6
$\begingroup$

I am studying the following proof for which an excerpt is provided below:

enter image description here

Update: I have written out a fully-detailed proof of an argument that seeks verify the claim that $\partial \psi$ is invertible. (1) I am unclear on what is special about the point $(x_0, 0)$ as the proof seems to goes through irrespective of the value of the particular point and (2) The author's logic at the end seems reversed to me. It would be helpful if someone could critique the proof below and indicate what step, if any, is incorrect. Also, I apologize for not including the actual TEX; it was formulated locally and I made use of many macros that mathjax wouldn't understand:

enter image description here enter image description here

$\endgroup$
  • $\begingroup$ You have a typo: you wrote that you may assume the first $m$ rows are non-zero. You meant "linearly independent" (as is clear in the next sentence). As you remark, the non-degeneracy is satisfied at all $(x_0, y)$, but the observation for $y\ne 0$ doesn't give us anything we care about. $\endgroup$ – Sam Lisi Apr 9 '12 at 21:43
  • $\begingroup$ @SamLisi Thanks for letting me know about that typo; I'll fix it. $\endgroup$ – ItsNotObvious Apr 10 '12 at 1:05
  • $\begingroup$ We are interested in the point $(x_0,0)$ (and not $(x_0,y)$, $y \neq 0$) because $f$ is the compositum of $X \xrightarrow{\alpha} X \times R^{n-m} \xrightarrow{\psi} R^n$, where $\alpha(x)=(x,0)$. You would have seen this you have completed the proof of the theorem. $\endgroup$ – user10676 Apr 10 '12 at 13:16
  • $\begingroup$ I found this on the Unanswered tab: is there still a question to be answered? I don't see anything wrong or illogical in the proof: the idea is simple and the execution is accurate. We must parametrize a neighborhood of $f(X_0)$ by $X_0$ times an open set, and this is done by augmenting $f$ with a linear map. $\endgroup$ – user31373 Jul 28 '12 at 22:08
  • $\begingroup$ @LeonidKovalev The question is still unanswered. Compare my proof with the excerpt given from the text; they can't both be right. $\endgroup$ – ItsNotObvious Jul 29 '12 at 14:14
1
$\begingroup$

(1) It is true that $\partial \psi(x_0,y)$ is invertible for all values of $y$, not just $0$. We only care about the invertibility of $\partial \psi(x_0,0)$ because we need a diffeomorphism onto a neighborhood of (a piece of) the given manifold, which corresponds to $y=0$.

(2) I agree that the logic of the last sentence in the first excerpt is backwards. Just ignore "Therefore" and replace "and thus" with "because".

$\endgroup$

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.