- #1
Frank Castle
- 580
- 23
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.
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.
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