I am studying about tangent bundle from the book "J. M lee", on page 66 of the book a map $\tilde{\phi} : \pi^{-1}(U) \to \mathbb{R}^{2n}$ is defined by $$v^i \frac{\partial}{\partial x^i}|_p \mapsto (x^1(p),\cdots, x^n(p),v^1,\cdots,v^n)$$ where $(x^1,\cdots,x^n)$ is the coordinate functions of $\phi$ and $(U,\phi)$ is a local coordinate chart of a smooth manifold $M$. It is written on page 252 that $\tilde{\phi}$ is a diffeomorphism but i am not able to prove it. Can someone explain it!



1 Answer 1


To see that the map $\tilde{\phi}$ is invertible note that $$\tilde{\phi}^{-1}(x^1(p), \dots , x^n(p),v^1, \dots , v^n) = v^i \frac{\partial}{\partial x^i}|_p$$ this map is well-defined as the components of a vector are uniquely specified given a choice of coordinates (recall $v^i = V[x^i]$ for $V=v^i \frac{\partial}{\partial x^i}|_p$). Also, the coordinate chart $x: M \rightarrow \mathbb{R}^n$ is injective so there exists just one point $p$ to which the vector $(x^1(p), \dots , x^n(p))$ corresponds. So, I hope I have convinced you it is a bijection in the $x$ coordinates. Of course, if there exists a coordinate system $y$ also containing $p$ then $\tilde{\phi}$ written as $$ \tilde{\phi}(w^i \frac{\partial}{\partial y^i}|_p) = (y^1(p),\dots, y^n(p), w^1, \dots , w^n) $$ is likewise a bijection. Indeed, we have two different local coordinate representatives. To show $\tilde{\phi}$ is smooth we should recall that as $x,y$ are compatible charts at $p$ the mappings $x \circ y^{-1}$ and $y \circ x^{-1}$ are smooth maps on $\mathbb{R}^n$ in the sense of multivariate real calculus on $\mathbb{R}$. Also, if the vector $V$ has components $v^i = V[x^i]$ in the $x$-coordinate system and $w^i=V[y^i]$ in the $y$-coordinate system then these are related due to the chain rule: $$ \frac{\partial}{\partial x^i} = \frac{\partial y^j}{\partial x^i}\frac{\partial}{\partial y^j}$$ Observe, $$ V=v^i\frac{\partial}{\partial x^i} = v^i\frac{\partial y^j}{\partial x^i}\frac{\partial}{\partial y^j} = w^j \frac{\partial}{\partial y^j} $$ thus, equating coefficients of the $y$-coordinate derivations, the components of the vector $V$ at $p$ are connected by the so-called contravariant transformation rule: $$ w^j = v^i\frac{\partial y^j}{\partial x^i}$$ Collecting our thoughts, we wish to argue $\tilde{\phi}$ is smooth by examining how two different coordinate representatives are related. Note: $$ (x^1(p), \dots , x^n(p),v^1, \dots , v^n) \mapsto (y^1(p),\dots, y^n(p), w^1, \dots , w^n)$$ is a mapping on $\mathbb{R}^{2n}$ for which the Jacobian matrix has the form: $$ \left[\begin{array}{cc} (y \circ x^{-1})' & 0 \\ 0 & [\frac{\partial y^j}{\partial x^i}] \end{array} \right]$$ Here we noted that $y = y \circ x^{-1} \circ x$ so differentiation of the mapping $x \rightarrow y$ at $p$ is in fact differentiation of $y \circ x^{-1}$ at $x(p)$. However, we know that derivative exists as the transition function $y \circ x^{-1}$ is smooth for charts containing a common point $p \in M$. Furthermore, just to be explicit, these two blocks in the Jacobian above are identical; $(y \circ x^{-1})' = [\frac{\partial y^j}{\partial x^i}]$. This is contravariance manifested, the components of a vector field change in the same fashion as the coordinates themselves. In any event, it should be evident by now that the Jacobian of the transition map for the $\tilde{\phi}$ map is nonsingular hence the transition map is smooth. Notice, if $M$ was a $C^k$ manifold then the transition maps $y \circ x^{-1}$ would just be $C^k$ and the tangent bundle would likewise be $C^k$. The map we considered in this discussion is sometimes called the adapted coordinate chart on the tangent bundle. A nice summary of it is that $\tilde{x} = (x,dx)$ in the sense that $\tilde{x}(p,v) = (x(p),dx_p(v))$. In that notation it is easier to argue $\tilde{y} = (y,dy)$ and so \begin{align} \tilde{y} &= (y,dy) \\ &= (y \circ x^{-1} \circ x, d(y \circ x^{-1} \circ x)) \\ &= (y \circ x^{-1} \circ x, d(y \circ x^{-1}) \circ dx)) \\ &= (y \circ x^{-1}, d(y \circ x^{-1})) \circ (x,dx) \\ &= (y \circ x^{-1}, d(y \circ x^{-1})) \circ \tilde{x} \\ \end{align} once again demonstrating that $\tilde{x}$ and $\tilde{y}$ are compatible charts on $TM$ near $p$.

  • $\begingroup$ Sorry I know this is an old question but I have a doubt... Once I have computed the preimage of an element in $\mathbb R^{2n}$, for example $(x^1,\dots,x^n,v^1,\dots v^n)$, which is $$\tilde{\varphi^{-1}}(x^1,\dots,x^n,v^1,\dots, v^n)=v^j\dfrac{\partial}{\partial x^i}\vert_{\varphi^{-1}(x^1,\dots,x^n)}$$ are $v^j\dfrac{\partial}{\partial x^i}\vert_{\varphi^{-1}(x^1,\dots,x^n)}$ and $(\varphi^{-1}(x^1,\dots,x^n),\textbf v)$ the same thing? Are those two equivalent things? $\endgroup$
    – Vajra
    Commented Jun 11, 2021 at 10:01

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