Suppose we take a piece of board with an origin marked at 0 and a drawing pin stuck in it at A say. We throw a piece of string on the board both of whose ends fall near 0. We fix both ends at 0 and then try to draw the string in to a bundle round 0. This may be possible (see Figure 89.1(a)) or impossible because the string always catches on A (Figure 89.1(b)). In the second case we say that the string is

tangled round A.

The result that follows may be stated precisely and proved by using the ideas of algebraic topology.

Plausible Lemma 89.1.If f:[a,b] → ℝ2is a continuous function with f(a) = f(b) = 0and f(t) ≠ x0for all t∊[a,b] then the point f(t) moves n times clockwise roundx0as t runs from a to b where n is an integer. The curve is tangled roundx0if and only n ≠ O.

The reader should convince herself that this lemma is plausible before continuing.

Does a Brownian motion in ℝ2 starting at Q tangle itself round a given point x0 or does it remain untangled? To make the question more precise let us consider a specific Brownian motion y(t) with y(0) = 0. Since Brownian motion in ℝ2 fails to pass through any given point with probability 1 we may assume that y(t) ≠ x0 for all t. Consider a small δ > 0 and a very small ε > 0 and times s1 < t1 < … defined Thus in some sense Sj marks the jth time y leaves the immediate vicinity of 0 and tj the jth time that y returns (see Figure 89.2).