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I am currently working through Guillemin & Pollack (slowly) and I've run into a bump quite early on. So far, we've recalled that maps from open sets to $\mathbb{R}^n$ are "well behaved". Their derivatives are linear maps between the copies of affine space that they are defined on and they satisfy the chain rule. We wish to extend derivatives to smooth maps between manifolds so that they act like normal derivatives when the manifolds are affine space, and so that they also satisfy the chain rule. The approach is this: Let $f:X\to Y$ be a smooth map of smooth $k$ and $l$ manifolds, respectively, with $x\mapsto y$. Let $\phi:(U,0)\to (X,x)$ parametrize $X$ around $x$ and let $\psi:(V,0)\to (Y,y)$ parametrize $Y$ around $y$ with $U\subset \mathbb{R}^k$ and $V\subset\mathbb{R}^l$ open sets. To get a commutative diagram, shrink $U$ and let $h$ complete the diagram, $h=\psi^{-1}\circ f\circ\phi$.

(To "shrink" properly, I believe we let $U'=U\cap\phi^{-1}(f^{-1}(\psi(V)))$)

So then we make the definition $df_x=d\psi_0\circ dh_0\circ d\phi_0^{-1}$. I'm having a hard time seeing why this definition is independent of the choice of parametrization. I believe it, but I can't lock it down set theoretically if you know what I mean. The book advises that we use a similar argument to showing that the tangent space of $X$ at $x$ is well defined. I've tried introducing alternative parametrizations $\eta$ and $\rho$ to parallel $\phi$ and $\psi$, but my diagrams are confusing and I am not sure which clever composition to make and then subsequently take a derivative of. Any help would be much appreciated. Thanks!

EDIT: Some more details on my progress. Suppose $\rho:V'\to Y$ is an alternative parametrization of $Y$ at $y$. Without loss of generality, let $\rho(V')=\psi(V)$. Then we have $h'=\rho^{-1}\circ f\circ \phi$ and the possibly different derivative map $df_x'=d\rho_0\circ dh_0'\circ d\phi_0^{-1}$. Then notice that $\rho\circ h'=\psi\circ h=f\circ\phi$ is a map from an open set in affince space to affine space. Then the chain rule applies and we get that $df_x'=d\rho_0\circ dh_0'\circ d\phi_0^{-1}=d\psi_0\circ dh_0\circ d\phi_0^{-1}=df_x$ as desired. I can get a symmetric result if we swap out $\phi$ for $\eta$, also. But I can't figure out how to change them both at once.

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  • $\begingroup$ It sounds like you pretty much have it! Can't you use your first computation to go from the pair $(\phi, \psi)$ to $(\phi, \rho)$, and then your second result to go from $(\phi, \rho)$ to $(\eta, \rho)$? That is, to replace both of the charts, replace them one at a time. That seems okay to me, or am I missing something? I think that what you wrote in your last paragraph (the "EDIT") is the most important part; this bit about changing parametrizations in sequence, one at a time, is just an add-on. $\endgroup$ – Zach Teitler May 11 '18 at 5:54
  • $\begingroup$ Yes, I see what you mean. However, I think that I was actually out of my mind when I wrote this originally. I was able to swap $\psi$ for $\rho$, but not $\eta$ for $\phi$. We are allowed, above, to write $h=\psi^{-1}\circ f\circ \eta$, put $\psi$ on the left hand side, and differentiate with the usual chain rule. When trying to swap out the other map, we are stuck: $h\circ \eta^{-1}=\psi^{-1}\circ f$ $\endgroup$ – Prototank May 11 '18 at 12:12
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Let's say we have a diagram like this: $$ \begin{array}{lcl} U' & \overset{h'}{\longrightarrow} & V' \\ \downarrow\varphi' & & \downarrow\psi' \\ X & \overset{f}{\longrightarrow} & Y \\ \uparrow\varphi & & \uparrow\psi \\ U & \overset{h}{\longrightarrow} & V \end{array} $$ Here $X,Y$ are manifolds; $U,U'$ are parametrizations of neighborhoods in $X$, with $\varphi(0)=x=\varphi'(0')$, and similarly for $V,V'$; the squares commute. To be explicit, say $1 \in V$, $1' \in V'$, with $\psi(1)=y=\psi'(1')=f(x)$.

From the bottom square, we propose to define $df_x$ to be $d\psi_1 \circ dh_0 \circ d\varphi_0^{-1}$. From the top square we propose to define $df_x$ to be $d\psi'_{1'} \circ dh'_{0'} \circ d{\varphi'}_{0'}^{-1}$. First of all, let's stop writing subscripts. The big rectangle commutes. Say we map from $U'$ to $V$: $$ h \circ \varphi^{-1} \circ \varphi' = \psi^{-1} \circ \psi' \circ h' . $$ Let $\Phi$ be a smooth extension of $\varphi^{-1}$ in a neighborhood of $x \in X \subseteq \mathbb{R}^N$. Let $\Psi$ be a smooth extensions of $\psi^{-1}$ in a neighborhood of $y \in Y \subseteq \mathbb{R}^M$. (Like Guillemin and Pollack do in the paragraph on page 9-10 where they show that $\dim T_x(X) = \dim X$.) We have $$ h \circ \Phi \circ \varphi' = \Psi \circ \psi' \circ h' , $$ since the image of $\varphi'$ is contained in $X$ and $\Psi|_X = \varphi^{-1}$, and the image of $\psi'$ is contained in $Y$, similarly. Every map in this equation is a smooth map between open sets in affine space. By the chain rule, $$ dh \circ d\Phi \circ d\varphi' = d\Psi \circ d\psi' \circ dh' . $$ The image of $d\varphi'$ is $T_x(X)$ (same as image of $d\varphi$; this is the well-definedness of the space $T_x(X)$, G&P page 9). So we can write $$ dh \circ d\Phi|_{T_x(X)} \circ d\varphi' = d\Psi|_{T_y(Y)} \circ d\psi' \circ h'. $$ As in G&P, since $\Phi \circ \varphi$ is the identity on a neighborhood of $0$ in $U$, we have $d\Phi \circ d\varphi$ is the identity map. So $d\Phi|_{T_x(X)} \circ d\varphi$ is the identity; now these are maps between $T_x(X)$ and $\mathbb{R}^n = T_0(U)$, which are spaces with the same dimension. So $d\Phi|_{T_x(X)} = (d\varphi)^{-1}$. We get $$ dh \circ d\varphi^{-1} \circ d\varphi' = d\psi^{-1} \circ d\psi' \circ h' . $$ Now all our problems are over; rearranging gives $$ d\psi \circ dh \circ d\varphi^{-1} = d\psi' \circ dh' \circ d{\varphi'}^{-1} $$ as desired.

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    $\begingroup$ Your bottomright most $Y$ I think is $V$. Thanks a lot, I can finally put this behind me. $\endgroup$ – Prototank May 11 '18 at 19:06
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    $\begingroup$ I'm glad I could help. :-) $\endgroup$ – Zach Teitler May 11 '18 at 19:11

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