My book is Connections, Curvature, and Characteristic Classes by Loring W. Tu (I'll call this Volume 3), a sequel to both Differential Forms in Algebraic Topology by Loring W. Tu and Raoul Bott (Volume 2) and An Introduction to Manifolds by Loring W. Tu (Volume 1).

Definition 1.5 gives the definition for Riemannian metric and Riemannian manifold. Example 1.9 says

If $F : N \to M$ is a diffeomorphism and $< , >$ is a Riemannian metric on $M$, then (1.3) defines an induced Riemannian metric $< , >'$ on $N$.

The following is my proof of Example 1.9.

  1. Question 1: Is this proof correct?

  2. Question 2:

    • If this proof is correct, then is there a way to do this without relying on pushforwards from Volume 1 or without injectivity of $F$?

      • I guess we can come up with a similar proof for an embedding, but embeddings are injective. So we'll have to go with investigating local diffeomorphisms, local diffeomorphisms onto image, immersions, etc.

      • I'm asking because the Example 1.10 seems to do similarly to Example 1.9 though the $F$ in Example 1.10 is not injective.

    • If this proof is incorrect, then why?


Notation from Volume 1 Section 2.4: For a smooth manifold $N$, let $\mathfrak X (N)$ be the set of smooth vector fields on $N$, and let $C^{\infty}N$ be the set of smooth functions on $N$ (not germs).

We must show that

A. (Not interested in proving this part, but I'm stating what is to be proven for completeness) For all $p \in N$, the mapping $\langle , \rangle'_p: (T_pN)^2 \to \mathbb R$ is an inner product on $T_pN$, where $\langle , \rangle'_p$ is given as follows:

  • Let $u,v \in T_pN$. Then $F_{*,p}u, F_{*,p}v \in T_{F(p)}M$.

  • Let $\langle , \rangle_{F(p)}: (T_{F(p)}M)^2 \to \mathbb R$ be the inner product on $T_{F(p)}M$ given by the Riemannian metric $\langle , \rangle$ on $M$, at the point $F(p) \in M$.

  • Then $(\langle , \rangle'_p)(u,v) = \langle u, v \rangle'_p = \langle F_{*,p}u, F_{*,p}v \rangle_{F(p)}$.

B. $\langle X,Y\rangle' \in C^{\infty}N$ for all $X,Y \in \mathfrak X (N)$, where $\langle X,Y\rangle': N \to \mathbb R$, $\langle X,Y \rangle'(p)=\langle X_p,Y_p\rangle'_p$ $=\langle F_{*,p}X_p,F_{*,p}Y_p\rangle_{F(p)}$.

To prove B:

  1. Let $X,Y \in \mathfrak X (N)$. Then, by Volume 1 Example 14.15, $F_{*}X$ and $F_{*}Y$ are defined vector fields on $M$.

  2. Hopefully, $F_{*}X$ and $F_{*}Y$ are smooth, i.e. $F_{*}X,F_{*}Y \in \mathfrak X (M)$. (I ask about this step here.)

  3. $\langle A, B \rangle \in C^{\infty} M$ for all $A,B \in \mathfrak X(M)$, by definition of $\langle , \rangle$ for $M$ (Definition 1.5).

  4. $\langle F_{*}X,F_{*}Y \rangle \in C^{\infty}M$, from (2) and (3).

  5. $\langle X,Y\rangle' = \langle F_{*}X,F_{*}Y \rangle \circ F$, i.e. $\langle X,Y\rangle'$ is the pullback by $F$ of $\langle F_{*}X,F_{*}Y \rangle$

  6. $\langle X,Y\rangle' \in C^{\infty}N$, by Volume 1 Proposition 6.9, by (4) and by smoothness of $F$.

