Different defns of homorphism
in [Math]
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From: Chip Eastham on 5 Jun 2010 22:47 On Jun 5, 11:22 am, Bill Dubuque <w...(a)nestle.csail.mit.edu> wrote: > Chip Eastham <hardm...(a)gmail.com> wrote: > > > Given the setting provided by Fred and Arturo, we can > > point out a special case in which a group homomorphism > > amounts to linear transformation of vector spaces. > > > Let A,B be vector spaces over the field Q of rational > > numbers. If F:A -> B is a homomorphism from A to B > > of the additive (abelian) groups of their vectors, > > then F is also a linear transformation (a homorphism > > of vector spaces). > > > [Sketch of proof: We know F "preserves" vector addition > > and identity for vector addition. It remains to show > > F preserves scalar multiplication, i.e. that for any > > rational scalar r and any vector v in A: > > > F(rv) = r F(v) > > > First we prove this for natural numbers r = n by > > expressing nv = v+...+v (n times) and applying > > the vector space properties of A and B and the > > assumption that F preserves vector addition. > > One then extends this to nonpositive integers by > > vector space properties of additive identity and > > inverses. Finally one extends it to arbitrary > > rational numbers r = p/q by exploiting: > > > pv = (p/q)v + (p/q)v + ... (p/q)v (q times) > > > and the vector space properties of A and B.] > > > This then motivates how one might construct an > > example of group homomorphism from one vector > > space to another that is not a vector space > > homomorphism (linear transformation), taking > > A,B to be vector spaces over a field larger > > than Q. > > > We can give a completely constructive example > > by taking the field A = B = Q(sqrt(2)) so A > > and B are both one-dimensional vector spaces > > over this field. Considered as vector spaces > > over Q, then A and B are two-dimensional, so > > we can define a Q-linear transformation (group > > homomorphism) by F(1) = 0, F(sqrt(2)) = 1. > > But this is not a linear transformation with > > respect to scalar field Q(sqrt(2)). > > Generally the torsion subgroup T of a divisible abelian group G > is divisible, G/T is a vector space over Q, and G = T (+) G/T > > --Bill Dubuque Thanks, Bill. However I may be missing how your example/construction fits into the discussion. Yes, the canonical quotient map G -> G/T is a group homomorphism that is not a homomorphism of vector spaces, but G here is not a vector space (although the torsion free part G/T turns out to be one). regards, chip
From: Bill Dubuque on 6 Jun 2010 11:30
Chip Eastham <hardmath(a)gmail.com> wrote: > On Jun 5, 11:22 am, Bill Dubuque <w...(a)nestle.csail.mit.edu> wrote: >> Chip Eastham <hardm...(a)gmail.com> wrote: >> >>> Given the setting provided by Fred and Arturo, we can >>> point out a special case in which a group homomorphism >>> amounts to linear transformation of vector spaces. >> >>> Let A,B be vector spaces over the field Q of rational >>> numbers. If F:A -> B is a homomorphism from A to B >>> of the additive (abelian) groups of their vectors, >>> then F is also a linear transformation (a homorphism >>> of vector spaces). >> >>> [Sketch of proof: We know F "preserves" vector addition >>> and identity for vector addition. It remains to show >>> F preserves scalar multiplication, i.e. that for any >>> rational scalar r and any vector v in A: >> >>> F(rv) = r F(v) >> >>> First we prove this for natural numbers r = n by >>> expressing nv = v+...+v (n times) and applying >>> the vector space properties of A and B and the >>> assumption that F preserves vector addition. >>> One then extends this to nonpositive integers by >>> vector space properties of additive identity and >>> inverses. Finally one extends it to arbitrary >>> rational numbers r = p/q by exploiting: >> >>> pv = (p/q)v + (p/q)v + ... (p/q)v (q times) >> >>> and the vector space properties of A and B.] >> >>> This then motivates how one might construct an >>> example of group homomorphism from one vector >>> space to another that is not a vector space >>> homomorphism (linear transformation), taking >>> A,B to be vector spaces over a field larger >>> than Q. >> >>> We can give a completely constructive example >>> by taking the field A = B = Q(sqrt(2)) so A >>> and B are both one-dimensional vector spaces >>> over this field. Considered as vector spaces >>> over Q, then A and B are two-dimensional, so >>> we can define a Q-linear transformation (group >>> homomorphism) by F(1) = 0, F(sqrt(2)) = 1. >>> But this is not a linear transformation with >>> respect to scalar field Q(sqrt(2)). >> >> Generally the torsion subgroup T of a divisible abelian group G >> is divisible, G/T is a vector space over Q, and G = T (+) G/T > > Thanks, Bill. However I may be missing how your example/construction > fits into the discussion. Yes, the canonical quotient map G -> G/T is > a group homomorphism that is not a homomorphism of vector spaces, > but G here is not a vector space (although the torsion free part G/T > turns out to be one). The point was to mention a more general perspective. Namely, how Q-vector spaces arise naturally from divisible abelian groups. --Bill Dubuque |