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Robert Bryant
Robert Bryant
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$\mathcal{T}$ is not a Lie group when $n>1$.

YouActually, the OP did not say whether youhe wanted $\mathcal{T}$ to be all possible sequences of compositions of these generating sets, but, if youhe did, then it is clear that $T$$\mathcal{T}$ is not a Lie group, in the sense that it is not defined as the set of solutions of some system of PDE for transformations of $\mathbb{R}^n$. For one thing, the group that they generate would properly contain the conformal group $\mathrm{O}(n{+}1,1)$ acting on $S^n$, which is known to be a maximal Lie group, i.e., there is no group (in Lie's sense) between the conformal group and the full diffeomorphism group. (NB: The group of analytic diffeomorphisms of $S^n$ is not a subgroup of the full diffeomorphims in Lie's sense because it is not defined as the set of solutions of some system of PDE.)

In particular, no group $G$ that contains $\mathcal{T}$ can preserve any geometric structures of the kind you mentionthe OP mentions because this would define a PDE that $\mathcal{T}$$G$ satisfies.

(By the way, note that $\mathcal{T}$, as the OP defined it, does not consist of smooth transformations of $S^n$ only when $n>1$, since the non-conformal affine transformations do not extend smoothly to $\infty$ except when $n=1$.)

$\mathcal{T}$ is not a Lie group when $n>1$.

You did not say whether you wanted $\mathcal{T}$ to be all possible sequences of compositions of these generating sets, but, if you did, then it is clear that $T$ is not a Lie group, in the sense that it is not defined as the set of solutions of some system of PDE for transformations of $\mathbb{R}^n$. For one thing, the group that they generate would properly contain the conformal group acting on $S^n$, which is known to be a maximal Lie group, i.e., there is no group (in Lie's sense) between the conformal group and the full diffeomorphism group. (NB: The group of analytic diffeomorphisms of $S^n$ is not a subgroup of the full diffeomorphims in Lie's sense because it is not defined as the set of solutions of some system of PDE.)

In particular, no group that contains $\mathcal{T}$ can preserve any geometric structures of the kind you mention because this would define a PDE that $\mathcal{T}$ satisfies.

$\mathcal{T}$ is not a Lie group when $n>1$.

Actually, the OP did not say whether he wanted $\mathcal{T}$ to be all possible sequences of compositions of these generating sets, but, if he did, then it is clear that $\mathcal{T}$ is not a Lie group, in the sense that it is not defined as the set of solutions of some system of PDE for transformations of $\mathbb{R}^n$. For one thing, the group that they generate would properly contain the conformal group $\mathrm{O}(n{+}1,1)$ acting on $S^n$, which is known to be a maximal Lie group, i.e., there is no group (in Lie's sense) between the conformal group and the full diffeomorphism group. (NB: The group of analytic diffeomorphisms of $S^n$ is not a subgroup of the full diffeomorphims in Lie's sense because it is not defined as the set of solutions of some system of PDE.)

In particular, no group $G$ that contains $\mathcal{T}$ can preserve any geometric structures of the kind the OP mentions because this would define a PDE that $G$ satisfies.

(By the way, note that $\mathcal{T}$, as the OP defined it, does not consist of smooth transformations of $S^n$ only when $n>1$, since the non-conformal affine transformations do not extend smoothly to $\infty$ except when $n=1$.)

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Robert Bryant
Robert Bryant
  • 108.4k
  • 8
  • 341
  • 453

$\mathcal{T}$ is not a Lie group when $n>1$.

You did not say whether you wanted $\mathcal{T}$ to be all possible sequences of compositions of these generating sets, but, if you did, then it is clear that $T$ is not a Lie group, in the sense that it is not defined as the set of solutions of some system of PDE for transformations of $\mathbb{R}^n$. For one thing, the group that they generate would properly contain the conformal group acting on $S^n$, which is known to be a maximal Lie group, i.e., there is no group (in Lie's sense) between the conformal group and the full diffeomorphism group. (NB: The group of analytic diffeomorphisms of $S^n$ is not a subgroup of the full diffeomorphims in Lie's sense because it is not defined as the set of solutions of some system of PDE.)

In particular, no group that contains $\mathcal{T}$ can preserve any geometric structures of the kind you mention because this would define a PDE that $\mathcal{T}$ satisfies.