7

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?

Proof:

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 $F$.

  • @user10354138 So, Example 1.10 is correct or incorrect? –  Jun 20 '19 at 12:01
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    ignore that. didn't see the unit circle – user10354138 Jun 20 '19 at 12:04
  • @user10354138 Thanks anyway. Are you sure Example 1.10 is not incorrect simply from $F$'s being not injective? (In other words, Example 1.10 either is correct or is incorrect for a different reason) –  Jun 20 '19 at 12:06
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    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. – user10354138 Jun 20 '19 at 12:09
  • @user10354138 Let $\tilde{F}$ be $F$ with restricted range $\tilde{F}: N \to F(N)$, $\tilde{F}(p)=F(p)$. Follow-up 1. Am I correct to say that that $F_{*,p}$ is injective at every $p \in N$ is equivalent to that $F$ is an immersion at every $p \in N$ and is weaker but not equivalent to that $\tilde{F}$ is a local diffeomorphism? Wikipedia: "A local diffeomorphism is a special case of an immersion". Follow-up 1.1. In this case, I think your parenthetical remark should be instead "(So $F$ is an immersion)" ? –  Jun 20 '19 at 12:14
  • @user10354138 Follow-up 2. So, $F: S^1 \to S^1, F(z)=z^2$ is local diffeo and thus immersion and therefore each $F_{*,p}$ is injective? –  Jun 20 '19 at 12:16
  • @user10354138 Follow-up 3. Actually if we have $F$ not injective even if $F$ immersion, I think the issue of the well-definedness of $F_X$ is that $(F_X)(q) = (F_X)q$ is not well-defined because $q$ has more than one element in its preimage. Are you saying that $F(p_1)=q=F(p_2),$ $(FX)q$ can be defined as both $(FX)q = F{,p_1}(X_{p_1})$ and $(F_X)q = F{,p_2}(X_{p_2})$ because we somehow have $F_{,p_1}(X_{p_1}) = F_{,p_2}(X_{p_2})$? –  Jun 20 '19 at 12:20
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    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. – user10354138 Jun 20 '19 at 12:27
  • @user10354138 Wow, it's getting late. I kind of misread. So, it's ok if $F$ is not injective or if the pushforward $F_{*}X$ is not defined (and I think not injective implies pushforward is not defined) as long as $F$ is an immersion? Thanks! –  Jun 20 '19 at 12:43
  • @user10354138 Wait where do I use $F$ local diffeo or at least $F$ immersion in my proof? So far I noticed only the part where I use $F$ injective. I guess either that I missed the part where I used $F$ local diffeo or at least $F$ immersion or that there is no such part, but merely that there exists another proof that does not use injectivity of $F$ –  Jun 20 '19 at 12:56
  • @user10354138 By the way, I think I proved for $M = N = S^1 \subseteq \mathbb R^2$ (rather than $\mathbb C$) that $F$ is local diffeo. A. Is this correct? B. Is this also true for $\mathbb C$ (under a different proof of course)? Proof: For each $p=(x_0,y_0) \in N$, $F_{,p}: T_pN \to T_{F(p)}M$ is represented by a matrix with determinant $(2x_0)^2-(-2y_0)(2y_0)=4$. Thus, each matrix is invertible. Equivalently, each $F_{,p}$ is an isomorphism. Equivalently, $F$ is local diffeo at each $p$. (I think the proof for $\mathbb C$ goes something like $F'(z_0) = 2z_0\frac{d}{dw}|_{F(z_0)} \ne 0$.) –  Jun 20 '19 at 13:46
  • @user10354138 Wait so how please would you prove Example 1.9 without using injectivity of $F$ but using that $F$ is a local diffeo or at least $F$ is an immersion? –  Jun 20 '19 at 13:47
  • @user10354138 A thought I had just now: If injectivity is not needed here, then is injectivity not needed there ? I think we could have there that $c$ is a local diffeo onto its image –  Jun 20 '19 at 15:20
  • @user10354138 About "onto image", is a surjective map a local diffeomorphism if and only if it is a local diffeomorphism onto its image? –  Jun 20 '19 at 15:25
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    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. – lEm Jun 21 '19 at 06:20
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    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.) – lEm Jun 21 '19 at 06:25
  • @lEm Thanks. I'll analyze this on Monday. For now, for question 1 please: is the proof correct with $F$ a diffeomorphism and using that $F$ is injective (even if you can make a different proof without using $F$ diffeo or injective) ? –  Jun 21 '19 at 13:02
