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The ever-reliable Wikipedia says:

... an adelic algebraic group is a semitopological group defined by...

No more details are given, and I was wondering if the multiplication only being separately continuous has any noticeable effect when working with adelic groups. I ask because there are some algebras which pop up in the study of von Neumann algebras where the multiplication is not jointly continuous, and this is mildly annoying.

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    $\begingroup$ If one does get a topological group the argument goes: Commutative linear group operations are continuous, so addition and multiplication are continuous, so polynomial maps between affine spaces are continuous, so algebraic maps between algebraic varieties are continuous, so algebraic group operations are continuous. I don't see any problem with this, but maybe one of the steps fails for subtle reasons? $\endgroup$
    – Will Sawin
    Commented Aug 16, 2013 at 3:26
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    $\begingroup$ Let $X$ be a separated finite type scheme over a global field $K$ with adele ring $A$. Choose a finite set $S_0$ of places of $K$ containing the archimedean ones so that $X$ extends to a separated flat $O_{K,S_0}$-scheme $X_0$ of finite type. Then $X(A)$ is the direct limit of the sets $X_0(A_K^S)$ topologized as direct products $\prod_{v\in S} X(K_v) \times \prod_{v\not\in S} X_0(O_v)$ for increasing $S$ containing $S_0$. The transition maps are open and the resulting locally compact Hausdorff topology is independent of $S_0$ and $X_0$. It is functorial in $X$ and respects fiber products. QED $\endgroup$
    – user36938
    Commented Aug 16, 2013 at 3:43
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    $\begingroup$ This limit construction coincides with the more familiar one in the affine case using the crutch of a closed immersion into an affine space. So there are no surprises if one is careful about the definitions. $\endgroup$
    – user36938
    Commented Aug 16, 2013 at 3:45
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    $\begingroup$ I just rolled back the latest (well-intentioned!) edit - I think that originally David's question was not asking for "intuition" but it was a "soft question" in the sense that he was deliberately not formulating a "sharply delineated, yes/no question", but something inviting more open-ended answers. $\endgroup$
    – Yemon Choi
    Commented Aug 16, 2013 at 22:55
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    $\begingroup$ @user36938 - would you mind expanding your comment into an answer? This makes it linkable, and won't get missed in a wash of comments. $\endgroup$
    – David Roberts ♦
    Commented Aug 17, 2013 at 0:04

1 Answer 1

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A locally compact semitopological semigroup which is also a group is in fact a topological group, i.e. the existence of inverses plus separate continuity actually forces joint continuity.

This is a theorem of Ellis, see

R. Ellis, Locally compact transformation groups. Duke Math. J. 24 (1957) no. 2, 119–125

Quoting from the MathReview:

Let G be a group of homeomorphisms on a locally compact space X. Suppose G has a Hausdorff topology such that multiplication is continuous in each variable separately. If the function π:G × X → X defined by π(g,x)=g(x) is continuous on the left, the author shows that π is jointly continuous.

EDIT: it's been noted in the comments that the part I quoted does not explain in any way why we can deduce continuity of inversion. In an earlier paper Ellis had shown that in a topological semigroup whose underlying semigroup is a group, inversion is continuous:

R. Ellis, A note on the continuity of the inverse. Proc. Amer. Math. Soc. 8 (1957), 372–373

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  • $\begingroup$ probably it is in Hindman-Strauss. $\endgroup$
    – Benjamin Steinberg
    Commented Aug 16, 2013 at 14:40
  • $\begingroup$ Fair enough, though after one has actually given a definition of the topology being considered in the first place (which we need for anything to make sense), the desired answer just drops right out by functoriality considerations, so to invoke Ellis' theorem (treating the adelic topology as a "black box") seems like invoking the inverse function theorem to prove the smoothness of matrix inversion. $\endgroup$
    – user36938
    Commented Aug 16, 2013 at 17:57
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    $\begingroup$ @user36938 Fair enough, but the point of my answer was that one doesn't even need to know any functoriality, or indeed anything about adeles. In other words, even a lowly analysis-type who is uncultured in number theory can give an answer:) I agree that if you do know the precise definition of the topology, then it is better to show joint continuity by looking at the topology rather than appealing to an oracle $\endgroup$
    – Yemon Choi
    Commented Aug 16, 2013 at 18:20
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    $\begingroup$ @Yemon: How does the "Ellis oracle" prove that inversion is continuous in the topology (so it is a topological group)? That is the basic reason for the topological complications when working with adelic constructions, namely that in the topological ring of adeles the inversion on units is not continuous for the subspace topology. $\endgroup$
    – user36938
    Commented Aug 16, 2013 at 22:08
  • $\begingroup$ @user36938 Automatic continuity of inversion is the bit that surprised me most when I learned of Ellis's theorem, although I don't claim it's the hardest part of the result. I've added some more information. $\endgroup$
    – Yemon Choi
    Commented Aug 16, 2013 at 22:14

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