the form $(x,x)$ in some way. The problem is that as points in $C_{2}(S^{1})$ approach points on the diagonal, the Gauss map has two limits— the forward and the backward sweeping tangents.

Compactifying the configuration space
-------------------------------------

Let us define $\overline{C_{2}(S^{1})}$, a compactification of $C_{2}(S^{1})$, by pasting two copies of the diagonal, $\Delta_{0}$ and $\Delta_{1}$ to its missing diagonal $\{(x,x)|x\in S^{1}\}$, as shown in the diagram below.

$$\xy 0;/r4pc/: (0,0)="a"; (1,1)="b" **@{--}; (0,1)="c" **@{-};
?(0.45)*{>>}; (0,1)="c"; "a" **@{-}; ?(0.45)*{\wedge}; (2,0)="a";
(3,1)="b" **@{-}; (2,1)="c" **@{-}; ?(0.45)*{>>}; (2,1)="c"; "a"
**@{-}; ?(0.45)*{\wedge}; (0,-.2)="a"; (1,.8)="b" **@{--};
(1,-.2)="c" **@{-}; ?(0.45)*{\wedge}; (1,-.2)="c"; "a" **@{-};
?(0.45)*{>>}; (2,-.2)="a"; (3,.8)="b" **@{-}; (3,-.2)="c" **@{-};
?(0.45)*{\wedge}; (3,-.2)="c"; "a" **@{-}; ?(0.45)*{>>};
(.5,1.3)*=0{\textbf{$C_{2}(S^{1})$}};
(2.5,1.3)*=0{\textbf{$\overline{C_{2}(S^{1})}$}};
\endxy$$  

Points on $\Delta_{0}$, which are limits of the form $\lim_{y\rightarrow x^{+}}(x,y)$, shall be denoted $(x,x^{+})$, with points on $\Delta_{1}$ correspondingly denoted $(x,x^{-})$. At these boundaries of $C_{2}(S^{1})$, the Gauss map converges to the tangent vector to the curve, sweeping either forwards or backwards depending on whether its input converges to a point in $\Delta_{0}$ or to a point in $\Delta_{1}$. This allows us to solve the problem of how to extend the Gauss map to the diagonal. $\Phi$ can be extended smoothly to a function $\overline{\Phi_{\gamma}}: \overline{C_{2}(S^{1})}\rightarrow
S^{2}$, 