Smooth maps on a manifold lie group

Question

$$ \operatorname{GL}_n(\mathbb R) = \{ A \in M_{n\times n} | \det A \ne 0 \} \\ \begin{align} &n = 1, \operatorname{GL}_n(\mathbb R) = \mathbb R - \{0\} \\ &n = 2, \operatorname{GL}_n(\mathbb R) = \left\{\begin{bmatrix}a&b\\c&d\end{bmatrix}\Bigg| ad-bc \ne 0\right\} \end{align} $$

$(\operatorname{GL}_n(\mathbb R),\cdot)$ is a group.

1. $AB$ is invertible if $A$ and $B$ are invertible.
2. $A(BC)=(AB)C$
3. $I=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
4. $A^{-1}$ is invertible if $A$ is invertible.

$$(\operatorname{GL}_n(\mathbb R) := \det{}^{-1}(\{0\}))$$ $\det{}^{-1}(\{0\})$ is open in $M_{n\times n}(\mathbb R)$.

$\det : M_{n\times n}(\mathbb R) \to \mathbb R$ is continuous, why?

$\dim \operatorname{GL}_n(\mathbb R) = n^2 - 1$, why?

$(\operatorname{GL}_n(\mathbb R),\cdot)$ is a Lie group if:

$$ \mu : G\times G \to G \\ \mu(A,B) = A\cdot B \:\text{ is smooth} \\ I(A) = A^{-1} \:\text{ is smooth} $$

How can I show this?

I want to show that general special linear group is Lie group. I could not show the my last step. How can I show that $AB$ and $A^{-1}$ are smooth? please help me I want to learn this. Thanks

Here the my handwritten notes https://i.stack.imgur.com/tkoMy.jpg

Answer 1

Well first you have to decide what exactly the "topology" on matrices is. Suppose we considered matrices just as vectors in $\mathbb{R}^{n^2}$, with the usual metric topology. Matrix multiplication by say $A$ take a matrix $B$ to another $\mathbb{R}^{n^2}$ vector where the entries are polynomials in the components of $B$.

The other bit is similar, if you know what crammer's rule is.

Answer 2

• $\det:M_n(\mathbb R)\to\mathbb R$ is continuous, since it is a polynomial map, if you regard it as a map $\det:\mathbb R^n\to\mathbb R$:

$$\det A=\sum\limits_{i+j=n}(-1)^{i+j}a_{ij}A_{ij}$$

where $A=(a_{ij}),$ and $A_{ij}$ is the determinant of the $(n-1)\times(n-1)$ matrix if we remove row $i$ and column $j$ from $A$.

• Now, $\dim GL_n(\mathbb R)=\dim M_n(\mathbb R)=\dim \mathbb R^n=n^2$.

  This is because $GL_n(\mathbb R)=\det ^{-1}(\mathbb R\setminus \{0\})$ is an open subset of $M_n(\mathbb R)$, since $\det$ is continuous and $\mathbb R\setminus \{0\}$ is open. Also, it is well-known that an open subset of an $n-$manifold is an $n-$manifold.

• Regarding the multiplication:

$$A\cdot B=(c_{ij}), \text{ where } c_{ij}=\sum\limits_{k=1}^na_{ik}b_{kj}$$ It is obvious that matrix multiplication is a vector function, with polynomial coordinate functions , therefore it is a smooth function.

• Finally, consider the map: $I:GL_n(\mathbb R)\to GL_n(\mathbb R)$ such that $I(A)=A^{-1}$. Then: $$I(A)=A^{-1}=\dfrac{1}{\det A}\mathrm {Adj}(A)$$ where $\det A$ is polynomial, and $\mathrm {Adj}(A) $, the adjoint of A, is easily seen as a vector function with polynomial coordinate functions. Therefore, $I$ is a vector function with rational coordinate functions, which implies that it is smooth.