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I have originally posted this question on math.SE, but it received little attention, so I repost it here.

Let $U\subset \mathbb{C}^{n}$ and $V\subset \mathbb{C}^{m}$ be open and connected. Let $\Phi:U\to V$ be a holomorphic map.

Is it true that there is a non-zero holomorphic function $w$ on a neighbourhood of $\Phi(U)$, which vanishes at every critical value of $\Phi$?

If this is wrong, is there any other refinement of Sard's theorem for the holomorphic maps between complex domains?

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    $\begingroup$ Neil Strickland's answer has me wondering if you should be looking for a local statement: every point in $U$ has a neighborhood $U'$ such that your statement is true when $\Phi$ is restricted to $U'$. $\endgroup$ – Tom Goodwillie Oct 5 '17 at 2:23
  • $\begingroup$ @TomGoodwillie Yes, that would be very nice, if true. In fact the local version would be sufficient for my needs. For one-dimensional case it is trivial (right?), but I'm afraid this might be one of the instances where the multi-dimensional case is quite different. $\endgroup$ – erz Oct 5 '17 at 5:20
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    $\begingroup$ I am thinking of the strategy that Milnor uses ito prove Sard's Theorem in the little book "Topology from the Differentiable Point of View". Maybe in the analytic case this strategy will show that the critical set is the union of finitely many analytic manifolds each of which has "small" image by induction on $m$ and $n$. Working locally is what would guarantee the finiteness. $\endgroup$ – Tom Goodwillie Oct 5 '17 at 14:09
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    $\begingroup$ You can show that the set of critical values is contained in a countable union of proper analytic subsets. See page 41 of E Chirka's book on Complex Analytic Sets $\endgroup$ – Mohan Ramachandran Oct 5 '17 at 20:35
  • $\begingroup$ @MohanRamachandran Thank you for the reference. $\endgroup$ – erz Oct 6 '17 at 8:12
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No. We can enumerate $\mathbb{Q}[i]$ as $\{a_0,a_1,a_2,\dotsc\}$ and then choose a holomorphic function $f\colon\mathbb{C}\to\mathbb{C}$ with $f(n)=a_n$ and $f'(n)=0$ for all $n$. This follows from a well-known interpolation theorem; some references are discussed at Which sequences can be extended to analytic functions? (e. g., Ackermann's function), for example. Now the set of critical values is dense, so you probably cannot do much better than Sard's conclusion that it has measure zero.

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    $\begingroup$ Well ... I'd say countable is better than measure zero. $\endgroup$ – Tom Goodwillie Oct 5 '17 at 0:43
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    $\begingroup$ Thank you! The result you probably refer to, which is quoted in one of the answers, is a variant of Mittag-Leffler and Weierstrass theorems, and can be found in 15.13 of Rudin's Real and Complex Analysis. $\endgroup$ – erz Oct 5 '17 at 5:15
  • $\begingroup$ (That was just for the completeness sake) $\endgroup$ – erz Oct 5 '17 at 5:28

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