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$$\text{T}^*U\to\varphi(U)\times(\mathbb{R}^m)^*,\space(x,\lambda)\mapsto\left(\varphi(x),(D_{\varphi(x)}\varphi^{-1})^*(\lambda)\right)$$ What does $D_{\varphi(x)}\varphi^{-1}$ mean? Because I know "$D$" of sth, normally sth should be a function between two manifolds. But how can this homeomorphism work here?

Upc
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1 Answers1

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So this question is (I believe) about finding a functoriality for the cotangent bundle construction.

Because it is more familiar, let us first examine the functorial natural of the tangent bundle. Let $\varphi:M\to N$ be a map of smooth manitolfds. Then, given a point $x\in M$, we have a linear map $D\varphi_x: T_xM\to T_{\varphi(x)}N$, namely the derivative of $\varphi$ at $x$. This can be computed by passing to a coordinate chart and then taking partial derivatives, if you so desire. Just as we can glue the tangent spaces at each point into the tangent bundle, we can glue these linear maps together to define a map between tangent bundles:

$$D\varphi:TM\to TN, (x,v)\mapsto (\varphi(x),D\varphi_x(v))$$

Similarly, if we can use $\varphi$ to produce a linear map on the fibers of the cotangent bundle in a natural way, we can glue them together to form a map between the cotangent bundles. Given a linear map $A:V\to W$, we have a dual map $A^*:W^*\to V^*$ by $A^*(f)(v)=f(A(v))$ for $f\in W^*$. Unfortunately, because this reverses direction, we can't define our map as an ordered pair $(\varphi, (D\varphi)^*)$

However, if $\varphi$ is locally invertible near a point $x$ (equiv to $D\varphi_x$ being invertible), then we can correct for the fact that taking duals reverses direction by reversing a second time with inversion, namely we define a map

$$(x,\lambda)\mapsto (\phi(x),((D\varphi_x)^*)^{-1}(\lambda)).$$

I assert that this is the same as the map you have. To see this, we note that if $\varphi(x)=y$ and $\varphi$ is locally invertible at $x$, then $D(\varphi^{-1})_y=(D\varphi_x)^{-1}$.

So what is the term in the question? It is the derivative of $\varphi^{-1}$ evaluated at $\varphi(x)$.

Don't feel bad that you couldn't figure out what the notation it meant. I had to come up with everything above before I could make sense of it. Parentheses around the $\varphi^{-1}$ would have made it significantly clearer to me.

Aaron
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  • A side remark: I prefer my version of things, because it gives a map of bundles just under the assumption that $\varphi$ is locally invertible, and so does not require a global inverse to exist. – Aaron Apr 13 '16 at 07:40
  • Now it is clearer. Thanks. But I still wanna ask a Little bit. So first we have to start from tangent space $T_xM$. Is $D(\cdot)$ of sth a matrix? So here does this start here also work like an adjoint-operator? And as I understood, if we want to calculate sth, we must end up with sth on $\mathbb{R}^m$, then we can do a normal calculus. But here, this $\varphi$ is a coordinate "evaluating" function from this manifold to its own coordinate system $\mathbb{R}^m$. I didn't see the Definition given by you: two manifolds, $TM \to TN$. – Upc Apr 13 '16 at 10:02
  • I do not know what you mean by sth. When you have a map between two manifolds, you can pass to coordinate charts, and that gives you a smooth map between open subsets of Euclidean spaces. At every point, the derivative of the map is a matrix. I had thought that $\varphi$ was a smooth function because you applied $D$ to it. You left a lot undefined in your question so I had to make assumptions about what everything was. It looked like a function from a subset of Euclidean space to a manifold? – Aaron Apr 13 '16 at 15:38
  • Yes, you are right... I didn't define them clearly. $\varphi$ is the homeomorphism(is it there a special term for it?) on a local chart $(U, \varphi)$. So actually, this is the reason why it made me feel confused... I dont know whether it is good or interesting to calculate the derivative of a homeomorphism of a chart. – Upc Apr 13 '16 at 18:33
  • It isn't really useful to calculate the derivative of the map from a chart, because the chart is by definition the thing that is giving the differentiable structure (unless you take a sheaf-theoretic approach), so you are sort of forced to use circular reasoning and are just calculating the derivative of the identity map from $U$ to itself. Which is why I was assuming that $\varphi$ was not just a coordinate map. I suppose that $D{\varphi}$ could be shorthand for the trivialization of the tangent bundle over a chart defined by $\varphi$? – Aaron Apr 13 '16 at 19:31