motivation of surgery - MathOverflow most recent 30 from http://mathoverflow.net2013-05-22T19:48:45Zhttp://mathoverflow.net/feeds/question/18359http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/18359/motivation-of-surgerymotivation of surgerystudent2010-03-16T11:25:34Z2011-10-02T19:09:19Z
<p>an $n$-surgery on m dim manifold M is to cut out $S^n\times D^{m-n}$and replace it by $D^{n+1}\times S^{m-n-1}$.
I want to know how this is invented?
I do know that the effect of passing a critical point of index $n$ in $m$-manifold is equivalent to attach an $n+1$-handle $D^{n+1}\times D^{m-n-1}$.Now the boundary of $D^n\times D^{m-n}$ is $S^n\times D^{m-n}\cup D^{n+1}\times S^{m-n-1}$,i think there must be some close relation between the special form of $n$-surgery and handle.can someone help make this clear? </p>
http://mathoverflow.net/questions/18359/motivation-of-surgery/18361#18361Answer by Petya for motivation of surgeryPetya2010-03-16T11:38:32Z2010-03-16T17:24:16Z<p>The best answer is to cite Tim Perutz answer to my question <a href="http://mathoverflow.net/questions/15647" rel="nofollow">Surgery and homology: a reference request</a>:</p>
<p>"To say that a smooth, closed manifold N is obtained by surgery along a (framed) sphere in M is to say that there is a cobordism P from M to N and a Morse function $f\colon P\to [0,1]$, with $f^{−1}(0)=M$, $f^{−1}(1)=N$, and exactly one critical point c."</p>
<p>Critical point corresponds to a handle, $S^n\times D^{m-n}$ lies in $M$, $D^{n+1}\times S^{m-n-1}$ lies in $N$ and spheres $S^n$ and $S^{m-n-1}$ are intersections of stable and unstable manifolds for $c$ with corresponding level sets.</p>
<p>In other words: to obtain $N$ from $M$ by a surgery one can consider $M\times [0,1]$, glue a handle using $S^n\times D^{m-n} \subset M=M\times 1$, smooth the resulted cobordism and take a component of a boundary. </p>
http://mathoverflow.net/questions/18359/motivation-of-surgery/21699#21699Answer by Andrew Ranicki for motivation of surgeryAndrew Ranicki2010-04-17T22:41:43Z2011-10-02T07:32:36Z<p><a href="http://en.wikipedia.org/wiki/Surgery_theory" rel="nofollow">The Wikipedia article on surgery theory</a> explains this! In addition, the Edinburgh <a href="http://www.maths.ed.ac.uk/~s1057008//surgerygroup/" rel="nofollow">Surgery Theory Study Group</a> provides all kinds of surgery-related materials, including <a href="http://www.youtube.com/user/SurgeryGroup" rel="nofollow">YouTube videos</a> (not for the squeamish). There is all kinds of surgery bric-a-brac on
<a href="http://www.maths.ed.ac.uk/~aar/surgery/ages.htm" rel="nofollow">Surgery Bits and Pieces</a> and also <a href="http://www.maths.ed.ac.uk/~aar/surgery/notes.htm" rel="nofollow">here</a>.</p>
http://mathoverflow.net/questions/18359/motivation-of-surgery/76981#76981Answer by Daniel Moskovich for motivation of surgeryDaniel Moskovich2011-10-02T13:53:55Z2011-10-02T13:53:55Z<p>This question has already been answered, but there's a tiny piece of intuition which I'd like to make explicit:</p>
<p>If you're thinking about a manifold in the PL world, surgery might look a bit arbitrary- why cut out and glue in those pieces and not others? Surgery's natural setting is the smooth world, where you're equipping a manifold with a Morse function $f\colon\, M\to \mathbb{R}$, and using information about critical points of $f$ to encode $M$.</p>
<p>It's actually a bit more involved than you might think it might be, but when you pass a critical point of $f$ you add a handle to $M$, and the boundary changes by surgery. For details, see answers to <a href="http://mathoverflow.net/questions/70248/searching-for-an-unabridged-proof-of-the-basic-theorem-of-morse-theory" rel="nofollow">this question</a>.</p>
<p>So really, surgery isn't an a-priori construction which somebody pulled from a hat- it is rather an operation which stems naturally and inevitably from Morse theory.</p>