Re: [math-fun] algebraic geometry query
[I sent this 1.5 hours ago, but through some glitch it still hasn't shown up in my Inbox.] CORRECTION: Where I wrote "In R^n, let W be a compact connected (n-1)-dimensional submanifold with connected boundary M = bd(W)," the correct dimension of W should be n (not n-1), as corrected below. Also, the affiliation mentioned is no longer current. --Dan ----- Hmm, Why wouldn't C^0 (just continuous) work. In R^n, let W be a compact connected n-dimensional submanifold with connected boundary M = bd(W). Claim: ------ If a continuous function f: W -> R^1 (= the reals) for which 0 lies in f(W) but not in f(M), then f must have an absolute extremum in int(W) = W - M. Proof à la Latto: ----------------- Since f is continuous on the compact set W, it must have both an absolute minimum and an absolute maximum on W. Since f is nonzero on the connected set M, f must take M into either (0,oo) or (-oo,0). If the former, f must have a local minimum in int(W); if the latter f must have a local maximum in int(W). [ ] Corollary: ---------- If f is also differentiable on int(W), then it has a critical point there (since any local extremum x of a function differentiable in a neighborhood of x must have a critical point at x). -----
Yesterday I proposed the intrinsic criterion for real 2-space zero-detection below. Nobody has yet shot it down --- which may well of course be because nobody has actually read it --- so I shall press on, making the optimistic assumption that it involves no grossly obvious errors. << Theorem: Given a continuously differentiable function f(x, y) , and a compact region R of the plane ( |R^2 ) with f nonzero on the boundary of R : f has a zero within R if and only if there is some (intrinsic critical) point (x, y) in R where *** df/dx = f = 0 ***. Proof (offered tentatively): By the implicit function theorem, the constraint f(x, y) = 0 defines a function x(y) single-valued in any interval of y where df/dx <> 0 . Such an interval meeting R is finite since R is compact, and its endpoints lie in the interior of R since f <> 0 on the boundary; therefore df/dx = 0 at those endpoints. The converse is trivial. QED.
On reflection, "region" is used rather loosely above: in fact, it seems that all that is required is that R is compact. If so, the condition cannot be relaxed in the same way that Dan did for the extrinsic version --- to f being merely C^0 continuous --- since that depended on connectivity. Which does smell just a little bit fishy ... Fred Lunnon On 6/1/14, Dan Asimov <dasimov@earthlink.net> wrote:
[I sent this 1.5 hours ago, but through some glitch it still hasn't shown up in my Inbox.]
CORRECTION:
Where I wrote "In R^n, let W be a compact connected (n-1)-dimensional submanifold with connected boundary M = bd(W)," the correct dimension of W should be n (not n-1), as corrected below.
Also, the affiliation mentioned is no longer current.
--Dan
----- Hmm, Why wouldn't C^0 (just continuous) work.
In R^n, let W be a compact connected n-dimensional submanifold with connected boundary M = bd(W).
Claim: ------ If a continuous function f: W -> R^1 (= the reals) for which 0 lies in f(W) but not in f(M), then f must have an absolute extremum in int(W) = W - M.
Proof à la Latto: ----------------- Since f is continuous on the compact set W, it must have both an absolute minimum and an absolute maximum on W. Since f is nonzero on the connected set M, f must take M into either (0,oo) or (-oo,0). If the former, f must have a local minimum in int(W); if the latter f must have a local maximum in int(W). [ ]
Corollary: ---------- If f is also differentiable on int(W), then it has a critical point there (since any local extremum x of a function differentiable in a neighborhood of x must have a critical point at x). -----
On 5/31/14, Andy Latto <andy.latto@pobox.com> wrote:
On Fri, May 30, 2014 at 4:28 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
In answer to Andy: it shouldn't matter which compactification is used --- so long as we don't forget about the necessity. "Or some other ..." --- trick question? There aren't any others!
Not true; the Stone-Cech compactification is much, much, larger than RP2.
Or without using such heavy machinery, consider the sequences (0,n) and (1, n). In a compactification, they each have to have a convergent subsequence. But there's no reason that these two convergent subsequences have to converge to the same point, as they do in the two compactifications you mention.
