Let $H$ be a real or complex Hilbert space. In the case where $H$ is infinite-dimensional, let us define a *half-dimensional* subspace as a subspace $W \subset H$ such that both $W$ and $W^\perp$ have infinite dimension.


Fix one half-dimensional subspace $W\_0$. The Grassmannian of $H$ is
$$\mathrm{Gr}(H, W\_0) = \{W \subset H ~|~ W \text{ is half-dimensional}, P\_W - P\_{W\_0} \text{ is Hilbert-Schmidt}\}. $$
Here for $W \subset H$ a subspace, $P\_W$ denotes the orthogonal projection onto $W$. $\mathrm{Gr}(H, W\_0)$ can be given the structure of a Hilbert manifold in a natural way (see e.g. the book "Loop Groups" of Pressley and Segal).


The space $\mathrm{Gr}(H, W\_0)$ has a tautological vector bundle $\tau$ over it, where the fiber is given by $\tau(L) = L$.



> 
> **Question:** Does $\tau$ have a nowhere vanishing section?
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I believe that in the case that $H$ is finite-dimensional (say of dimension $2n$), the answer is no, as one can show that the Euler class of $\tau$ is non-zero. But how would one proceed in the infinite-dimensional case?



Nonvanishing section of infinite-dimensional tautological bundle
Let $X$ be any paracompact space. Then Hilbert vector bundles over $X$ are classified by homotopy classes of maps $[X, BU(\mathcal H)]$. But when $\mathcal H$ is infinite-dimensional, the group $U(\mathcal H)$ is contractible (this is Kuiper's theorem), and hence every infinite-dimensional Hilbert bundle over a paracompact space is trivializable.


Hilbert manifolds modelled on a separable Hilbert space are metrizable. Separable metric spaces are paracompact. As long as your $H$ is a separable Hilbert space, the above argument implies that your bundle is trivializable. There is probably a straightforward extension of this argument in the non-separable case but I didn't think about it.


In particular, your bundle has infinitely many linearly independent sections.


