32
$\begingroup$

Now asked on MathOverflow.


Background: It seems that, after his groundbreaking work on quadratic forms and inventing Witt vectors, Ernst Witt developed the hobby of giving extremely short proofs to famous theorems. E.g. his collected works contain a one-page proof of the prime number theorem.

What interests me in this question is his six-line proof of the Gelfand-Mazur Theorem ("The only $\Bbb C$-Banach algebra $K$ which is a skew field is $\Bbb C$ itself") resp. its historic predecessor and now corollary, Ostrowski's Theorem ("The only complete Archimedean fields are $\Bbb R$ and $\Bbb C$"; not to be confused with the more famous Ostrowski's theorem which classifies the valuations on $\Bbb Q$). Witt's article is "Über einen Satz von Ostrowski", Arch. Math. 3 (1952), p. 334, reprinted on p. 404 of his Collected Papers. Unfortunately, I do not have access to either source right now. The best I could find is this (p. 245) English translation of its decisive three sentences.

I understand the first sentence, which says that w.l.o.g. we can assume $\dim_{\Bbb R}K>2$ and hence $K^\times$ simply connected. I also understand the third sentence which says that there cannot be an isomorphism between the additive group of $K$ and the multiplicative group $K^\times$ (obviously, as we are in characteristic $0$; there is a misprint in the translation, since of course it's the element $-1$ which is of order 2 in $K^\times$, and IIRC that's what Witt writes in the original). But the second sentence

The differential equation $x^{-1}dx = y$ then [i.e. assuming $K\setminus \lbrace 0\rbrace$ simply connected] engenders a global isomorphism between the multiplicative group $(x\neq 0$) and the additive group $(y)$.

seems to hide some details. I guess the idea is that by simply connectedness, path-integrating $f(x) = x^{-1}$ from the startpoint $1$ to any $x\neq 0$ gives a well defined "logarithm" function $F(x)$ which has the property $F(ab) = F(a) + F(b)$ and is bijective. (Which then, as said, I understand gives a contradiction and shows no such $K$ with $\Bbb R$-dimension $>2$ exists.)

Question 1: How to understand the highlighted sentence? In particular, is my interpretation correct and if yes, how exactly to show such an $F$ is a group isomorphism $K^\times \simeq (K,+)$?

Question 2: Does this prove Gelfand-Mazur, as the source of the above translation seems to imply, or merely Ostrowski's theorem, as Witt's own title claims? If it only proves Ostrowski's theorem, is there a way to upgrade this to a full proof of Gelfand-Mazur?

Note 1: As said I cannot check it right now, but I think in the original article Witt actually writes the differential equation $x^{-1}dx = dy$ which makes more sense to me.

Note 2: I am aware of the standard calculus proof that $F(x) = \int_1^x \frac{dt}{t}$ satisfies $F(ab) = F(a) +F(b)$ (actually, teaching that in my calculus class last week reminded me of this question), but it seems Witt is assuming a generalisation of this and maybe more differential calculus to a possibly infinite dimensional $\Bbb R$ resp. $\Bbb C$-vector space, which is a bit outside of my comfort zone.

$\endgroup$
5
  • $\begingroup$ How these arguments relies on essential features of $\mathbb{C}?$ It is unclear what part of the proof fails for $\mathbb{Q}.$ Re. integral. If you consider $\int_x^y t^{-1}\:dt$ along an interval connecting $x$ and $y$ then this integral is equal to $\int_0^1 ((1-t)x + ty)^{-1}(y-x)\:dt.$ In $\mathbb{Q},$ integral along triangle with vertices $i,j,k$ is not equal to 0. So $F$ is not a function on $\mathbb{Q}^\times.$ Existence of injective $F$ implies that $K^\times$ is commutative. $\endgroup$
    – dsh
    Sep 18, 2023 at 18:56
  • $\begingroup$ @dsh: Not sure what you mean. The proof talks about $\mathbb R$-algebras $K$, and $\mathbb Q$ is not an $\mathbb R$-algebra. We are showing by contradiction that certain such $K$-algebras cannot exist if their real dimension were $>2$; that excludes $\mathbb C$ already, so in a way this proof does not even talk about $\mathbb C$. What exact statement do you think is proved here for $\mathbb C$ which would also work for $\mathbb Q$? $\endgroup$ Sep 19, 2023 at 16:19
  • 1
    $\begingroup$ Sorry for confusion. I meant $\mathbb{H}$ instead of $\mathbb{Q}.$ I was wondering which part of the proof fails for $\mathbb{R}$ algebras of $\dim_{\mathbb{R}} K > 2.$ Integral in the division ring of quaternions is not zero along closed contour. $\endgroup$
    – dsh
    Sep 19, 2023 at 19:06
  • $\begingroup$ @dsh Oh, that makes much more sense! You raise a very good point there which I don't know the answer to. My only guess is that we somehow need a "complex structure" on $K$, which the quaternions lack. Btw, do you have a good reference for integration over the quaternions? $\endgroup$ Sep 19, 2023 at 20:10
  • 1
    $\begingroup$ You can check Bourbaki "Integration". Most of the time results re. integral of functions with values in $\mathbb{R}$ and Banach space are the same. For $t \mapsto t^{-1}$ Riemann's definition is sufficient. In the case of $\mathbb{H}$ definition of integral along the path is a problem due non-commutativity (need invariant wrt reparametrizations of path). Gentili and others "Regular functions of a quaternionic variable" provides Cauchy formula for quaternions for curves inside subspaces isomorphic to $\mathbb{C}$ (in this case integral along the path is easy to define). $\endgroup$
    – dsh
    Sep 19, 2023 at 23:22

0

You must log in to answer this question.