1
$\begingroup$

Let $O_2(\mathbb{R}) = \{A\in M_2(\mathbb{R})\mid AA^T=A^TA=I\}$ with the subspace topology induced by $M_2(\mathbb{R})$. Prove $O_2(\mathbb{R})$ is closed and disconnected. Then conclude from it that there is no continuous curve like $\rho:[0,1] \rightarrow O_2(\mathbb{R})$ such that $\rho(0)=I$ and $\rho(1)=\left( \begin{array}{ccc} 0 &-1 \\ -1&0\end{array}\right)$ .

  • 3
    Show that the function $A\to AA^T$ is continuous. – 2017-01-07
  • 3
    If $O_2(\mathbb{R})$ were connected, then the image of $\det: O_2(\mathbb{R}) \to \mathbb{R}$ would be connected, but this image is $\{-1,1\}$. Similar reasoning applies to showing the non-existence of such a continuous curve $\rho$. – 2017-01-07
  • 0
    @Hayden That also requires showing that the determinant is continuous. Not difficult, but it must be done. – 2017-01-07
  • 0
    @Arthur Yes, but that follows from the fact that it can be written as a polynomial of the entries of the matrix, and such polynomials are always continuous. – 2017-01-07
  • 0
    Cool, thx guys! – 2017-01-07

1 Answers 1

0

The inverse image of a closed set (e.g. the set containing only $I$) under a continuous function is closed.

As for "disconnected", do you know how to prove that the determinant of an orthogonal matrix is either $1$ or $-1$?

  • 0
    Yup, $det(A) = det(A^T)$ and the determinant of the product is the product of the determinants, it becomes a form of $x^2=1$ equation. – 2017-01-07
  • 0
    @AshkanRanjbar : So you can use that, and it quickly answers the part about disconnectedness and about the curve from on point to another. – 2017-01-07
  • 0
    And for the disconnectedness we have the continuous image being onto a discrete space, thus the domain itself must be disconnected? – 2017-01-07
  • 0
    Yes. The sets $\{a\}$ and $\{b\}$ are open subsets of $\{a,b\}$, so their inverse-images are open subsets of the domain. So the domain has disjoint complementary nonempty open sets. – 2017-01-07