90 THI'] ROYAL SOCIETY OF CANADA 



inside the circle. The map is a curvilinear triangle U (a, b, d) which 

 is everywhere convex to the outside. These two regions U (abc), 

 U (abd) have a region U (ab) in common; its contour passes through 

 A, B and is everywhere convex to the outside. Hence the straight 

 line A B lies within U (ab) and must therefore be the map of a line 

 within the region common to (abc), (abd). 



Having proved that there are in the x-plane curves AD, BD, CD 

 that map into straight lines in the u-plane, it follows that the map 

 of X is the rectilinear triangle D' D" D"^ with A, B, C at the middle 

 points of the sides. 



The problem of the inversion depends on the conformai repre- 

 sentation of the whole of a dissected Riemann surface T^ We may 



take this surface to be that in the 

 figure; the lines connecting d with a, 

 b, c being L (da), L (db), L (dc). 



Let the dissected sheet considered 

 by Schwarz carry the value u, and below 

 it place another sheet so that below each 

 place X, y is another place (x, -y). Let 

 the lower limit for u be placed at (d, o). 

 Then exactly corresponding paths on these two sheets give the values 

 u, and u^= -u. Now establish a connection of the two sheets along 

 the bridge L (dc) but maintain the cuts along L (da) , L (db) , treating 

 these as cuts along B, A, when B, A have been shrunk to L (da) and 

 L(db). We have now the dissected Riemann surface T. The effect 

 of the barriers on the upper sheet of T^ and absence of barriers on 

 the lower sheet is that a path from d on the lower sheet will have 

 above it a path from d that crosses A, B; hence the values of u, u^ 

 attached to th3 two covertical places will u, u^= - u4-2mûj + 2m'aji 

 where m, m^ are integers. The values of m, m* are shown to be 1 by 

 placing X at c, where u=u^ = w+a;i. The relation u +u^ =2(0 -\-2oA 

 shows that the join of u, u^ is bisected by (o+co^ and hence that the 

 aggregate of values of u^ forms a triangle which combines with the 

 u-triangle to give a rectilinear parallelogram of periods. By allowing 

 free passages over A, B, we see that the map of T is the network of 

 parallelograms of periods. 



IL Weierstrass's Method based on Abel's theorem. 



J 'au o 



, .., , dx /"x, y dx 



Letu=^ / * ,/TTT~r = / — , so that x takes one 



^^^^-^^ ya,. o y 



of the values a,, a,, a.,, ai for u=o. It has to be shown that x, y 

 are one-valued functions of u for all finite values of u, with no singula- 