  • @lEm Okay, I think I understand why we need $F$ to be an immersion (everywhere), but how do you do prove Example 1.9 without assuming injectivity of $F$ (question 2), assuming my proof for Example 1.9 is correct (question 1)? –  Jun 26 '19 at 08:42
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    I think you can use the same proof but applied locally. Given $p\in N$, pick a neighborhood $U\subset N$ of $p$ so that $U\cong F(U)$ (by assumption $F$ is a local diffeomorphism). Then you can pushforward vector fields on $U$ and check that $\langle, \rangle'$ is smooth on $U$. – lEm Jun 26 '19 at 08:57
  • @lEm Oh I think I get it. 1. $F$ immersion is used in Part A but not Part B of the proof? 2. You can pushforward vector fields on $U$ since each $F: U \to F(U)$ is injective? 3. Example 1.9 is still true if "$F: N \to M$ diffeomorphism of smooth manifolds" was replaced with "3A. $F: N \to M$ smooth map of manifolds, 3B. $N$ and $M$ of the same dimension, 3C. $F(N)$ manifold and $F: N \to F(N)$ local diffeomorphism" 4. "$F: N \to M$ diffeomorphism" = 3A + 3B + 3C + "F bijective" ? 5. You make use of the fact that local diffeomorphism is an immersion so immersion is actually not sufficient? –  Jun 26 '19 at 09:19
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    What I meant was that you don't need to require $F$ to be a diffeomorphism. You can always pull back metric from $M$ to $N$ if $F:N\to M$ is a local diffeomorphism (this guarantees that $F$ is an immersion, and $N, M$ are of the same dimension). – lEm Jun 26 '19 at 09:25
  • @lEm Oh I made a mistake. $F:N \to M$ is a local diffeomorphism implies $F:N \to F(N)$ is a local diffeomorphism but not conversely, so the chain goes like this? $$\text{local diffeomorphism} \implies \text{immersion} \implies \text{local diffeomorphism onto image}$$ –  Jun 26 '19 at 10:03
  • @user10354138 Apparently why immersion implies local diffeo onto image was asked in a comment but not answered unless local embedding is the same as local diffeomorphism onto image. I'm just about to investigate this equivalence. Why does immersion imply local diffeo onto image? –  Jun 27 '19 at 08:46
  • @lEm Apparently why immersion implies local diffeo onto image was asked in a comment but not answered unless local embedding is the same as local diffeomorphism onto image. I'm just about to investigate this equivalence. Why does immersion imply local diffeo onto image? –  Jun 27 '19 at 08:47
  • @lEm what is a local diffeomorphism onto image exactly please? I know what local diffeomorphisms, local homeomorphisms onto image and local homeomorphisms are. The issue with copying the definition of local homeomorphism onto image to local diffeomorphism onto image has the problem of the image possibly not being a submanifold or manifold while there is no such issue for local homeomorphism onto image since image can always be made into a subspace. –  Jul 22 '19 at 05:08
  • @user10354138 1. When you say immersions $F$ are local diffeomorphisms onto their images $F(N)$, if we assume that $F(N)$ is a (regular/an embedded) submanifold, then do you or do you not assume that $N$ and $F(N)$ have the same dimension (regardless of whether or not they actually have the same dimensions)? 2. If $F$ is an immersion with submanifold $F(N)$, then does $F(N)$ have the same dimension as $N$? –  Jul 23 '19 at 06:13
  • @user10354138 Edit: $F(N)$ and $N$ have the same dimension, or $N$ and $M$ have the same dimension or something. (I think $\dim N = \dim M$ implies $\dim N = \dim F(N)$) –  Jul 23 '19 at 06:33
  • @lEm I think I figured it out! Immersions are not local diffeomorphisms onto their images, but they are equivalent to local embeddings! For each $p \in N$ and $F$ immersion (everywhere), there exists a $U$ such that $F(U)$ is a submanifold of $M$ and $F|_{U} : U \to F(U)$ is a diffeomorphism. Local diffeomorphism onto image gives that $F(U)$ is both open in $F(N)$ and a submanifold of $M$, but we don't need both! A local embedding gives $F(U)$ is a submanifold, which is all we really need? –  Jul 26 '19 at 07:07
  • @user10354138 If I understand right both you and lEm have beautifully preserved my idea to use pushforwards for an immersion $F$. I ask about this here. You both can answer both this question and that question, and I can upvote both your answers to both questions. –  Jul 26 '19 at 08:01

