Lie Groups and Matrix Groups - Key Proof

We present a proof that $SL_n(\mathbb{R})$ is a smooth submanifold of $GL_n(\mathbb{R})$ , illustrating the general technique for establishing that matrix groups defined by polynomial equations are Lie groups.

Proof

Theorem: $SL_n(\mathbb{R}) = \{A \in GL_n(\mathbb{R}) : \det(A) = 1\}$ is a Lie group of dimension $n^2 - 1$ .

Proof:

We apply the implicit function theorem. Consider the determinant map: $\det: GL_n(\mathbb{R}) \to \mathbb{R}^*$ This is a smooth map, being a polynomial in the matrix entries. We want to show that $SL_n(\mathbb{R}) = \det^{-1}(1)$ is a smooth submanifold.

Step 1: Compute the derivative of the determinant map.

For $A \in GL_n(\mathbb{R})$ , the derivative $d(\det)_A: T_A GL_n(\mathbb{R}) \to \mathbb{R}$ can be computed using the formula: $\frac{d}{dt}\bigg|_{t=0} \det(A + tB) = \det(A) \cdot \text{tr}(A^{-1}B)$ for any matrix $B \in M_n(\mathbb{R}) = T_A GL_n(\mathbb{R})$ .

Step 2: Show the derivative is surjective.

At any point $A \in SL_n(\mathbb{R})$ (where $\det(A) = 1$ ), we have: $d(\det)_A(B) = \text{tr}(A^{-1}B)$

To show this is surjective as a map to $\mathbb{R}$ , we need to show we can achieve any value. Taking $B = A$ gives $\text{tr}(A^{-1}A) = \text{tr}(I) = n \neq 0$ . Thus the image contains a nonzero real number, and by linearity, the map is surjective.

Step 3: Apply the implicit function theorem.

Since $1 \in \mathbb{R}^*$ is a regular value of $\det$ (i.e., $d(\det)_A$ is surjective for all $A \in \det^{-1}(1)$ ), the preimage theorem (a consequence of the implicit function theorem) implies that $SL_n(\mathbb{R}) = \det^{-1}(1)$ is a smooth submanifold of $GL_n(\mathbb{R})$ of codimension 1.

Step 4: Compute the dimension.

Since $GL_n(\mathbb{R})$ has dimension $n^2$ and $SL_n(\mathbb{R})$ has codimension 1, we obtain: $\dim SL_n(\mathbb{R}) = n^2 - 1$

Step 5: Verify group operations are smooth.

The multiplication in $SL_n(\mathbb{R})$ is simply restriction of matrix multiplication from $GL_n(\mathbb{R})$ , which is smooth (polynomial in entries). Similarly, inversion is smooth on $GL_n(\mathbb{R})$ by Cramer's rule, and restriction to $SL_n(\mathbb{R})$ remains smooth. Moreover, since $\det(AB) = \det(A)\det(B)$ and $\det(A^{-1}) = (\det A)^{-1}$ , these operations preserve the condition $\det = 1$ .

Therefore, $SL_n(\mathbb{R})$ is a Lie group. □

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Remark

This proof technique is widely applicable. Any matrix group defined as the zero set of polynomial equations (e.g., $O(n)$ defined by $A^T A = I$ ) can be shown to be a Lie group using similar methods, provided one verifies that the defining equations have surjective differentials (regularity condition).

Example

For $O(n)$ , we define $F: GL_n(\mathbb{R}) \to \text{Sym}_n$ (symmetric matrices) by $F(A) = A^T A - I$ . The derivative at $A \in O(n)$ is $dF_A(B) = A^T B + B^T A$ . One can verify this is surjective onto the space of symmetric matrices, which has dimension $\frac{n(n+1)}{2}$ . Thus $O(n)$ has dimension $n^2 - \frac{n(n+1)}{2} = \frac{n(n-1)}{2}$ .

The key insight is that algebraic conditions (determinant equals 1) combined with smoothness of the defining functions naturally produce smooth manifold structures. This connection between algebra and geometry is fundamental to Lie theory.