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It's well-known that $SL(n, \mathbb{R})$ is diffeomorphic to $SO(n,\mathbb{R}) \times \mathbb{R}^{n(n+1)/2-1}$. The argument goes like this:

Let $T^+$ be the subset of $SL(n,\mathbb{R})$ consisting of the upper-triangular matrices with positive diagonal entries. Then QR factorization tells us that the map $$f = (Q,R) \mapsto QR : SO(n,\mathbb{R}) \times T^+ \to SL(n,\mathbb{R})$$ is a bijection. This map is also smooth: $SO(n,\mathbb{R})$ and $T^+$ are both closed submanifolds of $SL(n,\mathbb{R})$, and $f$ is simply the restriction of the multiplication operation $SL(n,\mathbb{R}) \times SL(n,\mathbb{R}) \to SL(n, \mathbb{R})$.

If we can somehow conclude that $f$ is a diffeomorphism, we are done, since $T^+$ is easily shown to be diffeomorphic to $\mathbb{R}^{n(n+1)/2-1}$! My problem is with this step: I only know how to do this for a few small values of $n$, where I understand the tangent spaces of $SO(n,\mathbb{R})$ and $T^+$ and can explicitly compute the differential $df$ in coordinates to show that $f$ is an immersion/submersion (either is sufficient by dimension-counting and the previously-established bijectivity). This is extremely tedious, though – there obviously must be a better way!

So, my question is this: what's the best/most elegant/easiest way (this is up to interpretation) to show that the map $f$ is actually a diffeomorphism? I'm happy to use theorems found in introductory differential topology/geometry books (Lee, Bott & Tu, Guillemin & Pollack, etc.) if a reference is given.

I'm aware of this post which is about the same topic, but does not explain why this map should be a diffeomorphism.

diracdeltafunk
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There is a direct argument why $f^{-1}$ is smooth (even though it is not elegant at all). There is an explicit formula for the $QR$ factorization using $Gram$-$Schmidt$. Since every $A\in SL(n,\mathbb{R})$ is invertible its columns $(a_1,\ldots,a_n)$ form a basis and hence Gram-Schmidt does not fail when starting with $(a_1,\ldots,a_n)$.

The one critical point where Gram-Schmidt might not be smooth is when we re-normalize the projected vectors, i.e. $$ v_j:= \frac{u_j}{\lVert u_j\rVert}, $$ where $u_j:= (Id-\sum_{k=1}^{j-1}\pi_{u_k})a_j$ and $\pi_{u_k}$ is the projection onto the subspace spanned by $u_k$. Yet for every $A\in SL(n,\mathbb{R})$ there is a small open neighborhood in $SL(n,\mathbb{R})$ where $\lVert u_j \rVert\geq L > 0$ for all $1\leq j \leq n$ and some $L\in\mathbb{R}$. Since the denominators of the derivatives at the critical point of the construction are given by powers of $\lVert u_j \rVert$ the QR-factorization is smooth.

dennis_s
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    Thank you! I don't fully understand the point of checking smoothness at the normalization step: the map $u \mapsto u/\Vert u \Vert : \mathbb{R}^n \setminus {0} \to \mathbb{R}^n \setminus {0}$ is smooth, and we certainly know the $u_j$'s will be nonzero, so I don't see what can go wrong? Also, correct me if I'm wrong but it's also important to check that the upper-triangular matrix we get after multiplying by the inverse of the orthogonal matrix produced by Gram-Schmidt has positive diagonal entries (to make sure this actually gives an inverse to the map $f$), although this isn't hard. – diracdeltafunk Jun 14 '20 at 02:00