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Question : How to show $SL(n,\mathbb R)$ diffeomorphic to $ SO(n) \times \mathbb R^{n(n+1)/2-1}$? Also, how to show $SL(n,\mathbb C)$ diffeomorphic to $ SU(n) \times \mathbb R^{n^2-1}$?

I have no idea about it. And, I just find a similar question that I guess can work here:

$GL^+ (n,\mathbb R)$ is diffeomorphic to $SO(n) \times T^+(n, \mathbb R)$ , where $T^+(n, \mathbb R)$ is the Lie group of all opper triangular real matrices with positive diagonal entries.

Simply, we can construct a diffeomorphism by QR decomposition, that is, $$A=QR \to (Q,R)$$ Then, we check that it's well-defined and smooth and construct an inverse.

However, for the origin question, I cannot construct a diffeomorphism just as above. Please help!

Hang
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Hint If the QR decomposition of a matrix in $GL^+(n, \Bbb R)$ is $A = QR$, then $$\det A = \det (QR) = \det Q \det R = \det R ,$$ so if $$\Phi : GL^+(n, \Bbb R) \to SO(n) \times T^+(n, \Bbb R)$$ is the diffeomorphism defined by that decomposition, $(Q, R)$ is in the image of $SL(n, \Bbb R) \subset GL^+(n, \Bbb R)$ iff $\det R = 1$, which implies that the restriction $$\Phi\vert_{SL(n, \Bbb R)} : SL(n, \Bbb R) \to SO(n) \times (T^+(n, \Bbb R) \cap SL(n, \Bbb R) )$$ is itself a diffeomorphism.

It remains to show that $T^+(n, \Bbb R) \cap SL(n, \Bbb R)$ is diffeomorphic to $\Bbb R^{n (n + 1) / 2 - 1}$. Any matrix $(t_{ij}) \in T^+(n, \Bbb R) \cap SL(n, \Bbb R)$ is upper-triangular and has determinant $1$, and so $$t_{11} \cdots t_{nn} = \det (t_{ij}) = 1 .$$ Thus, we can express $t_{nn}$ as a function of $t_{11}, \ldots, t_{n - 1, n - 1}$.

Travis Willse
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    Actually, $T^+(n, \Bbb R) \cong \Bbb R^{n(n+1)/2}$, no? But if $A = QR$, $\det A = 1$ together with $\det Q = \pm 1$ and $\det R>0$ implies that all of those determinants equal $1$. So $\dim( T^+(n,\Bbb R) \cap {\rm SL}(n, \Bbb R)) = n(n+1)/2 - 1$ because $\det R = 1$ kills one degree of freedom in the diagonal of $R$. If I'm missing something please point it out. – Ivo Terek Mar 31 '20 at 18:17
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    Thanks, @IvoTerek. The answer was written a long time ago, so I'm not sure what I was thinking at the time (and I'm surprised no one else had pointed out the issue in the last half-decade). Anyway, I've completely rewritten the answer. – Travis Willse Apr 03 '20 at 09:31
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    Wow, that was a major edit. I had just pointed out the detail to help people who might stumble upon the post later, but I didn't expect any feedback. Thank you for the effort and +1 – Ivo Terek Apr 03 '20 at 13:50
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    The original answer was garbled enough that leaving it might have done future readers more harm than good. Thanks again for your helpful comment. – Travis Willse Apr 03 '20 at 19:53