1. Oct 21, 2007

### Simfish

Hello there,

So I've noticed that at least out of the sources I've read, none of the point out the connection between additivity (a key operation that is emphasized on many texts) and homomorphisms. After all, a homomorphic function is merely a function wherein f(xy) = f(x)f(y). So a multiplicative function would be homomorphic under multiplication, and an additive function would be homomorphic under addition. Am I missing anything?

So is polynomial addition an isomorphism between the domain (the real line) and the range ONLY IF the function is monotonically increasing? (since bijectivity is a pre-req for isomorphism?) Is it a homomorphism otherwise?

Last edited: Oct 21, 2007
2. Oct 21, 2007

### matt grime

In what category are you thinking? Since your definitions don't seem quite right.

3. Oct 21, 2007

### CompuChip

You mean that if we have two sets A, B with operations $\circ_A, \circ_B$ on them (e.g. $\circ_A: A \times A \to A$ is function with appropriate restrictions, e.g. non-degenerate, bilinear) then you call a function homomorphic when
$$f(x \circ_A y) = f(x) \circ_B f(y)$$ for all x, y in A.
This is completely general and does not require to have an operation like addition or multiplication, if could be something completely different (composition of functions, or a commutator, or ...), but it is possible to have one or both. If you consider e.g. the group $(\mathbf{R}, +)$ or $(\mathbf{R}^* , \cdot)$ then a additive resp. multiplicative function will indeed be a homomorphism of the group to itself, but that is just a very specific example which is not special in any way (most functions are neither additive nor multiplicative), so why should you want to mention it explicitly (other than to demonstrate the idea?)

Also, if you add to monotonically increasing functions they stay monotonically increasing (for them to be bijective you should have strict monotonicity by the way, as $x \mapsto 0$ is also monotonic ). You ask if adding of polynomials is a isomorphism. I suppose you need a function
$f : \mathbf{R}[X] \to \mathbf{R}[X]$ (mapping polynomials to polynomials) and ask if $f(P(X) + Q(X)) = f(P(X)) + f(Q(X))$ iff P and Q are (strictly) monotonically increasing, but I'm not sure what you want to take for f. Or do you simply mean that (P + Q)(x) = P(x) + Q(x) (which is obviously true, you don't need to consider homomorphisms for this), or that P(x + y) = P(x) + P(y) (which is obviously not true, e.g. (x + y)^3 =/= x^3 + y^3).

Actually, I think either of us is mixing up some things, perhaps you could be more precise on what you mean?

4. Oct 23, 2007

### Simfish

Ah, it's just a way to demonstrate the idea (most abstract algebra texts have very few examples, so more examples helps me to understand the idea of a homomorphism, especially one applied to sets that aren't so small - those sets are usually in the examples we see in abstract algebra textbooks).

==
Or could it be a mapping from the values of the x-axis to the y values? After all, in the case of a quadratic polynomial, it would be f(x) = x^2. So the f here would map real numbers to real numbers (though here the map obviously is not bijective).

So, say, in the case of f(x) = x^2, f(x+y) = (x+y)^2 = f(x) + f(y) + 2xy (so here at least, we don't have a homomorphism).

So I think I'm thinking in terms of functional notation? In examples of abstract algebra texts, for example, we see that the exponential and logarithm functions are homomorphisms. After all, exp(x+y) = exp(x)exp(y) [OVER MULTIPLICATION] and log(x+y) = log(x) + log(y)[OVER ADDITION]. Actually, don't the homomorphic properties of the exponential and logarithmic functions make the functions so special? (since those homomorphic properties are precisely how we define such functions by, aside from the starting value?). In that case, the homomorphic property would actually be a significant one.

Polynomials probably weren't the best example here (since polynomials usually increase and decrease somewhere in the interval).

==
Oh, and sorry if my posts are confusing - I have a really weird learning style - so I initially make all sorts of propositions when I don't understand something very well - and those propositions are often wrong (but it's better than what I previously did - which was to stay silent - which certainly never helps)

Last edited: Oct 23, 2007
5. Oct 23, 2007

### CompuChip

OK, so I'm going to make a very bold statement...
There's this difference between abstract concepts and practical instances. Usually the abstract concepts are the powerful things which have beautiful and/or complicated proofs and it is usually these what mathematicians are after. Practical examples serve to illustrate the usefulness of the concept, and often also to provide ways to work with the abstract concepts more quickly in situations which occur more often. It is not uncommon though, that examples do not suffice to grasp the full idea, and you should just work with them until they become natural to appreciate their full power.

Sometimes I get confused as well, and start mixing things up. At least you just give it a try, and see where you end up (which is a good approach IMO) but has the downside of sometimes leaving people a little confused In any case, talking nonsense and showing people you don't understand and they should explain is always better than to keep silent and give them the impression that you understand it and let them find out after a long time. So just keep posting

So anyway, the polynomials were a bad choice, even strictly monotonous ones, as your counter example shows. As your other example shows, exp is a homomorphism (though you forget to mention between which groups, namely $(\mathbf{R}, +)$ and $(\mathbf{R}^*, \times)$, or actually $(\mathbf{R}^+, \times)$ -- where I used notations $\mathbf{R}^* = \mathbf{R} \setminus \{0\}, \mathbf{R}^+ = \{ x \in \mathbf{R} | x > 0 \}$).
Indeed, exp and log are two of the few functions which have this property, but I wouldn't say they are defined to have this property (rather, the exponential is the function that has itself as a derivative), in fact for any number r we have $r^{x+y} = r^x r^y$.

Just a question: where does your question come from? Are you learning group theory at the moment? Maybe I can give you more relevant information

Last edited: Oct 23, 2007
6. Oct 23, 2007

### Simfish

Oh I see. Yeah, as for the practical instances, I usually use them to at least understand the concept at first (they're the easiest way to start understanding a concept - or testing how general the concept is - when one asks questions such as "does this general concept apply to that practical example?")

Yeah, I'm learning group theory at the moment. I'm making questions up for myself as I go, since it seemed like additivity (but as you said, "(P + Q)(x) = P(x) + Q(x) " isn't a homomorphism! - it's just a notational confusion!). Thanks for your help!