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I have a set of objects $[x_1;x_2;\dots;x_n]$ and a permutation group $\pi$ that changes the position of these objects; this group can be represented as a matrix where only one nonvanishing entry (equal to 1) is located at each row (or column). Now I want to consider the case $n \mapsto \infty$. The group $\pi$ will be infinite-dimensional in this case and I want to promote this group to a Lie group. I can do this by defining a Lie algebra

$T_aT_b-T_bT_a = \sum_{d} c_{abd}T_d$

where I choose the generators $T_a$ to be transposition matrices that exchange only two points. The permutation group defined above will only be recovered if

$\pi = exp(\sum_d a_dT_d)$

with parameters $a_d$ satisfy certain conditions. This will create a special Lie group topology and makes this permutation group completely different from wellknown Lie groups like $SU(2)$. Which topological invariants I can compute?

I have heard that this group will be a discontinuous group. When I consider the principal bundle $E$ with base space $B$ and a fiber $F$ equipped with group actions of $\pi$ this fiber bundle will be not smooth, right?

Therefore the curvature form $\Omega$ may be not defined in some regions. But is it possible to compute something like the Pontryagin class $p=\int tr (\Omega \wedge \Omega)$?

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    This question is kind of all over the place. For starters, the infinite group you describe is usually denoted $S_{\Omega}$, where $\Omega$ is some countable set. $S_{\Omega}$ does not have any "dimension" since it is not a vector space. Moreover, why should it be a Lie group? Where's the smooth structure? – 2017-01-06

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