Frankel 1.1 Problems


The locus described by \(x^2 + y^2 - z^2 = c\) is a 2 dimensional submanifold of \(\mathbb{R}^3\) for \(c < 0\) and \(c = 0\) but not for \(c > 0\). This can be seen by investigating the functional dependence

$$z = z(x,y) = \sqrt{x^2+y^2-c}.$$

For \(c < 0, z(x,y)\) is differentiable over it's entire domain, by definition forming a 2D submanifold of \(\mathbb{R}^3\). For \(c = 0, z(x,y)\) is differentiable everywhere except at \(x = y = 0\), and therefore forms a submanifold if and only if the origin is omitted. For \(c > 0, z(x,y)\) is defined on the reals only for \(x^2+y^2 \ge c\) and differentiable only for \(x^2+y^2 > c\), so it only forms a submanifold for \(x^2+y^2 > c\).



SO(\(n\)) is defined to be the set of all orthogonal \(n \times n\) matrices \(x\) with \(\det x = 1\). The discussion on SO(3) in the book easily generalizes to show that SO(\(n\)) is an \(n\)-dimensional submanifold of \(\mathbb{R}^{n^2}\).



The special linear group is the set

$$\mathrm{SL}(n) := \{ n\times n \text{ real matrices } x | \det x = 1 \}.$$

Clearly SL(\(n\)) is a subset of the set of \(n \times n\) matrices, which itself is isomorphic to \(\mathbb{R}^{n^2}\). The condition on the determinant restricts SL(\(n\)) to be a space of dimension \(n^2 - 1\), but to show it's a submanifold we must demonstrate that \(\det x = 1\) defines a differentiable map. To do this we first expand the determinant around the \(j\)'th row,

$$\det x = \sum_{i=1}^n (-1)^{i+j} M_{ij}x_{ij}.$$

Here, \(M_{ij}\) is the minor.  Taking the derivative with respect to \(x_{kj}\),

$$\frac{\partial}{\partial x_{kj}}\det x = \sum_{i=1}^n (-1)^{i+j}M_{ij}\frac{\partial x_{ij}}{\partial x_{kj}} = (-1)^{k+j}M_{kj}.$$

As long a \(M_{kj}\) is non-zero, the implcit function theorem tells us that \(x_{kj}\) can be expressed as a differentiable function of the other components. Therefore, by definition, SL(\(n\)) is a \(n^2 - 1\) submanifold of \(\mathbb{R}^{n^2}\).