist's perspective so I apologise in advance for any lack of rigour.

As I understand it, given a base manifold $M$, a vector field $X:M\rightarrow TM$ is a section of the tangent bundle of the manifold $M$. Furthermore, a given vector field $X$ generates a local one-parameter group of diffeomorphisms $\sigma_{t}(p):=\sigma(t,p)$ that are the integral curves of X, i.e. $$\frac{d}{dt}\sigma^{\mu}_{t}(x)=X^{\mu}(\sigma_{t}(x))$$ with the initial condition $$\sigma^{\mu}_{0}(x)=x^{\mu}$$ where $\sigma^{\mu}_{t}(x)$ is the coordinate representation of the diffeomorphism $\sigma :\mathbb{R}\times M\rightarrow M$ and, similarly, $X^{\mu}(\sigma^{\mu}_{t}(x))$ are the coordinate components of $X$.

The first question I have, is that from reading Sean Carroll's GR notes, when introducing the Lie derivative, he states that "a single discrete diffeomorphism is insufficient; we require a one-parameter family of diffeomorphisms...". Is what is meant by this that in order to be able to compare vectors at different points on a manifold, we need a mapping between the two points that continuously connects the two points, such that one can meaningfully take a limit?! By this I mean, by varying the parameter $t$ the diffeomorphism $\sigma_{t}$ maps to different points, such that if we have some starting point $p\in M$ and some end point $q\in M$ then by varying $t$, $\sigma_{t}$ maps to a continuous set of points between $p$ and $q$, thus defining a continuous curve between $p$ and $q$. Such that, as we increase the value of the parameter $t$, $\sigma_{t}$ maps to different points(between $p$ and $q$), and for some value of $t$ it maps to the point $q$, such that we have a parametrised curve connecting the point $p$ to the point $q$, along which we can "drag" vectors from one (base) point to another.

Given this initial formalism, my second question pertains to the pullback and pushforward defined by the diffeomorphism $\sigma_{t}$ and calculating the Lie derivative of a vector:

Choosing a particular local coordinate chart, under the action of $\sigma_{t}$ (for infinitesimal $t$), a point $p$ whose coordinate is $x^{\mu}$, is mapped to $$x'^{\mu}:=\sigma^{\mu}_{t}(x)=\sigma^{\mu}(t,x)=x^{\mu}+tX^{\mu}(x)$$ Given a vector field $Y$, suppose we wish to evaluate its change as we move from the point $p$ to a nearby point $q$, whose coordinate is given by $x'^{\mu}=\sigma_{t}(x)$. To do this we must use the pullback map $\left(\sigma_{-t}\right)_{\ast}:T_{\sigma_{t}(x)}M\rightarrow T_{x}M$ to map the vector $Y\big\vert_{\sigma_{t}(x)}\in T_{\sigma_{t}(x)}M$ to $T_{x}M$ and then take the difference between this and the vector $Y\big\vert_{x}\in T_{x}M$. Let $e_{\mu}\big\vert_{x}$ and $e_{\mu}\big\vert_{x'}$ denote the coordinate bases for $T_{x}M$ and $T_{\sigma_{t}(x)}M$, respectively. Then, $$Y_{x'}:=Y\big\vert_{\sigma_{t}(x)}=Y^{\nu}(x')e_{\mu}\big\vert_{x'}=\left[Y^{\mu}(x)+tX^{\nu}(x)\frac{\partial Y^{\mu}(x)}{\partial x^{\nu}}+\mathcal{O}(t^{2})\right]e_{\mu}\big\vert_{x'}$$ Then, using the pullback map, $$\left(\sigma_{-t}\right)_{\ast}Y\big\vert_{x'}=\left[Y^{\mu}(x)+tX^{\nu}(x)\frac{\partial Y^{\mu}(x)}{\partial x^{\nu}}+\mathcal{O}(t^{2})\right]\left(\sigma_{-t}\right)_{\ast}e_{\mu}\big\vert_{x'}$$ Now this is the bit that I'm unsure about. How does one interpret $\left(\sigma_{-t}\right)_{\ast}e_{\mu}\big\vert_{x'}$? Does one simply use the chain rule, such that $$\left(\sigma_{-t}\right)_{\ast}e_{\mu}\big\vert_{x'}=\frac{\partial x^{\nu}}{\partial x'^{\mu}}e_{\nu}\big\vert_{x}=\left(\delta^{\nu}_{\,\mu}-t\frac{\partial X^{\nu}}{\partial x'^{\mu}}\right)e_{\nu}\big\vert_{x}=\left(\delta^{\nu}_{\,\mu}-t\frac{\partial X^{\nu}}{\partial x^{\lambda}}\frac{\partial x^{\lambda}}{\partial x'^{\nu}}\right)e_{\nu}\big\vert_{x}=\left(\delta^{\nu}_{\,\mu}-t\frac{\partial X^{\nu}}{\partial x{\mu}}+\mathcal{O}(t^{2})\right)e_{\nu}\big\vert_{x}$$ If this is the case, then I think I get what's going on, since then we have $$\left(\sigma_{-t}\right)_{\ast}Y\big\vert_{x'}=\left[Y^{\mu}(x)+tX^{\nu}(x)\frac{\partial Y^{\mu}(x)}{\partial x^{\nu}}+\mathcal{O}(t^{2})\right]\left[\delta^{\nu}_{\,\mu}-t\frac{\partial X^{\nu}}{\partial x{\mu}}+\mathcal{O}(t^{2})\right]e_{\nu}\big\vert_{x}\\ =\left[Y^{\mu}(x)+t\left(X^{\nu}(x)\frac{\partial Y^{\mu}(x)}{\partial x^{\nu}}-Y^{\nu}(x)\frac{\partial X^{\mu}(x)}{\partial x^{\nu}}\right)+\mathcal{O}(t^{2})\right]e_{\mu}\big\vert_{x}$$ and the Lie derivative of $Y$ with respect to $X$ is then $$\mathcal{L}_{X}Y=\lim_{t\rightarrow 0}\frac{\left(\sigma_{-t}\right)_{\ast}Y\big\vert_{x'}-Y\big\vert_{x}}{t}=\left[X^{\nu}(x)\frac{\partial Y^{\mu}(x)}{\partial x^{\nu}}-Y^{\nu}(x)\frac{\partial X^{\mu}(x)}{\partial x^{\nu}}\right)e_{\mu}\big\vert_{x}=\left[X,Y\right]$$ However, I'm quite unsure as to whether I've understood this all correctly. Any help and/or insight would be much appreciated.

