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One can define the linking number of disjointly embedded curves $K,L\subset S^{3}$ in a variety of ways, as is discussed in Chapter 5.D of Rolfsen's "Knots and Links". One way is the Gauss Integral $$\mathrm{lk}(K,L) = \frac{1}{4\pi}\int_{K\times L}\dfrac{\mathbf{x}-\mathbf{y}}{|\mathbf{x}-\mathbf{y}|^3}\cdot\mathrm{d}\mathbf{x}\times \mathrm{d}\mathbf{y}$$ (or symbols to that effect). This will be an integer when the curves are closed, and a real number in general.

The integral formula has been generalised to deal with the case of disjointly embedded closed manifolds $K^{k} ,L^{\ell}\subset S^{k+\ell+1}$ (see here and here for example). Presumably these formulas output real numbers when the manifolds $K$ and $L$ have boundaries.

My question concerns the situation of disjointly embedded submanifolds $K^k,L^\ell\subset S^n$, where $k+\ell >n-1$. Is there a useful notion of linking number in this case? For instance, take a surface and a curve in $3$ dimensions (so $k=2$, $\ell=1$ and $n=3$ in the above). Then we could try to define the linking number by somehow "integrating" $\mathrm{lk}(\gamma,L)$ over closed curves in $\gamma\subset K$.

This generalised linking number should be able to measure, say, how many times a curve passes through a length of tube.

Have such things been considered useful before? Or am I just talking nonsense?

Added: As Ryan points out in his comment, I'm not really looking for an isotopy invariant. Also (thanks to Kevin and Tom's answers) I'm slowly coming round to the idea that a single number won't really tell you much about the relative positions of the manifolds, but maybe a matrix valued function (with rows and columns indexed by homology bases for $K$ and $L$ in the appropriate dimensions) might be useful.

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The complement of $K$ has a fundamental group, so if $L$ was a circle you could think of conjugacy classes in $\pi_1 (S^n \setminus K)$ as being a type of "generalized linking number". This makes sense when $k+l=k+1>n-1$. So you're not talking nonsense. –  Ryan Budney Jul 20 '11 at 16:41
But given what Kevin and Daniel describe, any quest for such "generalized linking numbers" when $k+l>n-1$ pretty much has to be a more sensitive invariant than the "singular bordism of disjoint manifolds" level. i.e. if you modify $L$ or $K$ via a singular bordism keeping them disjoint, the "linking number" has no hope of being constant, unless it's trivial. –  Ryan Budney Jul 20 '11 at 16:48
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5 Answers

Sort of obvious, but: in general you get lots of linking numbers. If spaces $K$ and $L$ are mapped disjointly into $\mathbb R^n$ then for every $a\in H_i(K)$ and $b\in H_j(L)$ with $i+j=n-1$ you get a number. It can be obtained by pulling back a generator of $H^{n-1}(S^{n-1})$ via the evident map $K\times L\to S^{n-1}$ and evaluating the resulting element of $H^{n-1}(K\times L)$ on $a\times b\in H_{n-1}(K\times L)$. (If $a$ and $b$ are represented by smooth manifolds then this can be expressed as an integral.) This is a bilinear pairing $H_i(K)\times H_{n-i-1}(L)\to \mathbb Z$.

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I think the author wanted to explicitly rule-out this case -- his $i+j > n-1$. –  Ryan Budney Jul 20 '11 at 20:27
His $k+\ell>n-1$, you mean?. My $i$ and $j$ are not the dimensions of $K$ and $L$. –  Tom Goodwillie Jul 20 '11 at 20:30
@Tom: This is kind of what I had in mind, with my naive suggestion, and then trying to combine these into a single number. Although I have no ideas how to do this. –  Mark Grant Jul 20 '11 at 20:37
I don't think that a single number is the right thing to aim for, though. –  Tom Goodwillie Jul 20 '11 at 21:10
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The integral you listed computes the degree of the map $T^2\to S^2, (x,y)\to K(x)-L(y)/|K(x)-L(y)|$. Note that the degree completely determines the homotopy class of this map, i.e. $[T^2,S^2]\cong Z$.

This generalizes to consider the homotopy class of the map $[K\times L, S^{n-1}]$, for $K,L$ mapping disjointly to $R^n$. So, for example, if $K,L$ are 2-spheres and $n=4$, you can consider the homotopy class of the corresponding map in $[S^2\times S^2,S^3]\cong [S^4,S^3]=Z/2$, note that this wont be detected (directly) by the homological methods describe above. The subject of Link homotopy considers such invariants and generalizations, you can start with the old papers of Massey, Fenn-Rolfsen, Koschorke, etc.