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    $\begingroup$ ignore that. didn't see the unit circle $\endgroup$ – user10354138 Jun 20 '19 at 12:04
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    $\begingroup$ You can pull back the metric as long as $F_*$ is injective on the tangent space (so $F$ is a local diffeo onto image). Injectivity of $F$ only matters if you want global isometry. $\endgroup$ – user10354138 Jun 20 '19 at 12:09
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    $\begingroup$ For the inner product on $T_pN$, what matters is only the derivative of $F$ at $p$, not what $F$ does a million miles away. So injectivity of $F$ is irrelevant for pulling back the metric along an immersion. $\endgroup$ – user10354138 Jun 20 '19 at 12:27
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    $\begingroup$ If $F$ was not an immersion, then $F^*$ is not injective for some $p\in N$, so $F^*(u)=0$ for some non-zero $u\in T_p N$, then $\langle u, u\rangle'=0$. So $\langle, \rangle'$ does not define an inner product. Also please refrain from asking too many questions in a short period of time, it can get a bit overwhelming. $\endgroup$ – lEm Jun 21 '19 at 6:20
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    $\begingroup$ So at least we need $F$ to be an immersion everywhere, I believe this implies that $F$ is a local diffeomorphism onto $F(N)\subset M$. ($F(N)$ needs not be a manifold though.) $\endgroup$ – lEm Jun 21 '19 at 6:25

$\textbf{Question 1:}$ Yes, it is correct.

$\textbf{Question 2:}$ Yes, there is. Even though your proof is correct, it relies more on global properties than it needs to. The trick here is to do things locally, using coordinates.

Let $F\colon M\to N$ be a smooth map and $\left<\cdot\,,\cdot\right>$ be a metric on $N$. You can always define $\left<\cdot\,,\cdot\right>'$ on $M$ the way you did. Then $\left<\cdot\,,\cdot\right>'$ is easily seen to be bilinear and symmetric at each point (please tell me if this is not clear) and, in fact, we can show that it is also smooth (i.e., $\left<X,Y\right>'\colon N\to \mathbb{R}$ is smooth for any $X,Y\in\mathfrak{X}(N)$) without any further assumptions on $F$. After that, all that's left for it to be a metric is to be non-degenerate at each point, which you get by assuming that $(F_*)_p$ is injective at each point $p\in M$ (i.e., assuming $F$ is an immersion), as was already pointed out in the comments.

So let $U\subset M$ be a coordinate neighborhood in $M$ and $V\subset N$ a coordinate neighborhood in $N$ containing $F(U)$, with $\phi=(x^1,\ldots, x^m): U\to\mathbb{R}^m$ and $\psi=(y^1,\ldots, y^n):U\to\mathbb{R}^n$ the corresponding charts. Then for any vector field $\tilde{X}\in\mathfrak{X}(N)$, we have, for $q\in V$ $$\tilde{X}_q=\sum_{i=1}^n\tilde{X}^i\left(q\right)\left(\frac{\partial}{\partial y^i}\right)_q$$

for smooth functions $\tilde{X}^i:V\to\mathbb{R}$. Furthermore, since the $\frac{\partial}{\partial y^i}$'s form a basis for the tangent space at each point and $\left<\cdot\,,\cdot\right>$ is bilinear, you have functions $g_{ij}:U\to\mathbb{R}$,with $1\leq i,j\leq n$, such that, for any $\tilde{X},\tilde{Y}\in\mathfrak{X}(N)$ and $q\in V$


By assumption, this is smooth for every pair of vector fields, so the $g_{ij}$'s must be smooth.

Also, I'm not going to show this, as it's a basic fact of differential geometry (and an expected one too since $F_*$ is supposed to be a generalized derivative), but, for any vector field $X\in\mathfrak{X}(M)$ with

$$X_p=\sum_{i=1}^mX^i(p)\left(\frac{\partial}{\partial x^i}\right)_p$$ you have $$(F_*)_p(X_p)=\sum_{i=1}^m\sum_{j=1}^nX^i(p)\frac{\partial \tilde{F}^j}{\partial x^i}(p)\left(\frac{\partial}{\partial y^j}\right)_{f(p)}$$

where $\tilde{F}^j=y^j\circ F\circ \phi^{-1}:U\to \mathbb{R}$ for each $1\leq j\leq n$. Then, if $Y\in\mathfrak{X}(M)$ with

$$Y_p=\sum_{i=1}^mY^i(p)\left(\frac{\partial}{\partial x^i}\right)_p$$ you have $$\left<X,Y\right>'(p)=\sum_{i,j=1}^n\sum_{k,l=1}^mg_{ij}(f(p))X^k(p)\frac{\partial \tilde{F}^i}{\partial x^k}(p)Y^l(p)\frac{\partial \tilde{F}^j}{\partial x^l}(p)$$ which is smooth in $p$ since it's just a sum of products of smooth functions. Since the coordinate neighborhoods are arbitrary, we conclude that $\left<\cdot\,,\cdot\right>'$ is smooth.

