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L.S.,

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. I also understand the third equation I think.

These are pages 27 - 36 of the book.

My question is: How to show that the second equation holds? (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.

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?

Any help I would appreciate very much!

Thanks,

Willem

PS: earlier I posted the way to show the third equation, this is the link: Equivalent definitions of the tangent space

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