1
$\begingroup$

I'm having some difficulties understanding smooth maps between manifolds, and in particular I would like to know if the map $x \mapsto T_{x}M$ is a smooth map $M\to Gr(k,n)$? Here $x \in M$, where $M$ is a $k$-dimensional manifold in $\mathbb{R}^{n}$.

From Lee's Introduction to Smooth Manifolds, we have the definition:

Let $M, N$ be smooth manifolds and let $F : M \to N$ be any map. $F$ is smooth if for every $p \in M$, there exist smooth chart $(U, \phi)$ containing $p$ and $(V,\psi)$ containing $F(p)$ s.t. $F(U) \subseteq V$ and the composite map $\psi \circ F \circ \phi^{-1}$ is smooth from $\phi(U)$ to $\psi(V)$.

Is it possible to show that the map assigning $x$ to its tangent space is smooth directly from the definition?

$\endgroup$
  • $\begingroup$ Yes, the Grassmannian. I noticed that according to Wikipedia (Grassmannian), "The map assigning to x its tangent space defines a map from M to Gr(k, n)" $\endgroup$ – M. B. Jan 6 at 18:32
0
$\begingroup$

Sure, this is almost immediate from the definition of charts on the Grassmannian. Let $U\subseteq\mathbb{R}^{n\times k}$ be the open set consisting of $n\times k$ matrices of rank $k$. There is an obvious map $f:U\to Gr(k,n)$ taking a matrix to the span of its columns. The derivative of any local parametrization of $M$ defines a smooth map to $U$ whose composition with $f$ is exactly the map $x\mapsto T_xM$. So, it suffices to show that $f$ is smooth.

Now fix $i_1<i_2<\dots<i_k$ and let $V\subseteq U$ be the open subset consisting of matrices whose rows $i_1,\dots,i_k$ are linearly independent, and let $C\subseteq V$ be the set of matrices whose rows $i_1,\dots,i_k$ form the $k\times k$ identity matrix (which can be identified with $\mathbb{R}^{k(n-k)}$ since there are $n-k$ rows whose entries are unrestricted). By definition of the smooth structure on $Gr(k,n)$, $f$ restricted to $C$ is smooth (it is the inverse of one of the coordinate charts on $Gr(k,n)$). Now let $g:V\to C$ be the smooth map defined by $g(A)=AB^{-1}$ where $B$ is the $k\times k$ matrix formed by rows $i_1,\dots,i_k$ of $A$. The columns of $g(A)$ have the same span as the columns of $A$ since we just multiplied on the right by an invertible matrix, so $f=f\circ g$ on $V$. But $g$ maps to $C$ and $f$ is smooth on $C$, so we conclude that $f$ is smooth on $V$. Since open subsets $V$ of this form cover $U$, this proves $f:U\to Gr(k,n)$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ A couple of questions: 1. Why is $f$ restricted to $C$ smooth? Which coordinate chart is it the inverse of? 2. What do you mean by "precomposes"? I think I'm generally a little bit unsure about the sizes of the matrices. $A \in V$ is $k \times n$, but then what is "the inverse of the $k \times k$ matrix formed by its rows $i_{1},...,i_{k}$"? Do the rows $i_{i}$ not have $n$ entries each? $\endgroup$ – M. B. Jan 6 at 22:14
  • $\begingroup$ Oh, I can never remember which number is rows and which is columns in an "$a\times b$ matrix" and I think I got it backwards. $\endgroup$ – Eric Wofsey Jan 6 at 22:18
  • $\begingroup$ $f$ restricted to $C$ is the inverse of the standard coordinate chart on the Grassmannian corresponding to $(i_1,\dots,i_k)$, as described at en.wikipedia.org/wiki/…, for instance. $\endgroup$ – Eric Wofsey Jan 6 at 22:37

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.