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So I have a set P={ $p(\alpha,\beta,\gamma)=\pmatrix{1&\alpha&\beta\\0&1&\gamma\\0&0&1}$ $|$ $\alpha,\beta,\gamma$ $\in R$}. I needed to show P is a Lie group, which I have done.

I need to parametrise P, and I was asked to show that it is $R^3$, and, to show that the group multiplication is:

$p(\alpha,\beta,\gamma).p(x,y,z) = p(\alpha+x,\beta+\alpha z + y, \gamma+z)$

So I multiplied the matrices $\pmatrix{1&\alpha&\beta\\0&1&\gamma\\0&0&1}$. $\pmatrix{1&x&y\\0&1&z\\0&0&1}$ = $\pmatrix{1&\alpha+x&\beta+\alpha z + y\\0&1&\gamma+z\\0&0&1}$.

My question is, how would I parametrise the group above?

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There isn't "the" parameter space of $P$. You can parametrize $P$ in various ways, and here you have a parametrization by elements of $\mathbb R^3$. It's not clear to me what it means to ask whether this is "indeed" the parameter space. –  joriki Nov 15 '11 at 3:47
@Ray: The parametrization is the mapping $p:\mathbf{R}^3\rightarrow G\subset GL_3(\mathbf{R})$ given by mapping the triple $(\alpha,\beta,\gamma)$ to that upper triangular matrix. For example, your product formula may be written as $$p(\alpha,\beta,\gamma)p(x,y,z)=p(\alpha+x,\beta+\alpha z+y, \gamma+z).$$ Similary $p(\alpha,\beta,\gamma)^{-1}=p(-\alpha,-\beta+\alpha\gamma,-\gamma)$. If you drop the $p$ from these equations you get differentiable mappings from $R^3\times R^3\to R^3$ and $R^3\to R^3$ respectively, and that is what is needed to show that this is a Lie group (a single chart). –  Jyrki Lahtonen Nov 17 '11 at 14:31
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1 Answer

up vote 1 down vote accepted

You've already parametrized the group. The way you've written it establishes a bijection between the elements of $\mathbb R^3$ and the elements of $P$; that's what a parametrization is. The matrix multiplication you've carried out verifies the given multiplication law for the parameters, because the product of the matrices parametrized by the triples on the left-hand side is indeed the matrix parametrized by the triple on the right-hand side. There's nothing more to it than that.

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@Ray: It seems to me you may still be confused. What do you mean by "do the same as you did for $P$"? $P$ is the entire set; it's not a matter of doing the same as you did for $P$, but of applying what you did for $P$ to the identity and the inverses. All you need to do is write down the identity matrix and the inverse of $p(\alpha,\beta,\gamma)$ and read off the parameters according to the bijection you've already established. –  joriki Nov 15 '11 at 4:09
@Ray: I'm sorry, I'm having a hard time figuring out which part of this you don't understand. You seem to be labouring under a rather fundamental misunderstanding. Perhaps it might help to take a look at the Wikipedia article on parametrization. Your inverse is correct, but I don't understand what you mean by "so it is also in $\mathbb R^3$". This is a $3\times3$ matrix; it's in $\mathbb R^{3\times3}$, not $\mathbb R^3$; it is parametrized by a triple of real numbers, i.e. an element of $\mathbb R^3$, as specified in your definition of $P$. –  joriki Nov 15 '11 at 4:34
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