The Bott periodicity theorem states that $\pi_k(O(\infty))=\pi_{k+8}(O(\infty))$ and similarly for other classical Lie groups.
But his groups are defined as an inductive limit. For e.g. $GL(n,F)$, for $F=\mathbb{R}$ or $\mathbb{C}$, the inductive limit is not a priori the same as the group of all invertible transformations of $F^\infty$. Besides, it's certainly not isomorphic to the group of all invertible transformations of the (separable infinite dimensional) Hilbert space.

(1) What can we say about the group of all transformations of $F^\infty$ in the context of Bott periodicty?

(2) What are the Lie algebras of the aforementioned groups? Are they somehow related?

(3) Is it possible to represent some of these groups as an "infinite Dynkin diagram"? (For example, I would imagine $sl(\infty)=A_\infty$ being represented by an infinite row of connected circles).

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    $\begingroup$ I took the liberty to edit the question because, as it standed, it seemed to suggest that $F^\infty$ is Hilbert space (which is not). So, a priori, we're actually talking about three groups here, not two: $GL(\infty)=\lim_n GL(n)$, $GL(F^\infty)$, and $\mathrm{Aut}_{top}(\ell^2)$. (And same for the "orthogonal" versions). $\endgroup$
    – Qfwfq
    Feb 18, 2018 at 18:12
  • $\begingroup$ Is there any reason to think that there will be a sensible notion of Lie algebra for an arbitrary group? The automorphism groups are no longer differentiable manifolds, at least not modelled on finite-dimensional Euclidean spaces, nor even (as far as I can tell) on possibly-infinite-dimensional Banach spaces. $\endgroup$
    – LSpice
    Feb 21, 2018 at 14:27

1 Answer 1


Edit: I was not being precise and probably wrong. I thank Denis Nardin for that. I have therefore removed the remarks on $F^\infty$.

Let us consider the invertible bounded linear operators on a (real/complex) Hilbert space $H$. Kuiper's Theorem states that $GL(H)$ is contractible, so all higher homotopy groups vanish. If one considers the subgroup of all invertibles of the form $Id+K$ where $K$ is a compact operator one obtains another group $GL_C(H)$. The homotopy type of this space is $O(\infty)$ or $U(\infty)$ depending if the Hilbert space if real or complex.

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    $\begingroup$ Actually there are invertible linear operators on $\mathbb{R}^\infty$ not in $O(∞)$ (think about permutation operators for a permutation of $\mathbb{N}$ with infinite support). Dunno about the homotopy type. $\endgroup$ Feb 18, 2018 at 18:42
  • $\begingroup$ @DenisNardin: That is a very helpful comment. I always assumed this to be true without checking, but it is obviously false. There is another group lurking around on $F^\infty$ which is the space of invertible operators which differ of the identity by a finite rank operator. Let me think about it and see if I can say something correct in the next couple of days. $\endgroup$
    – Thomas Rot
    Feb 18, 2018 at 19:03
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    $\begingroup$ For what is worth, I think the reason why groups like $O(\infty)$ show up in homotopy theory is that $\mathbb{R}^\infty$ really should be thought of as an ind-space, not as a space (precisely, as an ind-object in the topological category of finite dimensional inner product spaces and isometric embeddings, and in fact the terminal object of the ind-category). $\endgroup$ Feb 18, 2018 at 20:18
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    $\begingroup$ For what it's worth, $F^\infty$ cannot be complete in any metric, by the Baire category theorem. (It is the union of countably many finite-dimensional subspaces, which are closed sets of empty interior). $\endgroup$ Feb 19, 2018 at 7:40
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    $\begingroup$ @LSpice In my previous comment I was only requiring the metric to define a vector space topology, not to be translation invariant (this is the usual convention in topological vector space theory). Your last parenthesized statement is not correct -- $F^\infty$ is actually complete in the unique uniformity defined by the direct sum topology (reference: Bourbaki's Topological Vector Spaces, III.21 Corollary 2). What my previous comment is saying, in this context, is that this uniformity is not metrizable. $\endgroup$ Feb 26, 2018 at 0:14

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