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Suppose that $M$ and $N$ are finite-dimensional vector spaces and therefore smooth manifolds and $F: M \to N$ is a smooth map. I have seen in my manifolds lecture that the Professor determines properties of the tangent map $F_{*p} : T_p M \to T_{F(p)}N$, and sometimes even its Jacobian matrix, simply by computing the directional derivative: $$ D_aF(r) = \lim_{t \to 0} \frac{1}{t}\left[ F(a + tr) - F(a) \right]. $$ In the case that $M = \mathbb{R}^m$ and $N = \mathbb{R}^n$, this is not surprising because then we have the formula $$ J_aF\cdot r = D_aF(r) \qquad \qquad (1) $$ So this formula (1) provides a convenient way to determine the Jacobian matrix from the directional derivative if $M$, $N$ are $\mathbb{R}^{m},\mathbb{R}^{n}$.

  1. What is the most convenient way to determine $F_*$ from the directional derivative in the case that $M$ and $N$ are any finite-dimensional vector space (and not necessarily $\mathbb{R}^{m},\mathbb{R}^{n}$)?
  2. Are there properties of $F_*$ (in particular rank or regular values) that are conveniently determined from the directional derivative whenever $M$ and $N$ are any finite-dimensional vector space? If so, how can I determine them?

EDIT: I am modelling the manifolds on $\mathbb{R}^d$, so I have coordinates $\phi : M \to \mathbb{R}^m$ on $M$ and $\psi : N \to \mathbb{R}^n$ on $N$. If $(b_i)_{i=1}^m$ is a basis for $M$$\phi(b_i) = e_i$ is the coordinate mapping of that basis ($e_i$ standard basis on $\mathbb{R}^d$). $\psi$ is defined similarly with whatever the basis is of $N$.

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In general, let $M$ be a smooth manifold modelled on a Banach space $X_1$, let $p \in M$. Then, a choice of chart $(U, \alpha)$ about $p$ allows us to identify the tangent space $T_pM$ with the underlying model space $X_1$, which means one can construct a linear isomorphism $\Phi_{\alpha,p}: T_pM \to X_1$ (in finite dimensions, this can be seen as the reason why the tangent space dimension and the manifold dimension are equal)

Next, suppose also that we have another smooth manifold $N$ modelled on a Banach space $X_2$, and a smooth map $F:M \to N$. Then, as above, a choice of chart $(W, \beta)$ about $F(p)$ allows us to construct a linear isomorphism $\Phi_{\beta, F(p)}: T_{F(p)}N \to X_2$.

Now, we employ a tool often used in linear algebra: we study the map $F_{*p}$ by studying the "isomorphism related" map $S := \Phi_{\beta,F(p)} \circ F_{*p} \circ \Phi_{\alpha,p}^{-1}: X_1 \to X_2$. To keep all these spaces and maps clear in your mind, consider the following diagram (which is commutative by definition of $S$):

$\require{AMScd}$ \begin{CD} T_pM @>{F_{*p}}>> T_{F(p)}N \\ @V{\Phi_{\alpha,p}}VV @VV{\Phi_{\beta, F(p)}}V \\ X_1 @>>{S}> X_2 \end{CD}

By unravelling the definitions of $F_{*p}$, it shouldn't be too hard to prove that \begin{equation} S = d(\beta \circ F \circ \alpha^{-1})_{\alpha(p)}. \end{equation} (here the notation $dG_{\xi}$ means the Frechet derivative of $G$ at $\xi$, where $G$ is a map between Banach Spaces). Hence, inverting the relation in the diagram, we can write \begin{align} F_{*p} &= \Phi_{\beta,F(p)}^{-1} \circ S \circ \Phi_{\alpha,p} \\ &= \Phi_{\beta,F(p)}^{-1} \circ d(\beta \circ F \circ \alpha^{-1})_{\alpha(p)} \circ \Phi_{\alpha,p} \end{align}

This idea should be familiar from linear algebra, where given a linear transformation between finite-dimensional spaces, a choice of basis on the domain and target space allow us to rephrase everything about the linear map in terms of its corresponding matrix relative to those bases.

Hence, if you want to study the linear map $F_{*p}$, you just choose charts in the domain and target space of $F$, and consider the derivative of the "chart-representative" map $\beta \circ F \circ \alpha^{-1}$. Usually, the spaces $X_1$ and $X_2$ will be "familiar" to us, whereas the tangent spaces will be "abstract" (their elements could be complicated objects like equivalence classes of curves, or derivations etc), hence we prefer to work with $S$ rather than $F_{*p}$ directly. So, any property of linear transformations which is preserved under isomorphism (such as rank) can be studied more easily via the "induced map" $S$ rather than the original map $F_{*p}$.