Andy
Compactness is all very well, but practical applications are liable to involve "regions" extending to infinity. Having earlier casually postponed consideration on the grounds that any old completion should solve the problem, I must now admit that it's not so simple. For a start, as Andy pointed out, there are quite a few choices. A principle which I elucidated early in life --- I believe it was on the occasion I was first exposed to the serial trumpet abuse of Miles Davis --- is that the potential for an activity does not automatically imply a recommendation to engage in it. Numerous other instances since observed include: dropping a 5lb hammer on one's foot; credit default swaps; and Stone–Čech c*mp*ct*f*c*t**n. Indeed, it's not clear that any of these alternatives has much to offer from the standpoint of computational geometry, besides the classical duo that in my innocence I was relying on. However, Andy's suggestion that (0, n) and (1, n) might be assigned different limits proves prescient, as this example involving a rectangular hyperbola demonstrates (it was time to get away from unit-circle counterexamples) --- Consider region R = { (x, y) | x >= 0 & y >= 0 } and functions (A) f = x*y + 1 ; or (B) f = x*y - 1 . In case (A) there are no zeros in R , in case (B) there are plenty; however neither projective nor complex boundary points at infinity can distinguish them. [Other examples might involve some complicated singularity at infinity.] However, I have an elementary workaround which looks intuitively plausible, at least in simple non-compact situations. Apply the (relaxed) intrinsic criterion with respect to THREE different coordinate frames, say: (x, y) ; (y, x) ; (x+y, x-y) . Courtesy of its asymptotes, an interior hyperbola may evade detection in two frames; but it cannot then also hide in the third. What constraints on a non-compact region & boundary would stand this up? And just how does one go about proving something like this anyway? Fred Lunnon On 6/2/14, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Yesterday I proposed the intrinsic criterion for real 2-space zero-detection below. Nobody has yet shot it down --- which may well of course be because nobody has actually read it --- so I shall press on, making the optimistic assumption that it involves no grossly obvious errors.
<< Theorem: Given a continuously differentiable function f(x, y) , and a compact region R of the plane ( |R^2 ) with f nonzero on the boundary of R : f has a zero within R if and only if there is some (intrinsic critical) point (x, y) in R where *** df/dx = f = 0 ***.
Proof (offered tentatively): By the implicit function theorem, the constraint f(x, y) = 0 defines a function x(y) single-valued in any interval of y where df/dx <> 0 . Such an interval meeting R is finite since R is compact, and its endpoints lie in the interior of R since f <> 0 on the boundary; therefore df/dx = 0 at those endpoints. The converse is trivial. QED.
On reflection, "region" is used rather loosely above: in fact, it seems that all that is required is that R is compact. If so, the condition cannot be relaxed in the same way that Dan did for the extrinsic version --- to f being merely C^0 continuous --- since that depended on connectivity. Which does smell just a little bit fishy ...
Fred Lunnon
On 6/1/14, Dan Asimov <dasimov@earthlink.net> wrote:
[I sent this 1.5 hours ago, but through some glitch it still hasn't shown up in my Inbox.]
CORRECTION:
Where I wrote "In R^n, let W be a compact connected (n-1)-dimensional submanifold with connected boundary M = bd(W)," the correct dimension of W should be n (not n-1), as corrected below.
Also, the affiliation mentioned is no longer current.
--Dan
----- Hmm, Why wouldn't C^0 (just continuous) work.
In R^n, let W be a compact connected n-dimensional submanifold with connected boundary M = bd(W).
Claim: ------ If a continuous function f: W -> R^1 (= the reals) for which 0 lies in f(W) but not in f(M), then f must have an absolute extremum in int(W) = W - M.
