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Let $V\subseteq \Bbb R^n$ be a vector subspace. If $V$ is rational, i.e. has a basis consisting of elements in $\Bbb Z^n$, then there’s a well-defined notion of the “volume of the lattice of V”: $$\operatorname{Vol}(L_V):=\sqrt{\det(B^\top B)},$$ where $B$ is any basis matrix of the lattice $L_V:=V\cap \Bbb Z^n$.

Question: Is it possible to extend this definition to non-rational $V$?

One idea I had is to approximate $V$ in the appropriate Grassmanian by rational subspaces. But one runs into the problem that the sequence of lattice volumes of the approximating subspaces might not converge, as can be seen from the sequence $V_j=\Bbb R (j, 1)$—this sequence converges to the $x$-axis, yet $\operatorname{Vol}(L_{V_j})= \sqrt{j^2+1}\to \infty$. In addition, I have no reason to think that given a sequence $V_j$ of rational subspaces converging to a rational subspace $V$ for which the sequence of lattice volumes does converge—say to a value $a$—that $a$ is necessarily $\operatorname{Vol}(L_V)$.

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  • $\begingroup$ Do you have even a single injective sequence $(V_j)$ for which the volume is bounded? (necessarily of dimension $\ge 2$ and codimension $\ge 1$, so $n\ge 3$) $\endgroup$
    – YCor
    Oct 30, 2018 at 21:40
  • $\begingroup$ @YCor I don’t have any example. I’m leaning toward that not being the correct definition, if such a definition even exists $\endgroup$ Oct 31, 2018 at 2:01
  • $\begingroup$ There's always a formal definition, namely embed the set of $k$-dimensional rational subspaces into $Gr(n,k)\times\mathbf{R}_+$, by $V\mapsto (V,\mathrm{Vol}(L_V))$ and consider the closure of the image. (There's no injective sequence with bounded volume iff this image is discrete; this notably holds for $k=1$.) $\endgroup$
    – YCor
    Oct 31, 2018 at 6:57
  • $\begingroup$ Doesn't if follow from Minkowski's second theorem that the image is discrete for all $k$? $\endgroup$ Nov 8, 2018 at 12:23

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