Page:American Journal of Mathematics Vol. 2 (1879).pdf/8

2 $$G$$ points whose notation differs in the particular that one group of three letters has suffered a change not cyclic are conjugate with respect to the conic $$S.$$ There are ten ways in which six letters can be divided into two groups of three each, hence there are ten pairs of points $$G.$$

Four $$G$$ points which lie on one Steiner-Plücker line are (Salmon, p. 362,) $$G\,(BDA\,.ECF),$$ $$G\,(EDF\,.BCA),$$ $$G\,(BCF\,.EDA),$$ $$G\,(BDF\,.ECA).$$ We shall call the Steiner-Plücker line on which they lie $$i\,(BE\,.CD\,.AF).$$ In the notation of an $$i$$ line the division into groups of two is important, but not the order of the letters in each group. The number of ways in which six things can be separated into three different groups of two things each is fifteen, hence there are fifteen lines $$i.$$ The $$G$$ points on one $$i$$ line are obtained by selecting one letter out of each group of two for the first group of three, and taking the remaining three letters, in the same order, for the other group of three. As this can be done in four different ways, there are four points $$G$$ on one line $$i.$$ Through one point $$G$$ pass three lines $$i;$$ viz., through $$G\,(ABC\,.DEF),$$ pass $$i\,(AD\,.BE\,.CF),$$ $$i\,(AE\,.BF\,.CD),$$ $$i\,(AF\,.BD\,.CE).$$ In writing the symbols for the $$i$$ lines through one $$G$$ point, it is necessary to observe that the cyclic order of the first letters of the three duads must be the same as that of the second letters; for instance, through $$G\,(ABC\,.DEF)$$ does not pass $$i\,(AD\,.BF\,.CE).$$

The Kirkman point which corresponds to the Pascal $$h\,(ABCDEF)$$ is (Salmon, p. 363,) the intersection of the Pascal lines $$h\,(ACEBFD),$$ $$h\,(CEADBF),$$ $$h\,(EACFDB).$$ We shall call this the Kirkman point $$H\,(ABCDEF).$$ The Pascal lines through a Kirkman point are obtained by taking the three odd letters in the order in which they stand, and then the three even letters, inverting the order of the last two, for the first Pascal; and then deriving the other two Pascals from this by a cyclic change of the first three letters in one direction and of the last three in the other direction. Similarly, the three Kirkman points on one Pascal, $$h\,(ACEBFD),$$ are $$H\,(AEFCDB),$$ $$H\,(EFABCD),$$ $$H\,(FAEDBC).$$ If we wish to know whether two given Pascals, as $$h\,(ABCDEF),$$ $$h\,(AEDBCF),$$ intersect in a Kirkman point or not, we have suffered opposite cyclic changes. The two lines just written are $$h\,(BCDEFA),$$ $$h\,(DBCFAE),$$ and they meet in $$H\,(BECADF).$$

The three $$H$$ points of one Cayley-Salmon line are (Salmon, p. 362,) $$H\,(ABCFED),$$ $$H\,(ADCBEF),$$ $$h\,(AFCDEB).$$ We shall call this Cayley-