So far everything I've said holds for general smooth manifolds $M,N$. But, in the case where $M$ and $N$ are finite dimensional vector spaces (not necessarily cartesian), we can make another simplification. Since they are vector spaces, the manifold structure is the one obtained by the identity chart, so in the discussion above, we can replace $\alpha = \text{id}_M$ and $\beta = \text{id}_N$ (and suppress the points $p$, $F(p)$ for convenience). Hence, our commutative diagram becomes $\require{AMScd}$ \begin{CD} T_pM @>{F_{*p}}>> T_{F(p)}N \\ @V{\Phi_{\text{id}_M}}VV @VV{\Phi_{\text{id}_N}}V \\ M @>>{S = dF_p}> N \end{CD}

Hence, we have that \begin{equation} F_{*p} = \Phi_{\text{id}_N}^{-1} \circ dF_p \circ \Phi_{\text{id}_M} \end{equation}

Hence, to determine the map $F_{*p}$ and its properties, all you need to do is compute $dF_p$ and its properties, and then "rephrase" all of that information into the language of tangent vectors via the isomorphisms.

Now, we have reduced the study of $F_{*p}$ to the study of $dF_p$. Note that for any $\xi \in M$, we have (by the chain rule) \begin{equation} dF_p(\xi) = \dfrac{d}{dt}\bigg|_{t=0} F(p + t \xi). \end{equation}


So as a conclusion: the bare minimum you need to do to determine $F_{*p}$ is the following:

  • Pick a convenient basis $\{\xi_1, \dots, \xi_n\}$ of $M$.
  • For each $\xi_i$ in the basis, compute \begin{equation} dF_p(\xi_i) = \dfrac{d}{dt}\bigg|_{t=0} F(p + t \xi_i) \end{equation} Since $dF_p$ is linear, by knowing its values on a basis, you know it everywhere on $M$.
  • Explicitly write down the isomorphisms $\Phi_{\text{id}_M}: T_pM \to M$, and likewise for $N$. (This of course depends on your actual construction of the tangent space; whether you constructed it as equivalence classes of curves/ derivations/ germs/ quotient by a certain ideal etc)
  • Perform the compositions mentioned above to calculate $F_{*p}$.

Reference: Loomis and Sternberg Advanced Calculus, section $9.4$ in particular, which is about tangent spaces (and also some earlier chapters for linear algebra)

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  • $\begingroup$ Thanks for writing up such a detailed answer! Would you consider adding the following? I have only seen manifolds where the spaces $\{X_i\}_{i=1}^2$ are $\mathbb{R}^m,\mathbb{R}^n$, therefore I have not seen the case where the coordinates $\alpha$, $\beta$ is the identity map. We have followed J.Lee and used the coordinate mapping $\tilde\alpha$, defined as follows: If $\{ b_i \}_{i=1}^m$ is a basis for $M$, then $\tilde\alpha$ sends $v^1b_1 + \cdots v^mb_m \mapsto (v^1,\cdots,v^m)$. If you could specify what the answer would be in this specific case, it would really help me out! $\endgroup$
    – Mikkel Rev
    Jun 15 '19 at 18:45
  • $\begingroup$ in that case, only consider the first half of what I said, and ignore the rest. So here, $M$ is the given vector space, $(M, \tilde{\alpha})$ is a chart on $M$, and so you should have an isomorphism $\Phi_{\tilde{\alpha},p}: T_pM \to \Bbb{R^m}$. This isomorphism is something you'd have to construct by yourself, using the definitions of tangent space you are used to (I hope you're atleast convinced that there exists an isomorphism). Now, do the same for $N$, and apply the first commutative diagram $\endgroup$
    – peek-a-boo
    Jun 15 '19 at 21:57
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    $\begingroup$ @MikkelRev yes it seems right. Also, are you using John M Lee's Introduction to smooth manifolds? Because if you are, then Proposition $3.8$ might be helpful in this context to see the isomorphism between $T_pM$ and $M$, when $M$ is a finite-dimensional vector space. $\endgroup$
    – peek-a-boo
    Jun 16 '19 at 14:07
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    $\begingroup$ It is good that you mentioned you only saw manifolds modeled on $\Bbb{R}^d$, and hence tried to relate what I said with everything you have learnt. But I hope you realise how inconvenient it is (mostly theoretically, but sometimes computationally) to insist on $\Bbb{R^d}$. As you can see, if you allow your model space to be any vector space, then you can allow for identity charts, which simplifies matters. However, if you (or your prof) insist on $\Bbb{R^d}$, then the diagram becomes bigger and you have a lot more maps to keep track of, and also you unnecessarily introduced bases for $M,N$. $\endgroup$
    – peek-a-boo
    Jun 16 '19 at 14:37
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    $\begingroup$ @EpsilonDelta yes I think it is a fantastic book, and I would without any doubt rate it as my favorite even above Spivak/Munkres (lol these are the only 3 books I used for advanced calc) especially chapter 3 on derivatives (even though its mainly a book on calculus, I found the introductory linear algebra sections very enlightening). However, do not mistake clarity for simplicity; this book is very clearly written, but is by no means easy. So, to decide if its suitable for you, it really depends on your background. Also, it depends on the topics of multivariable calc you're looking for. $\endgroup$
    – peek-a-boo
    Nov 10 '19 at 3:18

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