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Let $F$ be $\mathbb R$ or $\mathbb{C}$. Then $F^\infty$ is defined as the space of sequences in $F$ which eventually are zero. This space has some bad properties (for example it is not complete as a metric space with the natural metric), but it is nice from a homotopy point of view. I don't think it is hard to prove that the space of invertible linear operators of $F^\infty$ is homotopy equivalent to $O(\infty)$ or $U(\infty)$.

You also as about the invertible bounded linear operators on a (real/complex) Hilbert space $H$. Kuipers 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 compact one obtains another group $GL_C(H)$. The homotopy type of this space is again $O(\infty)$ or $U(\infty)$ depending on the choice of $F$.

I do not know anything about Lie Algebras, so maybe someone else can chime in.

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.

Let $F$ be $\mathbb R$ or $\mathbb{C}$. Then $F^\infty$ is defined as the space of sequences in $F$ which eventually are zero. This space has some bad properties (for example it is not complete as a metric space with the natural metric), but it is nice from a homotopy point of view. I don't think it is hard to prove that the space of invertible linear operators of $F^\infty$ is homotopy equivalent (isomorphic even probably) to $O(\infty)$ or $U(\infty)$.

You also as about the invertible bounded linear operators on a (real/complex) Hilbert space $H$. Kuipers 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 compact one obtains another group $GL_C(H)$. The homotopy type of this space is again $O(\infty)$ or $U(\infty)$ depending on the choice of $F$.

I do not know anything about Lie Algebras, so maybe someone else can chime in.

Let $F$ be $\mathbb R$ or $\mathbb{C}$. Then $F^\infty$ is defined as the space of sequences in $F$ which eventually are zero. This space has some bad properties (for example it is not complete as a metric space with the natural metric), but it is nice from a homotopy point of view. I don't think it is hard to prove that the space of invertible linear operators of $F^\infty$ is homotopy equivalent to $O(\infty)$ or $U(\infty)$.

You also as about the invertible bounded linear operators on a (real/complex) Hilbert space $H$. Kuipers 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 compact one obtains another group $GL_C(H)$. The homotopy type of this space is again $O(\infty)$ or $U(\infty)$ depending on the choice of $F$.

I do not know anything about Lie Algebras, so maybe someone else can chime in.

Let $F$ be $\mathbb R$ or $\mathbb{C}$. Then $F^\infty$ is defined as the space of sequences in $F$ which eventually are zero. This space has some bad properties (for example it is not complete as a metric space with the natural metric), but it is nice from a homotopy point of view. You can prove that the space of invertible linear operators is equal to $O(\infty)$ or $U(\infty)$ (depending if $F$ are the reals or the complex operators.

You also as about the invertible bounded linear operators on a (real/complex) Hilbert space $H$. Kuipers Theorem states that $GL(H)$ is contractible, so all higher homotopy groups vanish.

I do not know anything about Lie Algebras, so maybe someone else can chime in.

Let $F$ be $\mathbb R$ or $\mathbb{C}$. Then $F^\infty$ is defined as the space of sequences in $F$ which eventually are zero. This space has some bad properties (for example it is not complete as a metric space with the natural metric), but it is nice from a homotopy point of view. I don't think it is hard to prove that the space of invertible linear operators of $F^\infty$ is homotopy equivalent (isomorphic even probably) to $O(\infty)$ or $U(\infty)$.

You also as about the invertible bounded linear operators on a (real/complex) Hilbert space $H$. Kuipers 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 compact one obtains another group $GL_C(H)$. The homotopy type of this space is again $O(\infty)$ or $U(\infty)$ depending on the choice of $F$.

I do not know anything about Lie Algebras, so maybe someone else can chime in.