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There's been recent work on generalized linking numbers by Sinha, Walter, Munson and Volic. –  Ryan Budney Sep 13 '11 at 21:13
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You'd think the definition of linking number would be common knowledge, wouldn't you? Actually, it's surprising how few places treat the topic carefully. The best references I know are:

  • H. Schubert, Topology Allyn and Bacon, Inc. (1968; Zbl 0169.53802).
  • R. Hartley and K. Murasugi, Covering linkage invariants Can. J. Math. 29, 1312-1339 (1977; Zbl 0373.55004).
  • J. Lannes and F. Latour, Forme quadratique d'enlacement et applications Astérisque 26, 1-90 (1975; Zbl 0318.57017).

Ignoring the issue of integral formulae, which is an topic unto itself, a linking pairing between chains $a\in C_q(M)$ and $b\in C_{p}(M)$ where $\dim (M)=p+q+1$ should be the cup product of $[a]$ with the Bockstein of $[b]$, evaluated against the fundamental class of $M$. This is a rational number. Using Poincaré-Lefshetz duality and a little bit of homological algebra, this quantity can be re-interpreted as an intersection number, and (in favourable situations) explicitly calculated. The only places I know in which the details are worked out properly are the above references.

Getting to your question, the obvious algebraic generalization of the linking number (in arbitrary dimensions) is the cup product of $[L]$ with the Bockstein of $[K]$. There's nothing natural against which to evaluate this (is there?), so it's just a $p+q+1$-chain in $H_{p+q+1}(M;\mathbb{Q})$. So I would guess that the answer is no; you could define all kinds of things, but if the dimensions are wrong, then what you'd be getting would not be a linking number.

Update: Ryan's comment puts it very nicely, so I'm working it in:

...any quest for such "generalized linking numbers" when $k+l>n−1$ pretty much has to be a more sensitive invariant than the "singular bordism of disjoint manifolds" level. i.e. if you modify L or K via a singular bordism keeping them disjoint, the "linking number" has no hope of being constant, unless it's trivial.
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I'm a big fan of Rolfsen's "Knots and links" book for coverage of linking numbers. He doesn't do things in full-blown generality but IMO it's a plus that way. –  Ryan Budney Jul 20 '11 at 16:59
@Ryan If one is learning it for the first time then I agree. But if one actually needs to open up the box, to work with it, and to prove new results, then I think the technology really becomes necessary. At least, I never felt I had any real grasp on the linking number (I'm not sure I do yet!) until I thought hard for a very long time about the details of above constructions. –  Daniel Moskovich Jul 20 '11 at 17:20
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Just to mention here the ``linking numbers'' introduced by A.Haefliger [1]: for an embedding $f\colon S^k\sqcup S^l\to S^n$ he defines the linking number as certain element $$\lambda_1(f)\in\pi_k(S^{n-k-1}\vee S^{n-l-1}).$$ This invariant is stronger than all the homological invariants discussed above and than the $\alpha$-invariant suggested by Paul. However, it is still incomplete isotopy invariant.

[1] A. Haefliger, Enlacements de spheres en codimension superiure a 2, Comm. Math. Helv. 41 (1966-67), 51-72 (in French).

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By Alexander duality, $H_i(S^n \setminus K) \cong H_{n-1-i}(K)^*$ (I'm assuming coefficients in $\mathbb R$). Given an $i$-cycle $X$ in $S^n \setminus K$ (e.g. an $i$-cycle in $L$) and an $(n-1-i)$-cycle $Y$ in $K$, the pairing is given by integrating a formula similar to yours over $X\times Y$.

If $k+l = n-1$ we can take $X$ and $Y$ above to be the fundamental classes of $K$ and $L$ and we recover the usual linking number. When $k=2, l=1, n=3$, as in your example, we can take $Y$ to be a curve in $K$ which cuts the tube in question, and then the integral will measure the (signed) number of times $L$ goes through this tube.

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@Kevin: This seems like the right thing to do practically with the curve and tube example. But I kind of want the number to take into account all possible choices of $Y$ at once (as I tried to indicate in my comment to Tom's answer). Whether such a number could be interpreted in a meaningful way is open to debate. –  Mark Grant Jul 20 '11 at 20:42
As Tom Goodwillie said in another comment on this page, I don't think that trying to capture the pairings in my (or Tom's) answer with a single number is a good idea. At least, I don't see a way to do it usefully. –  Kevin Walker Jul 20 '11 at 22:07
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