Proof à la Latto: ----------------- Since f is continuous on the compact set W, it must have both an absolute minimum and an absolute maximum on W. Since f is nonzero on the connected set M, f must take M into either (0,oo) or (-oo,0). If the former, f must have a local minimum in int(W); if the latter f must have a local maximum in int(W). [ ]
Corollary: ---------- If f is also differentiable on int(W), then it has a critical point there (since any local extremum x of a function differentiable in a neighborhood of x must have a critical point at x). -----
The impression grows on me that this topic may have become mysteriously contentious for reasons unconnected to mathematics. So I'll risk one more version of this proof, expanded to fill gaps which have emerged meantime, and invite anyone still interested to email me off-list. Theorem: Given a continuously differentiable function f(x, y) , and a compact subset R of the Cartesian plane with f nonzero on the boundary of R : f has a zero in R if and only if there is some point (x, y) in R where df/dx = f = 0 . Proof: Denote by S the curve defined implicitly by f(x, y) = 0 . If f has a zero in R , some point lies both on S and in R ; and the entire connected component of S containing it lies in R by continuity, f being nonzero on the boundary of R . Let y' be an endpoint of the interval to which that component of S projects on the y-axis; y' is finite since it lies within the projection of the compact set R . By the implicit function theorem, constraint f(x, y) = 0 defines a function x = g(y) , single-valued in any interval of y where df/dx != 0 . Since g ceases to be single-valued at y' , we have df/dx = f = 0 at (x', y') where x' = g(y') ; also (x', y') in R . The converse is trivial. QED. Please note that the annulus & circle counterexample was retired -- ooh -- at least three days ago. While the worked example concerned is available on request, prospective consumers are warned that it does depend on the ability to sort the set {1,3,2} correctly into ascending order. Further extensions: non-compact R ; higher dimensional R and S . Fred Lunnon On 6/3/14, Fred Lunnon <fred.lunnon@gmail.com> wrote:
On 5/31/14, Andy Latto <andy.latto@pobox.com> wrote:
On Fri, May 30, 2014 at 4:28 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
In answer to Andy: it shouldn't matter which compactification is used --- so long as we don't forget about the necessity. "Or some other ..." --- trick question? There aren't any others!
Not true; the Stone-Cech compactification is much, much, larger than RP2.
Or without using such heavy machinery, consider the sequences (0,n) and (1, n). In a compactification, they each have to have a convergent subsequence. But there's no reason that these two convergent subsequences have to converge to the same point, as they do in the two compactifications you mention.
Andy
Compactness is all very well, but practical applications are liable to involve "regions" extending to infinity. Having earlier casually postponed consideration on the grounds that any old completion should solve the problem, I must now admit that it's not so simple. For a start, as Andy pointed out, there are quite a few choices.
A principle which I elucidated early in life --- I believe it was on the occasion I was first exposed to the serial trumpet abuse of Miles Davis --- is that the potential for an activity does not automatically imply a recommendation to engage in it. Numerous other instances since observed include: dropping a 5lb hammer on one's foot; credit default swaps; and Stone–Čech c*mp*ct*f*c*t**n.
Indeed, it's not clear that any of these alternatives has much to offer from the standpoint of computational geometry, besides the classical duo that in my innocence I was relying on. However, Andy's suggestion that (0, n) and (1, n) might be assigned different limits proves prescient, as this example involving a rectangular hyperbola demonstrates (it was time to get away from unit-circle counterexamples) ---
Consider region R = { (x, y) | x >= 0 & y >= 0 } and functions (A) f = x*y + 1 ; or (B) f = x*y - 1 . In case (A) there are no zeros in R , in case (B) there are plenty; however neither projective nor complex boundary points at infinity can distinguish them. [Other examples might involve some complicated singularity at infinity.]
However, I have an elementary workaround which looks intuitively plausible, at least in simple non-compact situations. Apply the (relaxed) intrinsic criterion with respect to THREE different coordinate frames, say: (x, y) ; (y, x) ; (x+y, x-y) . Courtesy of its asymptotes, an interior hyperbola may evade detection in two frames; but it cannot then also hide in the third.
What constraints on a non-compact region & boundary would stand this up? And just how does one go about proving something like this anyway?
Fred Lunnon
Isn't there an easy counterexample if the boundary of R is not connected? Let R be the (compact) annulus 1 <= |(x,y)| <= 3, with f(x,y) := x^2 + y^2 - 4. Then f is nonzero on bd(R) since f(bd(R)) = {-3,5}. f(x,y) = 0 on the circle in R where |(x,y)| = 2. But f has no critical point in R (i.e., no point where df/dx = df/dy = 0). Or am I misunderstanding the meaning of "intrinsic critical point (x,y)" ? * * * Conversely, this time let R = the unit disk D^2, and let f(x,y) = x^2 + y^2 + 1. Then f is clearly nonzero on bd(R), but has a critical point in int(R). * * * I must be misunderstanding this whole Theorem. --Dan
Theorem: Given a continuously differentiable function f(x, y) , and a compact region R of the plane ( |R^2 ) with f nonzero on the boundary of R : f has a zero within R if and only if there is some (intrinsic critical) point (x, y) in R where *** df/dx = f = 0
Arrrgh. Yet again I'm forced to post a correction: I omitted the last line, which should read: "And f is nonzero on int(R) as well." --Dan On Jun 2, 2014, at 6:15 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Isn't there an easy counterexample if the boundary of R is not connected?
