Exponential of Lie Groups.

Question

When the exponential map defined a bijection between the group G and their Lie algebra? The only example I know is the Heisenberg group.

Answer 1

The exponential map is a bijection implies that $G$ is contractible since the Lie algebra is a vector space, Let ${\cal G}$ be the Lie algebra of $G$, write ${\cal G}=V\oplus R$ where $R$ is the radical and $V$ is semi-simple, consider the subgroup $V_G$ of $G$ associated to $V$, the restriction of $exp$ to the Lie algebra of $V_G$ is injective if and only if $V_G$ is trivial, thus $V$ is trivial and $G$is solvable and simply connected. The converse is also true.

To show that $V$ is trivial, consider the restriction of $exp$ to the Lie algebra $D$ of the compact component $C$ of $V_G$ in its Iwasawa decomposition, this restriction is injective implies that $D$ is zero since for a compact Lie group, the exponential is surjective and if it is injective it is an homeomorphism and a vector space is compact if its dimension is zero. This shows that $C$ is trivial and henceforth $V$ is trivial since it is not solvable.

Answer 2

There is a really cool example that I love, given in a paper by Arsigny, et al. which shows that the symmetric positive definite matrices can be given a Lie group structure, and this is one such example where the exponential map is a bijection from the Lie group to its Lie algebra. Let $SPD(n)$ be the manifold of symmetric positive definite matrices. This can be realized as a manifold both as an open subset of the symmetric matrices $Symm(n)$, and also as a homogeneous space $SPD(n) \approx GL_n(\mathbb{R})/O(n)$ under the action $GL_n(\mathbb{R}) \times SPD(n) \to SPD(n)$ given by $(A, P) \to APA^T$. Using the spectral theorem we can diagonalize any $P \in SPD(n)$ as $P = QDQ^T$ with $D$ diagonal with positive elements and $Q \in O(n)$. Furthermore, the tangent space $T_PSPD(n) \cong Symm(n)$ at each point, hence the tangent vectors are symmetric matrices. The matrix exponential map $\exp:Symm(n) \to SPD(n)$ given by the Taylor series has the convenient computational property that

$$ \exp(P) \;\; =\;\; Q\exp(D) Q^T $$

and similarly the logarithm map $\log:SPD(n) \to Symm(n)$ defined by the usual Taylor expansion is defined for all matrices in $SPD(n)$ and serves as an inverse for $\exp$.

Now, the part that Arsigny and his collaborators introduced in their paper is how to turn $SPD(n)$ into a Lie group. This can be done by introducing the binary operation $\odot: SPD(n) \times SPD(n) \to SPD(n)$ given by

$$ A\odot B \;\; =\;\; \exp(\log(A) + \log(B)). $$

One can easily show that this operation is smooth and satisfies all of the conditions to turn $(SPD(n), \odot)$ into a Lie group. What's also exciting too is that in the paper, the authors introduce a scalar multiplication that turns $SPD(n)$ into a vector space.

Update 9/15/2019

I wanted to update this example with a couple more features that will make this analysis a little more complete.

• Notice that the Lie group product turns $SPD(n)$ into a commutative Lie group. This implies that the Lie algebra $Symm(n)$ is commutative, and rightfully so the bracket $[\cdot, \cdot]: Symm(n) \times Symm(n) \to Symm(n)$ is trivial. This fact is correctly reflected in the Campbell-Baker-Hausdorff formula:

$$ \log\left (\exp(A) \odot \exp(B)\right ) \;\; =\;\; A+B. $$

• Observe now that the Lie group structure $(SPD(n), \odot)$ is really just the pushforward of vector addition on $Symm(n)$ via $\exp$, and this is afforded to us by the fact that $\exp$ is a global diffeomorphism. We can extend this to a vector space structure on $SPD(n)$ by also pushing forward the scalar multiplication. Arsigny et al. defined the operation $\circledast:\mathbb{R} \times SPD(n) \to SPD(n)$ given by $$ \lambda\circledast P \;\; =\;\; \exp \left (\lambda \log(P)\right ). $$

One can show that $\circledast$ satisfies the distributive properties of scalar multiplication and in fact serves as a scalar multiplication on $SPD(n)$.

Update 2/4/2023

Another great example in this same vein was published recently that shows that correlation values themselves possess Lie group structure and this is another case where the group is isomorphic to the Lie algebra. Research has shown that correlation matrices have a quotient manifold structure

$$ Corr(n) \;\; =\;\; SPD(n)/Diag_+(n) $$

that arises from the group action $Diag_+(n)\times SPD(n) \to SPD(n)$ given by $(D,P) \to DPD$. A natural projection $\pi: SPD(n) \to Corr(n)$ is given by the usual map that takes a covariance matrix to the corresponding correlation matrix by dividing the $ij$ term by $\sigma_i\sigma_j$. Then in the special case $Corr(2)$, we see that ordinary matrix multiplication gives products that are elements of $SPD(2)$:

$$ \left [ \begin{array}{cc} 1 & x \\ x & 1 \\ \end{array} \right ]\left [ \begin{array}{cc} 1 & y \\ y & 1 \\ \end{array} \right ] \;\; =\;\; \left [ \begin{array}{cc} 1 + xy & x+y \\ x+y & 1+xy \\ \end{array} \right ]. $$

This can be used to define a product $C(x)*C(y) = \pi(C(x)C(y))$ where we find

$$ C(x)*C(y) \;\; =\;\; \left [ \begin{array}{cc} 1 & \frac{x+y}{1+xy} \\ \frac{x+y}{1+xy} & 1 \\ \end{array} \right ] $$

and $Corr(2)$ can be easily shown to be a commutative Lie group. Although this is not technically a matrix Lie group (the product is not purely from matrix multiplication), the paper goes on to subsequently show that left-invariant vector fields are of the form

$$ S_x \;\; =\;\; s\left (1-x^2\right ) $$

and therefore the exponential map, defined for arbitrary Lie groups as the map Exp$(S) = \gamma_S(1)$ (maximal integral curve of the vector field $S$), gives rise to the ODE

$$ \dot{\gamma} \;\; =\;\; s\left (1-\gamma^2\right ) $$

which has the straightforward solution $\gamma(t) = \tanh(st)$. Because $Corr(2)$ is essentially the interval $(-1,1)$ and $T_0Corr(2) = \mathbb{R}$, we find that $\tanh:\mathbb{R}\to Corr(2)$ is a bijective homomorphism with a smooth homomorphic inverse.