Important results with one or more than one proof [closed]

Question

Can you give examples of deep, important results that have only one known proof, and not just because the first proof is fairly recent, or because not many people really cared to think about it? How hard is the proof from the perspective of the non-expert in the field? In the opposite direction, can you give examples of important results, for which several genuinely different proofs were found? Are these proofs all considered equally hard, or some are much easier or, perhaps, even surprisingly easy?

For example, Carleson's theorem has more than one proof and, although the latest proof is much simplified, it is probably still quite technical. Poincare conjecture has one known proof, not easily accessible to non-experts. Kadison-Singer problem has one known proof, accessible to non-experts with a little effort. Capset problem has one known proof, fully accessible to non-experts. The last three examples are relatively recent results, and some of them are likely to remain the only known approaches for a long time, while maybe not others. Instead of trying to guess the future, what are interesting examples that are less recent?

Answer 1

I wonder whether 'deepness' is subjective or not. The Compactness Theorem of first-order logic has several proofs.

Theorem. (Gödel-Maltsev) Given a language $L$ and a set $S$ of first-order sentences in that language, if every finite subset of $S$ has a model, then $S$ has model.

(A language is just a set of relational, function and constant symbols, for example the language of ordered fields is $+,\times,=,\leqslant,0,1$; a first-order sentence in this language is a finite word using this language and $\exists,\forall,\land(and),\lor(or)$ and paranthesis for ease of reading only, for instance the sentence $\forall x\exists y(y^2=x)$; a model of a sentence is just a set where the sentence can be interpreted in a 'true' way. For instance $\bf R$ is not a natural model of the above sentence as $-1$ does not have a squareroot there, but $\bf C$ is a model of it.)

(For one immediate application, one can deduce in a straightforewared way that there is an ordered field containing $\bf R$ as well as infinitesimal/infinite elements while having the same first order properties as $\bf R$)

Some Proofs. (I believe there are many more)

(1) Gödel's original proof (for Gödel, in 1929, this Theorem was stated as a 'Remark') is from his Completeness Theorem, stated in the particular case where the language $L$ is countable, in which case the axiom of choice is not needed. Hard for me to say the nature of the proof. Grammatical maybe.

(2) I don't know precisely the nature of Maltsev's proof, published in German in 1936, and extending G.'s result to the case of an arbitrary signature $L$, using the axiom of choice.

(3) Los' proof via the 'explicit' construction of the model as ultraproduct of finite structures, using the 'axiom of the ultrafilter', which is weaker that the axiom of choice. I would say this proof is of a topological nature.

In the same (?) vein, Gromov's 'bounded version' of his Theorem stating that a finitely generated group of polynomial growth as a nilpotent subgroup of finite index has several (at least 2) proofs, apparently.

Theorem. (Gromov) For any positive integers $k$, $d$, $n$, there exists a positive integer $m$ such that any $n$-generated group, in which for all $r= 1,\dots,m$ the size of the ball of radius $r$ centered at the identity is at most $kr^d$, has a subgroup of index and nilpotency class at most $m$.

I read that Gromov's original proof is a Compactness argument. Van den Dries and Wilkie gave an alternative proof using Gödel's Compactness Theorem. Belegradek recently provided a third proof using yet another kind of compactness argument, using very little model theory.

Edit: To match the OP's Edit, I'm just adding that the field concerning the first Theorem should be 'model theory' which is currently classified (according to Bairwise, if I am not mistaken) as the latest of the 4 branches of 'mathematical logic'. It is central/important in the sense that it more or less gave birth to the 'field', and is used more or less tacitly in pretty much every Theorem of model theory. The second Theorem belongs to the field of 'geometric group theory', and I do not know about it's being central. It seems to be important in the sense that many mathematicians seem to be interested in it. It also has a wikipedia page in 3 major mathematical languages: https://fr.wikipedia.org/wiki/Th%C3%A9or%C3%A8me_de_Gromov_sur_les_groupes_%C3%A0_croissance_polynomiale

Answer 2

The first example that occurs to me is Hindman's theorem: If the set of positive integers is partitioned into finitely many pieces, then there is an infinite set $H$ such that all sums of finitely many (distinct) elements of $H$ lie in the same piece. Hindman's original proof is very complicated (Hindman himself has suggested that it could be used to torture graduate students) but it has the advantage of being elementary --- it can be formalized in a system only slightly stronger than $ACA_0$. A later, easier proof by Galvin and Glazer, now considered the standard proof, has the advantage that one can easily remember or reconstruct it, but it requires more powerful tools, including an application of Zorn's lemma to a collection of subsemigroups of a certain semigroup whose elements are ultrafilters. There's also an "intermediate" proof due to Baumgartner.

