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To show that $SL(2, \mathbb{R})$ is a manifold, we apply the Regular Value Theorem to the determinant function.

1. Define the Map

We consider the space of all $2 \times 2$ real matrices, $M_2(\mathbb{R}) \cong \mathbb{R}^4$. A general matrix $A$ is: $$A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$$

The determinant is a smooth map $f: M_2(\mathbb{R}) \to \mathbb{R}$ defined by: $$f(A) = ad - bc$$

The group $SL(2, \mathbb{R})$ is the level set $f^{-1}(1)$.

2. Check for Regularity

We must show that the gradient $\nabla f$ is non-zero for every matrix in $SL(2, \mathbb{R})$. $$\nabla f = \left( \frac{\partial f}{\partial a}, \frac{\partial f}{\partial b}, \frac{\partial f}{\partial c}, \frac{\partial f}{\partial d} \right) = (d, -c, -b, a)$$

The only way for $\nabla f$ to be the zero vector is if $a=b=c=d=0$. However, the zero matrix has a determinant of $0$. Since $0 \neq 1$, the zero matrix is not in $SL(2, \mathbb{R})$.

Consequently, for any $A \in SL(2, \mathbb{R})$, $\nabla f(A) \neq \mathbf{0}$, meaning $1$ is a regular value.

3. Conclusion

By the Regular Value Theorem, the dimension of the manifold is the dimension of the domain minus the dimension of the codomain: $$\dim(SL(2, \mathbb{R})) = 4 - 1 = 3$$ Thus, $SL(2, \mathbb{R})$ is a smooth 3-dimensional manifold.

This approach uses the Matrix Exponential to show that $SL(2, \mathbb{R})$ is a manifold by demonstrating that it looks locally like its tangent space at the identity.

1. The Key Identity

The relationship between the determinant and the trace is given by Jacobi’s formula: $$\det(e^X) = e^{\text{tr}(X)}$$

Let $\mathfrak{sl}(2, \mathbb{R})$ be the space of traceless $2 \times 2$ matrices: $$\mathfrak{sl}(2, \mathbb{R}) = \{ X \in M_2(\mathbb{R}) \mid \text{tr}(X) = 0 \}$$

If $X \in \mathfrak{sl}(2, \mathbb{R})$, then $\det(e^X) = e^0 = 1$. This confirms that the exponential map takes the linear subspace of traceless matrices into the group $SL(2, \mathbb{R})$.

2. Local Diffeomorphism

The derivative of $\exp$ at the origin is the identity map. By the Inverse Function Theorem, $\exp$ is a local diffeomorphism.

Specifically, there exists a neighborhood $U$ of $0$ in $\mathfrak{sl}(2, \mathbb{R})$ and a neighborhood $V$ of $I$ in $SL(2, \mathbb{R})$ such that: $$\exp: U \to V$$ is a smooth homeomorphism. Since $U$ is an open subset of the 3-dimensional vector space $\mathfrak{sl}(2, \mathbb{R})$, this provides a local coordinate chart for $SL(2, \mathbb{R})$ around the identity.

3. Conclusion

Because $SL(2, \mathbb{R})$ is a Lie group, we can translate this local chart to any point $A \in SL(2, \mathbb{R})$ using the smooth map $L_A(B) = AB$. This covers the entire group with smooth charts, proving it is a manifold of dimension 3.

The Basis of $\mathfrak{sl}(2, \mathbb{R})$

Any $2 \times 2$ matrix $X$ with $\text{tr}(X) = 0$ must be of the form: $$X = \begin{pmatrix} a & b \\ c & -a \end{pmatrix}$$

We can decompose this into three independent basis matrices:

1. $H = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$
2. $E = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$
3. $F = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$

Since any traceless matrix is a linear combination $aH + bE + cF$, the dimension is exactly 3.

Geometric Interpretation

The exponential map transforms these basis vectors into specific types of transformations within $SL(2, \mathbb{R})$:

• $\exp(tH)$: Represents a hyperbolic rotation (squeezing one axis while stretching the other).
• $\exp(tE)$ and $\exp(tF)$: Represent shear transformations.

Because $\exp$ is a local diffeomorphism near the origin, these three “directions” in the vector space $\mathfrak{sl}(2, \mathbb{R})$ translate directly into the three dimensions of the manifold $SL(2, \mathbb{R})$.

1. The Matrix $F-E$

The matrix $X = F - E$ is the standard generator for rotations: $$X = F - E = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$$ Since $X^2 = -I$, this matrix behaves like the imaginary unit $i$.

2. The Exponential Path

The exponential map $\exp(t(F-E))$ produces the standard rotation matrix: $$\exp(t(F-E)) = \begin{pmatrix} \cos t & -\sin t \\ \sin t & \cos t \end{pmatrix}$$ This is the group $SO(2)$, the subgroup of all orthogonal matrices with determinant 1.

3. Role in $SL(2, \mathbb{R})$ Topology

• Compactness: While most directions in $SL(2, \mathbb{R})$ (like those generated by $H$, $E$, or $F$) lead to non-compact, “unbounded” matrices, the direction $F-E$ is periodic.
• Topology: $SL(2, \mathbb{R})$ is topologically equivalent to $S^1 \times \mathbb{R}^2$. The $S^1$ factor (the circle) is exactly the path traced by $\exp(t(F-E))$.
• 1-Manifolds: This relates to your upcoming topic! The subgroup $SO(2)$ is a perfect example of a compact, connected 1-manifold embedded within a higher-dimensional manifold.

Annex Jacobi’s formula

1. Diagonalizable Case

Assume $X = P D P^{-1}$ where $D$ is diagonal with eigenvalues $\lambda_i$.

1. $\text{tr}(X) = \sum \lambda_i$.
2. $e^X = P e^D P^{-1}$, so $\det(e^X) = \det(e^D)$.
3. $\det(e^D) = \prod e^{\lambda_i} = e^{\sum \lambda_i} = e^{\text{tr}(X)}$.

2. Differential Equation Proof

Define $f(t) = \det(e^{tX})$.

• $f(0) = 1$.
• Using the chain rule and the derivative of the determinant at $I$: $$f'(t) = \det(e^{tX}) \cdot \text{tr}(X)$$
• Solving the ODE $\frac{df}{dt} = \text{tr}(X) f$: $$f(t) = e^{t \cdot \text{tr}(X)}$$
• Setting $t=1$ yields $\det(e^X) = e^{\text{tr}(X)}$.