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The title says it all. The motivation behind this question is that I'm stuck with a theorem in topology, which states that if $f,g:X \rightarrow \mathbb{K}$ (where $\mathbb{K}$ is either $\mathbb{R}$ or $\mathbb{C}$ - both endowed with the euclidean topology - and $X$ is some topological space) is continuous, then $f\cdot g$ is as well. Since the sketch of the proof goes like this: "We know that from that the multiplication $\textrm{mult}: \mathbb{K}^2 \rightarrow \mathbb{K}$ is continuous and $C:X \rightarrow \mathbb{K}^2, C(x)=(f(x),g(h))$ is as well, so must be $f \cdot g = C \circ \textrm{mult}$" I'm left with the problem (not having taken a full course in complex analysis yet) with which topology $\mathbb{C}^2$ traditionally is endowed, when one proves, that the multiplication of complex numbers is continuous.

(For this theorem I need the fact that this topology I'm looking for on $\mathbb{C}^2$ - viewed as the domain of $\textrm{mult}$ - is at least contained in the product topology obtained from $ \mathbb{C}^2 = \mathbb{C} \times \mathbb{C} $ , both equipped with the euclidean topology - $ \mathbb{C}^2$ viewed this time as the range of $C$ - because else the proof won't work. My guess is, that the topology I'm looking for on $\mathbb{C}^2$ - as the domain of $\textrm{mult}$ - is the product topology above described, but I couldn't quickly do a proof that using this topology $\textrm{mult}: \mathbb{C}^2 \rightarrow \mathbb{C}$ is continuous and since I've wasted enough time on this problem and also haven't taken a complex analysis course, I thought it is finally time to ask you guys about this, so that I can get a clear view of things.)

Side questions:

I'm also wondering why "euclidean topology" was mentioned, because as far as I knew, this makes sense only for $\mathbb{R}$ (I'm guessing with $\mathbb{C}$ it is meant, that the euclidean topology of $\mathbb{R}^2$ is used; could it be, that if the topology I'm looking for on $\mathbb{C}^2$ - as the domain of $\textrm{mult}$ - is the product topology, that it is the same as the euclidean one for $\mathbb{R }^4$ ?)

Does anybody know a different proof of the above theorem ? (One that especially doesn't use nets, because I've done a sketch of a proof thatis way and it is just painful - and it also doesn't adapt easily to other types of composing $f$ and $g$)

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What other topology is there on $\mathbb{C}^2$ to work with? –  scineram Sep 24 '11 at 14:35
    
@scineram I don't know; I was hoping (if there was another topology) for someone here to tell me... –  temo Sep 24 '11 at 18:10
    
@temo: There are many different topologies on any given point set. But only a very few topologies on $\mathbb{C}$ or $\mathbb{C}^2$ are "useful" (in the sense that the functions continuous under the topology form a class of functions that we care about). The Zariski topology is a weaker topology on $\mathbb{C}^n$ (weaker in the sense that there are less open sets), and the only functions $\mathbb{C}^n \to \mathbb{C}^m$ that are continuous under the Zariski topology (on both domain and codomain) are regular maps (polynomials). –  Shaun Ault Sep 25 '11 at 22:50

1 Answer 1

up vote 2 down vote accepted

Yes, the Euclidean topology on $\mathbb{C}$ is the same as the Euclidean topology on $\mathbb{R}^2$, meaning they are both induced by the Euclidean distance metric. The product topology on $X \times Y$ is generated by finite intersections of the open strips $p_1^{-1}(U)$, $p_2^{-2}(V)$, where $U$ is open in $X$ and $V$ is open in $Y$, and $p_1, p_2$ are the projection maps onto the $X$ and $Y$ components, respectively. For $\mathbb{R} \times \mathbb{R}$, it suffices to take $U$ and $V$ to be open intervals, so it is easy to see that the open "boxes" $(a,b)\times (c,d)$ are open in $\mathbb{R} \times \mathbb{R}$ and form a base. But in any given open metric ball of $\mathbb{R}^2$, each point in the ball can be surrounded by an open "box" of small enough dimensions. Conversely, every open box has a small open metric ball surrounding each of its points. This shows $\mathbb{R}^2$ (Euclidean) $\cong \mathbb{R} \times \mathbb{R}$. Moreover, "Euclidean" $\mathbb{C}^2$ and Euclidean $\mathbb{R}^4$ have the same topology.

About continuity, you mentioned in your title "complex addition", but your question had to deal with "complex multiplication," so I'll assume multiplication is what you really meant. Consider the multiplication map: $\mu : \mathbb{C}^2 \to \mathbb{C}$, written in terms of real and imaginary components:

$$\mu : \mathbb{R}^4 \to \mathbb{R}^2,$$

$$\mu(a, b, c, d) = (ac-bd, bc+ad).$$

On the left, we interpret $(a, b, c, d)$ as $(a + bi, c + di) \in \mathbb{C}^2$. On the right, interpret $(ac-bd, bc+ad)$ as $(ac-bd) + (bc+ad)i \in \mathbb{C}$. Then clearly $\mu$ is continuous in all four variables, because the component functions $ac-bd$ and $bc+ad$ are continuous (being polynomial in the arguments).

Hope this helps!

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