# Is there a group between $SO(2,\mathbb{R})$ and $SL(2,\mathbb{R})$?

Is there a non-trivial subgroup $H \subset SL(2,\mathbb{R})$ such that $H \supset SO(2,\mathbb{R})$ ?

My intuition is that, since $\dim SO(2)=1$ and $\dim SL(2)=3$, there should be some group between, but I can't point out one.

Note : in the complex case, $H:= SO(2,\mathbb{C}) \cup \left\{ \left( \matrix{ a&b \\ b&-a} \right) ~|~ a^2+b^2=-1 \right\}$ is an example.

• @KCd Can you be more specific? I've thought about Iwasawa decomposition and polar decomposition, but the problem is that the decomposition is not a group morphism. Commented Sep 25, 2014 at 21:45
• I made an error in my previous comment, so I deleted it.
– KCd
Commented Sep 25, 2014 at 22:10
• Wikipedia's page on $SL(2,\mathbb{R})$ ( en.wikipedia.org/wiki/SL2%28R%29 ) says that 'the circle group $SO(2)$ is a maximal compact subgroup of $SL(2, \mathbb{R})$.' Commented Sep 26, 2014 at 0:10
• @StevenStadnicki : Yes, but I don't think this helps. Commented Sep 26, 2014 at 0:20
• Your example over the complex numbers is not a group. Commented Sep 26, 2014 at 0:23

I think there's a fairly simple bare hands proof, even when you don't require $H$ to be a Lie group.

Consider the image of the unit circle under an element of $SL(2,\mathbb{R})$. This is always an ellipse, centred on the origin, where the product of the lengths of the major axis and minor axis is equal to $4$. The elements of $SO(2,\mathbb{R})$ are precisely those giving a circle (i.e., a major axis of length $2$).

If $H$ contains $SO(2,\mathbb{R})$ and also another element $A$ of $SL(2,\mathbb{R})$ which gives an ellipse with major axis $2a$, where $a>1$, then by pre-composing and post-composing $A$ with suitable rotations, we get all elements of $SL(2,\mathbb{R})$ giving an ellipse with major axis $2a$. So we can assume $A=\begin{pmatrix}a&0\\0&\frac{1}{a}\end{pmatrix}$.

Then $H$ also contains $AR_\theta A$ for every $\theta$, where $R_\theta$ is a rotation through an angle of $\theta$. For $\theta=0$ this element maps the unit circle to an ellipse with major axis $2a^2$, and for $\theta=\frac{\pi}{2}$ to a circle. By the Intermediate Value Theorem, for intermediate values of $\theta$ we can get any major axis length between $2$ and $2a^2$, and by the argument above, $H$ must then contain all elements of $SL(2,\mathbb{R})$ that give an ellipse with major axis between $2$ and $2a^2$.

Repeating, we can get any length of major axis.

• When you compose with rotations, you are not going to change the length of the minor axis of an ellipse. So your statement "we get all such elements" seems false to me. Am I missing something?
– Seub
Commented Sep 26, 2014 at 4:29
• @Seub The length of the minor axis is determined by the length of the major axis, since we are only concerned with elements of $SL(2,\mathbb{R})$, which preserve the area of the ellipse. I'll edit to clarify what I mean. Commented Sep 26, 2014 at 6:30
• @Seub He means all elements with the same major axis length $2a$ (and minor axis $\frac2a$) but different directions. Commented Sep 26, 2014 at 6:30
• Of course, thank you.
– Seub
Commented Sep 26, 2014 at 12:08
• Now I think this is convincing. If it really works, congrats, that's very nice. Still, I think you should edit to clarify at least your argument using the intermediate value theorem. Maybe something like this: let $f \in SL(2,\mathbb{R})$ giving an ellipse with major axis $2a$. Then the map $\Psi : f \,SO(2) \to SL(2,\mathbb{R})$, $\Psi(g) = g \, f$ gives elements with major axes $2$ (take $g = ?$) and $2a^2$ (take $g = f$), therefore also any major axis $\in [2, 2a^2]$ because $f SO(2)$ is connected. [NB: in the first case, maybe take $g = f \circ r_{\pi/2}$, I'm not sure]
– Seub
Commented Sep 26, 2014 at 13:46

Apparently not. There is a nontrivial 2 dimensional real Lie algebra, it can even be written as 2 by 2 matrices, $$\left( \begin{array}{rr} a & b \\ 0 & 0 \end{array} \right)$$ see http://en.wikipedia.org/wiki/Table_of_Lie_groups#Real_Lie_groups_and_their_algebras

I get that the exponential of this is $$\left( \begin{array}{cc} e^a & b \left( \frac{e^a - 1}{a} \right) \\ 0 & 1 \end{array} \right)$$ Note $$\frac{e^a - 1}{a} = 1 + \frac{a}{2} + \frac{a^2}{6} + \cdots$$ is analytic, and becomes $1$ when $a=0.$

The trouble is that you need, as a Lie algebra for something in between, matrices that are both trace-free and anti-trace free, as $$\left( \begin{array}{rr} a & b \\ -b & -a \end{array} \right)$$ which do not make a Lie algebra.

Oh, the trivial Lie group of dimension 2 can be realized in $SO_4,$ maximal torus

• What if the intermediate group isn't a Lie group? Commented Sep 25, 2014 at 23:48
• @StevenStadnicki, dunno. Commented Sep 25, 2014 at 23:52