Manifolds and their dimension

Question

Why are the following two sets manifolds and what are their dimensions?

1. The set of all 2 by 2 matrices with determinant 1.
2. The set of all inner products on $\mathbb R^3$.

I have difficulty in understanding the two sets in the first place. For example, in the second set, I know that inner products are simply real numbers. Can I then say that the set of all inner products is simply like the real line, i.e. $\mathbb R$. Hence, its dimension is 1? Thank you!

Answer 1

A 2 by 2 matrix is simply an array of real numbers written:

$$\left[ \begin{matrix}a & b \\ c & d\end{matrix} \right]$$

"2 by 2" refers to the number of rows and columns. The determinant is given by $ad - bc$, hence the set of 2 by 2 matrices with determinant 1 is the same as the subset of $\mathbb{R}^4$ cut out by the equation $ad - bc - 1 = 0$. Now, there is a nice theorem which states:

Thm: If $f:\mathbb{R}^n \rightarrow \mathbb{R}$ is a smooth function, then the set $\{\mathbf{x} \mid f(\mathbf{x}) = 0\}$ is a smooth $n-1$ dimensional manifold, provided that whenever $f(\mathbf{x}) = 0$ the following vector is not the zero vector.

$f'(\mathbf{x}) := \left[\begin{array}{c} \frac{\partial f}{\partial x_1}(\mathbf{x}) \\ \frac{\partial f}{\partial x_2}(\mathbf{x}) \\ \vdots \\ \frac{\partial f}{\partial x_n}(\mathbf{x})\end{array}\right]$

$\square$

Now, if you accept this theorem, it follows that the set of 2 by 2 matrices is a 3-dimensional smooth manifold, using the function $f(a,b,c,d) = ad - bc - 1$. The derivative of $f$ is given by:

$f'(a,b,c,d) = \left[\begin{array}{c} d \\ -c \\ -b \\ a \end{array}\right]$

Now, if $ad - bc = 1$, then $f'(a,b,c,d)$ is not the zero vector, and so the theorem applies.

Answer 2

Let $\operatorname{SL}(2,\Bbb R)$ be the collection of all $2\times 2$ real matrices with determinant one and let $$ M=\left\{(w,x,y,z)\in\Bbb R^4: wz-xy=1 \right\} $$ Note that there is an obvious bijection $$ \phi:\operatorname{SL}(2,\Bbb R)\to M $$ I'll leave it to you to define $\phi$.

Now, let $f:\Bbb R^4\to \Bbb R$ be $f(w,x,y,z)=wz-xy-1$ and note that $M=f^{-1}(\{0\})$. Moreover, $$\nabla f(w,x,y,z)=\langle z,-y,-x,w\rangle$$ so the only critical point of $f$ is $\mathbf 0\notin M$. One may then use the implicit function theorem to show that $M$ is a smooth hypersurface in $\Bbb R^4$. That is, $M$ is a smooth manifold of dimension $4-1=3$.

The above is either a straightforward application of a few basic theorems covered in any differential topology course or a nice exercise on the IFT. Once we know that $M$ is a manifold it is an easy exercise to endow $\operatorname{SL}(2,\Bbb R)$ with the structure of a smooth three-dimensional manifold using the bijection $\phi$.

Answer 3

Brian and Alex's answers are more correct and more to the point than what I'm going to say, but maybe this hand-wavy discussion is helpful for getting intuition.

The collection of all $2 \times 2$ matrices with determinant 1, denoted $\mathrm{SL}(2, \mathbb{R})$, is just a collection of four numbers, call them $(a, b, c, d)$, subject to the condition that $ad - bc = 1$. This is a subset of $\mathbb{R}^4$, and if it were an open subset you'd be done (it would then be a 4-dimensional open submanifold of $\mathbb{R}^4$). However, this is not an open subset. It is a closed set since it is the preimage of the closed set $\{1\} \subseteq \mathbb{R}$ under the determinant map, which is continuous.

Now, why is it a manifold? You could use some theorems to say this is a hypersurface of $\mathbb{R}^4$, but let me give you another point of view. Think about letting $U_a \subseteq \mathrm{SL}(2, \mathbb{R})$ be the collection of those matrices where the $a$-entry is not zero. Define $U_b$, $U_c$, and $U_d$ similarly. As we can't have all four entries $(a, b, c, d)$ be zero and have determinant $1$, these four sets cover $\mathrm{SL}(2, \mathbb{R})$.

Inside $U_a$, you know that $d = \frac{1 + bc}{a}$, and so $d$ is determined by your other three values $a$, $b$, and $c$. You have three variables you can "control" and so $U_a$ should be three-dimensional: think of the $a$, $b$, and $c$ as being coordinates inside of $U_a$. You can do something similar for the other charts.

Now of course there are details to check about these charts fitting together nicely so that you get a smooth manifold, but maybe that gives some intuition about why $\mathrm{SL}(2, \mathbb{R})$ should be a 3-dimensional submanifold sitting inside of $\mathbb{R}^4$.