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I am an undergraduate self-studying Differential Geometry from Lang's book "Fundamentals of Differential Geometry". I think there will be a series of doubts from my side, starting with this one.

In the chapter on MANIFOLDS, Lang has said a lot of things which I do not understand why are true.

Background: Consider $X$ to be a set and $(U_i,\phi_i)$ to be a chart, where $U_i\subset X$ and $\phi_i:U_i\to E_i$ where $E_i$ is a Banach space, and $\phi_i$ is a bijection to an open set $\phi_i(U_i)$ of $E_i$.

The statements that I do not understand:

If two charts $(U_i,\phi_i)$ and $(U_j,\phi_j)$ are such that $U_i\cap U_j$ is non-null, then by considering the derivative of $\phi_j\phi_i^{-1}:\phi_i(U_i\cap U_j)\to \phi_j(U_i\cap U_j)$, we see that $E_i$ and $E_j$ are toplinearly isomorphic i.e. there exists a continuous linear isomorphism between $E_i$ and $E_j$.

Why is this true? I know that $\phi_j\phi_i^{-1}$ is an isomorphism from $\phi_i(U_i\cap U_j)$ to $\phi_j(U_i\cap U_j)$. Why then will the two Banach spaces $E_i$ and $E_j$ be top-linearly isomorphic?

The set of points $x\in X$ such that there exists a chart $(U_i,\phi_i)$ at $x$ such that $E_i$ is toplinearly isomorphic to a given space $E$ is both open and closed.

Why is this true?

Therefore, on each connected component of $X$, we can assume that we have an $E-$atlas i.e. collection of charts, for some fixed $E$.

Why is this true, then? I think this has more to do with topology than differential geometry, but want to clarify.

Landon Carter
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2 Answers2

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1) The derivative is a continuous linear map between tangent spaces. Because the map is a diffeomorphism, its derivative must be an isomorphism. The tangent space to any point of a Banach space is the Banach space itself.

2) Clearly it's open. If $x_n \to x$, and there is an $E$-chart around each $x_n$, then pick an $F$-chart around $x$. Necessarily this contains the $x_n$ for large $n$. Thus the tangent space at $x_n$ is isomorphic both to $E$ and $F$, and so $E \simeq F$ as desired.

3) A set that is both open and closed is a union of connected components.

  • Sorry, I have not yet encountered tangent spaces in the book. Should I then accept 1) as a statement I cannot prove and move on, or is there a different explanation? – Landon Carter Aug 23 '16 at 15:51
  • For 2), you say $x$ is trivially open because of the existence of the chart $(U_i,\phi_i)$ where $U_i$ is an open neighborhood of $x$, am I correct? – Landon Carter Aug 23 '16 at 15:52
  • No, ignore my talk of tangent spaces. Do you understand why the derivative of a map at a single point is a map between $E_i$ and $E_j$, and why it must be an isomorphism? 2) Yes.
    • –  Aug 23 '16 at 15:53
  • It is a linear map from $E_i$ to $E_j$ is all I know. but why isomorphism? Sorry if I am irritating you but I am a beginner. – Landon Carter Aug 23 '16 at 16:14
  • Also your explanation to 2) depends on "tangent space" again. – Landon Carter Aug 23 '16 at 16:16
  • @LandonCarter "Tangent space" in this context can just be taken to mean "The Banach space the chart is an open subset of", by definition. For (2), I'm invoking the result of (1), that two different compatible charts at the same point must be based on isomorphic Banach spaces. For (1): You have a map $df_x: E_i \to E_j$, but you also have an inverse map, $dg_{f(x)}: E_j \to E_i$. Because $fg = gf = \text{Id}$, the chain rule says that the composition of these maps is the identity, whence they are isomorphisms. –  Aug 23 '16 at 16:21
  • Thank you for clearing things up. I understood everything except "but you also have an inverse map". Why is the existence of this inverse map guaranteed? Is this a property of a Banach space? – Landon Carter Aug 23 '16 at 16:37
  • @LandonCarter That's the definition of a diffeomorphism! A smooth map with a smooth inverse. The inverse to your map $\phi_j \phi_i^{-1}$ is just $\phi_i \phi_j^{-1}$. –  Aug 23 '16 at 16:39
  • I did not know that the derivative map is a diffeomorphism.I just knew it is continuous linear. I shall read this up then. Thanks for your efforts. – Landon Carter Aug 23 '16 at 16:41
  • @LandonCarter You misunderstand me. The map $\phi_j \phi_i^{-1}$ is a diffeomorphism. The argument I gave showed that if $f$ is a diffeomorphism with smooth inverse $g$, then $df_x$ is an isomorphism $E_i \to E_j$. –  Aug 23 '16 at 16:42
  • Oh! Yeah now this all makes sense. Indeed I misunderstood you, forgive me! And thank you very much!! – Landon Carter Aug 23 '16 at 16:44