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I have seen in many textbooks that the pullback of an arbitrary tensor field of type (r,s) under the diffeomorphism $\phi:M \rightarrow N$ is defined as

$\phi^* T(\eta_1,\dots, \eta_r, X_1, \dots, X_s) = T( (\phi^{-1})^*(\eta_1), \dots, (\phi^{-1})^*(\eta_r), \phi_* X_1, \dots, \phi_* X_s)$

where $\eta_i \in T_p^*(M)$ is a covector and $X_j \in T_p(M)$ is a vector. So, in the case of the metric tensor this would reduce to the following:

$\phi^*g(X,Y) = g(\phi_*X, \phi_*Y)$

where $X,Y$ are vectors.

Now, at the same time, Wikipedia suggests that we could find the pullback of such a tensor as

$\phi^*g(X,Y) = g(\phi X, \phi Y)$

and my question is how do they get $\phi_*$ to become $\phi$?

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    $\begingroup$ I'm not sure which part of the wikipedia artice you're reading but it should be $(\phi^*g)(X,Y) = g(\phi_*X, \phi_*Y)$ (which, apart from notation, is pretty much what's written in this section). $\endgroup$
    – peek-a-boo
    Jun 21, 2020 at 17:29
  • $\begingroup$ @peek-a-boo On the page you link to, I am trying to understand the second equation in the section "Pullback of multilinear forms" where they have $g(\phi(X), \phi(Y))$ $\endgroup$
    – user11128
    Jun 21, 2020 at 18:27

1 Answer 1

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If $\phi:V \to W$ is a linear map, then for any $p\in V$, the tangent mapping/pushforward mapping $T\phi_p$ or $d\phi_p$ or $\phi_{*,p}$ (however you want to use the notation) is a linear map $T_pV \to T_{\phi(p)}W$. But for a vector space, the tangent space can be canonically identified with itself: $T_pV \cong V$ and $T_{\phi(p)}W\cong W$. Because of this, you can "think of" the tangent mapping as a map $V \to W$. This is simply the derivative of a linear transformation $\phi:V \to W$ at the point $p \in V$. But a linear transformation is its own derivative.


If you want a more precise formulation of what I said above, here it is: on any (say finite-dimensional) vector space $V$, and any $p \in V$, there is a canonical isomorphism $\xi_{V,p}:T_pV \to V$. Note that the exact construction of this isomorphism will depend on which definition of tangent space you're using, but in any case, it is a good idea to prove this yourself. Similarly we have an isomorphism $\xi_{W,\phi(p)}:T_{\phi(p)}W \to W$. If you unwind the definitions of everything, you'll see that the following diagram commutes:

$\require{AMScd}$ \begin{CD} T_pV @>{\phi_{*p}}>> T_{\phi(p)}W \\ @V{\xi_{V,p}}VV @VV{\xi_{W,\phi(p)}}V \\ V @>>{D\phi_p = \phi}> W \end{CD} In other words, $\phi = \xi_{W,\phi(p)} \circ \phi_{*,p} \circ (\xi_{V,p})^{-1}$, or said differently once again, up to isomorphisms, for each $p \in V$, we have $\phi_{*,p} = \phi$. But all of this is only because $\phi$ is a linear transformation.


But in the general case, if you have smooth manifolds $M,N$, and you have a metric tensor $g$ on $N$ and a diffeomorphism $\phi:M \to N$, there is no reason to even expect that $M,N$ have vector space structures, so it doesn't even make sense to talk about $\phi$ being linear. This is why we have to use the push-forward map, and there is no sense in which we can "identify" the push-forward with the original map itself.

See this for a more general perspective of everything I mentioned here (with slightly different notation).


Edit: In response to comment.

The author DOES NOT say $(s^{-1})^*g(x,y) = g(s^{-1}(x), s^{-1}(y))$. He says \begin{align} d_{B^n}(x,y) &= [(s^{-1})^*d_{\mathcal{H}^n}](x,y) = d_{\mathcal{H}^n}(s^{-1}(x), s^{-1}(y)) \end{align} These are completely different statements. Note that if you have two (let's for simplicity say simply connected) Riemannian manifolds $(M,g)$ and $(N,h)$. Then, the metric tensors $g$ and $h$ give rise to distance functions $d_g$ and $d_h$ respectively (in the article, the author refers to these as $d_{B^n}$ and $d_{\mathcal{H}^n}$). Now, suppose we have a diffeomorphism $\phi:M \to N$. Then, we can consider the following objects:

