Topological zero divisors of $\mathbb C^n$

Question

We know that every zero divisor is a topological zero divisor but not every topological zero divisor is a zero divisor. First I define the terms: Zero divisor: In a Banach Algebra $A$ an element $x\in A$ is said to be a zero divisor if there exist $0\neq y\in A$ such that $xy=0$. Topological divisor of zero: An element $x\in A$ is said to be a topological divisor of zero if there exist a sequence ${x_n}$ with $||x_n||=1$ such that $xx_n\rightarrow 0$ as $n\rightarrow\infty$.

I was trying to find topological zero divisors in $\mathbb C^n$. Suppose $x=(x_1,x_2,...,x_n)$ is a topological zero divisor. Now every topological zero divisor is a singular element, so $x$ is also a singular element. Then we must have at least one $1\leq i \leq n$ such that $x_i=0$. Take $y=(0,\dots,0,1,0,\ldots,0)$ where $1$ is at $i^{\rm th}$ place. Then $y\neq 0$ and $xy=0$. So $x$ is a zero divisor.

So in $\Bbb C^n$ all topological zero divisors are in fact zero divisors. Am I correct?

Answer 1

Yes, you are correct (I interpret singular elements as being "non-invertible" elements) . To supply some details : if $x$ is a topological zero divisor , and $x_n \in \mathbb C^n$ are such that $\|x_n\| = 1$, $xx_n \to 0$, then we claim that $x$ is singular , for if there is $y$ with $yx=xy = \mathbf 1$ (where $\mathbf 1$ is the unit element) then $yxx_n = (yx)x_n = x_n$, however $yxx_n = y(xx_n) \to y\mathbf 0 = \mathbf 0$ by continuity of left multiplication by a fixed element.

It follows that $x_n \to \mathbf 0$, a contradiction since $\|x_n - \mathbf 0\| = 1$ for all $n$. Hence, $x$ is singular.

Write $x = (x_1,...,x_n)$ (now $x_n$ are complex numbers, thus we move away from the previous notation). Since $x$ is singular, at least one component $x_j \neq 0$, since otherwise $\left(\frac 1{x_j}\right)$ is an inverse of $x$. Now, if $x_j = 0$ then $y$ given by $y_i = 0$, $i \neq j$ and $y_j =1$ is an element such that $yx = xy=\mathbf 0$, completing the argument. $\blacksquare$

---

Can we do better, though? Like, this is just $\mathbb C^n$ that we talked about. So some more work might give us a little more benefit.

Indeed, we made heavy use of the structure of $\mathbb C^n$ above, because we knew precisely what the set of zero divisors was. However, there's a different way we can capture this result, in more generality.

Let $B$ be any finite dimensional Banach algebra. Then, any (left/right alone will suffice as well) topological divisor of zero is a usual divisor of zero.

Proof : Suppose $x \in B$ and $\|x_n\| =1$, $xx_n \to 0$. The key point is that the unit ball of a finite dimensional Banach Algebra is compact (for a hint, you can write each element in terms of the finitely many basis elements and work with the component sequences, using the fact that these are bounded and Bolzano-Weierstrass holds), and hence there is a subsequence $x_{n_k} \to z$ for some $z \in B$,but then $xx_{n_k} \to xz = 0$ so $x$ is a zero divisor. $\blacksquare$

---

In infinite dimensions, topological zero divisors have other characterizations. This one is nice :

Let $G \subset B$ be the set of invertible elements of a Banach algebra. Then $\partial G$, the topological boundary of $G$, consists purely of topological divisors of zero.

Proof : Suppose that $x \in \partial G$. Since $G$ is open, $\partial G \cap G = \emptyset$ so $x \notin G$. Now, let $x_n$ be invertible, $x_n \to x$. We claim that $\|x_{n}^{-1}\| \to \infty$ as $n \to \infty$.

Indeed, let $\|x_{n_k}\|^{-1} \leq M$ for all $k$ and some $M$. Then , use the identity : $$ \|x_{n_k}^{-1} - x_{n_l}^{-1}\| = \|x_{n_k}^{-1}(x_{n_l} - x_{n_k})x_{n_l}^{-1}\| \leq \|x_{n_k}\|^{-1} \|x_{n_l}\|^{-1}\|x_{n_k} - x_{n_l}\| $$

To see that if $x_{n_k}$ is convergent, then so must $x_{n_k}^{-1}$ be. If this converges to some $y \in B$ then $yx=xy=1$ can be verified, a contradiction. Hence the assumption is false.

Now, if $\|x_n^{-1}\| \to \infty$ then of course $xx_n^{-1} \to 0$ but $x_n^{-1} \not \to 0$, so normalizing $x_n^{-1}$ to have norm $1$ we conclude that $x$ is a topological zero divisor. $\blacksquare$

---

Finally, we must touch on permanent singularity, and a nice characterization theorem of Arens, proved here.

Definition : A singular element $a \in B$ is called permanently singular if for every Banach algebra $Y$ that extends $B$ (i.e. the norm of $Y$ restricts to the norm of $B$ on $B$ , and they share the same unit), $a$ is still singular in $Y$.

The remarkable theorem of Arens is the following :

Let $B$ be a unital commutative Banach algebra. Then, $a \in B$ is a topological divisor if and only if it is permanently singular.

Another nice theorem of Zelazko eliminates topological zero divisors by passing through an isomorphism :

Let $A$ be a Banach algebra, then there exists an extension algebra $B$ of $A$ and a homomorphism $\Theta : A \to B$ such that $\Theta$ , restricting the codomain to its image is an isometric isomorphism, and $x \in A$ is a topological zero divisor if and only if $\Theta(x)$ is a "usual" zero divisor (in their respective domains).

These results help us understand topological and usual zero divisors better.