# Does the pullback of the metric tensor have the form $\phi^* g(X,Y)=g(\phi X,\phi Y)$?

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 $$\phi$$?

• 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). Jun 21, 2020 at 17:29
• @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 $g(\phi(X), \phi(Y))$ Jun 21, 2020 at 18:27

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."

• 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! Jun 21, 2020 at 18:55
• @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 Jun 21, 2020 at 19:40
• 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. Jun 21, 2020 at 19:42
• 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 $(\phi^*g)(X,Y) = g(\phi_*X, \phi_*Y)$. Jun 21, 2020 at 19:59
• 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? Jun 21, 2020 at 20:47