Gruppi cancellabili

Un gruppo $K$ si dice cancellabile se, per ogni coppia di gruppi $G$, $H$, 

$G \times K \cong H × K$ implica $G \cong H.$ 

Si noti che non si può in generale passare al quoziente per $K$ in ambo i termini di $G \times K \cong H × K$: il motivo è che non è detto che il dato isomorfismo mandi il sottogruppo $\{1\} \times K$ di $G \times K$ nel sottogruppo $\{1\} \times K$ di $H \times K$. 

Infatti, esistono gruppi che non sono cancellabili. La più semplice famiglia di controesempi si ottiene prendendo come $G$ un gruppo arbitrario e ponendo $H=G \times G$ e $K=$prodotto numerabile infinito di copie di $G$. 

In questo esempio $K$ non è finitamente generato, ma esistono anche esempi finitamente presentati: addirittura, si dimostra che $\mathbb{Z}$ non è cancellabile in generale [1].

Invece, ogni gruppo finito $K$ è cancellabile. Questo risultato è dovuto a Hirshon [2], si veda anche la dimostrazione elementare data in [3].

Riferimenti.

[1]  https://math.stackexchange.com/questions/349826/counterexample-g-times-k-cong-h-times-k-implies-g-cong-h?noredirect=1&lq=1

[2] R. Hirshon, On Cancellation in Groups, The American Mathematical Monthly, Vol. 76, No. 9 (1969), pp. 1037-1039

[3] https://math.stackexchange.com/questions/349855/does-g-times-k-cong-h-times-k-imply-g-cong-h?noredirect=1&lq=1

Trasposte

 

Happy hour

Un'infinità numerabile di matematici entra in un bar, dove c'è l'happy hour "Drink a 1,20 euro". Il primo ordina un drink, il secondo due drink, il terzo tre drink e così via. Sospirando, il barista si rivolge a uno di loro e gli fa: "Ramanujan, te l'ho già spiegato: dovete pagare come tutti, non te li do 'sti 10 centesimi!"

Hully Gully

E se prima eravamo ℵ₀ a ballare l'Hully Gully, adesso siamo ℵ₀ a ballare l'Hully Gully.

- Popular joke, attribution unknown -

Technique

 A technique is a trick that works twice.

- Attribution unknown -

The Foias constant

The Foias constant [1] is the unique initial value $x_1= \alpha$ such that the sequence defined by the recurrence $$x_{n+1} = \left(1+\frac{1}{\, x_n} \right)^n$$ diverges to $+\infty$. Numerically, it is 

$$\alpha=1.18745235112650 \ldots$$

No closed form for the constant is known, and its transcendence has not been proven so far.

The constant is named after Ciprian Ilie Foias (1933-2020), a Romanian-American mathematician famous for contributing to PDE, Operator Theory, and Control Theory [2, 3].

The history of its discovery is a curious example of serendipity. In the mid-seventies, when Foias was teaching at the University of Bucharest, an error of a typist changed an easy basic exercise to a very challenging one. Foias took the challenge and eventually solved the accidentally invented difficult problem [4].

References.
[1] https://en.wikipedia.org/wiki/Foias_constant
[2] https://en.wikipedia.org/wiki/Ciprian_Foias
[3] Remembrances of Ciprian Ilie Foias, Notices AMS 69 (9), October 2022.
[4] J. Ewing, C. Foias: An Interesting Serendipitous Real Number. In Finite versus Infinite: Contributions to an Eternal Dilemma (Ed. C. Caluse and G. Păun). London: Springer-Verlag, pp. 119–126, 2000.

Watch your back

 

Whitney approximation theorem: $\mathbb{S}^n$ is simply connected for $n \geq 2$

Let me give a proof of the fact that the $n$-sphere $\mathbb{S}^n \subset \mathbb{R}^{n+1}$ is simply connected for $n \geq 2$, which is based on differential topology methods (instead than on the Seifert-Van Kampen theorem). We will use the following two results:

(1) Whitney Approximation Theorem: Every continuous map $f \colon M \to N$ between smooth manifolds is homotopic to a smooth map.

(2) Sard Theorem: Let $f \colon  M \to N$ be a smooth map between smooth manifolds, and let $\operatorname{Crit}(f)$ be the locus of points of $M$ where the differential df has rank $< \dim(N)$. Then the image of $\operatorname{Crit}(f)$ has zero Lebesgue measure in $N$. In particular, if $\dim(M) < \dim(N)$ then $f$ cannot be surjective.

Given a continuous loop $c \colon [0, \, 1] \to \mathbb{S}^n$, we want to show that $c$ is homotopically trivial. By (1), it is not restrictive to assume that $c$ is piecewise smooth. Then, by (2), $c$ cannot be surjective (recall that we are assuming $n \geq 2$).

Now, take a point $p$ not in the image of $c$. Then we can see $c$ as a loop in $\mathbb{S}^n-\{p\}$, which is a contractible space (it is homeomorphic to $\mathbb{R}^n$ via the stereographic projection). So $c$ is homotopically trivial.

Remark. The regularization provided by (1) is essential here. In fact, there exist continuous, non-smooth loops "of Peano type", whose image is the whole sphere.