[Ref. 'Core Principles of Special and General Relativity' by Luscombe]

Let $M,\tilde M$ respectively be $n$ and $n'$-dimensional manifolds and $\psi:M\to\tilde M$ be a smooth map. $p\in M$ such that $q=\psi(p)\in\tilde M$. Suppose that $f:\tilde M\to\mathbb{R}$ is a smooth function. The pullback $f\circ\psi:M\to\mathbb{R}$ will also be smooth. So far so good.

Now if we have a vector $\mathbf{t}\in T_p(M)$, the text I'm reading states that $(\psi_*\mathbf{t})f\equiv\mathbf{t}(f\circ\psi)\in T_q(\tilde M)$ without further justification, so I'm trying to prove it.

Definition of vector that's mentioned in the book and the one I'm using: a vector $\mathbf{t}\in T_p(M)$ is an operator $t^i\frac{\partial}{\partial x^i}$ on smooth functions $f:M\to\mathbb{R}$ such that $$\mathbf{t}(f)=t^i\frac{\partial f}{\partial x^i}\ \bigg|_p=t^i\frac{\partial}{\partial x^i}(f\circ\phi^{-1})\ \bigg|_{\phi(p)}$$ where $\{x^i\}$ are the coordinates of $p$ induced by a chart $(U,\phi)$.

I realize if the statement $(\psi_*\mathbf{t}_i)f\equiv\mathbf{t}_i(f\circ\psi)\in T_q(\tilde M)$ holds for basis vectors $\{\mathbf{t}_i\}$ of $T_p(M)$, then I'll be done.

Let a coordinate system $\{x^i\}$ for point $p$ correspond to a chart $(U,\phi)$ in the atlas of $M$. Then effectively I have to prove that (Einstein summation convention) $$\mathbf{t}_i(f\circ\psi)=\frac{\partial}{\partial x^i}(f\circ\psi\circ\phi^{-1})\ \bigg|_{\phi(p)}=\tilde t^j\frac{\partial}{\partial y^j}(f\circ\tilde\phi^{-1})\ \bigg|_{\tilde\phi(q)}$$

for some coordinate system $\{y^i\}$ for $q$ corresponding to some chart $(\tilde U,\tilde\phi)$ in the atlas of $\tilde M$.

I'm struggling to prove this (or maybe I've formulated the problem wrong). I'm guessing the smoothness of $\psi$, i.e. the smoothness of $\tilde\phi\circ\psi\circ\phi^{-1}:\mathbb{R}^n\to\mathbb{R}^{n'}$ for all $\phi,\tilde\phi$ (chart homeomorphisms in $M,\tilde M$ respectively) will come into play. I can express $f\circ\psi\circ\phi^{-1}:\mathbb{R}^n\to\mathbb{R}$ as $(f\circ\tilde\phi^{-1})\circ(\tilde\phi\circ\psi\circ\phi^{-1})$ and maybe proceed along those lines, but I don't know how the denominator will switch to $y^j$ or how the $(\tilde\phi\circ\psi\circ\phi^{-1})$ part will go away.

Note that I don't want to use transformation properties relating $x^i$ and $y^j$ across different manifolds, because in the book, those transformation properties are derived from the main result that I'm trying to prove in this question. Any help appreciated as always!

  • $\begingroup$ What is your definition of $\psi_*$ and hence $\psi_*\mathbf{t}$? $\endgroup$ Commented Jun 22, 2020 at 21:45
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    $\begingroup$ I'm not entirely sure what you're trying to show. I'm going to assume that you agree with the definition of $(\psi_\ast \textbf{t})f \equiv \textbf{t}(f \circ \psi)$ and you want to show that because of this definition, $(\psi_\ast \textbf{t}) \in T_q(M')$. Well, what does it mean to be a tangent vector in $T_q(M')?$ Use your favourite definition of tangent vector (derivation of functions, velocity curves, even germs of smooth functions) and show that those properties hold for $\psi_\ast(\textbf{t})$. $\endgroup$ Commented Jun 22, 2020 at 21:47
  • $\begingroup$ Your equality then follows since any vector in $T_q(M')$ will be expressible in any chosen basis of $T_q(M')$, but it's not really required. $\endgroup$ Commented Jun 22, 2020 at 21:49
  • $\begingroup$ @MichaelAlbanese: Most generally speaking, my interpretation of $\psi_*$ is that it's an operator that acts on vectors in $T_p(M)$ to give an operator $\psi_*\mathbf{t}$ $\endgroup$
    – Shirish
    Commented Jun 22, 2020 at 21:52
  • $\begingroup$ @OsamaGhani: I think I should've explicitly mentioned the definition of a tangent vector that I'm using. Thanks for pointing it out! But I'm trying to do pretty much what you mentioned in the first comment. Have edited the question to include the tangent vector definition I'm using. $\endgroup$
    – Shirish
    Commented Jun 22, 2020 at 22:00

1 Answer 1


You're basically one step away from completion! Consider, as you pointed out

$$\frac{\partial}{\partial x^i} (f \circ \psi \circ \phi^{-1}) = \frac{\partial}{\partial x^i} (f \circ \tilde{\phi}^{-1} \circ \tilde{\phi} \circ \psi \circ \phi^{-1})$$

Denote $\tilde{\phi} \circ \psi \circ \phi^{-1}(x)$ by $y(x)$ i.e. it parametrizes points $y$ as functions of $x$ (check the domains and ranges to see these works out). Then you get by the chain rule (I'm omitting points of evaluation but everything makes sense) $$\frac{\partial}{\partial x^i} (f \circ \tilde{\phi}^{-1} \circ \tilde{\phi} \circ \psi \circ \phi^{-1}) = \frac{\partial (f \circ \tilde{\phi}^{-1})}{\partial y} \frac{\partial y}{\partial x^i} = \tilde{t}^j \frac{\partial}{\partial y^j} (f \circ \tilde{\phi}^{-1})$$

I may have swept some details under the rug but this is the gist of the argument. From here you can actually see how to derive transformations between tangent vectors in different coordinates too.


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