Answer 3

I believe the Atiyah-Singer Index Theorem qualifies as deep. There are several different approaches to proving it, see here.

Answer 4

The Mordell conjecture (proved by Faltings (2x), Vojta and Bombieri).

The Weil conjectures (proved by Deligne (2x)).

The Theorem of Roth (proved by Roth and Faltings).

References:

• Faltings, Gerd (1983). "Endlichkeitssätze für abelsche Varietäten über Zahlkörpern" [Finiteness theorems for abelian varieties over number fields]. Inventiones Mathematicae (in German) 73 (3): 349–366. doi:10.1007/BF01388432. MR 0718935.

• Faltings, Gerd (1984). "Erratum: Endlichkeitssätze für abelsche Varietäten über Zahlkörpern". Inventiones Mathematicae (in German) 75 (2): 381. doi:10.1007/BF01388572. MR 0732554.

• Faltings, Gerd (1991). "Diophantine approximation on abelian varieties". Ann. of Math. 133 (3): 549–576. doi:10.2307/2944319. MR 1109353.

• Faltings, Gerd (1994). "The general case of S. Lang's conjecture". In Cristante, Valentino; Messing, William. Barsotti Symposium in Algebraic Geometry. Papers from the symposium held in Abano Terme, June 24–27, 1991. Perspectives in Mathematics. San Diego, CA: Academic Press, Inc. ISBN 0-12-197270-4. MR 1307396.

• Bombieri, Enrico (1990). "The Mordell conjecture revisited". Ann. Scuola Norm. Sup. Pisa Cl. Sci. 17 (4): 615–640. MR 1093712.

• Vojta, Paul (1991). "Siegel's theorem in the compact case". Ann. of Math. 133 (3): 509–548. doi:10.2307/2944318. MR 1109352.

• Deligne, Pierre (1974), "La conjecture de Weil. I", Publications Mathématiques de l'IHÉS (43): 273–307, ISSN 1618-1913, MR 0340258

• Deligne, Pierre (1980), "La conjecture de Weil. II", Publications Mathématiques de l'IHÉS (52): 137–252, ISSN 1618-1913, MR 601520

• Bombieri, Gubler, Heights in Diophantine Geometry, http://ebooks.cambridge.org/ebook.jsf?bid=CBO9780511542879

• Faltings, Wüstholz, Diophantine approximations on projective spaces, http://gdz.sub.uni-goettingen.de/dms/load/img/?PID=GDZPPN002111748

Answer 5

Whether you consider it a deep, very deep or shallow result:

Theorem. The symplectic group ${\bf Sp}_{2n}({\mathbb R})$ is included in ${\bf SL}_{2n}({\mathbb R})$.

One proof is purely algebraic and uses the Pfaffian (as a matter of fact, the same result is true when one replaces ${\mathbb R}$ by another field).

The other proof is more familiar. Using the polar decomposition, one proves that ${\bf Sp}_{2n}({\mathbb R})$ is diffeomorphic to ${\mathbb R}^\ell\times{\bf U}_n$. Because the unitary group is connected, one obtains the connectedness of the symplectic group. We conclude with the fact that the determinant can take only the dicrete values $\pm1$.

Edit. Here is the Pfaffian proof. Recall that the Pfaffian is a polynomial $Pf$ in the entries of the $2n\times2n$ alternate matrix, such that $$Pf(A)^2=\det A,\qquad Pf(J_n)=1,\quad J_n:=\begin{pmatrix} 0_n & I_n \\\\ -I_n & 0_n \end{pmatrix}.$$ It has the fundamental property that if $P\in{\bf M}_{2n}$, then $$Pf(M^TAM)=Pf(A)\cdot\det M.$$ Apply this identity to $A=J_n$ and to $M$ a symplectic matrix, you obtain $1=1\cdot\det M$. This proof is in my book Matrices (Springer GTM 216).