2 Answers2

2

$\textbf{Question 1:}$ 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 $q\in V$

$$\left<\tilde{X},\tilde{Y}\right>(q)=\sum_{i,j=1}^ng_{ij}(q)\tilde{X}^i(q)\tilde{Y}^j(q)$$

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

$$\omega'_p(v_1,\ldots,v_k)=\omega_{f(p)}\left(\left(F_*\right)_pv_1,\ldots,\left(F_*\right)_pv_k\right)$$

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.

Paulo Mourão
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  • Paulo Mourão, thanks. In short, $F$ immersion alone suffices? –  Jul 21 '19 at 06:21
  • Is the "basic fact" you mention based on Proposition 8.11 (but now we do $F_{*,p}$ on a linear combination of the basis elements rather than on just a single basis element)? –  Jul 21 '19 at 06:23
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    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. – Paulo Mourão Jul 21 '19 at 08:58
  • Thanks. (Again please) is $F$ immersion alone sufficient to induce a Riemannian metric? –  Jul 21 '19 at 09:59
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    Yes definitely (sorry I had missed your first comment). It is called the $\textbf{pullback metric}$ apparently, according to this Wikipedia article. – Paulo Mourão Jul 21 '19 at 10:21
  • Oh that's nice. Thanks! No need to be sorry. And to clarify, what you do in this long answer is prove $F$ immersion implies metric, without assuming anything stronger than $F$ immersion (eg $F$ diffeo, $F$ local diffeo, etc)? –  Jul 21 '19 at 10:27
  • Yes exactly, but not just that. Smoothness follows even without it being an immersion (it's true for any smooth map). You only need the immersion condition (and it suffices) for it to be nondegenerate. – Paulo Mourão Jul 21 '19 at 10:35
  • Wait wait the immersion is ONLY for Part A (the inner product part) because the immersion is NOT for Part B (the smoothness part) i.e. Part B can be proved from just $F$ smooth without $F$ being an immersion? –  Jul 21 '19 at 10:36
  • Wait do I actually indeed prove Part B for smooth $F$ even if $F$ is not an immersion? –  Jul 21 '19 at 10:42
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    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. – Paulo Mourão Jul 21 '19 at 11:34
  • Paulo Mourão thanks! –  Jul 21 '19 at 11:36
  • Paulo Mourão what is a local diffeomorphism onto image exactly please? I know what local diffeomorphisms, local homeomorphisms onto image and local homeomorphisms are. The issue with copying the definition of local homeomorphism onto image to local diffeomorphism onto image has the problem of the image possibly not being a submanifold or manifold while there is no such issue for local homeomorphism onto image since image can always be made into a subspace. –  Jul 22 '19 at 05:08
  • Paulo Mourão, actually about the immersion being sufficient for part A and any smooth map works for part B, does this assume or not assume anything like 1. N and M have equal dimensions or 2. $F(N)$ is a (regular/an embedded) submanifold and $N$ and $F(N)$ have equal dimensions (you might recall our previous discussion about dimensions)? –  Jul 23 '19 at 06:33
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    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). – Paulo Mourão Jul 23 '19 at 12:14
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    Let us continue this discussion in chat. – Paulo Mourão Jul 23 '19 at 12:17
0

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.

Alekos Robotis
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  • Thanks. Please clarify: This answers question 1 but not question 2? –  Jul 22 '19 at 05:09