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Why are the sphere $$S=\lbrace |x|=1\rbrace$$ and the Stieffel manifolds of orthonormal $n$-frames $$V_n=\lbrace (x_1,\dots,x_n)\in S^n\mid i\neq j\Rightarrow\langle x_i|x_j\rangle=0\rbrace$$ of a infinite dimensional separable Hilbert space $\mathscr{H}$ contractible? I've read a proof of this about a year ago, but I can't find it, and I don't remember the argument.

Chris Brooks
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Olivier Bégassat
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  • Does this article help? –  Sep 19 '12 at 22:27
  • see http://math.ucr.edu/home/baez/week151.html for the sphere. I suppose the Stieffel's are bundles with contractible fiber over a contractible base which would make them contractible, but I'm not sure in infinite dimensions. – Bob Terrell Sep 20 '12 at 00:18
  • @BobTerrell Thank you both, I have found a proof for the contractibility of the sphere online, and it easily generalises to a proof of the contractibility of the Stiefel Manifolds. I will post it tomorow if I find the time. – Olivier Bégassat Sep 20 '12 at 00:22

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Let $\mathscr H$ be an infinite dimensional Hilbert space. Let us write $$\mathscr{V}_p=\{\text{all linearly independent }x_\bullet=(x_1,\dots,x_p)\in\mathscr{H}^p\}$$ and $$\mathscr{V}_p^\perp=\{\text{all orthonormal }x_\bullet=(x_1,\dots,x_p)\in\mathscr{H}^p\}$$ The Gram-Schmidt orthonormalization process induces a retraction of $\mathscr{V}_p^\perp\hookrightarrow \mathscr{V}_p$. Affine interpolation between $x_\bullet$ and its Gram-Schmidtization $\widetilde{x_\bullet}=\mathrm{GS}(x_\bullet)$ produces for every $t\in[0,1]$ a linearly independent family $(1-t)x_\bullet+t\widetilde{x_\bullet}$. Thus, the inclusion $\mathscr{V}_p^\perp\hookrightarrow \mathscr{V}_p$ is a deformation retraction and it is enough to prove that $\mathscr{V}_p$ is contractible.

Let $(e_n)_{n \in \mathbb N}$ be an orthonormal family in $\mathscr{H}$. Define a linear isometry $T$ of $\mathscr H$ by setting $$\begin{cases} \forall n\in\mathbb N,&Te_n=e_{n+1}\\ \forall v\in V,& Tv=v \end{cases}$$ where we put $V=\overline{\langle e_n\rangle}^\perp_{n\in\Bbb{N}}$.

Fact. Let $x_\bullet\in\mathscr{V}_p$ be free, $a,b\in \mathbb R$ with $a+b\neq0$, then $ax_\bullet +bT^px_\bullet$ is also free.

This follows immediately from the fact that $x$ and $T^p x$ are collinear iff $x\in V$ in which case $T^p x=x$.

We define a homotopy $H$ from $\mathrm{id}_{\mathscr{V}_p}$ to the constant map equal to $e_\bullet:=(e_1,\dots ,e_p)$. $$ H: \left\{ \begin{array}{ccc} [0,2]\times \mathscr{V}_p & \longrightarrow & \mathscr{V}_p,\\ (t,x_\bullet) & \longmapsto & \begin{cases} 0\leq t\leq 1:& (1-t)x_\bullet+tT^px_\bullet\\ 1\leq t\leq 2:& (2-t)T^px_\bullet+(t-1)e_\bullet \end{cases} \end{array} \right.$$ The fact quoted above means that $H$ is well-defined for $0\leq t\leq 1$. This map is still well-defined for $1 < t \leq 2$ since for $t>1$ the orthogonal projection of $(2-t)T^px_\bullet+(t-1)e_\bullet$ onto the span of $e_1,\dots,e_p$ is $(t-1)e_\bullet$ thus free.

Thus $\mathscr{V}_p$ is contractible and so is $\mathscr{V}_p^\perp$.

Olivier Bégassat
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  • Nice proof. What's more, I think that if we consider instead $T : \begin{cases} H &\to H \ e_n &\mapsto e_{n+1} \text{ for } n \in [0,p-1] \ e_n &\mapsto e_n \end{cases}$, we can prove the same result for every Hilbert space with $dim(H) \geq p$ (we don't need $T$ to be an isometry). – Jeyrome Sapin Dec 22 '20 at 14:10
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    Thank you @Vandrin : ) I doubt you can do anything like this in finite dimension since the finite dimensional Stieffel manifolds aren't contractible. – Olivier Bégassat Dec 22 '20 at 14:12
  • I agree. My mistake was to consider that the projection of $(2-t)T^px_\bullet+(t-1)e_\bullet$ onto the span of $e_1,\dots,e_p$ is still $(t-1)e_\bullet$ in this case. – Jeyrome Sapin Dec 22 '20 at 14:46