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Suppose $f$ and $g$ are two embeddings of $S^1$ in $R^3$ or in $S^3$. How do I show whether they form a link or not?

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Well... it is a difficult problem! :) You can try to evaluate some linking invariants, which may tell you there two circles are linked, like the linking number (wikipedia will tell you about it) and others, but in general pretty much everything can fail. – Mariano Suárez-Alvarez Mar 28 '11 at 5:21
Do you have any further information about f and g? – Bey Mar 28 '11 at 5:47
@Mariano: there are computable complete linking invariants. See Zare's response. – Ryan Budney Mar 30 '11 at 15:51

I think the common interpretation of your question is to determine whether you can pull the components apart with an ambient isotopy. This is a harder problem than you might expect, but it is doable. The key is that being able to pull the knots apart is equivalent to whether there is a topological sphere in the complement in $S^3$ which is not contractible. (A really large sphere containing everything is contractible in $S^3$, but not $\mathbb{R}^3$.) Normal surface theory lets us check for interesting $S^2$s algorithmically.

You can triangulate the complement of the link (already not a trivial operation) and get a finite-dimensional space of normal surfaces. Then you can construct a polytope such that if there is any incompressible $S^2$, there is one which is a vertex of the polytope. So, you find all of the vertices and check them.

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Gauss's linking integral will give the linking number: $$N = \frac{1}{4\pi}\int\int \frac{\vec{f}(\sigma)-\vec{g}(\tau)}{|\vec{f}(\sigma)-\vec{g}(\tau)|^3}\cdot \left(\frac{d\vec{f}}{d\sigma}\times \frac{d\vec{g}}{d\tau}\right)\;d\sigma\;d\tau$$

Per comment by Alon Amit, the linking number can be zero for maps that can't be teased apart such as the Whitehead link, but if it's nonzero, they definitely can't be separated. (And by the way, even an "unlink" is a "link" so I'm not certain this is what you're looking for.)

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The Whitehead Link has linking number 0, but I'd be hard pressed to say that its components "don't link together", since you obviously can't move them apart. It's indeed unclear what the OP means by "forms a link or not". – Alon Amit Mar 28 '11 at 8:39
@Alon; Thanks I'll edit accordingly. – Carl Brannen Mar 28 '11 at 8:41

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