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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 $\overline{S}$.

I've looked for the origin of the use of the symbol $\partial$ in topology without luck. I have only found references for its use in calculus. I've searched through History of Topology (Ioan Mackenzie James) online without success (but this may be my poor searching). Just visually scanning the 1935 Topologie von Alexandroff und Hopf, I do not see $\partial$ employed.

I have two questions:

Q1. Is there a sense in which the boundary operator $\partial$ is analogous to a derivative?

Q2. What is the historical origin for the use of the symbol $\partial$ in topology?

Thanks!

Addendum. Although Q2 has not been addressed, it seems appropriate to accept one among the wealth of insightful responses to Q1. Thanks to all!

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Vague answer that needs to be explained in more detail: The boundary operator $\partial$ is analogous to the exterior derivative $d$ acting on differential forms. I think they can be viewed as adjoints of each other, which act on complexes (submanifolds versus differential forms) that are dual to each other (via integration). –  Deane Yang Nov 16 '10 at 16:36
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As Deane says, it's Stoke's theorem: $$\int_M d\omega = \int_{\partial M} \omega$$ If you write integration as a pairing, then $$\langle d\omega, M\rangle = \langle \omega, \partial M\rangle$$ shows that $\partial$ is the adjoint of the de Rham $d$. –  José Figueroa-O'Farrill Nov 16 '10 at 16:49
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I think the analogy between the boundary operator and differentiation is a lot stronger in the context of singular homology or cell complexes. In that setting you can map a cell to its boundary taken with orientation which ensures that the boundary of a boundary vanishes. And of course in this setting the boundary is really just the differential of the complex computing the homology of your space, which shows the connection to the classical "d", since you could also have computed the cohomology of the space using the de Rham complex where the differential is just the ordinary differential. –  Dan Petersen Nov 16 '10 at 16:54
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Oops -- I should have written Stokes's Theorem. (For the Nth time, I wish I could edit comments.) –  José Figueroa-O'Farrill Nov 16 '10 at 17:04
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@Willie: such as ... this one? Seriously, that's "Qn2" in the original post! –  Loop Space Nov 16 '10 at 18:40

5 Answers 5

up vote 63 down vote accepted

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 $r=0$:

$$ |\partial S| = \frac{d}{dr} |S_r| |_{r=0}.$$

Thus, for instance, the boundary $\partial D_r$ of the disk $D_r$ of radius $r$ has circumference $\frac{d}{dr} (\pi r^2) = 2\pi r$.

More generally, one intuitively has the Newton quotient-like formula

$$ \partial S = \lim_{h \to 0^+} \frac{S_h \backslash S}{h};$$

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 $S$).

In a similar spirit, one informally has the "chain rule"

$$ {\mathcal L}_X S "=" (X \cdot n) \partial S $$

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).

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.

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.

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I really like your definition of the boundary as the limit of a "difference quotient". This is a good example of what "morally correct" means to me. –  Deane Yang Nov 16 '10 at 21:10
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I second Deane's comment. This erudite and enlightening response is more than my question deserves, and I am appreciative! In fact all the comments and answers are illuminating. Thanks! –  Joseph O'Rourke Nov 17 '10 at 0:21

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 $\partial(A\times B) = ((\partial A)\times \overline{B}) \cup (\overline{A} \times (\partial B))$.

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Very nice observation!!! –  Joseph O'Rourke Nov 17 '10 at 21:44
    
For CAT(0) spaces, there is a compactification also denoted by $\partial$ and this rule only holds non-equivariantly in this context. This is a bit of a flaw of the analogy. –  HenrikRüping Nov 17 '10 at 22:22

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 $+$.

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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 currents. 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.

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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 calculate the absolute value of the gradient of $u$ and consider all points where this value is large enough as an edge point. If your image is a (smoothed) characteristic function, then this will give you an approximation to the boundary of the set by thresholding the absolute value of the gradient of the image.

Similarly, the morphological gradient also gives the boundary of an object. Here, you consider the object as the support of an indication function. For this function you calculate the so-called closing with some parameter $\epsilon>0$ (which is nothing else as the indicator function of the $\epsilon$-enlaged object). The morphological gradient is then the difference of this closing and the function itself and encodes the boundary of the object.

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