The Unapologetic Mathematician

Mathematics for the interested outsider

Some Topological Fields

Okay, hopefully now that I’m back in Kentucky and I’ve gotten the first day of all three of my classes under my belt, I can get back into the groove.

In a little while we’re going to want to talk about “evaluating” a power series like we did when we considered polynomials as functions. But when we try to map into our base field, we get a sequence of values and we ask questions about convergence. That means we need to have a topology on our field! Luckily there are a few hanging around.

The real numbers have a topology. In fact, that’s really their main characteristic. The rational numbers have a topology too, but the whole motivation of the construction of the real numbers is that rational numbers have a lot of holes that sequence limits fall through. Since we’re talking about sequences here we really need the tighter weave that the real numbers provide.

What else can we use? The complex numbers form a two-dimensional vector space over the reals, which means that as a vector space we have the isomorphism \mathbb{C}\cong\mathbb{R}\times\mathbb{R}. So let’s just use the product metric, along with the topology it gives on the complex numbers.

Let’s be a little explicit here: the product metric depends on picking the basis \{1,i\} for \mathbb{C} as a vector space over \mathbb{R}. We get the “size” of a complex number |a+bi|=\sqrt{a^2+b^2}, and then we define the distance between two complex numbers as the size of their difference.

I said before that there may be many different distance functions that give the same topology, so why do we really want to use this one? Well, it turns out that this formula can actually be written in a really neat way in complex number language. If we have a complex number z=a+bi we also have its conjugate \bar{z}=a-bi. Then we can calculate the product

z\bar{z}=(a+bi)(a-bi)=(a^2+b^2)+0i

This is just the square of our distance function, written as a complex number! But also notice that this formula is symmetric between a complex number and its conjugate. That is, conjugation preserves the size of complex numbers, as we should expect because there’s no a priori difference between the two.

Now we need to prove that the field operations are continuous. For multiplication, for example, we need to ask that if z_2 is close to z_1 and w_2 is close to w_1, then z_2w_2 is close to z_1w_1. We write z_2=z_1+\Delta z and w_2=w_1+\Delta w, and multiply out

z_2w_2=z_1w_1+z_1\Delta w+\Delta zw_1+\Delta z\Delta w

The condition is that for any real \epsilon>0 we can find real \delta_z and \delta_w so that \delta_z>|\Delta z| and \delta_w>|\Delta w| together imply that |z_1\Delta w+\Delta zw_1+\Delta z\Delta w|<\epsilon. From here it’s a bunch of mucking around with our formula, but none of it is very difficult.

At the end of the day, we’ve got two topological fields to work with — the real numbers \mathbb{R} and the complex numbers \mathbb{C} — and we can talk about evaluating power series from \mathbb{R}[[X]] or \mathbb{C}[[X]].

UPDATE: I forgot to mention that it’s also easy to see that the norm is multiplicative. That is, |zw|=\sqrt{zw\overline{zw}}=\sqrt{z\bar{z}}\sqrt{w\bar{w}}=|z||w|.

August 26, 2008 Posted by | Algebra, Ring theory | 5 Comments

   

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