In my book Vector Analysis by Klaus Jänich,

Three different 'versions' of the Tangent space of a point $p$ at a differentiable variety are being discussed.

The 'geometrical': the set of differentiable curves in which pass through $p$ at $t=0$. (with two curves equivalent when they have the same derivative in 0 on chart)

The 'algebraic': the set of al derivations on the ring of germs at $p$, that satisfy the product rule.

The 'physical' the set of al vectors that send al the charts around $p$ to a subspace of $\mathbb R^n$, with the property that for any two charts the associated vectors in $\mathbb R^n$ are mapped to eachother by the differential of the transition map.

Then three maps are given,

$\phi_1$ : geometric $ \rightarrow $ algebraic by $f\mapsto(f\circ\alpha$)$'$$(0)$

$\phi_2$ : algebraic $ \rightarrow $ physical by $(U,h)$ $\mapsto$ $(v(h_1), ..., v(h_n))$

$\phi_3$ : physical $ \rightarrow $ geometric by $(\alpha(t) := h^{-1}(h(p) + tv(U,h))$

and then is stated that

$\phi_1 \circ\phi_2 \circ\phi_3 = Id$

$\phi_2 \circ\phi_3 \circ\phi_1 = Id$

$\phi_3 \circ\phi_1 \circ\phi_2 = Id$

And it is shown for the first equation that it holds. These are pages 27 - 36 of the book.

My question is: How to show that the other equations also hold? (this is not a homework question) The book says it goes in a similar way as the first, but I get stuck at a point. For instance the third equation, what I have tried is this:

You have a 'physical' tangent vector $v$, which sends al the charts around $p$ to a subspace of $\mathbb R^n$. Now you want to make it a geometric tangent vector, so we define the curve $(\alpha(t) := h^{-1}(h(p) + tv(U,h))$. Now we want to make it a algebraic tangent vector, so we define the vector $v'$ that maps every $f$ in the ring of germs around $p$ to $(f\circ\alpha)'(0)$, so that's now $(f\circ h^{-1}(h(p) + tv(U,h)))'(0)$.

Then we want to make it a physical vector again, so it now becomes $v''$ which sends every $(U,h)$ around $p$ to $(v'(h))$, which then becomes $(h$ 0 ($h^{-1}(h(p) + tv(U,h))))'(0)$

but how do we know now that that is the same as the original $v$? I thought it should be possible to somehow use the chainrule for derivatives, and then maybe you could throw away the $h(p)$ term since you're taking the derivative with respect to $t$! But then still I don't now how this would be done and if it would help.

Any help I would appreciate very much!



  • $\begingroup$ Oke I'm terribly sorry, you can just say $h o h^{-1}$ cancels, and then if you take the derivative with respect to $t$, you just get $v(U,h)$ $\endgroup$ Oct 20, 2013 at 13:58
  • $\begingroup$ Only a remark concernig $\LaTeX$: The composition symbol $\circ$ is \circ and you do not need to add dollar signs between each symbol, but one dollar at the beginning an one in the end suffice for the whole formula. $\endgroup$
    – gofvonx
    Oct 20, 2013 at 16:55
  • $\begingroup$ thank you gofvonx. Should I delete my question now? $\endgroup$ Oct 20, 2013 at 16:57
  • 1
    $\begingroup$ Also, since you can answer the question yourself you might want to post an answer on your own and accept it... $\endgroup$
    – gofvonx
    Oct 20, 2013 at 16:57
  • $\begingroup$ I have done that! But now I still wonder about the second equation. Then if you take the composite, you'll get in the end a derivation that sends any $f$ in the ring of germs at $p$ to $f \circ h^{-1} \circ (h(p) + tv(U,h))'(0)$. If you apply the chainrule twice you end up with $f'(p)*(h^{-1})'(p)*v(U,h)$! But it should be something like $v(f)$, which totally doesn't look like it.. could you please help me with this last equation? I thank you! $\endgroup$ Oct 21, 2013 at 7:44

1 Answer 1


In $(h \circ $($h^{-1}(h(p) + tv(U,h))))'(0)$, the $h \circ h^{-1}$ cancel, what's then left is $(h(p) + tv(U,h))'(0)$.

Since you take the derivative with respect to $t$, this now becomes a constant: $v(U,h)$. This is the same as $v(U,h)$


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