More generally, a multilinear map $\omega_q:\left(T_qN\right)^k\to\mathbb{R}$, for each $q\in N$, that varies smoothly with $q$, in the sense that $\omega(X_1,\ldots,X_k):N\to\mathbb{R}$ is smooth for any $X_1,\ldots,X_k\in\mathfrak{X}(N)$, is called a $k$-covariant tensor field and you can show, similarly to what I did above, that $\omega'_p:\left(T_pM\right)^k\to\mathbb{R}$ given by


varies smoothly with $p$. $\omega'$ is called the pullback of $\omega$ and is usually written $F^*\omega$. What this shows is that, unlike the pushforward, the pullback is always smooth and well-defined without any further assumptions on $F$, other than being smooth.

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    $\begingroup$ It is exactly that Proposition $8.11$, I've just written out what they are saying, there is no difference. Im not sure I understand what you mean but, in general, you can't represent the $F_{*,p}$ in terms of a single basis element, that is not what they are doing there, $i$ is not fixed. $\endgroup$ – Paulo Mourão Jul 21 '19 at 8:58
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    $\begingroup$ Yes definitely (sorry I had missed your first comment). It is called the $\textbf{pullback metric}$ apparently, according to this Wikipedia article. $\endgroup$ – Paulo Mourão Jul 21 '19 at 10:21
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    $\begingroup$ Yes, you can prove part B without assuming that $F$ is an immersion. As for the question you linked, I haven't had a chance to look at it yet. Hopefully, I will soon. $\endgroup$ – Paulo Mourão Jul 21 '19 at 11:34
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    $\begingroup$ You don't need to assume anything, except smoothness of the map, because you do $\textbf{not}$ need pushforwards of vector fields to be well-defined (you're taking pushforwards of vectors pointwise). Your $\left<\cdot,\cdot\right>$ is a particular instance of what is called a tensor field on your manifold and $\left<\cdot,\cdot\right>'$ is called the pullback of that tensor field and it is always smooth if the map is smooth (i.e. B always holds). $\endgroup$ – Paulo Mourão Jul 23 '19 at 12:14
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    $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – Paulo Mourão Jul 23 '19 at 12:17

Yes, I think you are basically correct.

Because $F$ is a diffeomorphism, $F:N\to M$ it induces an isomorphism of tangent spaces $F_{*,p}:T_pN\to T_{F(p)}M$. This allows us to define (as you did) an inner product pointwise on $T_pN$ by $\langle u,v\rangle_p=\langle F_* u, F_*v\rangle_{F(p)}$ for any $u,v\in T_pN$. We just need to check that these definitions of inner products $\langle \:\cdot,\cdot\:\rangle$ vary smoothly with $p$ in the sense necessary to define a Riemannian metric.

To do this, let $X,Y\in \mathfrak{X}(N)$ be given, and notice that $F$ pushes forward smooth vector fields to smooth vector fields (being a $\mathscr{C}^\infty$ diffeomorphism). So, $F_* X ,F_* Y\in \mathfrak{X}(M)$. Then on $N$, $$ \langle X,Y\rangle:N\to \mathbb{R}$$ given by $p\mapsto \langle F_*X_p,F_*Y_p\rangle_{F(p)}$ is smooth, being a composition of $p\mapsto F(p)\mapsto \langle F_* X_p, F_* Y_p\rangle_{F(p)}.$ The second map is smooth by one of the characterizations of smoothness of a Riemannian metric, and smoothness of the pushforward vector fields.

  • $\begingroup$ Thanks. Please clarify: This answers question 1 but not question 2? $\endgroup$ – user636532 Jul 22 '19 at 5:09

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