Is the boundary $\partial S$ analogous to a derivative? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T01:47:03Z http://mathoverflow.net/feeds/question/46252 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/46252/is-the-boundary-partial-s-analogous-to-a-derivative Is the boundary $\partial S$ analogous to a derivative? Joseph O'Rourke 2010-11-16T16:26:53Z 2011-03-30T20:28:17Z <p>Without prethought, I mentioned in class once that the reason the symbol $\partial$ is used to represent the boundary operator in topology is that its behavior is akin to a derivative. But after reflection and some research, I find little support for my unpremeditated claim. Just sticking to the topological boundary (as opposed to the boundary of a manifold or of a simplicial chain), $\partial^3 S = \partial^2 S$ for any set $S$. So there seems to be no possible analogy to Taylor series. Nor can I see an analogy with the fundamental theorem of calculus. The only tenuous sense in which I can see the boundary as a derivative is that $\partial S$ is a transition between $S$ and the "background" complement $\overline{S}$.</p> <p>I've looked for the origin of the use of the symbol $\partial$ in topology without luck. I have only found <a href="http://en.wikipedia.org/wiki/%E2%88%82" rel="nofollow">references</a> for its use in calculus. I've searched through <em>History of Topology</em> (Ioan Mackenzie James) online without success (but this may be my poor searching). Just visually scanning the 1935 <em>Topologie</em> von Alexandroff und Hopf, I do not see $\partial$ employed.</p> <p>I have two questions:</p> <blockquote> <p><b>Q1</b>. Is there a sense in which the boundary operator $\partial$ is analogous to a derivative?</p> <p><b>Q2</b>. What is the historical origin for the use of the symbol $\partial$ in topology?</p> </blockquote> <p>Thanks!</p> <p><b>Addendum</b>. Although <b>Q2</b> has not been addressed, it seems appropriate to accept one among the wealth of insightful responses to <b>Q1</b>. Thanks to all!</p> http://mathoverflow.net/questions/46252/is-the-boundary-partial-s-analogous-to-a-derivative/46285#46285 Answer by Terry Tao for Is the boundary $\partial S$ analogous to a derivative? Terry Tao 2010-11-16T20:46:49Z 2010-11-16T21:50:28Z <p>The surface area $|\partial S|$ of a (bounded, smooth) body $S$ is the derivative of the volume $|S_r|$ of the $r$-neighbourhoods $S_r$ of $S$ at $S=0$:</p> <p>$$|\partial S| = \frac{d}{dr} |S_r| |_{r=0}.$$</p> <p>Thus, for instance, the boundary $\partial D_r$ of the disk $D_r$ of radius $r$ has circumference $\frac{d}{dr} (\pi r^2) = 2\pi r$.</p> <p>More generally, one intuitively has the Newton quotient-like formula</p> <p>$$\partial S = \lim_{h \to 0^+} \frac{S_h \backslash S}{h};$$</p> <p>the right-hand side does not really make formal sense, but certainly one can view $S_h \backslash S$ as a $[0,h]$-bundle over $\partial S$ for $h$ sufficiently small (in particular, smaller than the radius of curvature of $S$).</p> <p>In a similar spirit, one informally has the "chain rule"</p> <p>$${\mathcal L}_X S "=" (X \cdot n) \partial S$$</p> <p>for the "Lie derivative" of $S$ along a vector field $X$, where $n$ is the outward normal. (There may also be a divergence term, depending on whether one is viewing $S$ as a set, a measure, or a volume form.) Again, this does not really make formal sense, although Stokes' theorem already captures most of the above intuition rigorously (and, as noted in the comments, Stokes' theorem is probably the clearest way to link boundaries and derivatives together).</p> <p>EDIT: A more rigorous way to link boundaries with derivatives proceeds via the theory of distributions. The weak derivative $\nabla 1_S$ of the indicator function of a smooth body $S$ is equal to $-n d\sigma$, where $n$ is the outward normal and $d\sigma$ is the surface measure on $\partial S$. (This is really just a fancy way of restating Stokes' theorem, after one unpacks all the definitions.) This can be used, for instance, to link the Sobolev inequality with the isoperimetric inequality.