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?

  • $\begingroup$ 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

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 $f:U\to Gr(k,n)$.

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  • $\begingroup$ 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

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