Garrett 03-30-2012 1 • Interlude: Calculus on spheres: invariant integrals, invariant
... Theorem: (instance of Schur’s Lemma) For a finite-dimensional irreducible representation V of a group G, any G-intertwining ϕ : V → V of V to itself is scalar. Proof: First, claim that the collection HomG (V, V ) of all Gintertwinings of finite-dimensional V to itself is a division ring. Indeed, giv ...
... Theorem: (instance of Schur’s Lemma) For a finite-dimensional irreducible representation V of a group G, any G-intertwining ϕ : V → V of V to itself is scalar. Proof: First, claim that the collection HomG (V, V ) of all Gintertwinings of finite-dimensional V to itself is a division ring. Indeed, giv ...
SUM AND PRODUCT OF DIFFERENT SETS 1 Mei
... point out that the geometric approach does not distinguish between sets of integers and sets of real numbers. On the other hand, it does not provide nontrivial lower bounds on |A + B| + |AB|, if the set B is much smaller that A. It is also not enough for showing that |AB| > (|A||B|)1−² for all A, B ...
... point out that the geometric approach does not distinguish between sets of integers and sets of real numbers. On the other hand, it does not provide nontrivial lower bounds on |A + B| + |AB|, if the set B is much smaller that A. It is also not enough for showing that |AB| > (|A||B|)1−² for all A, B ...
SOME ALGEBRAIC DEFINITIONS AND CONSTRUCTIONS
... Proposition 13. If f : R −→ S is any homomorphism of rings, then its kernel is an ideal I and its image is isomorphic to R/I. Proposition 14. R/I is an integral domain if and only if I is prime. R/I is a field if and only if I is maximal. A maximal ideal is prime, but not conversely. Exercise 15. Le ...
... Proposition 13. If f : R −→ S is any homomorphism of rings, then its kernel is an ideal I and its image is isomorphic to R/I. Proposition 14. R/I is an integral domain if and only if I is prime. R/I is a field if and only if I is maximal. A maximal ideal is prime, but not conversely. Exercise 15. Le ...
