Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm reading Fractals Everywhere by Michael Barnsley. On pp. 6-8 [1] he defines a linear space which, he says, "is also called a vector space." However, his definition of a linear space only requires closure under vector addition and scalar multiplication. There is no mention, for example, of additive inverses. (I realize other properties are required.)

He also defines the Riemann Sphere (see section 1.5), and later, in an exercise, asks the reader to show that the Riemann Sphere is not a vector space.

I can see that the Riemann Sphere is not a vector space in the traditional sense because $\infty$ has no additive inverse. But it seems that using Barnsley's definition, it would be a vector space, if we adopt the convention that $\infty + x = \infty$ for all $x \in \mathbb{C}$ and that $a \cdot \infty = \infty$ for all $a \in \mathbb{R}$.

Am I missing something? Thanks.

[1] http://goo.gl/d0NzN

share|improve this question
add comment

2 Answers 2

up vote 6 down vote accepted

You might call this thing (vector addition without inverses and scalar multiplication) a semi-module (wikipedia) over a field. The definition is as follows (I assume that you have an additive identity, even though you don't specify one):

A semimodule over a field $k$ is a set $M$ together with a function $+:M\times M \to M$ (vector addition) which defines an associative and commutative operation, an element $0\in M$ which is an identity element for $+$, and an action $k\times M \to M$ (scalar multiplication) which distributes over $+$.

The problem is, once you have scalar multiplication distributing over vector addition, you have $v + (-1)v = (1-1)v = 0$, thus $(-1)v$ is an additive inverse! Thus a semimodule over a field is the same as a module over a field which is just another name for a vector space. Thus the definition of a 'linear space' implies the remaining vector space axioms.

You can get around this by forgetting about some of the structure on the field, namely the additive inverse, but most often people drop both the additive and the multiplicative inverse to get a semiring = rig (wikipedia). Then semimodules over a rig are very interesting objects, and feature in an area called tropical geometry. A basic example in tropical geometry is the rig $\mathbb{R}\cup \{\infty\}$ (similar to your example), but where 'addition' is $x\oplus y = min\{x,y\}$ and 'multiplication' is $x\otimes y = x+y$.

share|improve this answer
    
I have never heard of this. Thanks. –  Kerry Mar 11 '11 at 6:06
    
Thanks for the response, @David. How do we know that 0v = 0? That is, how do we know that for any vector v in M, the product of the additive identity for the field k and v is equal to the additive identity (a vector) for the set M? I don't think TeX works in comments, so to be clear: the first '0' in "0v = 0" is a scalar, and the second '0' is a vector. –  David Mar 11 '11 at 18:40
    
Never mind, I see it now: v = 1*v = (1+0)*v = 1*v + 0*v = v + 0*v, so 0*v is an (really the) additive identity for M. –  David Mar 11 '11 at 19:50
add comment

I don't feel I really contributed anything. I think the usually we treat $C^{*}$ is to treat it as $\mathbb{C}\mathbb{P}_{1}$, in other words the complex "lines" passing the complex "plane". Hence for two elements in the plane we have $(c_{1},c_{2})\approx (\frac{c_{1}}{c_{2}},1)$ for $c_{2}\not=0$ and vector addition just follows. At $\infty$ we have $c_{2}=0$ and the whole line $(c_{1},0)$ was identified as $\infty$.

Now if you add two points in $C^{*}$, we are essentially adding two lines, and we may define $(a,b)+(c,d)=(ad+bc,bd)=(\frac{a}{b}+\frac{c}{d},1)$ if $b,d$ are not both zero. In case $d$ ended up being $0$ we should have $(a,b)+(c,d)=(ad,0)$, which is identified to the infinity point. This kind of justified $a+\infty=\infty$ in the Riemann Sphere.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.