4
$\begingroup$

Let $M$ be a smooth manifold and consider the Lee's definition of the tangent space $T_pM$ (so $T_pM$ is the vector space of derivations at $p$). The canonical definition of tangent bundle (as set) of $M$ is: $$TM=\bigcup_{p\in M}\{ p\}\times T_pM$$ so it is the disjoint union of all tangent spaces; but L.W.Tu in his "Introduction to Manifolds" says that the tangent spaces are already disjoint and for this reason he defines $$TM=\bigcup_{p\in M} T_pM$$

Why we can't find a common derivation between $T_pM$ and $T_qM$ if $q\neq p$? I think that Tu's statement is not true.

$\endgroup$
  • 2
    $\begingroup$ How did Lee and Tu respectively define $T_pM$? $\endgroup$ – Neal Nov 4 '12 at 12:32
  • $\begingroup$ This should also be differential-topology, not differential-geometry. $\endgroup$ – M.B. Nov 4 '12 at 12:33
  • $\begingroup$ I just noticed this question. I'll add that with the definition of $T_pM$ that I give in my book (derivations of $C^\infty(M)$ at $p$), the zero derivation is a derivation at $p$ for every $p$, so defining the tangent bundle as a simple union would not work. I understand the advantages of defining $T_pM$ as the set of derivations of the space of germs (and I often think of it that way myself), but for pedagogical reasons I made the decision to use the conceptually simpler definition involving $C^\infty(M)$, which I thought would be a little easier for novices to wrap their heads around. $\endgroup$ – Jack Lee Jun 9 '17 at 18:42
  • $\begingroup$ @JackLee math.stackexchange.com/questions/2982860/… $\endgroup$ – An old man in the sea. Nov 3 '18 at 13:48
3
$\begingroup$

A derivation at $p\in M$ is in particular a linear map $\partial: C^\infty_p \to \mathbb R$ defined on the set of germs of smooth functions at $p$, and similarly for $q$.
So the sets of derivations at $p$ and $q$ are disjoint simply because they consist of maps with different domains (namely $C^\infty_p$ and $C^\infty_q$). And maps with different domains cannot be equal, as follows from the set-theoretical definition of "map".

$\endgroup$
  • $\begingroup$ Tu's definition of derivation is different from that in Lee's book. The first, uses germs instead the latter says that a derivation is a linear function with domain $C ^{\infty} (M)$ $\endgroup$ – Dubious Nov 4 '12 at 15:58
  • $\begingroup$ As an algebraic/analytic geometer I much prefer the definition with germs. The other definition works in differential geometry because of the specific result that the canonical map $\mathcal C^\infty(M) \to \mathcal C^\infty_p$ is surjective. This is completely false in the algebraic/analytic category. $\endgroup$ – Georges Elencwajg Nov 4 '12 at 17:20
  • $\begingroup$ But how are we sure that we always have $C^\infty_p \neq C^\infty_q$ for $q\neq p$? $\endgroup$ – An old man in the sea. Nov 3 '18 at 12:50
  • $\begingroup$ in a hausdorff space p and q are separated by open neighborhoods which i suspect make up the domains of the elements of $C_p^\infty$ and $C_q^\infty$. $\endgroup$ – peter Feb 13 at 10:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.