Different defns of homorphism
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From: Peter on 2 Jun 2010 10:44 Hi, Excuse my ignorance, but I wanted to find out if there are different defns of what a "homomorphism" is. From abstract algebra, I recall that f is a homomorphism from groups (A,*) to (B,&) iff for every a and b in A f(a*b) = f(a)&f(b) I was trying to get a clear understanding of the relationship between a homomorphism and a linear transformation (on a vector space V over the field F, denoted V(F)). The linearity condition implies that all linear transformations are homomorphisms, so I started to wonder if all homomorphisms on vector spaces also linear transformations. The reason why this is not clear to me is that I am not sure whether you could construct a homomorphism on a vector space that would not satisfy the second property that a linear transformation must satisfy, homogeneity. When I look at texts however that deal with this question things become somewhat confusing. In "Abstract Algebra" by W. E. Deskins, pg. 494, section 12.1 "Homomorphisms and Linear Transformations", we find the following definition of a homorphism: " Definition 12.1 The mapping A from vector space V(F) to vector space W(F) is a homomorphism or linear transformation from V(F) to W(F) iff (a u + b v)A = a (uA) + b (vA) for every a,b in F and u,v in V" while another text I looked at talks about "homomorphisms of vector spaces" (which is the same as Deskins "homomorphism") Sorry for the stupid question, but I just want to get a clear understanding of this concept. Thank you Peter
From: Peter on 2 Jun 2010 11:00 It seems that there is a homophism between vector spaces that is not a linear transformation: Let V be the space of mxn matrices having complex entries and f be complex conjugation. If the field F is the set of complex numbers, then V(F) is a vector space. Under addition, f is a homomorphism from V to V since f(M + N) = (M+N)* = M* + N* = f(M) + f(N) for every M,N in V with * denoting complex conjugation. f however is not a linear transformation as it is not homogeneous since f(aM) = (aM)* = a* M* not equal to a M* = a f(M) for a in the set of complex numbers when Im(a) is non-zero. So, I'm still confused why they use the same terms interchangeably. Peter
From: FredJeffries on 2 Jun 2010 11:36 On Jun 2, 8:00 am, Peter <pwoly...(a)yahoo.com> wrote: > It seems that there is a homophism between vector spaces that is not a > linear transformation: > > Let V be the space of mxn matrices having complex entries and f be > complex conjugation. > If the field F is the set of complex numbers, then V(F) is a vector > space. > Under addition, f is a homomorphism from V to V since > f(M + N) = (M+N)* = M* + N* = f(M) + f(N) for every M,N in V with * > denoting complex conjugation. > f however is not a linear transformation as it is not homogeneous > since > f(aM) = (aM)* = a* M* not equal to a M* = a f(M) for a in the set of > complex numbers when Im(a) is non-zero. > > So, I'm still confused why they use the same terms interchangeably. You're leaving off the second part of the terms. Your example is a homomorphism OF GROUPS. An linear transformation is a homomorphism OF VECTOR SPACES. We do not speak of generally homomorphisms as such. It is a homomorphism of some particular algebraic structure -- a homomorphism of groups, a homomorphism of rings, a homomorphism of vector spaces... As your example shows, it is possible for a mapping to be a homomorphism with respect to one particular structure but not a homomorphism with respect to another structure on the same sets. See http://en.wikipedia.org/wiki/Homomorphism > > Peter
From: Peter on 2 Jun 2010 11:54 On Jun 2, 11:36 am, FredJeffries <fredjeffr...(a)gmail.com> wrote: > You're leaving off the second part of the terms. Your example is a > homomorphism OF GROUPS. An linear transformation is a homomorphism OF > VECTOR SPACES. > > We do not speak of generally homomorphisms as such. It is a > homomorphism of some particular algebraic structure -- a homomorphism > of groups, a homomorphism of rings, a homomorphism of vector spaces... > > As your example shows, it is possible for a mapping to be a > homomorphism with respect to one particular structure but not a > homomorphism with respect to another structure on the same sets. > I suspected that this was the case, but was not sure. Thank-you very much! Peter
From: Bill Dubuque on 5 Jun 2010 11:22
Chip Eastham <hardmath(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 |