Consider the flag variety (or flag manifold, depending on who you are) $V=\mathrm {Fl} (3,\mathbb C)$ of complete flags of subspaces of $\mathbb C^3$. That is, an element of M is a tuple (L , P) consisting of line, and plane in $\mathbb C^3$, so that the line lies in the plane.

Suppose that we have a fixed choice of basis for every plane in $\mathbb C^3$. Then we have a consistent isomorphism $P\to\mathbb C^2$ for every plane. Then choice of a flag breaks apart into choice of a plane (an element of $\mathrm{Gr}(2,3)\cong\mathrm{Gr}(1,3)\cong\mathbb P^2 (\mathbb C)$), and then a choice of a line in any plane (an element of $\mathrm{Gr}(1,2)\cong\mathbb P^1 (\mathbb C)$) which can be isomorphically embedded into that plane. In other words, every choice of basis for all of our planes (map sending points of $\mathrm{Gr}(2,3)$ to pairs of linearly independent vectors lying in that plane) gives us a bijection

$\mathrm {Fl} (3,\mathbb C) \to \mathbb P^1\times \mathbb P^2$,

or in fact of any flag variety (over any field) to a product of projective spaces (over that field). (We are using the relation that $\mathrm{Gr}(k,n)\cong\mathrm{Gr}(n-k,n)$, by sending k-dimensional subspaces to their orthogonal complements.)

The natural next question is whether we can make a choice of basis so that this bijection is in fact a homeomorphism (or hopefully an isomorphism of varieties/diffeomorphism of manifolds.) It would appear that this is possible if and only if we can make a "continuous choice of basis" for the planes which make up $\mathrm{Gr}(2,3)$: that is a continuous map $\mathrm{Gr}(2,3)\to\mathbb C^3\times \mathbb C^3$ so that the two resulting vectors are a basis for the plane we started with. Such a map is necessarily injective.

Thinking about this spatially, many "natural-looking" choices of basis will fail to be continuous. (Hold your hand out with thumb and forefinger at a right-angle. Then rotate their spanned plane about the axisyour thumb. After rotating 180$^{\circ}$, you get the same plane, but now you've negated one of your basis vectors: this choice of basis is not continuous.)

Hence, my question is this: does there exist a continuous choice of basis for planes in a finite-dimensional $\mathbb C$-vector space in general? Over $\mathbb R$ (/other topological fields)? For particular dimensions (even/odd)? This problem appears to be a bit algebraic-topological in nature, and I'm not sure how to approach it. (Gaussian elimination, for instance, fails to give a continuous choice.)

Ultimately, of course, the question I am interested in is whether this bijection (and hopefully homeomorphism) can be an isomorphism of varieties. There are obviously other ways to approach that question, but continuous choice of basis seems an interesting question in its own right.

  • $\begingroup$ Just check that the flat variety and the product of projective spaces have different homotopy type, for example. What you do have is a fibration with total space the flag variety and fiber and base projective spaces ---this is in fact a locally trivial fiber bundle, but not a trivial one. $\endgroup$ – Mariano Suárez-Álvarez Aug 17 '12 at 22:01
  • 2
    $\begingroup$ The world would be much more boring if flag varieties were products of projective spaces! $\endgroup$ – Mariano Suárez-Álvarez Aug 17 '12 at 22:02
  • $\begingroup$ Taking orthogonal complements requires fixing a choice of inner product. Instead of taking orthogonal complements, you can take annihilators in the dual space - this gives an isomorphism between Grassmannians over an arbitrary field. $\endgroup$ – Qiaochu Yuan Aug 17 '12 at 23:51
  • $\begingroup$ Thank you very much all! I always find it really satisfying when an alternative method of proof can imply something nontrivial about the original method. (In this case, checking the homotopy types of these varieties lets us easily see that linear subspaces don't have a continuous choice of basis.) $\endgroup$ – Xander Flood Aug 20 '12 at 12:52
  • $\begingroup$ @MarianoSuárez-Alvarez Please consider converting your comment into an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. $\endgroup$ – Julian Kuelshammer Jun 23 '13 at 8:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.