We have the following definition of orientability for finite dimensional vector spaces:
Definition: Let $\Bbb{E}$ be a finite dimension vector space and $\mathscr{B}(\Bbb{E})$ be the set of all ordered basis of $\Bbb{E}$, i.e. if $\dim \Bbb{E}=n$, then $\scr{B}(\Bbb{E})=\{(x_1,\dots,x_n)\in \Bbb{E}^n\,:\, \{x_1,\dots,x_n\} \text{ is a basis for $\Bbb{E}$}\}$. We define an equivalence relation on $\scr{B}(\Bbb{E})$ as follows: given $B_1=(x_1,\dots,x_n)$ and $B_2=(y_1,\dots,y_n)\in \scr{B}(\Bbb{E})$ we write $B_1\sim B_2$ if $\det A>0$, where $A$ is the matrix of the isomorphism "change of basis" $T:\Bbb{E}\to\Bbb{E}$, such that $T(x_i)=y_i$, $i=1,\dots,n$. If $B_1\sim B_2$ we say that $B_1$ has the same orientation of $B_2$.
With this relation, $\scr{B}(\Bbb{E})/\sim$ has two equivalence classes, one of them (usually that one containing the standard basis in the case $\Bbb{E}=\Bbb{R}^n$) is called positive orientation and the other one negative orientation. We also have a definition of orientability for differentiable manifolds using this notion.
My question is quite simple: is possible to generalize the notion of orientability for infinite dimensional vector spaces? If we try to generalize the notion directly from the finite dimensional case we face the problems of saying what is an ordered basis and the lack of matrices for linear isomorphisms, so we cannot state the condition of the "positive determinant of the matrix change of basis".