Let $E$ be a Banach space over the field $\mathbb K$, where $\mathbb K$ is $\mathbb R$ or $\mathbb C$. Let $\mathcal{B}(E)=\{\text{bounded operators on }E\}$ and $\mathrm{GL}(E)=\{T\in\mathcal B(E)\mid \exists S\in\mathcal B(E):TS=ST=\mathrm {id}_E\}$. Can we prove or disprove that $\mathrm{GL}(E)$ is connected when $\mathbb K=\mathbb C$ and dis-connected when $\mathbb K=\mathbb R$?
Case $\mathbb K=\mathbb C$: I find this solution to the finite dimensional case still works, since we have holomorphic functional calculus.
Case $\mathbb K=\mathbb R$: the result is false for $\mathbb R^n$, where $n\in\mathbb N$, because $\det$ is continuous and $\det(\mathrm{GL}(\mathbb R^n))=\mathbb R\setminus\{0\}$ is not connected. But $\det$ tools seem not avaliable for the infinite dimensional case. It should be simpler if we consider real Hilbert space. We also lose holomorphic functional calculus. I don't know about spectral theorem for real Hilbert space of infinite dimension...
Even, for example, $E=\ell_2=\{x\colon\mathbb N\to\mathbb R\mid\sum_{n} x_n^2<\infty\}$ is a $\mathbb R$-Hilbert space. Can we prove that $\mathrm{id}$ and $T$, reprsentated by matrix $\mathbf{diag}(-1,1,\ldots,1,\ldots)$ are not in the same connected componment of $\mathrm{GL}(\ell_2)$?