Last edited: Jan 23, 2017
2. Feb 8, 2017

### TeethWhitener

3. Feb 20, 2017

### xaos

Second Bump.

This seems to be a good calculation. One thing I don't see is how the high order terms behave under the diffeomorphisms, since this gives only a first order condition on the expansion. They would have to fall off to zero faster than the region collapses, but I don't see what information allows this. Perhaps my eyes are blurred...

4. Mar 16, 2017

### Frank Castle

If anyone has any ideas on this it'd be much appreciated. I'd really like to understand the intuition behind "one-parameter family of diffeomorphisms".

5. Mar 16, 2017

### Staff: Mentor

Your calculation looks good to me, i.e. without drawing some pictures of which parts happen actually where. As far as I understood it, the role of one-parameter (connected) subgroups is only to establish a correspondence to one-dimensional Lie subalgebras as the basis for the correspondence subgroups $\leftrightarrow$ subalgebras. All I've found was an example (torus) where the author finishes:
It's a book about Lie groups, so he works directly with the exponential map from the beginning, which makes things a little bit less abstract.

6. Mar 17, 2017

### zwierz

then try to define and calculate the Lie derivative of 1-form $\omega=\omega_i(x)dx^i,\quad \mathcal L_X\omega=?$
another useful exercise show that $g_X^s\circ g_Y^t= g_Y^t\circ g_X^s\Longleftrightarrow [X,Y]=0$, here $g_X^t$ is one parametric group generated by the vector field $X$

7. Mar 17, 2017

### Frank Castle

Is the idea behind the need for a one-parameter family of diffeomorphisms that a single diffeomorphism $\phi$ with map a given point $p$ to one (and only one) "new" point $q=\phi(p)$. As such this is not enough if we wish to connect two points by a curve that we can subsequently parallel transport a vector between their respective tangent spaces. To do so we need a "family" of diffeomorphisms - one for each value of some parameter $t$, since then for different values of $t$, $\phi_{t}$ will map the initial point $p$ to different points between $p$ and $q$ (i.e. by varying $t$, $\phi_{t}(p)$ corresponds to different points between $p$ and $q$). Thus, $\phi_{t}(p)$ describes a curve, parametrised by $t$ with the constraints that $\phi_{0}(p)=p$ and $\phi_{\tau}(p)=q$ for some value of $t=\tau$. Would this be a correct understanding of the intuition behind it at all?!

8. Mar 17, 2017

### TeethWhitener

This is my understanding of it. A single diffeomorphism $\phi$ from $M \rightarrow M$ basically just relabels the points on $M$. A family of diffeomorphisms at a fixed point $p$ denoted $\phi_t(p)$ maps $\mathbb{R} \rightarrow M$, so it's equivalent to a curve on $M$ parameterized by $t \in \mathbb{R}$.

Edit: one other thing Wald talks about, which I didn't see in Carroll, is that $\phi_{s+t} = \phi_s \circ \phi_t$ and $\phi_0(p) = p$, which means that the family of diffeomorphisms is actually an Abelian group.

9. Mar 17, 2017

### Frank Castle

Ok cool, this makes intuitive sense to me.

10. Mar 17, 2017

### Staff: Mentor

... which are exactly the defining equations of flows of the vector field.