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In class we saw A 1-1 homomorphism from $\operatorname{Iso}(\mathbb{R})$ to $GL(2,\mathbb{R})$

$$\operatorname{Iso}(\mathbb{R})\cong \left\{ \begin{pmatrix}\pm1 & x\\ 0 & 1 \end{pmatrix}|x\in\mathbb{R}\right\}. $$

How can I get this result ? (It works, but it's probably not a guess, whats the idea ?)

Seeing this I think that there is a 1-1 homomorphism from $\operatorname{Iso}(\mathbb{R}^2)$ to $GL(3,\mathbb{R})$ , how can we find it ? (that's why I'm trying to gain a better understanding of $\operatorname{Iso}(\mathbb{R})$)

I could use some help with this.

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$\text{Iso}(\mathbb{R}^n)$ embeds into $\text{GL}_{n+1}(\mathbb{R})$. The idea is to consider $(n+1) \times (n+1)$ matrices whose top left $n \times n$ block is a linear isometry, whose last row is $0, 0, ... 1$, and whose last column describes a translation; these matrices act on the set of column vectors which end in $1$, which gives the relevant copy of $\mathbb{R}^n$. – Qiaochu Yuan Apr 27 '12 at 22:23
@QiaochuYuan can you please go into more details ? I can't see how this fits the case I gave, the last row is what you said and all the information needed for the translation is in the last column - but why it is $x 1$ and not another number ? why we take $+1,-1$ as $a_{11}$ ? I can't see how to generalize this even to $n=2$... – Belgi Apr 27 '12 at 22:26
It's the semidirect product of the orthogonal group $O(n)$ (rotating) and $\mathbf R^n$ (translating). $O(1)$ is just $\pm1$. – Dylan Moreland Apr 27 '12 at 22:29
Look at the matrix form below. Perhaps it will answer some of your questions about why $Iso(\mathbb{R})$ is put together the way it is. I'd recommend testing out a few points, especially in the case where $(x,y)=0$ or the orthogonal submatrix in the upper left corner is the identity. – Brett Frankel Apr 27 '12 at 22:34
up vote 2 down vote accepted

How about: $$\operatorname{Iso}(\mathbb{R}^2)\cong \left\{ \begin{pmatrix}a & b & x\\ c & d & y\\ 0 & 0 & 1\end{pmatrix}\right\}. $$

where $$ \begin{pmatrix} a & b\\ c & d \end{pmatrix}\in O_2(\mathbb{R})$$

and $(x,y)$ is a vector in $\mathbb{R}^2$. Represent the point $(x_0, y_0)$ by the vector $(x_0,y_0, 1)$.

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Indeed, multiplying the above matrix with $(x_0,y_0,1)$ really makes me see this, thanks! – Belgi Apr 27 '12 at 22:40

The Mazur-Ulam theorem gives us that any isometry of $\mathbb R^n$ is of the form $f(x)=Ax+b$ where $A\in GL(n,\mathbb R)$ and $b\in \mathbb R^n$. Intuitively, this is just saying that every such isometry is a linear map composed with a translation. We can embed $GL(n,\mathbb R)$ in $GL(n+1,\mathbb R)$ by simply adding a $1$ in the bottom right corner and zeroes elsewhere on the last row and column. We can embed $\mathbb R^n$ in $GL(n+1,\mathbb R)$ by the map $$\begin{pmatrix} x_1\\\vdots\\ x_n\end{pmatrix}\mapsto \begin{pmatrix} 1 & 0 & \cdots & x_1\\ & \ddots & \ddots & \vdots\\ 0 & \cdots & 1 & x_n\\ 0 &\cdots & 0 &1\\ \end{pmatrix}$$ which can be fairly easily verified by matrix multiplication. Putting these two together by multiplying the two matrices, we get an embedding $\mathrm{Iso}(\mathbb R^n)\to GL(n+1,\mathbb R)$. Note that combining two embeddings like this does not always work, but in this case it does because the images of the two embeddings intersect only at the identity $I\in GL(n+1,\mathbb R)$.

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I am trying to understand the answer but I don't see where the rotations are coming into this matrix in the way you presented it as two embedings... – Belgi Apr 27 '12 at 22:55
@Belgi Rotations are in $GL(n,\mathbb R)$, which we can embedd in $GL(n+1,\mathbb R)$ and then combine this embedding with the embedding of $\mathbb R^n$ by multiplying the results of the two embeddings. In the case $n=2$, this is precisely the embedding Brett gave. – Alex Becker Apr 27 '12 at 22:57
@AlexBecker I must be confused: why is it $GL_n$ and not $O_n$? It seems to me that something like $\begin{pmatrix}2 & \\ & 1\end{pmatrix}$ won't preserve the metric. (Even with my confusion this was still a useful read: I didn't know that this theorem had a name!) – Dylan Moreland Apr 28 '12 at 0:19
@DylanMoreland Isometries are in $O_n$ of course. However, since we just need an embedding and $O_n\subset GL_n$, we might as well use an embedding of $GL_n$. – Alex Becker Apr 28 '12 at 0:55

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