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Show that $SL(n , \mathbb{R})$ is a regular submanifold of $GL(n, \mathbb{R})$ of codimension $1$.

I know that I have to use Regular level set theorem but don't know how to proceed with that. I suppose I have to construct a map $F : SL(n , \mathbb{R}) \rightarrow GL(n, \mathbb{R}) $ and then proceed somehow but I don't see any way. I have looked Special linear group as a submanifold of $M(n, \mathbb R)$ but I haven't studied Lie groups so far and so couldn't understand the solution. So, can someone please explain this to me using regular level set theorem?

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user383517
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1 Answers1

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To show that $SL_n(\mathbb{R})$ is a regular submanifold of $GL_n(\mathbb{R})$, we let $f\colon GL_n(\mathbb{R})\rightarrow \mathbb{R}^{\times}$ be the determinant map $f(A)=\det(A)$, and apply the regular level set theorem to $f^{-1}(1)=SL_n(\mathbb{R})$. We need to verify that $1$ is a regular value of $f$.This follows by expanding the determinant along any row or any column, and we see that a matrix $A\in GL_n(\mathbb{R})$ is a critical point of $f$ if and only if all the $(n-1)\times(n-1)$-minors in the expansion are zero. From this determinant expansion we see that such a matrix has determinant zero. Since every matrix in $SL_n(\mathbb{R})$ has determinant $1$, all matrices of determinant $1$ are regular points of $\det$. So we are done.

Dietrich Burde
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