609 
NOTES AND REFERENCES. 
445, 451, 454. We have the two papers by K. Rohn, “Die Flachen vierter Ordnung 
hinsichtlich ihrer Knotenpunkte und ihrer Gestalten,” Preisschr. der F. J. Gesell. zu 
Leipzig (Leipzig, 1886, pp. 1—58), and same title Math. Ann. t. xxix. (1887), pp. 81—97. 
I have not been able to examine the conclusions arrived at in these papers with as 
much care as would have been desirable. 
I call to mind that for a &-nodal quartic surface the tangent cone from any node 
is a sextic cone with (k — 1) nodal lines, breaking up it may be into cones of lower 
orders—see table p. 265: and that we distinguish the quartic surfaces according to 
the forms of the sextic cones corresponding to the k nodes respectively. It will be 
recollected that (6) denotes a sextic cone, (6 X ) a sextic cone with one nodal line, 
(5i, 1) a sextic cone breaking up into a quintic cone with one nodal line and a plane ; 
and so in other cases. 
There is a sort of break in the theory ; in fact when the number of nodes is 
not greater than 7 these may be any given points whatever, and taking the 7 points 
at pleasure we have surfaces with 8 nodes, and 9 nodes, but not with any greater 
number of nodes, viz. for a surface with 10 or more nodes, it is not permissible to 
take 7 of these as points at pleasure, so that the theory of the surfaces with 10 
or more nodes is so to speak separated off from that of the surfaces with a smaller 
number of nodes. For the case of 10 nodes we have the symmetroid 10(3, 3) and 
other forms, for 11 nodes Rohn finds 3 or ? 4 forms; for 12 nodes he has four forms, 
viz. my 3 forms and a fourth form 12 rf ; for 13 nodes he has two forms, 13& agreeing 
with my 13«, and 13« which replaces my non-existent form 13p ; for 14 nodes, 15 nodes 
and 16 nodes he has in each case a single form, agreeing with my results. Without 
endeavouring to complete the theory, I write down a table as follows: 
No. of 
Nodes 
Form of Cones 
Remarks 
16 
16(1, 1, 1, 1, 1, 1) 
15 
15(2, 1, 1, 1, 1) 
14 
8 (3 15 1, 1, 1) + 6 (2, 2, 1, 1) 
13 
13 6 = 13« 
3(4„ 1, 1) + 1(3, 1, 1, 1) + 9 (3„ 2, 1) 
5) 
13« 
1(2, 2, 2) + 12 (4„ 1, 1) 
13« replaces my non-existent 
12 
12 & =12 a 
12(4„ 2) 
13„, =13(2, 2, 2) 
55 
12«= 12, 
.2(5,, l) + 6(3„ 3,) + 4(3, 2, 1) 
C. VIT. 
77