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?

  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ What is $\gamma'(1)$? $\endgroup$ – Daniel Fischer Jul 11 '15 at 18:15
  • 1
    $\begingroup$ 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

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 $\mathbb{R}^{n+m}$.


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.


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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.