By William Fulton
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This booklet relies on a lecture direction that I gave on the collage of Regensburg. the aim of those lectures used to be to provide an explanation for the function of Kahler differential varieties in ring concept, to organize the line for his or her program in algebraic geometry, and to guide as much as a little analysis difficulties The textual content discusses nearly completely neighborhood questions and is for this reason written within the language of commutative alge- algebra.
This path was once learn within the division of arithmetic on the collage of Washington in spring and fall 1999.
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Additional resources for Algebraic Curves: An Introduction to Algebraic Geometry
Thus M\ll = ¢. P 46 Let us then define 111 by 111 : ~p E M~t Sp ~ 0}. First, we assume that 111 is not empty. Then 111 is a non empty open part of M and in each point p of Ill, we know that (Vh)p = 0. The classical Pick-Berwald theorem then implies that 111 is an open part of a nondegenerate ellipsoid or hyperboloid. Thus detS is a constant different from zero on 111" The continuity of detS then implies that fit = M. Finally, we may assume that S = 0 on the whole of M. Thus by Proposition 2, we can suppose that M is given by the equation z -- P(x,y), where P is a polynomial of degree at most k + 1, and that the canonical affine normal vector field is given by (0,0,1).
Amer. Math. Soc. 210, 75-106 (1975). 2. , Rigoli, M. and Woodward, minimal immersions of S 2 into CP n. 3. M. : On conformal Math. Ann. 279, 599-620 L. M. (1988). : Minimal immersions of S 2 and RP 2 into CP n with few higher order singularities. To appear in Math. Proc. Camb. Phil. Soc. 4. Bolton, J. and Woodward, forms. M. : On immersions of surfaces into space Soochow J. of Mathematics 5. Bolton, J. M. immersions with St-symmetry. 6. Bolton, J. M. 14, 11-31 (1988). : On the Simon conjecture for minimal Math.
Then it follows from Lemma 5 that a = 0 and ,11 = ,l 2. Thus, we have (i). Therefore, we may assume that detS = 0. But then we know that there exists an eigenvector u of S with eigenvalue zero. Again there are two possibilities. a. h(u,u) -- 0. In this case, we can find a vector v, such that h(v,v) = 0 and h(u,v) = 1. Using the equation of Ricci, we then obtain that h(Sv,u) = h(v,Su) = 0. Hence Sv has no component in the direction of v. Thus, we have (iii). b. h(u,u) $ 0. Here, we may assume, by taking - ~ as normal, that h(u,u) = 1.