Let R be the (compact) annulus 1 <= |(x,y)| <= 3, with f(x,y) := x^2 + y^2 - 4.
Then f is nonzero on bd(R) since f(bd(R)) = {-3,5}.
f(x,y) = 0 on the circle in R where |(x,y)| = 2.
But f has no critical point in R (i.e., no point where df/dx = df/dy = 0).
Or am I misunderstanding the meaning of "intrinsic critical point (x,y)" ?
* * *
Conversely, this time let R = the unit disk D^2, and let f(x,y) = x^2 + y^2 + 1.
Then f is clearly nonzero on bd(R), but has a critical point in int(R).
----- Theorem: Given a continuously differentiable function f(x, y) , and a compact region R of the plane ( |R^2 ) with f nonzero on the boundary of R : f has a zero within R if and only if there is some (intrinsic critical) point (x, y) in R where *** df/dx = f = 0. -----
"Intrinsic" == *********************** *** df/dx = f = 0 *** *********************** WFL On 6/3/14, Dan Asimov <dasimov@earthlink.net> wrote:
Arrrgh. Yet again I'm forced to post a correction:
I omitted the last line, which should read:
"And f is nonzero on int(R) as well."
--Dan
On Jun 2, 2014, at 6:15 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Isn't there an easy counterexample if the boundary of R is not connected?
Let R be the (compact) annulus 1 <= |(x,y)| <= 3, with f(x,y) := x^2 + y^2 - 4.
Then f is nonzero on bd(R) since f(bd(R)) = {-3,5}.
f(x,y) = 0 on the circle in R where |(x,y)| = 2.
But f has no critical point in R (i.e., no point where df/dx = df/dy = 0).
Or am I misunderstanding the meaning of "intrinsic critical point (x,y)" ?
* * *
Conversely, this time let R = the unit disk D^2, and let f(x,y) = x^2 + y^2 + 1.
Then f is clearly nonzero on bd(R), but has a critical point in int(R).
----- Theorem: Given a continuously differentiable function f(x, y) , and a compact region R of the plane ( |R^2 ) with f nonzero on the boundary of R : f has a zero within R if and only if there is some (intrinsic critical) point (x, y) in R where *** df/dx = f = 0. ----- _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Dear Fred, Thank you for the clarification. Did any of my two (corrected) proofs make sense? I'm puzzled by the deafening silence. Despite my not understanding this new terminology of yours -- and I really wish you would write clearly -- my argument should be 100% relevant nonetheless, in both directions. Regards, Dan On Jun 3, 2014, at 3:41 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
"Intrinsic" ==
*********************** *** df/dx = f = 0 *** *********************** WFL
On 6/3/14, Dan Asimov <dasimov@earthlink.net> wrote:
Arrrgh. Yet again I'm forced to post a correction:
I omitted the last line, which should read:
"And f is nonzero on int(R) as well."
--Dan
On Jun 2, 2014, at 6:15 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Isn't there an easy counterexample if the boundary of R is not connected?
Let R be the (compact) annulus 1 <= |(x,y)| <= 3, with f(x,y) := x^2 + y^2 - 4.
Then f is nonzero on bd(R) since f(bd(R)) = {-3,5}.
f(x,y) = 0 on the circle in R where |(x,y)| = 2.
But f has no critical point in R (i.e., no point where df/dx = df/dy = 0).
Or am I misunderstanding the meaning of "intrinsic critical point (x,y)" ?
* * *
Conversely, this time let R = the unit disk D^2, and let f(x,y) = x^2 + y^2 + 1.
Then f is clearly nonzero on bd(R), but has a critical point in int(R).
----- Theorem: Given a continuously differentiable function f(x, y) , and a compact region R of the plane ( |R^2 ) with f nonzero on the boundary of R : f has a zero within R if and only if there is some (intrinsic critical) point (x, y) in R where *** df/dx = f = 0. ----- _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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