</p> <p>In a similar spirit, $\frac{1_{S_h} - 1_S}{h}$ converges in the sense of distributions as $h \to 0$ to surface measure $d\sigma$ on $\partial S$, thus providing a rigorous version of the intuitive difference formula given previously.</p> http://mathoverflow.net/questions/46252/is-the-boundary-partial-s-analogous-to-a-derivative/46288#46288 Answer by BS for Is the boundary $\partial S$ analogous to a derivative? BS 2010-11-16T21:39:34Z 2010-11-16T21:39:34Z <p>I think that there are (at least) two occurences of boundary in your question. One is the notion of frontier in general topology (closure minus interior) the other in differential (or geometric) topology, namely the set of points where the space under consideration is like $\mathbb{R}_+\times\mathbb{R}^{n-1}$ rather than like $\mathbb{R}^n$. The second is the one related to derivative in de Rham's theory of <a href="http://en.wikipedia.org/wiki/Homological_current" rel="nofollow"><em>currents</em></a>. These are functionals (continuous in a suitable topology) on differential forms (maybe twisted) on a smooth manifold $M$. The boundary of a current $T$ is then defined by Stokes formula as the adjoint of exterior differential : $\partial T$ on $\alpha$ is $T$ on $d\alpha$. And that's it ! For example, a current might be integration on some submanifold with boundary inside $M$, and its $\partial$ is then integration on the boundary of the submanifold. De Rham himself wrote a book on the subject, but I suspect that the ideas have now evolved to much higher depths (so to speak), with all this higher categories stuff, which I'm not competent to discuss.</p> http://mathoverflow.net/questions/46252/is-the-boundary-partial-s-analogous-to-a-derivative/46392#46392 Answer by Dirk for Is the boundary $\partial S$ analogous to a derivative? Dirk 2010-11-17T19:04:36Z 2010-11-17T19:04:36Z <p>Another, not too mathematical, analogy comes from image processing. There you can consider an image $u$ as a real valued function on a rectangle, say. A basic method for edge detection is to caluclate the absolute value of the gradient of $u$ and consider all points where this value is large enough as an edge point. If you image is a (smoothed) characteristic function, then this will give you an approximation to the boundary of the set.</p> <p>Similarly, the <a href="http://en.wikipedia.org/wiki/Morphological_Gradient" rel="nofollow">morphological gradient</a> also gives the boundary of an object.</p> http://mathoverflow.net/questions/46252/is-the-boundary-partial-s-analogous-to-a-derivative/46410#46410 Answer by Vaughn Climenhaga for Is the boundary $\partial S$ analogous to a derivative? Vaughn Climenhaga 2010-11-17T21:27:08Z 2010-11-17T21:27:08Z <p>A partial answer to Q1 -- apologies if this is obvious, but I don't see it written here yet, and this is the thing that made me sit up and take notice of the fact that there's some sort of connection between the boundary operator $\partial$ and differentiation. If $X$ and $Y$ are two topological spaces and $A \subset X$, $B\subset Y$ are closed, then they satisfy a product rule of sorts: $$\partial(A\times B) = ((\partial A)\times B) \cup (A \times (\partial B)).$$ This also works without the assumption that $A$ and $B$ are closed if you're willing to weaken the analogy a bit by replacing the right-hand side with $\partial(A\times B) = ((\partial A)\times \overline{B}) \cup (\overline{A} \times (\partial B))$.</p> http://mathoverflow.net/questions/46252/is-the-boundary-partial-s-analogous-to-a-derivative/60120#60120 Answer by Denis Serre for Is the boundary $\partial S$ analogous to a derivative? Denis Serre 2011-03-30T20:28:17Z 2011-03-30T20:28:17Z <p>I like Vaughn's answer, but it seems to me that another form of product rule holds. If $A$ and $B$ are reasonnable subsets of $\mathbb R^n$ (perhaps one needs only that they are the closure of their interior), then $$\partial(A\cap B)=(\partial A\cap B)\cup(A\cap\partial B).$$ Here, $\cap$ and $\cup$ are the boolean operators, which are the analogues of $\times$ and $+$.</p>