  • first is the pullback tensor field $(\phi^{-1})^*g$ on $N$ (as defined above).
  • second is the pull-back distance function $(\phi^{-1})^*d_g$ on $N$, which is DEFINED as \begin{align} [(\phi^{-1})^*d_g](x,y) &= d_g\left( \phi^{-1}(x), \phi^{-1}(y)\right) \qquad \text{for all $x,y \in N$} \end{align}

Note that although we are using the same notation $(\phi^{-1})^*$, and calling both of them "pullbacks", these are completely different things. The first is a pullback of tensor field, while the second is a pull-back of a distance function. The word "pull-back" should be thought of literally as the name suggests: you have a certain object defined on one space (eg. a tensor field or distance function), and you have a invertible map between two spaces. Then, you can use this map to "transport" this object to the new space.

Now, here is a theorem which you should try to prove (it is really just an exercise in unwinding all the definitions).

Theorem.

Let $(M,g), (N,h)$, $\phi:M \to N$, and $d_g,d_h$ all have the same definitions as above. If $h = (\phi^{-1})^*g$ then $d_h = (\phi^{-1})^*d_g$.

What this says is that if your metric tensors are related to each other by a pullback, then so are the associated distance functions. Note that this is precisely what the author is saying in the first sentence of his proof:

"Since the stereographic projection $s: \mathcal{H}^+ \to B^n$ is a Riemannian isometry, it is also a metric isometry for the induced distances."

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  • $\begingroup$ Great answer. A couple of follow ups: (1) would the isomorphism $\xi_{V, p} $ be the exponential map? (2) I'm interested in applying this to the case where $\phi$ is the stereo graphic projection from the one sheeted hyperboloid to the poincare disk and since these will both be vector spaces, I think I can use the result! $\endgroup$
    – user11128
    Jun 21, 2020 at 18:55
  • $\begingroup$ @user11128 no it has nothing to do with the exponential map. In general on any manifold $M$ modeled on a vector space $V$ (typically $\Bbb{R}^n$), if you take a chart $(U, \alpha)$, then for each $p \in M$, you can construct an isomorphism $\xi_{\alpha,p}:T_pM \to V$ (i.e the tangent bundle is a locally trivial and linear on each fiber). For the case where $M = V$ is itself a vector space, we can trivially use the identity chart to induce an isomorphism. If you use the definition of tangent space using equivalence classes of smooth curves, then $\endgroup$
    – peek-a-boo
    Jun 21, 2020 at 19:40
  • $\begingroup$ in the general case, the isomorphism $\xi_{\alpha,p}:T_pM \to V$ is given as taking an equivalence class of curves $[\gamma]_p \in T_pM$, with $\gamma(0) = p$ and mapping it to the vector $(\alpha \circ \gamma)'(0) \in V$. In the case of $M=V$, a vector space, using the identity chart $(V, \text{id}_V)$, we get an induced isomorphism $\xi_{\text{id}_V,p}:T_pV \to V$, $[\gamma] \mapsto (\text{id}_V \circ \gamma)'(0) = \gamma'(0)$. i.e each curve gets mapped to its tangent vector. What I refer to as $\xi_{\text{id}_V,p}$ here is what I simply referred to as $\xi_{V,p}$ in my answer above. $\endgroup$
    – peek-a-boo
    Jun 21, 2020 at 19:42
  • $\begingroup$ as for point (2), I'm not actually familiar with these spaces (I've only had an introductory course in differential geometry), but from a brief google search, these spaces are only subsets of vector spaces, they aren't vector spaces themselves. Also, stereographic projection isn't a linear map, so you have to keep the pushforwards everywhere; i.e you have to use $(\phi^*g)(X,Y) = g(\phi_*X, \phi_*Y)$. $\endgroup$
    – peek-a-boo
    Jun 21, 2020 at 19:59
  • $\begingroup$ that's peculiar. i'm looking at proposition 8.6 of arxiv.org/pdf/2003.11180.pdf. in this case, $s^{-1}$ is the inverse stereographic projection and he does a pullback with respect to this, you will notice he has $(s^{-1})^*g(x,y) = g(s^{-1}x, s^{-1}y)$. do you understand why? $\endgroup$
    – user11128
    Jun 21, 2020 at 20:47

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