Elements of the coset $G/H$, where $G=\text{GL}^+(2)$ and $H=\text{SO}(2)$

Question

In this paper, in section $2$, a method to write the elements of the coset of $G/H$ is provided for $\text{GL}(4)$, but I am interested in $\text{GL}^+(2)$.

My matrix representation of $\mathfrak{gl}(2)$ is

$$ \begin{bmatrix} a+x& -b+y\\ b+y & a-x \end{bmatrix} $$

My matrix representation of $\mathfrak{so}(2)$ is

$$ \begin{bmatrix} 0& -b\\ b & 0 \end{bmatrix} $$

---

Reading the paper, it states that the element $g$ of $G$ can be decomposed as $g=\gamma h$, where

$$ \gamma = \exp \left( \begin{bmatrix} a+x& +y\\ y & a-x \end{bmatrix} \right) \in G/H $$

and where

$$ h = \exp \left( \begin{bmatrix} 0& -b\\ b & 0 \end{bmatrix} \right) \in H $$

Is this correct, or no?

I am suspicious of the argument, because to me

$$ \exp \left( \begin{bmatrix} a+x& y\\ y & a-x \end{bmatrix} \right)\exp \left( \begin{bmatrix} 0& -b\\ b & 0 \end{bmatrix} \right) \neq \exp \left( \begin{bmatrix} a+x& -b+y\\ b+y & a-x \end{bmatrix} \right) $$

Thus, $\gamma h$ does not appear to realize all elements of $G$. Or do we not care about some missing elements for cosets?

Answer 1

In the $GL_2(\mathbb R)$ case, the author's conclusion is correct, but it doesn't really come from Lie algebra manipulations. For $GL_n(\mathbb R)$, and for other reductive groups, there is the Cartan decomposition $G=\mathrm{exp}(\mathfrak s)\cdot K$, where $K$ is the maximal compact obtained as exponentiating the $+1$ eigenspace for a chosen Cartan involution (e.g., the standard $\gamma\to -\gamma^\top$), and $\mathfrak s$ is the $-1$ eigenspace. The latter is not a Lie subalgebra. In the standard case, it is symmetric matrices, while the Lie algebra $\mathfrak k$ of $K$ is skew-symmetric matrices.

For $GL_n(\mathbb R)$, the proof of this decomposition is just about the spectral theorem for self-adjoint operators on $\mathbb R^n$...

Answer 2

Even if $\exp(X+Y)\ne\exp(X)\exp(Y)$ in general, that doesn't stop elements of the form $\exp(X_1+Y_1)$ from being expressible as $\exp(X_2)\exp(Y_2)$ for a pair $X_2,Y_2$ which is different from $X_1,Y_1$. So your observation about lie algebra elements not commuting, and therefore $\exp$ not being a homomorphism, is not at all helpful for your conclusion about the author being wrong.

The situation we have here is a Lie group $G$ and subgroup $K$, with lie algebras $\mathfrak{g}$ and $\mathfrak{k}$, with a complementary vector subspace $\mathfrak{p}$ to $\mathfrak{k}$ within $\mathfrak{g}$ (i.e. $\mathfrak{g}=\mathfrak{k}\oplus\mathfrak{p}$ as vector spaces) which is not necessarily itself a lie subalgebra, and the question is if $\exp(\mathfrak{g})=\exp(\mathfrak{k})\exp(\mathfrak{p})$, despite the fact that $\exp(X+Y)\ne\exp(X)\exp(Y)$.

In particular, this situation has $\mathfrak{g}=\mathfrak{k}\oplus\mathfrak{p}$ as part of the Cartan decomposition, as Garrett points out. For illustration, let's look at a different situation: knit products, where the complementary vector subspace is in fact a lie subalgebra of $\mathfrak{g}$ just like $\mathfrak{k}$. Let me change letters to be more consistent with the literature. We have subgroups $H,K\le G$ for which $G=HK$ (i.e. all $g\in G$ are expressible as $g=hk$ with $h\in H,k\in K$) and $H\cap K$ is trivial (i.e. if $g=hk$ then $h\in H,k\in K$ are unique). If we added the condition that elements of $H$ and $K$ commute with each other, then we could say $G=H\times K$ is an internal direct product, but a knit product has no such condition. In other words, the multiplication map $H\times K\to G$ given by $(h,k)\mapsto hk$ is a diffeomorphism but not necessarily a homomorphism, if $G$ is a knit product of $H$ and $K$.

Here is perhaps the simplest example: the (orientation-preserving component of the) affine group $\mathrm{Aff}_0(1,\mathbb{R})$, consisting of maps $x\mapsto ax+b$ ($a>0$). We can model such maps as matrices $[\begin{smallmatrix} a&b\\0&1\end{smallmatrix}]$ acting on vectors $[\begin{smallmatrix} x\\1 \end{smallmatrix}]$. This group is a knit product of subgroups $K=\{[\begin{smallmatrix} a&0\\0&1\end{smallmatrix}]\}\cong\mathbb{R}^\times$ and $H=\{[\begin{smallmatrix}1&h\\0&1\end{smallmatrix}]\}\cong\mathbb{R}$. Indeed, a little bit of work shows:

$$ \exp\left(\begin{bmatrix}x&y\\0&0\end{bmatrix}\right)= \exp\left(\begin{bmatrix}x&0\\0&0\end{bmatrix}\right) \exp\left(\begin{bmatrix}0&\frac{1}{x}(1-e^{-x})y\\0&0\end{bmatrix}\right). $$

For $\mathrm{GL}_2\mathbb{R}$ in particular, it is a good exercise to verify (a) any rotation matrix is an exponential of an antisymmetric matrix, (b) any symmetric matrix is an exponential of a symmetric matrix, and (c) every invertible matrix is a symmetric matrix times a rotation matrix.