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I solved the following exercise. Could someone please check my work?

Prove that $\operatorname{Isom_n}{\mathbb R^n}$ is a matrix group. Is it compact?

My work:

We recall that

$$\operatorname{Isom_n}{\mathbb R^n} = \left \{ \left ( \begin{array}{cc} A & 0 \\ V & 1 \end{array} \right ) \mid A \in O(n) (\mathbb K), V \in \mathbb R^n \right \} $$

By considering blockwise multiplication, blockwise inversion and the multiplicativity of the determinant it is clear that $\operatorname{Isom_n}{\mathbb R^n} $ is a subgroup of $GL_{n+1}$.

Since $|\det|$ is continuous and

$$ \det \left ( \begin{array}{cc} A & 0 \\ V & 1 \end{array} \right ) = \det A$$

it is clear that if $A_n \to A$ and $|\det A_n| = 1$ then also $|\det A| = 1$ hence $\operatorname{Isom_n}{\mathbb R^n} $ is closed in $GL_{n+1}$.

It is however not closed in $M_{n+1}$. To see this note that

$$ X_n = \left ( \begin{array}{cc} D_n & 0 \\ V & 1 \end{array} \right ) $$

where $D_n$ is the diagonal matrix with $d_1 = {1\over n}$, $d_2 = n$ and $d_k = 1$ otherwise is a sequece in $\operatorname{Isom_n}{\mathbb R^n}$ that converges to a matrix of zero determinant.

Similarly we see that $\operatorname{Isom_n}{\mathbb R^n}$ is not compact: the norm of $X_n$ is not bounded hence $\operatorname{Isom_n}{\mathbb R^n}$ is not bounded and therefore not compact.

learner
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1 Answers1

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The question doesn't reduce to just showing that $O(n)$ is closed in $GL_n(\mathbb{R})$. In the formula for $Isom_n(\mathbb{R}^n)$, you have not only an $A \in O(n)$, but also $V \in \mathbb{R}^n$, which you never considered.

Ted
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  • Yes, I didn't but $V$ can be arbitrary in both $\mathrm{Aff} $ and $\mathrm{Isom}$ so the limit of a sequence in $\mathrm{Isom}$ will be in $\mathrm{Aff}$. And in the previous exercise I showed that $\mathrm{Aff}$ is a matrix group so I thought I could use it in this exercise... am I wrong? – learner Dec 28 '14 at 01:39
  • Perhpas you could elaborate just a little bit more or in different words about what my mistake is please? I don't yet understand it. – learner Dec 28 '14 at 01:41
  • I have now posted my other proof here. – learner Dec 28 '14 at 01:49
  • @learner If your argument was intended to show that Isom is closed in $GL_{n+1}$, then your argument is correct. I thought you were addressing the second part of the question, whether Isom is compact. – Ted Dec 31 '14 at 07:59
  • No, I was trying to show that it is closed. Just the closed part... Please can you edit your answer accordingly? Then I'll accept it... – learner Jan 22 '15 at 06:13
  • I also added my new insights hoping my approach is now correct. – learner Jan 22 '15 at 06:21