0
$\begingroup$

First of all I am sorry for the bad notation, but I shall denote the smooth manifold of real symmetric $n \times n$ matrices by $S_n(\mathbb{R})$. I shall denote the transpose of a matrix $A$ by $A^T$

Define the map $f:GL_n(\mathbb{R}) \to S_n(\mathbb{R})$ by $f(A) \to A^T A$. I want to show that this map is a smooth map.

I am using the definition from the book An Introduction to Manifolds by Loring W. Tu. For every point $A \in GL_n(\mathbb{R})$, I need charts $(U,U',\phi)$ and $(V,V',\psi)$ such that the map $$\psi \circ f \circ\phi :U' \to V'$$ is smooth. But I am unable to proceed this way. I am unable to get charts and then showing smoothness of the composite function might get complicated.

How should I solve it? Is there some other way for the problem?

$\endgroup$
3
  • 1
    $\begingroup$ What do charts for these matrices look like? (Hint: Just "flatten" the matrix, this is a chart for it. Restricting the flattening to $GL_n(\mathbb{R})$ and $S_n(\mathbb{R})$ will still provide a chart). In these coordinates, $A^TA$ just has polynomial entries! (Also you need $\psi \circ f \circ \phi^{-1}$ not $\phi$) $\endgroup$ Commented Apr 10, 2021 at 20:45
  • $\begingroup$ What do you mean by 'flatten the matrix'? $\endgroup$ Commented Apr 10, 2021 at 21:09
  • $\begingroup$ Use the "obvious" isomorphism between $M_(\mathbb{R})$ and $\mathbb{R}^{n^2}$ $\endgroup$ Commented Apr 11, 2021 at 1:39

1 Answer 1

0
$\begingroup$

Let $M_n(\mathbb R) \approx \mathbb R^{n^2}$ denote the manifold of all $n\times n$-matrices and define $$F : M_n(\mathbb R) \to M_n(\mathbb R), F(A) = A^TA .$$ This is a smooth map because each coordinate function $F_{ij} : M_n(\mathbb R) \to \mathbb R$ is smooth. Clearly $F(M_n(\mathbb R)) \subset S_n(\mathbb R)$. Since $GL_n(\mathbb R)$ is open in $M_n(\mathbb R)$ and $S_n(\mathbb R)$ is a submanifold of $M_n(\mathbb R)$, we see that that the restriction $$f : GL_n(\mathbb R) \stackrel{F}{\to} S_n(\mathbb R)$$ is smooth.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .