Proving a submanifold of $SL_2(\mathbb{R})$

Question

I already showed that $SL_2(\mathbb{R})$ is a 3-dimensional manifold.

Now I want to show that the subspace $E$ of symmetric matrices whose eigenvalues are positive in $ SL_2( \mathbb{R})$ is a submanifold of dimension 2.

I tried showing that it is closed, but even if I did, it doesn't help me show that $E$ is of dimension 2.

I also tried using the level set theorem; but I couldn't figure out which map to use.

Answer 1

EDITED

Hint: Start over. What are the equations that dictate symmetry and $\det=1$? Show that this gives you a submersion. This manifold has two components: The symmetry condition dictates that the eigenvalues must be real, so having their product be $1$ forces them both to have the same sign.

Final comment: In fact, you can write this subset of $\Bbb R^4$ explicitly as a graph of a smooth function (which proves directly that it's a submanifold): If $A=\begin{bmatrix}x_1&x_2\\x_3&x_4\end{bmatrix}$, then we have $x_1x_4-x_2^2=1$, so $x_4=(1+x_2^2)/x_1$. This clearly has two components.

Answer 2

Let's see, $A = \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right]$. You know $ad-bc=1$ (this is $det(A)=1$) and $b=c$ ( this is $A^T=A$). Looks like 2 equations and 4 unknowns. Now, work out the details.