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Problem

Let $Sl(n, \mathbb R)$ be the set of $n \times n$ matrices with determinant equal to $1$. Show that $Sl(n, \mathbb R)$ as a subset of ${\mathbb R}^{n^2}$ has measure zero.

I'll write what I did up to now:

If $A=\begin{pmatrix} a_{11} & ...& a_{1n}\\ ... & ...& ...\\ a_{n1}& ...& a_{nn}\end{pmatrix}$, then I can identify $A$ with a$=(a_{11},...,a_{1n},...,a_{n1},...,a_{nn}) \in {\mathbb R}^{n^2}$; at the same time, if x$=(x_1,...,x_{n^2})$, I can identify this vector with the matrix that has the first nth coordinates of x as its first row, the second nth coordinates as its second row and so on, notice that xcan also be written as x$=(x_{11},...,x_{1n},...,x_{n1},...,x_{nn})$.

If I define $\phi:{\mathbb R}^{n^2} \to \mathbb R$ as $$\phi((x_{11},...,x_{1n},...,x_{n1},...,x_{nn}))=\sum_{\sigma \in S_n}sgn(\sigma)\prod_{i=1}^{n} x_{\sigma(i),i}$$

then $\phi$(x)=$det(A_x)$, where $A_x$ is the matrix identified with x.

It is clear that $\phi$ is continuous, what I want to prove is that $m(\phi^{-1}(\{1\}))=0$. I have no idea what to do next, all I know is that the preimage of $\{1\}$ is a closed set and that $m(\{1\})=0$.

I would really appreciate any suggestions to help me with this problem.

user16924
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    \You may try to show that $SL(n, \mathbb R)$ is an $n^2-1$ dimensional submanifold by using implicit function theorem on $\phi$. To differentiate $\phi$ see http://math.stackexchange.com/questions/1447692/why-the-gradient-of-log-detx-is-x-1-and-where-did-trace-tr-go –  Sep 24 '15 at 21:14
  • I haven't studied manifolds yet, is there another way of solving the problem without having to go through manifolds? – user16924 Sep 24 '15 at 21:48
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    The complement of $SL(n, {\Bbb R})$ is open and dense in $M(n, {\Bbb R})$. – Rob Arthan Sep 24 '15 at 21:52
  • Suppose I could prove that, I don't know how could I deduce from there that $SL(n,\mathbb R)$ has measure zero. For example, if a set $S \subset \mathbb R^k$ has measure zero, then its complement has to be dense in $\mathbb R^k$, because if not we could find a ball $B(x,\epsilon) \subset S$, which is absurd.So density is an absolute condition for the complement of a null set, but I don't see why the reverse implication works for a closed null set. – user16924 Sep 26 '15 at 17:53

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