Download Algebraic Topology Aarhus 1982. Proc. conf. Aarhus, 1982 by I. Madsen, B. Oliver PDF

By I. Madsen, B. Oliver

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Extra resources for Algebraic Topology Aarhus 1982. Proc. conf. Aarhus, 1982

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A point z0 is a removable singularity if the limit of the function at this point exists and, therefore, we may remove the singularity taking the limit as the value of the function at z0. A point z0 is a pole of a function f(z) if it is a zero of the function 1/f(z). Finally, z0 is an essential singularity if both limits of f(z) and 1/f(z) at z0 do not exist. The Lauren series centred at a removable singularity has not powers with negative exponent. The series centred at a pole has a finite number of powers with negative exponent, and that centred at an essential singularity has an infinite number of powers with negative exponent.

Let z and t be complex numbers and a, b, c and d vectors fulfilling: z=ab t=cd Then the following equalities are equivalent: zt=tz ⇔ abcd=cdab 18 RAMON GONZALEZ CALVET A complex number z and a vector c do not commute, but they can be permuted by conjugating the complex number: z c = a b c = c b a = c z* Every real number commute with any vector. However every imaginary number anticommute with any vector, because the imaginary unity e12 anticommute with e1 as well as with e2: zc=−cz z imaginary The complex plane In the complex plane, the complex numbers are represented taking the real component as the abscissa and the imaginary component as the ordinate.

13 O = c C + ( 1 − c ) C' with a, b and c being real coefficients. Equating the first and second equations we obtain: a A + ( 1 − a ) A' = b B + ( 1 − b ) B' which can be rearranged as: a A − b B = − ( 1 − a ) A' + ( 1 − b ) B' Dividing by a − b the sum of the coefficients becomes the unity, and then the equation represents the line passing through the points AB and A'B' (belonging to both pencils), which is the line P: P= a b a −1 b −1 A− = A' − B' a−b a−b a−b a−b RAMON GONZALEZ CALVET 50 By equating the second and third equations for the line O, and the third and first ones, analogous equations for Q and R are obtained: Q= b c b −1 c −1 B− C= B' − C' b−c b−c b−c b−c R= c a c −1 a −1 C− A= C' − A' c−a c−a c−a c−a Now we must prove that the lines P, Q and R belong to the same pencil, that is, fulfil the equation: R=dP+(1−d)Q with d real With the substitution of the former equations into the last one we have: c a b c  b   a  C− A=d  A− B + ( 1 − d )  B− C − − c−a c−a a − b b − a b c b c     Arranging all the terms at the left hand side, the expression obtained must be identical to zero because A, B and C are independent lines (not belonging to the same pencil): b (1 − d )  c (1 − d )  a   da  db  c − a − b − c − a  A +  a − b − b − c  B +  c − a + b − c  C ≡ 0       This implies that the three coefficients must be null simultaneously yielding a unique value for d, d= a−b a−c fact which proves that P, Q and R are lines of the same pencil.

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