# Is the map assigning $x$ to its tangent space smooth?

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?

• 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)" – 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 $$i_1 and let $$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)$$.
• 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? – M. B. Jan 6 at 22:14
• 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. – Eric Wofsey Jan 6 at 22:18
• $f$ 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. – Eric Wofsey Jan 6 at 22:37