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I need to show for $n=3$ that $SL(n,\mathbb{R})=\{A \in M(n, \mathbb{R}) : detA=1 \}$ is a $(n^2 -1)$ dimensional smooth submanifold of the vector space $M(n,\mathbb{R})$ of all real $n \times n$ matrices. I would assume that I need to use the regular value theorem and use the determinant map to get the result, but I'm a bit unsure on how to set this up correctly. Any help would be appreciated.

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You want to consider the smooth map $\det\colon M_n(\mathbb{R})\to\mathbb{R}$. If you can show $1$ is a regular value of $\det$, then $SL(n,\mathbb{R})=\det^{-1}(1)$ is a smooth manifold of dimension $$\dim(M_n(\mathbb{R}))-\dim(\mathbb{R})=n^2-1$$ by the regular value theorem.

To show $1$ is a regular value of $\det$, first note that the domain and codomain are vector spaces, so they may be identified with their own tangent spaces.

If $A\in M_n(\mathbb{R})$, then $d(\det)_A(A)$ can be computed as

$$\lim_{t\to 0}\frac{\det(A+tA)-\det(A)}{t}=\lim_{t\to 0}\frac{(1+t)^n\det(A)-\det(A)}{t}=n\cdot\det(A) $$

This shows $d(\det)_A$ is nonzero linear functional, hence surjective, for invertible $A$. This implies in particular that $1$ is a regular value of $\det$.

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    $\begingroup$ Awesome. This is so much better than what I had in mind. $\endgroup$ – Braindead Mar 25 '16 at 19:19
  • $\begingroup$ Can you clarify something for me? If you have chosen another point $B$ (instead of $A$), calculated $d(\det)_A(B)$ and verified it was nonzero, it would also work. You just needed to find one vector for which the derivative is nonzero. Right? $\endgroup$ – Soap Feb 27 '17 at 1:15
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    $\begingroup$ @Simoes Right, you just need to exhibit one vector where $d(\det)_A$ is nonzero, so that $d(\det)_A$ is not identically zero. Since the codomain is 1 dimensional, this forces it to be surjective. $\endgroup$ – Ben West Feb 27 '17 at 1:35
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Ben West's solution is extremely clean and simple, and it should be accepted as the answer.

This is the solution that I was thinking about.

$GL_n = \det^{-1}(\mathbb{R}\setminus\{0\})$

Since $\det$ is continuous, $GL_n$ is an open subset of $M_n$, meaning it has dimension $n^2$.

Restrict $\det$ to $GL_n$. Then, for any $A\in GL_n$ and any $B\in M_n \sim T_A GL_n$, we have:

\begin{align} d(\det)_A(B) &= \lim_{t\to0} \frac{\det(A+tB) - \det(A)}{t} \\ & =\det(A) \lim_{t\to0} \frac{\det(I+tA^{-1}B)-1}{t}\\ &= \det(A)\mathrm{tr}(A^{-1}B) \end{align}

Since $\det$ maps into a 1-dimensional space, we just need one $B$ that makes $\mathrm{tr}(A^{-1}B)≠0$. So, take $B=A$.

This shows that every non-zero number is a regular value of $\det$.

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