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Exercise 1457Topology › Topology of Mn(ℝ) › Applications and the case n = 2L3Added 10 Nov 2026

The convex hull of SO(2) is a flat disc

#M2(R)#convex hull#rotations#complex coordinates

Statement

Identify R2\mathbb{R}^2 with C\mathbb{C} and write A∈M2(R)A\in M_2(\mathbb{R}) as Ax=zx+wxˉAx=zx+w\bar x, where z=(a+d)+i(c−b)2z=\frac{(a+d)+i(c-b)}2 and w=(a−d)+i(b+c)2w=\frac{(a-d)+i(b+c)}2 for A=(abcd)A=\begin{pmatrix}a&b\\c&d\end{pmatrix} (A↦(z,w)A\mapsto(z,w) is a linear bijection onto C2\mathbb{C}^2). Let Rα=(cos⁡α−sin⁡αsin⁡αcos⁡α)R_\alpha=\begin{pmatrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{pmatrix} and SO(2)={Rα}SO(2)=\{R_\alpha\}. Show that
conv⁡SO(2)={A: w=0, ∣z∣≤1},\operatorname{conv}SO(2)=\{A:\ w=0,\ |z|\le1\},
a closed disc (of dimension 22) in the 44-dimensional space M2(R)M_2(\mathbb{R}).

Hint

The rotation RαR_\alpha is multiplication by eiαe^{i\alpha} on C\mathbb{C}, so it has (z,w)=(eiα,0)(z,w)=(e^{i\alpha},0). A linear bijection preserves convex hulls. Then write any zz with ∣z∣≤1|z|\le1 as a convex combination of two opposite points of the unit circle.

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Solution

Idea. Under the linear isomorphism A↦(z,w)A\mapsto(z,w), SO(2)SO(2) becomes the unit circle {(eiα,0)}\{(e^{i\alpha},0)\} in the plane {w=0}\{w=0\}, and the convex hull of a circle is the closed disc.

Step 1: SO(2)SO(2) in coordinates. Rαx=eiαxR_\alpha x=e^{i\alpha}x for x∈Cx\in\mathbb{C}. From the entries: a=d=cos⁡αa=d=\cos\alpha, c=sin⁡αc=\sin\alpha, b=−sin⁡αb=-\sin\alpha, so z=2cos⁡α+2isin⁡α2=eiαz=\frac{2\cos\alpha+2i\sin\alpha}2=e^{i\alpha} and w=0w=0. Hence
SO(2)⟷{(eiα,0):α∈R}.SO(2)\longleftrightarrow\{(e^{i\alpha},0):\alpha\in\mathbb{R}\}.

Step 2: linear bijections preserve convex hulls. Let T:M2(R)→C2T:M_2(\mathbb{R})\to\mathbb{C}^2 be A↦(z,w)A\mapsto(z,w). TT is linear and bijective, so it maps convex combinations to convex combinations, and T(conv⁡X)=conv⁡T(X)T(\operatorname{conv}X)=\operatorname{conv}T(X). So we must show conv⁡{(eiα,0)}={(z,0):∣z∣≤1}\operatorname{conv}\{(e^{i\alpha},0)\}=\{(z,0):|z|\le1\}.

Step 3: ⊆\subseteq. The set D={(z,0):∣z∣≤1}D=\{(z,0):|z|\le1\} is convex (the closed unit disc is convex, by the triangle inequality ∣λz1+(1−λ)z2∣≤λ∣z1∣+(1−λ)∣z2∣≤1|\lambda z_1+(1-\lambda)z_2|\le\lambda|z_1|+(1-\lambda)|z_2|\le1) and contains the circle. So conv⁡{(eiα,0)}⊆D\operatorname{conv}\{(e^{i\alpha},0)\}\subseteq D.

Step 4: ⊇\supseteq. Let ∣z∣≤1|z|\le1 and write z=reiφz=re^{i\varphi} with r=∣z∣∈[0,1]r=|z|\in[0,1]. Then
z=1+r2 eiφ+1−r2 (−eiφ).z=\frac{1+r}2\,e^{i\varphi}+\frac{1-r}2\,\bigl(-e^{i\varphi}\bigr).
Indeed 1+r2−1−r2=r\frac{1+r}2-\frac{1-r}2=r. The two weights are ≥0\ge0 and sum to 11, and eiφe^{i\varphi}, −eiφ=ei(φ+π)-e^{i\varphi}=e^{i(\varphi+\pi)} lie on the unit circle. So (z,0)(z,0) is a convex combination of two points of the circle.

Back in matrices. In real terms: conv⁡SO(2)={(p−qqp):p2+q2≤1}\operatorname{conv}SO(2)=\left\{\begin{pmatrix}p&-q\\q&p\end{pmatrix}:p^2+q^2\le1\right\}, and the matrix with (p,q)=r(cos⁡φ,sin⁡φ)(p,q)=r(\cos\varphi,\sin\varphi) equals 1+r2Rφ+1−r2Rφ+π\frac{1+r}2R_\varphi+\frac{1-r}2R_{\varphi+\pi}.

Remark. This hull is only 22-dimensional. In contrast, conv⁡O(2)\operatorname{conv}O(2) is the whole operator-norm ball {∥A∥≤1}={∣z∣+∣w∣≤1}\{\|A\|\le1\}=\{|z|+|w|\le1\}, which is 44-dimensional: the improper rotations {z=0,∣w∣=1}\{z=0,|w|=1\} are needed to fill it. In particular the 2×22\times2 matrices of operator norm ≤1\le1 cannot be written as convex combinations of rotations alone.

Lesson. Convex hulls are invariant under linear isomorphisms, so choose coordinates in which the set is a familiar one (here a circle).