# Show that $\text{SO}(2)$ is compact.

Let $$\text{Mat}_2(\mathbb{R})$$ be the set of $$2\times 2$$ real matrices with the topology obtained by regarding $$\text{Mat}_2(\mathbb{R})$$ as $$\mathbb{R}^4$$. Let $$\text{SO}(2)=\{A\in\text{Mat}_2(\mathbb{R}); A^TA=I_2, \det A=1\}$$ where $$A^T$$ denotes the transpose of $$A$$, and $$I_2$$ is the $$2\times 2$$ identity matrix.

The subspace topology of $$\text{SO}(2)$$ is obtained from $$\mathbb{R}^4$$, where we identify a $$2\times 2$$ matrix with a point in $$\mathbb{R}^4$$ by using the matrix entries as coordinates. Viewing $$\text{SO}(2)$$ as a subset of $$\mathbb{R}^4$$, it is enough to show that $$\text{SO}(2)$$ is bounded and closed in $$\mathbb{R}^4$$.

I was able to show that $$\text{SO}(2)$$ is bounded. For any matrix $$A\in\text{SO}(2)$$, we have that $$|A|=\sqrt{2}$$, using the Euclidean metric of $$\mathbb{R}^4$$.

I want to show that $$\text{SO}(2)$$ is closed by showing that $$\mathbb{R}^4\setminus\text{SO}(2)$$ is open. I am not sure how to start this.

In addition we have: \begin{align*} A^T A\\ \\ =\begin{pmatrix} a_1 & a_3 \\ a_2 & a_4\end{pmatrix}\begin{pmatrix} a_1 & a_2 \\ a_3 & a_4\end{pmatrix}\\ \\ =\begin{pmatrix} (a_1)^2+ (a_3)^2 & a_1a_2+a_3a_4 \\ a_1a_2+a_3a_4 & (a_2)^2+(a_4)^2\end{pmatrix}\\ \\ =\begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix} \end{align*} and \begin{align*} \det{\begin{pmatrix} a_1 & a_2 \\ a_3 & a_4\end{pmatrix}}\\ \\ =a_1a_4-a_2a_3=1. \end{align*}

• Continuous maps on topological spaces are defined as maps f: A-> B in which all closed sets in B are images of closed sets in A. In metric spaces this is an equivalent characterisation of continuity and used all the time to show that something is closed. In 90% of examples there is a way to rewrite a given set as reverse image of some continuous function. – Martin Erhardt Sep 24 '18 at 23:07
• Thank you for this general tip @MartinErhardt – user475040 Sep 25 '18 at 18:09
• Just to clarify: Different continuous maps were given as examples below. As long as I can show that $\text{SO}(2)$ is the preimage of closed set, the "specific" function can be different, just as long as it is continuous and the $\text{SO}(2)$ is a subset of the domain. @MartinErhardt – user475040 Sep 25 '18 at 18:29

A continuous function $$f : \text{Mat}_2(\mathbb{R}) \to \mathbb{R}^4$$ is defined by $$f(A) = (a_1^2 + a_3^2,a_1a_2 + a_3a_4,a_2^2 + a_4^2,a_1a_4 - a_2a_3)$$. Then

$$\text{SO}(2) = f^{-1}(1,0,1,1)$$

which is closed by the continuity of $$f$$.

• You mean $\text{Mat}_2(\mathbb{R})$? It should take the topology of Euclidean space, so an open set in $\text{Mat}_2(\mathbb{R})$ is an open ball? – user475040 Sep 25 '18 at 18:39
• As you said in your question, $\text{Mat}_2(\mathbb{R})$ is isomorphic to $\mathbb{R}^4$. Whether you arrange the four components in a $2 \times 2$- matrix or regard them as coordinates of a vector in $\mathbb{R}^4$ is topologically irrelevant. In fact, each finite-dimensional real vector space $V$ can be endowed with a norm, and all norms on $V$ are equivalent. Therefore $V$ carries a canonical topology induced by any norm. Thus $\text{Mat}_2(\mathbb{R})$ in fact takes the topology of a four-dimensional Euclidean space. – Paul Frost Sep 25 '18 at 22:07
• The equation $f(A) = (1,0,1,1)$ is equivalent to $A^TA = I_2$ (first three components) and $\text{det}(A) = 1$ (last component). Considering only $\text{det}(A) = 1$ is not enough. – Paul Frost Sep 25 '18 at 22:23
• Ok and being continuous in each component is enough to say $f$ is continuous? – user475040 Sep 25 '18 at 22:37
• Yes. A function $g : X \to \mathbb{R}^n$ is continuous of and only if all $n$ component functions $g_i : X \to \mathbb{R}$ are continuous. – Paul Frost Sep 25 '18 at 22:43

It is simpler to note that$$SO(2)=\left\{\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}\,\middle|\,\theta\in\mathbb{R}\right\}.$$Therefore, the map$$\begin{array}{rccc}f\colon&[0,2\pi]&\longrightarrow&SO(2)\\&\theta&\mapsto&\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}\end{array}$$is surjective. Since it is also continuous and $$[0,2\pi]$$ is compact, $$SO(2)$$ is compact too.

• I don't necessarily think that it's very easy to notice the equality. Of course once it's seen the argument is simple – leibnewtz Sep 24 '18 at 22:55

Quick answer: $$\text{det}$$ is continuous and $$Y = \{1\} \subset \mathbb{R}$$ is closed, so $$\text{SO}_{2} = \det^{-1}(Y)$$ is closed too.

If you prefer to show $$M_{2,2} \setminus \text{SO}_{4}$$ is open, you need: $$\forall A \in M_{2,2} \setminus \text{SO}_{2}, \,\, \exists \varepsilon > 0, \,\, \exists B \in M_{2,2} \setminus \text{SO}_{2}\,\, \text{s.t.} \,\,|B-A| < \varepsilon$$ So let $$A \in M_{2,2} \setminus \text{SO}_{2}$$. $$\det(A) \neq 1$$. Since $$\text{det}$$ is continuous, (take $$\varepsilon = |1-\det(A)|/2 := \eta$$ in the definition): $$\exists \delta > 0 ,\,\,\forall B, |B-A| < \delta \implies |\text{det}(B) - \text{det}(A)| < \eta$$ Therefore, in particular, $$|B-A| < \delta \implies \text{det}(B) \neq 1$$

Then $$\delta$$ is your required $$\varepsilon$$

• $\text{SO}(2) \subsetneqq det^{-1}(Y)$. Consider the diagonal matrix with entries $2$ and $1/2$. – Paul Frost Sep 25 '18 at 22:15
• Oh yeah, that's embarrassing – preferred_anon Sep 26 '18 at 7:47

For groups such as these, the most effective method is to simply realize that they're given by algebraic equations. Note that matrix multiplication is a map $$\mathbb{R}^{n^2} \times \mathbb{R}^{n^2} \to \mathbb{R}^{n^2}$$ given by a polynomial in each variable. By definition, the map $$det: \mathbb{R}^{n^2} \to \mathbb{R}$$ is also given by a polynomial.

Hence the conditions $$AA^T-I=0$$ and $$det(A)-1=0$$ realize $$SO(2)$$ as the zero set of a collection of polynomials, and therefore the group is closed.

One may use this argument to show that $$SO(n)$$, $$O(n)$$, $$SL(n)$$, $$SP(n)$$, etc are all closed, even if the underlying field is $$\mathbb{C}$$