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Prove that $SL(n)=\{A\in \Bbb{R}^{n\times n}:\det(A)=1\}$ is a differentiable submanifold.

The determinant function is smooth since it's a polynomial, and we have $\det^{-1}(1)=SL(n)$. So it suffices to prove that $d \det(A)$ has constant nonzero rank for all $A\in SL(n)$.

By the Laplace's formula, we can write $$\det A = \sum_{j=1}^n (-1)^{i+j} a_{ij} \det (A^{*})$$ where $A^*$ is $A$ with the i-th row and j-th column removed. Then $$\frac{\partial}{\partial a_{ij}} \det A = (-1)^{i+j}\det(A^*).$$

So we need to show that $$\left( (-1)^{i+j}\det(A^*) \right)_{ij}$$ has constant nonzero rank.

How can we proceed?

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    $\begingroup$ Another approach might be to show that it is a Lie group: It's easy to show that $SL_n(\mathbb{R}) \subseteq GL_n(\mathbb{R})$ is a group under multiplication. The restriction of the multiplication and inversion maps from $GL_n(\mathbb{R})$ are still smooth on $SL_n(\mathbb{R})$. $\endgroup$ – Mnifldz Jul 11 '15 at 17:46
  • $\begingroup$ I'm afraid I haven't learned about Lie Groups yet :( $\endgroup$ – iwriteonbananas Jul 11 '15 at 17:47
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    $\begingroup$ Since the codomain of $\det$ is one-dimensional, you want to show that $d\det$ is not the zero functional at any point of $SL(n)$. If you look at $\gamma(t) = \det (t\cdot A)$, what do you see? $\endgroup$ – Daniel Fischer Jul 11 '15 at 17:49
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    $\begingroup$ What is $\gamma'(1)$? $\endgroup$ – Daniel Fischer Jul 11 '15 at 18:15
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    $\begingroup$ Along the same lines @DanielFischer is suggesting, you might try to apply Euler's Theorem on Homogeneous Functions. $\endgroup$ – Ted Shifrin Jul 12 '15 at 18:16
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First, we need a theorem:

Suppose $U\subset \mathbb{R}^{n+m}$ is an open set and $f \colon U \to \mathbb{R}^m$ is a $\mathcal{C}^{\infty}$ map. Let $q \in \mathbb{R}^m$ and $M = f^{-1}(q)$. If $\mathrm{D}f(x)$ has rank $m$ for all $x \in M$, then $M$ is an $n$-dim. submanifold of $\mathbb{R}^{n+m}$.

Statement:

Let $SL_n(\mathbb{R}) = \{ A \in \mathrm{Mat}_{n \times n}\ \lvert \mathrm{det} A = 1\}$. Then:

  1. If $A \in SL_n(\mathbb{R})$, then $\mathrm{D \ det}(A)$ has rank $1$.
  2. $SL_n(\mathbb{R})$ is an $(n^2 -1) -$ manifold.

Proof:

Note, that $\mathrm{det} \colon \mathbb{R}^{n^2} \to \mathbb{R}$ is polynomial and so $\mathcal{C}^{\infty}$. To show $(1)$ it is only necessary to show that some directional derivative $\mathrm{D_{B}det}(A)$ is nonzero. We compute for $A \in SL_n(\mathbb{R})$,

$$ \mathrm{D_{A}det}(A) = \frac{d}{dt} \mathrm{det}(A + tA)\lvert_{t=0} = \frac{d}{dt} (1 + t)^n \mathrm{det}(A)\lvert_{t=0} = \frac{d}{dt}(1+t)^n \lvert_{t=0} = n. $$ Hence, $(1)$ is shown. To see $(2)$, we use the above theorem. The set $GL_{n}(\mathbb{R}) \subset \mathbb{R}^{n^2 -1} \times \mathbb{R}$ is an open set and $\mathrm{det}\colon GL_{n}(\mathbb{R}) \to \mathbb{R}$ is a $\mathcal{C}^{\infty}$- map. The set $\mathrm{det}^{-1}(1) = SL_{n}(\mathbb{R})$, and $\mathrm{D \ det}(A)$ has rank $1$ for each $A \in SL_n(\mathbb{R})$. Therefore, by the above theorem, $SL_{n}(\mathbb{R})$ is an $(n^2 - 1)$-submanifold.

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Since $SL(n) = \det^{-1}(1)$ and the determinant $\det : {\mathbb R}^{n^2} \to {\mathbb R}$ is smooth, you just need to show that the gradient $\nabla \det (A)$ is never zero at a unimodular matrix $A$ (ie, $A \in SL(n)$).

As you computed, the $n^2$ entries of $\nabla \det (A)$ are precisely:

$$ \left(\nabla \det (A)\right)_{i,j} = (-1)^{i+j} \det(A^*_{i,j}) $$

Here, $A^*_{i,j}$ is the $(n-1)\times (n-1)$-matrix obtained by deleting row $i$ and column $j$. So, denoting $C_{i,j}:=(-1)^{i+j} \det(A^*_{i,j})$, the matrix $C$ is precisely what is called the cofactor (or "adjugate") matrix of $A$: https://en.wikipedia.org/wiki/Adjugate_matrix, which verifies:

$$ C = (\det A) A^{-1} $$

This means that $C$ is also unimodular ($\det C=1$). Hence, $\nabla \det (A)$ (with entries $C_{i,j}$) has, at least, one non-zero entry, and can not be the zero vector.

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