Does $\mathrm{SL}(2, \mathbb{R})$ have a non-empty boundary?

Question

Does $\mathrm{SL}(2, \mathbb{R})$ have a non-empty boundary?

Edit:

I've shown that $1$ is a regular value, and hence $\mathrm{SL}(2, \mathbb{R})$ is a three-dimensional manifold as Ted's hint. But I am not sure about what to do next. One more step please?

Answer 1

Hint: Prove it is a $3$-dimensional manifold.

Answer 2

Since $1$ is a regular value of the determinant function, $\det^{-1}\{1\} = \mathrm{SL}(2;\mathbb{R})$ is a submanifold of the two-by-two matrices. In particular, since it is a manifold, it has no boundary.