$\DeclareMathOperator\Gr{Gr}$Let $E$ be a real, finite dimensional vector space of dimension $n$. Let $\Gr(k)$ be the set of linear subspaces of dimension $k$ of $E$. I am wondering what structures the manifold $\Gr(k)$ inherits from $E$, beyond differentiability. When $E$ is Euclidean, $\Gr(k)$ gets a Riemannian metric. When $E$ is not, I guess $\Gr(k)$ should still inherit a structure from the affine structure of $E$ but I can't clarify my ideas.

In particular, I am wondering if there is a notion of straight path between subspaces.

Can anyone help? Thanks.

  • $\begingroup$ It inherits the structure of real (or complex) analytic manifold, and also has a standard structure of a compact projective variety. $\endgroup$ Sep 6, 2021 at 12:57
  • 1
    $\begingroup$ These things go by the name of Grassmannian structures. Besides the description in Ben's answer below you can look at the works of Goncharov (Generalized conformal structures on manifolds, 1987) and the works of Akivis and Goldberg (e.g., Conformal and Grassmann structures, 1998). $\endgroup$ Sep 10, 2021 at 11:44

1 Answer 1


$\DeclareMathOperator\Gr{Gr}$I discussed this in my thesis.

Lemma: Every tangent space of the Grassmannian is a tensor product $T_P \Gr(k)=P^* \otimes (E/P)$; these isomorphisms are invariant under the general linear group $\operatorname{GL}_E$.

Proof: Take a smooth path $P(t)\in \Gr(k)$. Write each $P(t)$ as the kernel of a linear map $\varphi(t)$, smoothly varying in $t$, say $\varphi(t)\colon E\to Q$, for fixed vector spaces $E,Q$, with $\varphi(t)$ onto. This $\varphi(t)$ is uniquely determined by its kernel $P(t)$ up to changing to $g(t)\varphi(t)$, where $g(t)$ is a path in $\operatorname{GL}_Q$. Let $\bar\varphi(t)$ be the quotient isomorphism $E/P(t)\cong Q$ by quotienting out the kernel of $\varphi(t)$. Then check easily that $\eta(t)=\left.\bar\varphi(t)^{-1}\varphi'(t)\right|_{P(t)}$ is well defined (i.e. unchanged if we replace $\varphi(t)$ by $g(t)\varphi(t)$). Check that to each tangent vector $v=P'(0)\in T\Gr(k)$, there is a unique well defined value of $\eta(0) \in P(0)^*\otimes (E/P(0))$, so that $v\mapsto \eta(0)$ is a linear isomorphism $T_P \Gr(k)\cong P^*\otimes (E/P)$. $\Box$

In particular each tangent space contains a cone of tangent vectors which are rank one, i.e. pure tensors $\xi\otimes v$ for some $\xi\in P^*$ and $v\in E/P$, called the bad cone; it was named by Gluck, Warner and Yang.

H. Gluck, F. Warner, C. T. Yang, Division algebras, fibrations of spheres by great spheres and the topological determination of a space by the gross behaviour of its geodesics, Duke Mathematical Journal, vol. 50, no. 4, December 1983.

You can recover the Grassmannian from this field of cones in some sense; see papers of Hwang and Mok.

J.M. Hwang, N. Mok, Uniruled projective manifolds with irreducible reductive $G$-structures, Journal fur die reine und angewandte Mathematik, vol. 490, 1997, pp. 55-64.

A tensor product structure on each tangent space is equivalent to that field of cones. A tensor product structure has local invariants, which one can uncover using Cartan's method of equivalence, and the case of maximal dimensional Lie algebra of local symmetry vector field is that of open subsets of the Grassmannian. In particular, a compact and simply connected manifold with a tensor product structure in its tangent spaces, with maximal dimensional symmetry Lie algebra, is diffeomorphic to the universal covering space of the Grassmannian with its usual tensor product structure. The proof is involved: you use the tensor product structure to reduce structure group of the frame bundle, and then arrive at the Cartan geometry of the Grassmannian. The complete proof, for a much broader class of geometric structures on manifolds (in the real and complex cases), appears in lemma 12 of my paper

B. McKay, Holomorphic geometric structures on Kaehler-Einstein manifolds, Manuscripta Math., vol 153 (no. 1-2), 2017, pp.1--34.

To answer your more recent question: The automorphism group of the Grassmannian does not preserve a projective connection or a path geometry, as I proved in

McKay, Benjamin, Complete complex parabolic geometries. Int. Math. Res. Not. 2006.

In other words there is no family of paths on the Grassmannian which are, in local coordinates, precisely the solutions of a second order system of ordinary differential equations, and which is invariant under the action of the projective general linear group. But I suspect that there is an invariant family of paths everywhere tangent to the bad cone and defined by some ordinary differential equations of second or third order, so that is what I would suggest you might look for. The idea is to look at the representation theory of the projective group: in the Lie algebra $\mathfrak{sl}_{n+1}$ of the projective group, take the noncompact simple root $\alpha$, and the root vectors $e_{\alpha},e_{-\alpha}$ and their bracket; together these three vectors span a copy of $\mathfrak{sl}_2$ inside $\mathfrak{sl}_{n+1}$. This $\mathfrak{sl}_2$ has an orbit in the Grassmannian, which is a rational curve tangent to the bad cone. These curves form a small dimensional family, so solve some lower order ordinary differential equation. See

McKay, Benjamin Rigid geometry on projective varieties. Math. Z. 272 (2012), no. 3-4, 761–791

for a discussion of these on arbitrary flag manifolds and on manifolds with parabolic geometries.

  • $\begingroup$ Thank you for your detailed answer! In am also interested in a possible notion of geodesics. I will edit the question. $\endgroup$
    – Chevallier
    Sep 9, 2021 at 8:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.