I recently read about Grassmannian manifolds.

The following question naturally comes to mind.

Let $GR_k(\mathbf R^n)$ is the grassmannian manifold of $k$ dimensional linear subspaces of $\mathbf R^n$. Let $M$ be a smooth $k$-manifold in $\mathbf R^n$. Define a function $f:M\to M\times GR_k(\mathbf R^n)$ as $$f(\mathbf p)=(\mathbf p,T_{\mathbf p}M)$$ for all $\mathbf p\in M$. Then is $f$ continuous? Is it also smooth?

(All our tangent spaces pass through the origin. Usually the point at which we talk about the tangent space is also important. But since we are working in a Euclidean space, we can translate all of them to the origin).

I have no idea how to go about proving or disproving the above. Does anybody know if the above is true? If not, then are there some extra conditions under which the above is true?


Think of a particle moving in $3$-space. We can ask if the particle is moving ``continuously" in space. If $f:\mathbf R\to\mathbf R^3$ is a function such that at time $t$ the particle is at $f(t)$, then we say that the particle is moving continuously with time if the function $f$ is continuous.

Similary, we ave an intuitive notion of an infinite rigid rod (basically a line) moving continuously in $2$ or $3$-space. To make this notion precise we can use the concept of the real projective space. Suppose at time $t$, the line be written as $L(t)$. We say that the line is moving continuously with time if the function $f:\mathbf R\to \mathbf RP^3$ defined as $f(t)=L(t)$ is continuous.

The Grassmann manifolds generalize this notions to ``planes moving continuously (or smoothly)" in Euclidean spaces.

  • 3
    $\begingroup$ Yes! Google "Gauss map". $\endgroup$ – user64687 Jan 14 '15 at 11:28
  • 1
    $\begingroup$ Yes, you can prove this directly by using the standard "graph" coordinate charts on the Grassmannian. $\endgroup$ – Travis Jan 14 '15 at 11:42
  • $\begingroup$ Googling "Gauss map" I found a wikipedia article which says that this is true in a oriented manifold. I haven't read oriented manifolds yet, but I know that they are require a stronger definition. Does this mean the result isn't true for a usual manifold? $\endgroup$ – caffeinemachine Jan 14 '15 at 11:44
  • $\begingroup$ Alternatively, you could use the Stiefel manifold en.wikipedia.org/wiki/Stiefel_manifold $\endgroup$ – Orest Bucicovschi Jan 14 '15 at 11:44
  • 1
    $\begingroup$ @caffeinemachine: Still it would be nice to have understand in a synthetic way why the tangent space varies continuously. This is certainly a challenge if $X$ is say the boundary of a convex set with all boundary points "smooth" (en.wikipedia.org/wiki/Supporting_hyperplane). We should also think about the case of curves, and the case of hypersurfaces, since these seem easier. $\endgroup$ – Orest Bucicovschi Jan 14 '15 at 14:39

For a parametrization $\phi\colon U \to X$

$$\phi(u) = ( \phi_1, \ldots, \phi_n)(u_1, \ldots, u_k)$$

the tangent space at any point $\phi(u)$ has a basis $$\{\frac{\partial \phi}{\partial u_1}, \ldots, \frac{\partial \phi}{\partial u_k}\}$$

(like the good old way when you calculate $E$, $F$, $G$ for a surface)

This give a smooth map from $U$ to the manifold of frames, that further projects to a map of the Grassmanian.

Now the map $U\to X$ is a local diffeomorphism. So you get $f_{|\phi(U)}\colon \phi(U) \to \mathbb{R}^n \times G_{n,k}$ smooth. Locally smooth implies smooth.

Alternatively, if $X$ is described locally by $f_1=0, \ldots f_{n-k}=0$ with the jacobian $(\frac{\partial f_i}{\partial x_l})$ of rank $n-k$ around some point then the tangent space at a point $x$ is given by all the vectors $v =(v_1, \ldots v_n)$ so that $\nabla f_j \cdot v = 0$ , $\ j = 1, \ldots n-k$

Obs: To get tangent (and cotangent) spaces you need partial derivatives.

Come to think about it, if we have a local parametrization of the manifold, the Grassmann coordinates of the tangent planes are the big minors of the jacobian ( $\binom{n}{k}$ of them).

  • $\begingroup$ Thank you. This certainly helped. I am trying to digest it better. $\endgroup$ – caffeinemachine Jan 14 '15 at 12:14
  • $\begingroup$ No worries. I suggest doing some concrete examples. $\endgroup$ – Orest Bucicovschi Jan 14 '15 at 12:17
  • $\begingroup$ Compare for manifold with a parametrization and for manifolds as zeroes of some functions. $\endgroup$ – Orest Bucicovschi Jan 14 '15 at 12:23

Your Answer

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

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