Let $(\mathbb G,*)$ be a Carnot group. Thus, by definition, $\mathbb G$ is a connected and simply connected Lie group whose Lie algebra $\mathfrak g$ admits a stratification, that is $$\mathfrak g=V_1\oplus\dots\oplus V_r$$ such that $$[V_1,V_{i}]=V_{i+1}$$ for $1\leq i\leq r-1$ and every other commutator of lenght > $r$ is zero (see here).

We set $m_i:=dim(V_i)$ and $h_0=0$, $h_{i+1}=h_i+m_{i+1}$ for $0\leq i\leq r-1$, thus $h_1+\dots+h_r=n$, where $n$ is the (topological) dimension of $\mathbb G$.

For these groups, we know that the $Exp\colon (\mathfrak g, \diamond)\to (\mathbb G,*)$ is an isomorphism of Lie groups, where with $\diamond$ I have denoted the Hausdorff-Campbell operation on $\mathfrak g$. Moreover, given a basis $\{Z_1,\dots,Z_n\}$ of $\mathfrak g$ adapted to the stratification, that is $\{Z_{h_i+1},\dots, Z_{h_{i+1}}\}$ is a basis for $V_{i+1}$, we can identify $\mathfrak g$ with $\mathbb R^n$ setting $$\mathfrak g\ni\sum_{i=1}^nx_iZ_i\mapsto (x_1,\dots,x_n)\in \mathbb R^n. $$ If we call the above map $\pi$, we get that $Exp\circ \pi^{-1}\colon \mathbb R^n\to \mathbb G$ is an isomorphism of Lie group if we endow $\mathbb R^n$ with the composition law $\bullet$ defined as $$x\bullet y:=\pi(\sum x_i Z_i\diamond \sum_{i=1}^n y_i Z_i) \quad (*).$$

Now my doubts are:

  1. In papers, I often read sentences like "Let $\mathfrak g$ be the Lie algebra generated, as a vector space by three vectors $X_1,X_2,X_3$ with the only non trivial commutation relation being $X_3=[X_1,X_2]$. Thus $\mathfrak g$ is a nilpotent algebra stratified by the strata $V_1=span\{X_1,X_2\}$ and $V_2=span\{X_3\}$. Let $(\mathbb G,*)$ be the only connected and simply connected Lie group whose Lie algebra is $\mathfrak g$." Which theorem ensures us that there exists a connected and simply connected Lie group with $\mathfrak g$ as a Lie algebra and it is unique?
  2. With the notation as in the previous question, suppose that there exists a change of basis $\phi$ that sends the basis $\{X_1,X_2,X_3\}$ in $\{Y_1,Y_2,Y_3\}$. A change of basis is not necessarily a Lie algebras isomorphism, thus $Y_1,Y_2,Y_3$ may have different commutation relations from just $[Y_1,Y_2]=Y_3$ and all the possible others zero. Nevertheless $\mathfrak g$ is spanned also by $\{Y_1,Y_2,Y_3\}$. Thus, the only connected and simply connected Lie group that has Lie algebra spanned by $\{Y_1,Y_2,Y_3\}$ is again $(\mathbb G,*)$, right?
  3. Suppose that a Carnot group $(\mathbb G,*)$ is given. Suppose we have fixed a basis of its Lie algebra $\mathfrak g$ adapted to the stratification. Thus, if we use this basis, suppose $\bullet$ is the group law on $\mathbb R^n$ as in $(*)$. If we now use another basis adapted to the stratification, that may have different commutation relation from the former, we get a composition law $\odot$ on $\mathbb R^n$ that might be different from $\bullet$. Thus the groups $(\mathbb R^n,\bullet)$ and $(\mathbb R^n, \odot)$ are different Lie groups that are the representation in exponential coordinate of the same group $(\mathbb G,*)$. Is this also true?

Thanks a lot.


1 Answer 1

  1. Cartan-Lie theorem. Also known as Lie's third theorem - but more correctly: Cartan extended Lie's theorem to prove the result that you're looking for. Some links:

Lie's third theorem, wikipedia

Lie's three theorems at nLab

  1. Commutation relations are definitely preserved under change of basis! Try it.

  2. Answered by the answer to 2.


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